Most frequent k chars distance
The string distance function (SDF) that is the subject of this entry was proposed in a paper published in 2014 as a method for quickly estimating how similar two ordered sets or strings are. This SDF has also been proposed as being suitable in bioinformatics, e.g. for comparing sequences in the wp:fasta format.
Unfortunately, the original paper by Sadi Evren Seker et al. (arXiv:1401.6596 [cs.DS]) [1] has a number of internal inconsistencies and for this and other reasons is not entirely clear in several key respects, but the paper does give some worked examples and several of these (notably the two given under the "Worked Examples" heading below) agree with the interpretation used by the authors of the Python entry herein, so that interpretation is used in the present task description.
- The Task
(a) Show a time-efficient implementation of the SDF as if by a function mostFreqKSDF(s1, s2, k, n), where s1 and s2 are arbitrary strings, and k and n are non-negative integers as explained below;
(b) Show the values produced by the SDF for the following values of s1 and s2, with k = 2 and n = 10:
| String Inputs | "Top Two" | SDF Output (k=2, n=10) |
|---|---|---|
| 'night'
'nacht' |
n:1 i:1
n:1 a:1 |
8 |
| 'my'
'a' |
m:1 y:1
a:1 |
10 |
| ‘research’
‘research’ |
r:2 e:2
r:2 e:2 |
2 |
| ‘significant’
‘capabilities’ |
i:3 n:2
i:3 a:2 |
4 |
(c) Show the value produced by the SDF for k = 2, n = 100,
and the two strings:
"LCLYTHIGRNIYYGSYLYSETWNTGIMLLLITMATAFMGYVLPWGQMSFWGATVITNLFSAIPYIGTNLV" "EWIWGGFSVDKATLNRFFAFHFILPFTMVALAGVHLTFLHETGSNNPLGLTSDSDKIPFHPYYTIKDFLG"
- Reference Algorithm
The original paper employs an unusual scheme for encoding the frequencies of single letters as a string of letters and numbers. This scheme has several disadvantages, and so in the following we take a more abstract view and simply assume, solely for the purposes of this description, that two functions are available:
1) MostFreqK(s, k) returns (in decreasing order of their frequency of occurrence) the k most frequently occurring letters in s, with any ties broken by the position of the first occurrence in s;
2) For any letter c in MostFreqK(s, k), Freq(s,c) returns the frequency of c in s.
- MFKsimilarity
Next define MFKsimilarlity(s1, s2, k) as if by the following pseudo-code:
int function MFKsimilarity(string s1, string s2, int k):
let similarity = 0
let mfk1 = MostFreqK(s1, k)
let mfk2 = MostFreqK(s2, k)
for each c in mfk1
if (c occurs in mfk2)
then similarity += Freq(s1, c) + Freq(s2, c)
end
return similarity
- String Distance Wrapper Function
Now define MostFreqKSDF as if by:
int function MostFreqKSDF(string s1, string s2, int k, int n)
return n - MFKsimilarity(s1, s2, k)
- Worked Examples
The original paper gives several worked examples, including these two:
(i) MostFreqKSDF("night", "nacht", 2, 10)
For "night", the "top two" letters with their frequencies are: n:1 and i:1.
For "nacht", the "top two" letters with their frequencies are: n:1 and a:1.
Since "n" is the only shared letter amongst these candidates, the SDF is 10 - 1 - 1 = 8, as per the original paper.
(2) MostFreqKSDF( "LCLYTHIGRNIYYGSYLYSETWNTGIMLLLITMATAFMGYVLPWGQMSFWGATVITNLFSAIPYIGTNLV", "EWIWGGFSVDKATLNRFFAFHFILPFTMVALAGVHLTFLHETGSNNPLGLTSDSDKIPFHPYYTIKDFLG", 2, 100)
For the first string, the "top two" letters with their frequencies are: L:9 and T:8; and for the second, the "top two" are F:9 and L:8.
Since "L" is the only shared letter amongst these top-k characters, and since the sum of the corresponding frequencies is 9+8 = 17, the SDF is 83.
11l
<lang 11l>F most_freq_khashing(inputString, K)
DefaultDict[Char, Int] occuDict
L(c) inputString
occuDict[c]++
V occuList = sorted(occuDict.items(), key' x -> x[1], reverse' 1B)
V outputDict = Dict(occuList[0 .< K])
R outputDict
F most_freq_ksimilarity(inputStr1, inputStr2)
V similarity = 0
L(c, cnt1) inputStr1
I c C inputStr2
V cnt2 = inputStr2[c]
similarity += cnt1 + cnt2
R similarity
F most_freq_ksdf(inputStr1, inputStr2, K, maxDistance)
R maxDistance - most_freq_ksimilarity(most_freq_khashing(inputStr1, K), most_freq_khashing(inputStr2, K))
V str1 = ‘LCLYTHIGRNIYYGSYLYSETWNTGIMLLLITMATAFMGYVLPWGQMSFWGATVITNLFSAIPYIGTNLV’ V str2 = ‘EWIWGGFSVDKATLNRFFAFHFILPFTMVALAGVHLTFLHETGSNNPLGLTSDSDKIPFHPYYTIKDFLG’ V K = 2 V maxDistance = 100 V dict1 = most_freq_khashing(str1, 2) print(dict1, end' ":\n") print(dict1.map((c, cnt) -> c‘’String(cnt)).join(‘’)) V dict2 = most_freq_khashing(str2, 2) print(dict2, end' ":\n") print(dict2.map((c, cnt) -> c‘’String(cnt)).join(‘’)) print(most_freq_ksdf(str1, str2, K, maxDistance))</lang>
- Output:
[L = 9, T = 8]: L9T8 [F = 9, L = 8]: F9L8 83
C++
<lang cpp>#include <string>
- include <vector>
- include <map>
- include <iostream>
- include <algorithm>
- include <utility>
- include <sstream>
std::string mostFreqKHashing ( const std::string & input , int k ) {
std::ostringstream oss ;
std::map<char, int> frequencies ;
for ( char c : input ) {
frequencies[ c ] = std::count ( input.begin( ) , input.end( ) , c ) ;
}
std::vector<std::pair<char , int>> letters ( frequencies.begin( ) , frequencies.end( ) ) ;
std::sort ( letters.begin( ) , letters.end( ) , [input] ( std::pair<char, int> a ,
std::pair<char, int> b ) { char fc = std::get<0>( a ) ; char fs = std::get<0>( b ) ; int o = std::get<1>( a ) ; int p = std::get<1>( b ) ; if ( o != p ) { return o > p ; } else { return input.find_first_of( fc ) < input.find_first_of ( fs ) ; } } ) ;
for ( int i = 0 ; i < letters.size( ) ; i++ ) {
oss << std::get<0>( letters[ i ] ) ;
oss << std::get<1>( letters[ i ] ) ;
}
std::string output ( oss.str( ).substr( 0 , 2 * k ) ) ;
if ( letters.size( ) >= k ) {
return output ;
}
else {
return output.append( "NULL0" ) ;
}
}
int mostFreqKSimilarity ( const std::string & first , const std::string & second ) {
int i = 0 ;
while ( i < first.length( ) - 1 ) {
auto found = second.find_first_of( first.substr( i , 2 ) ) ;
if ( found != std::string::npos )
return std::stoi ( first.substr( i , 2 )) ;
else
i += 2 ;
} return 0 ;
}
int mostFreqKSDF ( const std::string & firstSeq , const std::string & secondSeq , int num ) {
return mostFreqKSimilarity ( mostFreqKHashing( firstSeq , num ) , mostFreqKHashing( secondSeq , num ) ) ;
}
int main( ) {
std::string s1("LCLYTHIGRNIYYGSYLYSETWNTGIMLLLITMATAFMGYVLPWGQMSFWGATVITNLFSAIPYIGTNLV" ) ;
std::string s2( "EWIWGGFSVDKATLNRFFAFHFILPFTMVALAGVHLTFLHETGSNNPLGLTSDSDKIPFHPYYTIKDFLG" ) ;
std::cout << "MostFreqKHashing( s1 , 2 ) = " << mostFreqKHashing( s1 , 2 ) << '\n' ;
std::cout << "MostFreqKHashing( s2 , 2 ) = " << mostFreqKHashing( s2 , 2 ) << '\n' ;
return 0 ;
} </lang>
- Output:
MostFreqKHashing( s1 , 2 ) = L9T8 MostFreqKHashing( s2 , 2 ) = F9L8
Go
<lang go>package main
import (
"fmt" "sort"
)
type cf struct {
c rune f int
}
func reverseStr(s string) string {
runes := []rune(s)
for i, j := 0, len(runes)-1; i < j; i, j = i+1, j-1 {
runes[i], runes[j] = runes[j], runes[i]
}
return string(runes)
}
func indexOfCf(cfs []cf, r rune) int {
for i, cf := range cfs {
if cf.c == r {
return i
}
}
return -1
}
func minOf(i, j int) int {
if i < j {
return i
}
return j
}
func mostFreqKHashing(input string, k int) string {
var cfs []cf
for _, r := range input {
ix := indexOfCf(cfs, r)
if ix >= 0 {
cfs[ix].f++
} else {
cfs = append(cfs, cf{r, 1})
}
}
sort.SliceStable(cfs, func(i, j int) bool {
return cfs[i].f > cfs[j].f // descending, preserves order when equal
})
acc := ""
min := minOf(len(cfs), k)
for _, cf := range cfs[:min] {
acc += fmt.Sprintf("%c%c", cf.c, cf.f)
}
return acc
}
func mostFreqKSimilarity(input1, input2 string) int {
similarity := 0
runes1, runes2 := []rune(input1), []rune(input2)
for i := 0; i < len(runes1); i += 2 {
for j := 0; j < len(runes2); j += 2 {
if runes1[i] == runes2[j] {
freq1, freq2 := runes1[i+1], runes2[j+1]
if freq1 != freq2 {
continue // assuming here that frequencies need to match
}
similarity += int(freq1)
}
}
}
return similarity
}
func mostFreqKSDF(input1, input2 string, k, maxDistance int) {
fmt.Println("input1 :", input1)
fmt.Println("input2 :", input2)
s1 := mostFreqKHashing(input1, k)
s2 := mostFreqKHashing(input2, k)
fmt.Printf("mfkh(input1, %d) = ", k)
for i, c := range s1 {
if i%2 == 0 {
fmt.Printf("%c", c)
} else {
fmt.Printf("%d", c)
}
}
fmt.Printf("\nmfkh(input2, %d) = ", k)
for i, c := range s2 {
if i%2 == 0 {
fmt.Printf("%c", c)
} else {
fmt.Printf("%d", c)
}
}
result := maxDistance - mostFreqKSimilarity(s1, s2)
fmt.Printf("\nSDF(input1, input2, %d, %d) = %d\n\n", k, maxDistance, result)
}
func main() {
pairs := [][2]string{
{"research", "seeking"},
{"night", "nacht"},
{"my", "a"},
{"research", "research"},
{"aaaaabbbb", "ababababa"},
{"significant", "capabilities"},
}
for _, pair := range pairs {
mostFreqKSDF(pair[0], pair[1], 2, 10)
}
s1 := "LCLYTHIGRNIYYGSYLYSETWNTGIMLLLITMATAFMGYVLPWGQMSFWGATVITNLFSAIPYIGTNLV" s2 := "EWIWGGFSVDKATLNRFFAFHFILPFTMVALAGVHLTFLHETGSNNPLGLTSDSDKIPFHPYYTIKDFLG" mostFreqKSDF(s1, s2, 2, 100) s1 = "abracadabra12121212121abracadabra12121212121" s2 = reverseStr(s1) mostFreqKSDF(s1, s2, 2, 100)
}</lang>
- Output:
input1 : research input2 : seeking mfkh(input1, 2) = r2e2 mfkh(input2, 2) = e2s1 SDF(input1, input2, 2, 10) = 8 input1 : night input2 : nacht mfkh(input1, 2) = n1i1 mfkh(input2, 2) = n1a1 SDF(input1, input2, 2, 10) = 9 input1 : my input2 : a mfkh(input1, 2) = m1y1 mfkh(input2, 2) = a1 SDF(input1, input2, 2, 10) = 10 input1 : research input2 : research mfkh(input1, 2) = r2e2 mfkh(input2, 2) = r2e2 SDF(input1, input2, 2, 10) = 6 input1 : aaaaabbbb input2 : ababababa mfkh(input1, 2) = a5b4 mfkh(input2, 2) = a5b4 SDF(input1, input2, 2, 10) = 1 input1 : significant input2 : capabilities mfkh(input1, 2) = i3n2 mfkh(input2, 2) = i3a2 SDF(input1, input2, 2, 10) = 7 input1 : LCLYTHIGRNIYYGSYLYSETWNTGIMLLLITMATAFMGYVLPWGQMSFWGATVITNLFSAIPYIGTNLV input2 : EWIWGGFSVDKATLNRFFAFHFILPFTMVALAGVHLTFLHETGSNNPLGLTSDSDKIPFHPYYTIKDFLG mfkh(input1, 2) = L9T8 mfkh(input2, 2) = F9L8 SDF(input1, input2, 2, 100) = 100 input1 : abracadabra12121212121abracadabra12121212121 input2 : 12121212121arbadacarba12121212121arbadacarba mfkh(input1, 2) = 112a10 mfkh(input2, 2) = 112210 SDF(input1, input2, 2, 100) = 88
Haskell
<lang Haskell>module MostFrequentK
where
import Data.List ( nub , sortBy ) import qualified Data.Set as S
count :: Eq a => [a] -> a -> Int count [] x = 0 count ( x:xs ) k
|x == k = 1 + count xs k |otherwise = count xs k
orderedStatistics :: String -> [(Char , Int)] orderedStatistics s = sortBy myCriterion $ nub $ zip s ( map (\c -> count s c ) s )
where
myCriterion :: (Char , Int) -> (Char , Int) -> Ordering
myCriterion (c1 , n1) (c2, n2)
|n1 > n2 = LT |n1 < n2 = GT |n1 == n2 = compare ( found c1 s ) ( found c2 s )
found :: Char -> String -> Int
found e s = length $ takeWhile (/= e ) s
mostFreqKHashing :: String -> Int -> String mostFreqKHashing s n = foldl ((++)) [] $ map toString $ take n $ orderedStatistics s
where
toString :: (Char , Int) -> String
toString ( c , i ) = c : show i
mostFreqKSimilarity :: String -> String -> Int mostFreqKSimilarity s t = snd $ head $ S.toList $ S.fromList ( doublets s ) `S.intersection`
S.fromList ( doublets t )
where
toPair :: String -> (Char , Int)
toPair s = ( head s , fromEnum ( head $ tail s ) - 48 )
doublets :: String -> [(Char , Int)]
doublets str = map toPair [take 2 $ drop start str | start <- [0 , 2 ..length str - 2]]
mostFreqKSDF :: String -> String -> Int ->Int mostFreqKSDF s t n = mostFreqKSimilarity ( mostFreqKHashing s n ) (mostFreqKHashing t n ) </lang>
- Output:
mostFrequentKHashing "LCLYTHIGRNIYYGSYLYSETWNTGIMLLLITMATAFMGYVLPWGQMSFWGATVITNLFSAIPYIGTNLV" 2 "L9T8" *MostFrequentK> mostFrequentKHashing "EWIWGGFSVDKATLNRFFAFHFILPFTMVALAGVHLTFLHETGSNNPLGLTSDSDKIPFHPYYTIKDFLG" 2 "F9L8"
J
Solution:<lang j>NB. String Distance Wrapper Function mfksDF =: {:@:[ - (mfks@:(mfkh&.>)~ {.)~
NB. Most Frequent K Distance mfks =: score@:(charMap@:[ {"1 charVals@:])/@:kHashes
score =. ([ +/ .* =)/ NB. (+ +/ .* *.&:*)/ for sum += freq_in_left + freq_in_right
charMap =. (,&< i.&> <@:~.@:,)&;/
charVals =. (; , 0:)"1
kHashes =. 0 1 |: ({.&>~ [: <./ #&>)
NB. Most Frequent K Hashing mfkh =: _&$: : (takeK freqHash) NB. Default LHA of _ means "return complete frequency table"
takeK =. (<.#) {. ]
freqHash =. ~. (] \:~ ,.&:(<"0)) #/.~
NB. No need to fix mfksDF mfkh =: mfkh f. mfks =: mfks f.</lang>
Examples:<lang j>verb define fkh =. ;@:,@:(":&.>) NB. format k hash
assert. 'r2e2 e2s1' (-: [: fkh 2&mfkh)&>&;: 'research seeking' assert. 2 = mfks 2 mfkh&.> 'research';'seeking'
assert. 'n1i1 n1a1' (-: [: fkh 2&mfkh)&>&;: 'night nacht' assert. 9 = 2 10 mfksDF 'night';'nacht'
assert. 'm1y1 a1' (-: [: fkh 2&mfkh)&>&;: 'my a' assert. 10 = 2 10 mfksDF 'my';'a'
assert. 'r2e2' -: fkh 2 mfkh 'research' assert. 6 = 2 10 mfksDF 'research';'research' NB. task says 8; right answer is 6
assert. 'a5b4 a5b4' (-: [: fkh 2&mfkh)&>&;: 'aaaaabbbb ababababa' assert. 1 = 2 10 mfksDF 'aaaaabbbb';'ababababa'
assert. 'i3n2 i3a2' (-: [: fkh 2&mfkh)&>&;: 'significant capabilities' assert. 7 = 2 10 mfksDF 'significant';'capabilities' NB. task says 5; right answer is 7
assert. 'L9T8 F9L8' (-: [: fkh 2&mfkh)&>&;: 'LCLYTHIGRNIYYGSYLYSETWNTGIMLLLITMATAFMGYVLPWGQMSFWGATVITNLFSAIPYIGTNLV EWIWGGFSVDKATLNRFFAFHFILPFTMVALAGVHLTFLHETGSNNPLGLTSDSDKIPFHPYYTIKDFLG' assert. 100 = 2 100 mfksDF 'LCLYTHIGRNIYYGSYLYSETWNTGIMLLLITMATAFMGYVLPWGQMSFWGATVITNLFSAIPYIGTNLV';'EWIWGGFSVDKATLNRFFAFHFILPFTMVALAGVHLTFLHETGSNNPLGLTSDSDKIPFHPYYTIKDFLG' NB. task says 83; right answer is 100
'pass' )
pass</lang>
Notes: As of press time, there are significant discrepancies between the task description, its pseudocode, the test cases provided, and the two other existing implementations. See the talk page for the assumptions made in this implementation to reconcile these discrepancies (in particular, in the scoring function).
Java
Translation of the pseudo-code of the Wikipedia article wp:Most frequent k characters to wp:java implementation of three functions given in the definition section are given below with wp:JavaDoc comments:
<lang java>import java.util.Collections; import java.util.Comparator; import java.util.HashMap; import java.util.LinkedHashMap; import java.util.ArrayList; import java.util.List; import java.util.Map;
public class SDF {
/** Counting the number of occurrences of each character
* @param character array
* @return hashmap : Key = char, Value = num of occurrence
*/
public static HashMap<Character, Integer> countElementOcurrences(char[] array) {
HashMap<Character, Integer> countMap = new HashMap<Character, Integer>();
for (char element : array) {
Integer count = countMap.get(element);
count = (count == null) ? 1 : count + 1;
countMap.put(element, count);
}
return countMap;
}
/**
* Sorts the counted numbers of characters (keys, values) by java Collection List
* @param HashMap (with key as character, value as number of occurrences)
* @return sorted HashMap
*/
private static <K, V extends Comparable<? super V>>
HashMap<K, V> descendingSortByValues(HashMap<K, V> map) {
List<Map.Entry<K, V>> list = new ArrayList<Map.Entry<K, V>>(map.entrySet()); // Defined Custom Comparator here Collections.sort(list, new Comparator<Map.Entry<K, V>>() { public int compare(Map.Entry<K, V> o1, Map.Entry<K, V> o2) { return o2.getValue().compareTo(o1.getValue()); } });
// Here I am copying the sorted list in HashMap // using LinkedHashMap to preserve the insertion order HashMap<K, V> sortedHashMap = new LinkedHashMap<K, V>(); for (Map.Entry<K, V> entry : list) { sortedHashMap.put(entry.getKey(), entry.getValue()); } return sortedHashMap;
}
/**
* get most frequent k characters
* @param array of characters
* @param limit of k
* @return hashed String
*/
public static String mostOcurrencesElement(char[] array, int k) {
HashMap<Character, Integer> countMap = countElementOcurrences(array);
System.out.println(countMap);
Map<Character, Integer> map = descendingSortByValues(countMap);
System.out.println(map);
int i = 0;
String output = "";
for (Map.Entry<Character, Integer> pairs : map.entrySet()) {
if (i++ >= k) break;
output += "" + pairs.getKey() + pairs.getValue();
}
return output;
}
/**
* Calculates the similarity between two input strings
* @param input string 1
* @param input string 2
* @param maximum possible limit value
* @return distance as integer
*/
public static int getDiff(String str1, String str2, int limit) {
int similarity = 0;
int k = 0; for (int i = 0; i < str1.length() ; i = k) { k ++; if (Character.isLetter(str1.charAt(i))) { int pos = str2.indexOf(str1.charAt(i));
if (pos >= 0) { String digitStr1 = ""; while ( k < str1.length() && !Character.isLetter(str1.charAt(k))) { digitStr1 += str1.charAt(k); k++; }
int k2 = pos+1; String digitStr2 = ""; while (k2 < str2.length() && !Character.isLetter(str2.charAt(k2)) ) { digitStr2 += str2.charAt(k2); k2++; }
similarity += Integer.parseInt(digitStr2) + Integer.parseInt(digitStr1);
} } } return Math.abs(limit - similarity);
}
/**
* Wrapper function
* @param input string 1
* @param input string 2
* @param maximum possible limit value
* @return distance as integer
*/
public static int SDFfunc(String str1, String str2, int limit) {
return getDiff(mostOcurrencesElement(str1.toCharArray(), 2), mostOcurrencesElement(str2.toCharArray(), 2), limit);
}
public static void main(String[] args) {
String input1 = "LCLYTHIGRNIYYGSYLYSETWNTGIMLLLITMATAFMGYVLPWGQMSFWGATVITNLFSAIPYIGTNLV";
String input2 = "EWIWGGFSVDKATLNRFFAFHFILPFTMVALAGVHLTFLHETGSNNPLGLTSDSDKIPFHPYYTIKDFLG";
System.out.println(SDF.SDFfunc(input1,input2,100));
}
}</lang>
JavaScript
Defines the functions for counting and frequency outside of the hashing function, as it is easier to calculate the comparison from a structured result rather than parsing a hash.
<lang javascript>//returns an object of counts keyed by character const kCounts = str => {
const counts = {};
for (let char of str) {
counts[char] = counts[char] ? counts[char] + 1 : 1;
}
return counts;
};
//returns an array of length k containing the characters with the highest counts const frequentK = (counts, k) => {
//note that this is written for clarity rather than speed, //as it sorts all of the counts when only the top k are needed return Object.keys(counts) .sort((a, b) => counts[b] - counts[a]) .slice(0, k);
};
//returns a hashed string of the most frequent k characters and their frequencies const mostFreqKHashing = (str, k) => {
const counts = kCounts(str);
return frequentK(counts, k)
.map(char => char + counts[char])
.join("");
};
//numeric score of similarity based on the sum of counts of characters appearing in the top k of both strings const mostFreqKSimilarity = (str1, str2, k) => {
const counts1 = kCounts(str1);
const counts2 = kCounts(str2);
const freq1 = frequentK(counts1, k);
const freq2 = frequentK(counts2, k);
let similarity = 0;
for (let char of freq1) {
//only considers a character if it is in the top k of both strings
if (freq2.includes(char)) {
//the examples on this page are inconsistent in whether the similarity
//should be the sum of the two counts or only the shared count (the minimum of the two)
//this code uses the sum
similarity += counts1[char] + counts2[char];
}
}
return similarity;
};
//subtracts the similarity score from the maxDifference const mostFreqKSDF = (str1, str2, k, maxDistance) => {
return maxDistance - mostFreqKSimilarity(str1, str2, k);
};</lang>
Testing with Jest <lang javascript>test('hash of "LCLYTHIGRNIYYGSYLYSETWNTGIMLLLITMATAFMGYVLPWGQMSFWGATVITNLFSAIPYIGTNLV" is "L9T8"', () => {
expect(mostFreqKHashing(str1, 2)).toBe("L9T8");
});
test('hash of "EWIWGGFSVDKATLNRFFAFHFILPFTMVALAGVHLTFLHETGSNNPLGLTSDSDKIPFHPYYTIKDFLG" is "F9L8"', () => {
expect(mostFreqKHashing(str2, 2)).toBe("F9L8");
});
test("SDF of strings 1 and 2 with k=2 and max=100 is 83", () => {
expect(mostFreqKSDF(str1, str2, 2, 100)).toBe(83);
});</lang>
Testing in Console <lang javascript>const str1 = "LCLYTHIGRNIYYGSYLYSETWNTGIMLLLITMATAFMGYVLPWGQMSFWGATVITNLFSAIPYIGTNLV"; const str2 = "EWIWGGFSVDKATLNRFFAFHFILPFTMVALAGVHLTFLHETGSNNPLGLTSDSDKIPFHPYYTIKDFLG"; const K = 2; const maxDistance = 100; console.log(mostFreqKHashing(str1, K)); console.log(mostFreqKHashing(str2, K)); console.log(mostFreqKSDF(str1, str2, K, maxDistance));</lang>
- Output:
L9T8 F9L8 83
jq
Works with gojq, the Go implementation of jq
Preliminaries <lang jq># bag of words def bow(stream):
reduce stream as $word ({}; .[($word|tostring)] += 1);
- Like sort_by(f) but for items that compare equal, retain the original order
def sort_by_decreasing(f):
# Add $n-$i def enumerate: . as $in | length as $n | reduce range(0;$n) as $i ([]; . + [ [$n-$i, $in[$i] ] ]); enumerate | sort_by((.[1]|f), .[0]) | reverse | map(.[1]);</lang>
The Tasks <lang jq># Output: { characters: array_of_characters_in_decreasing_order_of_frequency, frequency: object} def MostFreqKHashing($K):
def chars: explode | range(0;length) as $i | [.[$i]] | implode;
. as $in
| bow(tostring | chars) as $bow
| $bow
| to_entries
| sort_by_decreasing(.value) # if two chars have same frequency than get the first occurrence in $in
| (reduce .[0:$K][] as $kv ({}; .[$kv.key] = $kv.value) ) as $frequency
| {characters: map(.key)[:$K], $frequency};
def MostFreqKSimilarity($in1; $in2; $K):
[$in1, $in2] | map( MostFreqKHashing($K)) as [$s1, $s2]
| reduce $s1.characters[] as $c (0;
$s2.frequency[$c] as $f
| if $f then . + $s1.frequency[$c] + $f
else . end) ;
def MostFreqKSDF($inputStr1; $inputStr2; $K; $maxDistance):
$maxDistance - MostFreqKSimilarity($inputStr1; $inputStr2; $K);
def MostFreqKSDF($K; $maxDistance):
. as [$inputStr1, $inputStr2] | $maxDistance - MostFreqKSimilarity($inputStr1; $inputStr2; $K);
def task2:
["night", "nacht"], ["my", "a"], ["research", "research"], ["research", "seeking"], ["significant", "capabilities"] | MostFreqKSDF(2; 10) as $sdk | [., $sdk] ;
def task100:
["LCLYTHIGRNIYYGSYLYSETWNTGIMLLLITMATAFMGYVLPWGQMSFWGATVITNLFSAIPYIGTNLV", "EWIWGGFSVDKATLNRFFAFHFILPFTMVALAGVHLTFLHETGSNNPLGLTSDSDKIPFHPYYTIKDFLG"] | MostFreqKSDF(2; 100);
task2, task100</lang>
- Output:
[["night","nacht"],8] [["my","a"],10] [["research","research"],2] [["research","seeking"],6] [["significant","capabilities"],4] 83
Kotlin
The code for the MostFreqKSimilarity function differs in this task from that in the associated Wikipedia article. Also the description is inconsistent with the code in both cases.
What I've decided to assume is that the frequency for commonly occurring characters must be the same in both strings for it to be added to the 'similarity' variable which seems to be the implication of both descriptions. This gives the same results as the Wikipedia article for all except the last example where it gives 100 rather than 83.
It's also evident that you can't just add the frequency of each character to the output string of the MostFreqKHashing function and then expect to be able to parse it afterwards. This is because, in the general case, any printing characters (including digits) could occur in the input string and the frequency could be more than 9. I've therefore encoded the frequency as the character with the same unicode value rather than the frequency itself.
I have no idea what NULL0 is supposed to mean so I've ignored that. <lang scala>// version 1.1.51
fun mostFreqKHashing(input: String, k: Int): String =
input.groupBy { it }.map { Pair(it.key, it.value.size) }
.sortedByDescending { it.second } // preserves original order when equal
.take(k)
.fold("") { acc, v -> acc + "${v.first}${v.second.toChar()}" }
fun mostFreqKSimilarity(input1: String, input2: String): Int {
var similarity = 0
for (i in 0 until input1.length step 2) {
for (j in 0 until input2.length step 2) {
if (input1[i] == input2[j]) {
val freq1 = input1[i + 1].toInt()
val freq2 = input2[j + 1].toInt()
if (freq1 != freq2) continue // assuming here that frequencies need to match
similarity += freq1
}
}
}
return similarity
}
fun mostFreqKSDF(input1: String, input2: String, k: Int, maxDistance: Int) {
println("input1 : $input1")
println("input2 : $input2")
val s1 = mostFreqKHashing(input1, k)
val s2 = mostFreqKHashing(input2, k)
print("mfkh(input1, $k) = ")
for ((i, c) in s1.withIndex()) print(if (i % 2 == 0) c.toString() else c.toInt().toString())
print("\nmfkh(input2, $k) = ")
for ((i, c) in s2.withIndex()) print(if (i % 2 == 0) c.toString() else c.toInt().toString())
val result = maxDistance - mostFreqKSimilarity(s1, s2)
println("\nSDF(input1, input2, $k, $maxDistance) = $result\n")
}
fun main(args: Array<String>) {
val pairs = listOf(
Pair("research", "seeking"),
Pair("night", "nacht"),
Pair("my", "a"),
Pair("research", "research"),
Pair("aaaaabbbb", "ababababa"),
Pair("significant", "capabilities")
)
for (pair in pairs) mostFreqKSDF(pair.first, pair.second, 2, 10)
var s1 = "LCLYTHIGRNIYYGSYLYSETWNTGIMLLLITMATAFMGYVLPWGQMSFWGATVITNLFSAIPYIGTNLV" var s2 = "EWIWGGFSVDKATLNRFFAFHFILPFTMVALAGVHLTFLHETGSNNPLGLTSDSDKIPFHPYYTIKDFLG" mostFreqKSDF(s1, s2, 2, 100) s1 = "abracadabra12121212121abracadabra12121212121" s2 = s1.reversed() mostFreqKSDF(s1, s2, 2, 100)
}</lang>
- Output:
input1 : research input2 : seeking mfkh(input1, 2) = r2e2 mfkh(input2, 2) = e2s1 SDF(input1, input2, 2, 10) = 8 input1 : night input2 : nacht mfkh(input1, 2) = n1i1 mfkh(input2, 2) = n1a1 SDF(input1, input2, 2, 10) = 9 input1 : my input2 : a mfkh(input1, 2) = m1y1 mfkh(input2, 2) = a1 SDF(input1, input2, 2, 10) = 10 input1 : research input2 : research mfkh(input1, 2) = r2e2 mfkh(input2, 2) = r2e2 SDF(input1, input2, 2, 10) = 6 input1 : aaaaabbbb input2 : ababababa mfkh(input1, 2) = a5b4 mfkh(input2, 2) = a5b4 SDF(input1, input2, 2, 10) = 1 input1 : significant input2 : capabilities mfkh(input1, 2) = i3n2 mfkh(input2, 2) = i3a2 SDF(input1, input2, 2, 10) = 7 input1 : LCLYTHIGRNIYYGSYLYSETWNTGIMLLLITMATAFMGYVLPWGQMSFWGATVITNLFSAIPYIGTNLV input2 : EWIWGGFSVDKATLNRFFAFHFILPFTMVALAGVHLTFLHETGSNNPLGLTSDSDKIPFHPYYTIKDFLG mfkh(input1, 2) = L9T8 mfkh(input2, 2) = F9L8 SDF(input1, input2, 2, 100) = 100 input1 : abracadabra12121212121abracadabra12121212121 input2 : 12121212121arbadacarba12121212121arbadacarba mfkh(input1, 2) = 112a10 mfkh(input2, 2) = 112210 SDF(input1, input2, 2, 100) = 88
Nim
It is impossible to get consistent results. We chose to implement Wikipedia algorithm (note that the page is no longer available, except in Internet Archive). With this algorithm, we get the same results as those of Wikipedia except for the last example where we get 91 instead of 83.
In order to avoid the limitation to 10 occurrences, we chose to use an ordered table (equivalent to Python OrderedDict). We could have used Kotlin solution or used a sequence of tuples (char, count) but the ordered table gives better performance.
<lang Nim>import algorithm, sugar, tables
type CharCounts = OrderedTable[char, int]
func `$`(counts: CharCounts): string =
for (c, count) in counts.pairs: result.add $c & $count
func mostFreqKHashing(str: string; k: Positive): CharCounts =
var counts: CharCounts # To get the counts in apparition order. for c in str: inc counts.mgetOrPut(c, 0) counts.sort((x, y) => cmp(x[1], y[1]), Descending) # Note that sort is stable. var count = 0 for c, val in counts.pairs: inc count result[c] = val if count == k: break
func mostFreqKSimilarity(input1, input2: CharCounts): int =
for c, count in input1.pairs:
if c in input2:
result += count
func mostFreqKSDF(str1, str2: string; k, maxDistance: Positive): int =
maxDistance - mostFreqKSimilarity(mostFreqKHashing(str1, k), mostFreqKHashing(str2, k))
const
Pairs = [("night", "nacht"),
("my", "a"),
("research", "research"),
("aaaaabbbb", "ababababa"),
("significant", "capabilities")]
for (str1, str2) in Pairs:
echo "str1: ", str1 echo "str2: ", str2 echo "mostFreqKHashing(str1, 2) = ", mostFreqKHashing(str1, 2) echo "mostFreqKHashing(str2, 2) = ", mostFreqKHashing(str2, 2) echo "mostFreqKSDF(str1, str2, 2, 10) = ", mostFreqKSDF(str1, str2, 2, 10) echo()
const
S1 = "LCLYTHIGRNIYYGSYLYSETWNTGIMLLLITMATAFMGYVLPWGQMSFWGATVITNLFSAIPYIGTNLV" S2 = "EWIWGGFSVDKATLNRFFAFHFILPFTMVALAGVHLTFLHETGSNNPLGLTSDSDKIPFHPYYTIKDFLG"
echo "str1: ", S1 echo "str2: ", S2 echo "mostFreqKHashing(str1, 2) = ", mostFreqKHashing(S1, 2) echo "mostFreqKHashing(str2, 2) = ", mostFreqKHashing(S2, 2) echo "mostFreqKSDF(str1, str2, 2, 100) = ", mostFreqKSDF(S1, S2, 2, 100)</lang>
- Output:
str1: night str2: nacht mostFreqKHashing(str1, 2) = n1i1 mostFreqKHashing(str2, 2) = n1a1 mostFreqKSDF(str1, str2, 2, 10) = 9 str1: my str2: a mostFreqKHashing(str1, 2) = m1y1 mostFreqKHashing(str2, 2) = a1 mostFreqKSDF(str1, str2, 2, 10) = 10 str1: research str2: research mostFreqKHashing(str1, 2) = r2e2 mostFreqKHashing(str2, 2) = r2e2 mostFreqKSDF(str1, str2, 2, 10) = 6 str1: aaaaabbbb str2: ababababa mostFreqKHashing(str1, 2) = a5b4 mostFreqKHashing(str2, 2) = a5b4 mostFreqKSDF(str1, str2, 2, 10) = 1 str1: significant str2: capabilities mostFreqKHashing(str1, 2) = i3n2 mostFreqKHashing(str2, 2) = i3a2 mostFreqKSDF(str1, str2, 2, 10) = 7 str1: LCLYTHIGRNIYYGSYLYSETWNTGIMLLLITMATAFMGYVLPWGQMSFWGATVITNLFSAIPYIGTNLV str2: EWIWGGFSVDKATLNRFFAFHFILPFTMVALAGVHLTFLHETGSNNPLGLTSDSDKIPFHPYYTIKDFLG mostFreqKHashing(str1, 2) = L9T8 mostFreqKHashing(str2, 2) = F9L8 mostFreqKSDF(str1, str2, 2, 100) = 91
Perl
<lang Perl>#!/usr/bin/perl use strict ; use warnings ;
sub mostFreqHashing {
my $inputstring = shift ;
my $howmany = shift ;
my $outputstring ;
my %letterfrequencies = findFrequencies ( $inputstring ) ;
my @orderedChars = sort { $letterfrequencies{$b} <=> $letterfrequencies{$a} ||
index( $inputstring , $a ) <=> index ( $inputstring , $b ) } keys %letterfrequencies ;
for my $i ( 0..$howmany - 1 ) {
$outputstring .= ( $orderedChars[ $i ] . $letterfrequencies{$orderedChars[ $i ]} ) ;
}
return $outputstring ;
}
sub findFrequencies {
my $input = shift ;
my %letterfrequencies ;
for my $i ( 0..length( $input ) - 1 ) {
$letterfrequencies{substr( $input , $i , 1 ) }++ ;
}
return %letterfrequencies ;
}
sub mostFreqKSimilarity {
my $first = shift ;
my $second = shift ;
my $similarity = 0 ;
my %frequencies_first = findFrequencies( $first ) ;
my %frequencies_second = findFrequencies( $second ) ;
foreach my $letter ( keys %frequencies_first ) {
if ( exists ( $frequencies_second{$letter} ) ) {
$similarity += ( $frequencies_second{$letter} + $frequencies_first{$letter} ) ;
} } return $similarity ;
}
sub mostFreqKSDF {
(my $input1 , my $input2 , my $k , my $maxdistance ) = @_ ; return $maxdistance - mostFreqKSimilarity( mostFreqHashing( $input1 , $k) ,
mostFreqHashing( $input2 , $k) ) ; }
my $firststring = "LCLYTHIGRNIYYGSYLYSETWNTGIMLLLITMATAFMGYVLPWGQMSFWGATVITNLFSAIPYIGTNLV" ; my $secondstring = "EWIWGGFSVDKATLNRFFAFHFILPFTMVALAGVHLTFLHETGSNNPLGLTSDSDKIPFHPYYTIKDFLG" ; print "MostFreqKHashing ( " . '$firststring , 2)' . " is " . mostFreqHashing( $firststring , 2 ) . "\n" ; print "MostFreqKHashing ( " . '$secondstring , 2)' . " is " . mostFreqHashing( $secondstring , 2 ) . "\n" ; </lang>
- Output:
MostFreqKHashing ( $firststring , 2) is L9T8 MostFreqKHashing ( $secondstring , 2) is F9L8
Phix
with javascript_semantics function mostFreqKHashing(string input, integer k) string cfs = "" sequence cfsnx = {} for i=1 to length(input) do integer r = input[i], ix = find(r,cfs) if ix>0 then cfsnx[ix][1] -= 1 else cfs &= r cfsnx = append(cfsnx,{-1,length(cfs)}) end if end for cfsnx = sort(cfsnx) -- (aside: the idx forces stable sort) sequence acc := {} for i=1 to min(length(cfs),k) do integer {n,ix} = cfsnx[i] acc &= {cfs[ix], -n} end for return acc end function function mostFreqKSimilarity(sequence input1, input2) integer similarity := 0 for i=1 to length(input1) by 2 do for j=1 to length(input2) by 2 do if input1[i] == input2[j] then integer freq1 = input1[i+1], freq2 = input2[j+1] if freq1=freq2 then similarity += freq1 end if end if end for end for return similarity end function function flat(sequence s) string res = "" for i=1 to length(s) by 2 do res &= sprintf("%c%d",s[i..i+1]) end for return res end function procedure mostFreqKSDF(string input1, input2, integer k, maxDistance) printf(1,"input1 : %s\n", {input1}) printf(1,"input2 : %s\n", {input2}) sequence s1 := mostFreqKHashing(input1, k), s2 := mostFreqKHashing(input2, k) printf(1,"mfkh(input1, %d) = %s\n", {k,flat(s1)}) printf(1,"mfkh(input2, %d) = %s\n", {k,flat(s2)}) integer result := maxDistance - mostFreqKSimilarity(s1, s2) printf(1,"SDF(input1, input2, %d, %d) = %d\n\n", {k, maxDistance, result}) end procedure constant tests = {{"research", "seeking"}, {"night", "nacht"}, {"my", "a"}, {"research", "research"}, {"aaaaabbbb", "ababababa"}, {"significant", "capabilities"}} for i=1 to length(tests) do string {t1,t2} = tests[i] mostFreqKSDF(t1, t2, 2, 10) end for string s1 := "LCLYTHIGRNIYYGSYLYSETWNTGIMLLLITMATAFMGYVLPWGQMSFWGATVITNLFSAIPYIGTNLV", s2 := "EWIWGGFSVDKATLNRFFAFHFILPFTMVALAGVHLTFLHETGSNNPLGLTSDSDKIPFHPYYTIKDFLG" mostFreqKSDF(s1, s2, 2, 100) s1 = "abracadabra12121212121abracadabra12121212121" s2 = reverse(s1) mostFreqKSDF(s1, s2, 2, 100)
Output matches Go and Kotlin.
Python
unoptimized and limited <lang python>import collections def MostFreqKHashing(inputString, K):
occuDict = collections.defaultdict(int)
for c in inputString:
occuDict[c] += 1
occuList = sorted(occuDict.items(), key = lambda x: x[1], reverse = True)
outputStr = .join(c + str(cnt) for c, cnt in occuList[:K])
return outputStr
- If number of occurrence of the character is not more than 9
def MostFreqKSimilarity(inputStr1, inputStr2):
similarity = 0
for i in range(0, len(inputStr1), 2):
c = inputStr1[i]
cnt1 = int(inputStr1[i + 1])
for j in range(0, len(inputStr2), 2):
if inputStr2[j] == c:
cnt2 = int(inputStr2[j + 1])
similarity += cnt1 + cnt2
break
return similarity
def MostFreqKSDF(inputStr1, inputStr2, K, maxDistance):
return maxDistance - MostFreqKSimilarity(MostFreqKHashing(inputStr1,K), MostFreqKHashing(inputStr2,K))</lang>
optimized
A version that replaces the intermediate string with OrderedDict to reduce the time complexity of lookup operation: <lang python>import collections def MostFreqKHashing(inputString, K):
occuDict = collections.defaultdict(int)
for c in inputString:
occuDict[c] += 1
occuList = sorted(occuDict.items(), key = lambda x: x[1], reverse = True)
outputDict = collections.OrderedDict(occuList[:K])
#Return OrdredDict instead of string for faster lookup.
return outputDict
def MostFreqKSimilarity(inputStr1, inputStr2):
similarity = 0
for c, cnt1 in inputStr1.items():
#Reduce the time complexity of lookup operation to about O(1).
if c in inputStr2:
cnt2 = inputStr2[c]
similarity += cnt1 + cnt2
return similarity
def MostFreqKSDF(inputStr1, inputStr2, K, maxDistance):
return maxDistance - MostFreqKSimilarity(MostFreqKHashing(inputStr1,K), MostFreqKHashing(inputStr2,K))</lang>
Test: <lang python>str1 = "LCLYTHIGRNIYYGSYLYSETWNTGIMLLLITMATAFMGYVLPWGQMSFWGATVITNLFSAIPYIGTNLV" str2 = "EWIWGGFSVDKATLNRFFAFHFILPFTMVALAGVHLTFLHETGSNNPLGLTSDSDKIPFHPYYTIKDFLG" K = 2 maxDistance = 100 dict1 = MostFreqKHashing(str1, 2) print("%s:"%dict1) print(.join(c + str(cnt) for c, cnt in dict1.items())) dict2 = MostFreqKHashing(str2, 2) print("%s:"%dict2) print(.join(c + str(cnt) for c, cnt in dict2.items())) print(MostFreqKSDF(str1, str2, K, maxDistance))</lang>
- Output:
OrderedDict([('L', 9), ('T', 8)]):
L9T8
OrderedDict([('F', 9), ('L', 8)]):
F9L8
83
Racket
<lang Racket>#lang racket
(define (MostFreqKHashing inputString K)
(define t (make-hash)) (for ([c (in-string inputString)] [i (in-naturals)]) (define b (cdr (hash-ref! t c (λ() (cons i (box 0)))))) (set-box! b (add1 (unbox b)))) (define l (for/list ([(k v) (in-hash t)]) (list (car v) k (unbox (cdr v))))) (map cdr (take (sort (sort l < #:key car) > #:key caddr) K)))
(define (MostFreqKSimilarity inputStr1 inputStr2) ; not strings in this impl.
(for*/sum ([c1 (in-list inputStr1)] [c2 (in-value (assq (car c1) inputStr2))]
#:when c2)
(+ (cadr c1) (cadr c2))))
(define (MostFreqKSDF inputStr1 inputStr2 K maxDistance)
(- maxDistance (MostFreqKSimilarity (MostFreqKHashing inputStr1 K)
(MostFreqKHashing inputStr2 K))))
(MostFreqKSDF
"LCLYTHIGRNIYYGSYLYSETWNTGIMLLLITMATAFMGYVLPWGQMSFWGATVITNLFSAIPYIGTNLV" "EWIWGGFSVDKATLNRFFAFHFILPFTMVALAGVHLTFLHETGSNNPLGLTSDSDKIPFHPYYTIKDFLG" 2 100)
- => 83
- (Should add more tests, but it looks like there's a bunch of mistakes
- in the given tests...)
</lang>
Raku
(formerly Perl 6)
My initial impressions of this task are registered on the discussion page under "Prank Page?".
The "most frequent k characters" hashing function is straightforward enough to implement. The distance function is incomprehensible though. The description doesn't match the pseudo-code and neither of them match the examples. I'm not going to bother trying to figure it out unless there is some possible value.
Maybe I am too hasty though. Lets give it a try. Implement a MFKC routine and run an assortment of words through it to get a feel for how it hashes different words.
<lang perl6># Fairly straightforward implementation, actually returns a list of pairs
- which can be joined to make a string or manipulated further.
sub mfkh ($string, \K = 2) {
my %h;
$string.comb.kv.map: { %h{$^v}[1] //= $^k; %h{$^v}[0]++ };
%h.sort( { -$_.value[0], $_.value[1] } ).head(K).map( { $_.key => $_.value[0] } );
}
- lets try running 150 or so words from unixdic.txt through it to see
- how many unique hash values it comes up with.
my @test-words = <
aminobenzoic arginine asinine biennial biennium brigantine brinkmanship britannic chinquapin clinging corinthian declination dickinson dimension dinnertime dionysian diophantine dominican financial financier finessing fingernail fingerprint finnish giovanni hopkinsian inaction inalienable inanimate incaution incendiary incentive inception incident incidental incinerate incline inclusion incommunicable incompletion inconceivable inconclusive incongruity inconsiderable inconsiderate inconspicuous incontrovertible inconvertible incurring incursion indefinable indemnify indemnity indeterminacy indian indiana indicant indifferent indigene indigenous indigent indispensable indochina indochinese indoctrinate indonesia inequivalent inexplainable infantile inferential inferring infestation inflammation inflationary influential information infringe infusion ingenious ingenuity ingestion ingredient inhabitant inhalation inharmonious inheritance inholding inhomogeneity inkling inoffensive inordinate inorganic inputting inseminate insensible insincere insinuate insistent insomnia insomniac insouciant installation instinct instinctual insubordinate insubstantial insulin insurrection intangible intelligent intensify intensive interception interruption intestinal intestine intoxicant introduction introversion intrusion invariant invasion inventive inversion involution justinian kidnapping kingpin lineprinter liniment livingston mainline mcginnis minion minneapolis minnie pigmentation pincushion pinion quinine quintessential resignation ruination seminarian triennial wilkinson wilmington wilsonian wineskin winnie winnipeg
>;
say @test-words.map( { join , mfkh($_)».kv } ).Bag;</lang>
- Output:
Bag(i2n2(151))
One... Nope, I was right, it is pretty much worthless.
Ring
<lang ring>
- Project : Most frequent k chars distance
load "stdlib.ring" str1 = "LCLYTHIGRNIYYGSYLYSETWNTGIMLLLITMATAFMGYVLPWGQMSFWGATVITNLFSAIPYIGTNLV" str2 = "EWIWGGFSVDKATLNRFFAFHFILPFTMVALAGVHLTFLHETGSNNPLGLTSDSDKIPFHPYYTIKDFLG" see "Str1 = " + str1 + nl see "Str2 = " + str2 + nl
MostFreqKHashing(str1,"str1") MostFreqKHashing(str2,"str2")
func MostFreqKHashing(str3,strp)
chr = newlist(26,2)
for n = 1 to 26
str = char(n+64)
cstr = count(str3,str)
chr[n][1] = str
chr[n][2] = cstr
next
chr = sortsecond(chr)
chr = reverse(chr)
see "MostFreqKHashing(" + strp + ",2) = "
see chr[1][1] + chr[1][2] + chr[2][1] + chr[2][2] + nl
func count(cString,dString)
sum = 0
while substr(cString,dString) > 0
sum++
cString = substr(cString,substr(cString,dString)+len(string(sum)))
end
return sum
func sortsecond(alist)
aList = sort(alist,2)
for n=1 to len(alist)-1
for m=n to len(aList)-1
if alist[m+1][2] = alist[m][2] and alist[m+1][1] < alist[m][1]
temp = alist[m+1]
alist[m+1] = alist[m]
alist[m] = temp ok
next
next
return aList
</lang> Output:
Str1 = LCLYTHIGRNIYYGSYLYSETWNTGIMLLLITMATAFMGYVLPWGQMSFWGATVITNLFSAIPYIGTNLV Str2 = EWIWGGFSVDKATLNRFFAFHFILPFTMVALAGVHLTFLHETGSNNPLGLTSDSDKIPFHPYYTIKDFLG MostFreqKHashing(str1,2) = L9T8 MostFreqKHashing(str2,2) = F9L8
Sidef
<lang ruby>func _MostFreqKHashing(string, k) {
var seen = Hash()
var chars = string.chars
var freq = chars.freq
var schars = freq.keys.sort_by {|c| -freq{c} }
var mfkh = []
for i in ^k {
chars.each { |c|
seen{c} && next
if (freq{c} == freq{schars[i]}) {
seen{c} = true
mfkh << Hash(c => c, f => freq{c})
break
}
}
}
mfkh << (k-seen.len -> of { Hash(c => :NULL, f => 0) }...)
mfkh
}
func MostFreqKSDF(a, b, k, d) {
var mfkh_a = _MostFreqKHashing(a, k); var mfkh_b = _MostFreqKHashing(b, k);
d - gather {
mfkh_a.each { |s|
s{:c} == :NULL && next
mfkh_b.each { |t|
s{:c} == t{:c} &&
take(s{:f} + (s{:f} == t{:f} ? 0 : t{:f}))
}
}
}.sum
}
func MostFreqKHashing(string, k) {
gather {
_MostFreqKHashing(string, k).each { |h|
take("%s%d" % (h{:c}, h{:f}))
}
}.join
}
var str1 = "LCLYTHIGRNIYYGSYLYSETWNTGIMLLLITMATAFMGYVLPWGQMSFWGATVITNLFSAIPYIGTNLV"
var str2 = "EWIWGGFSVDKATLNRFFAFHFILPFTMVALAGVHLTFLHETGSNNPLGLTSDSDKIPFHPYYTIKDFLG"
say "str1 = #{str1.dump}" say "str2 = #{str2.dump}"
say
say("MostFreqKHashing(str1, 2) = ", MostFreqKHashing(str1, 2)) say("MostFreqKHashing(str2, 2) = ", MostFreqKHashing(str2, 2)) say("MostFreqKSDF(str1, str2, 2, 100) = ", MostFreqKSDF(str1, str2, 2, 100))
say
var arr = [
%w(night nacht), %w(my a), %w(research research), %w(aaaaabbbb ababababa), %w(significant capabilities),
]
var k = 2 var limit = 10
for s,t in arr {
"mfkh(%s, %s, #{k}) = [%s, %s] (SDF: %d)\n".printf(
s.dump, t.dump,
MostFreqKHashing(s, k).dump,
MostFreqKHashing(t, k).dump,
MostFreqKSDF(s, t, k, limit),
)
}</lang>
- Output:
str1 = "LCLYTHIGRNIYYGSYLYSETWNTGIMLLLITMATAFMGYVLPWGQMSFWGATVITNLFSAIPYIGTNLV"
str2 = "EWIWGGFSVDKATLNRFFAFHFILPFTMVALAGVHLTFLHETGSNNPLGLTSDSDKIPFHPYYTIKDFLG"
MostFreqKHashing(str1, 2) = L9T8
MostFreqKHashing(str2, 2) = F9L8
MostFreqKSDF(str1, str2, 2, 100) = 83
mfkh("night", "nacht", 2) = ["n1i1", "n1a1"] (SDF: 9)
mfkh("my", "a", 2) = ["m1y1", "a1NULL0"] (SDF: 10)
mfkh("research", "research", 2) = ["r2e2", "r2e2"] (SDF: 6)
mfkh("aaaaabbbb", "ababababa", 2) = ["a5b4", "a5b4"] (SDF: 1)
mfkh("significant", "capabilities", 2) = ["i3n2", "i3a2"] (SDF: 7)
Tcl
<lang tcl>package require Tcl 8.6
proc MostFreqKHashing {inputString k} {
foreach ch [split $inputString ""] {dict incr count $ch}
join [lrange [lsort -stride 2 -index 1 -integer -decreasing $count] 0 [expr {$k*2-1}]] ""
} proc MostFreqKSimilarity {hashStr1 hashStr2} {
while {$hashStr2 ne ""} {
regexp {^(.)(\d+)(.*)$} $hashStr2 -> ch n hashStr2 set lookup($ch) $n
}
set similarity 0
while {$hashStr1 ne ""} {
regexp {^(.)(\d+)(.*)$} $hashStr1 -> ch n hashStr1 if {[info exist lookup($ch)]} { incr similarity $n incr similarity $lookup($ch) }
} return $similarity
} proc MostFreqKSDF {inputStr1 inputStr2 k limit} {
set h1 [MostFreqKHashing $inputStr1 $k]
set h2 [MostFreqKHashing $inputStr2 $k]
expr {$limit - [MostFreqKSimilarity $h1 $h2]}
}</lang> Demonstrating: <lang tcl>set str1 "LCLYTHIGRNIYYGSYLYSETWNTGIMLLLITMATAFMGYVLPWGQMSFWGATVITNLFSAIPYIGTNLV" set str2 "EWIWGGFSVDKATLNRFFAFHFILPFTMVALAGVHLTFLHETGSNNPLGLTSDSDKIPFHPYYTIKDFLG" puts [MostFreqKHashing $str1 2] puts [MostFreqKHashing $str2 2] puts [MostFreqKSDF $str1 $str2 2 100]</lang>
- Output:
L9T8 F9L8 83
- A more efficient metric calculator
This version is appreciably more efficient because it does not compute the intermediate string representation “hash”, instead working directly on the intermediate dictionaries and lists: <lang tcl>proc MostFreqKSDF {inputStr1 inputStr2 k limit} {
set c1 [set c2 {}]
foreach ch [split $inputStr1 ""] {dict incr c1 $ch}
foreach ch [split $inputStr2 ""] {dict incr c2 $ch}
set c2 [lrange [lsort -stride 2 -index 1 -integer -decreasing $c2[set c2 {}]] 0 [expr {$k*2-1}]]
set s 0
foreach {ch n} [lrange [lsort -stride 2 -index 1 -integer -decreasing $c1[set c1 {}]] 0 [expr {$k*2-1}]] {
if {[dict exists $c2 $ch]} { incr s [expr {$n + [dict get $c2 $ch]}] }
}
return [expr {$limit - $s}]
}</lang> It computes the identical value on the identical inputs.
Typescript
with added type annotations
<lang typescript>//returns an object of counts keyed by character const kCounts = (str: string): Record<string, number> => {
const counts: Record<string, number> = {};
for (let char of str) {
counts[char] = counts[char] ? counts[char] + 1 : 1;
}
return counts;
};
//returns an array of length k containing the characters with the highest counts const frequentK = ( counts: Record<string, number>, k: number ): string[] => {
//note that this is written for clarity rather than speed, //as it sorts all of the counts when only the top k are needed return Object.keys(counts) .sort((a, b) => counts[b] - counts[a]) .slice(0, k);
};
//returns a hashed string of the most frequent k characters and their frequencies const mostFreqKHashing = (str: string, k: number): string => {
const counts = kCounts(str);
return frequentK(counts, k)
.map(char => char + counts[char])
.join("");
};
//numeric score of similarity based on the sum of counts of characters appearing in the top k of both strings const mostFreqKSimilarity = ( str1: string, str2: string, k: number ): number => {
const counts1 = kCounts(str1);
const counts2 = kCounts(str2);
const freq1 = frequentK(counts1, k);
const freq2 = frequentK(counts2, k);
let similarity = 0;
for (let char of freq1) {
//only considers a character if it is in the top k of both strings
if (freq2.includes(char)) {
//the examples on this page are inconsistent in whether the similarity
//should be the sum of the two counts or only the shared count (the minimum of the two)
//this code uses the sum
similarity += counts1[char] + counts2[char];
}
}
return similarity;
};
//subtracts the similarity score from the maxDifference const mostFreqKSDF = ( str1: string, str2: string, k: number, maxDistance: number ): number => {
return maxDistance - mostFreqKSimilarity(str1, str2, k);
};</lang>
Wren
<lang ecmascript>import "/seq" for Lst import "/sort" for Sort import "/trait" for Stepped
var mostFreqKHashing = Fn.new { |input, k|
var indivs = Lst.individuals(input.toList).map { |indiv| [indiv[0], indiv[1]] }.toList
var cmp = Fn.new { |p1, p2| (p2[1] - p1[1]).sign }
Sort.insertion(indivs, cmp)
return indivs.take(k).reduce("") { |acc, p| acc + "%(p[0])%(String.fromByte(p[1]))" }
}
var mostFreqKSimilarity = Fn.new { |input1, input2|
var similarity = 0
for (i in Stepped.new(0...input1.count, 2)) {
for (j in Stepped.new(0...input2.count, 2)) {
if (input1[i] == input2[j]) {
var freq1 = input1[i + 1].bytes[0]
var freq2 = input2[j + 1].bytes[0]
if (freq1 == freq2) similarity = similarity + freq1
}
}
}
return similarity
}
var mostFreqKSDF = Fn.new { |input1, input2, k, maxDistance|
System.print("input1 : %(input1)")
System.print("input2 : %(input2)")
var s1 = mostFreqKHashing.call(input1, k)
var s2 = mostFreqKHashing.call(input2, k)
System.write("mfkh(input1, %(k)) = ")
var i = 0
for (c in s1) {
System.write((i % 2 == 0) ? c : c.bytes[0])
i = i + 1
}
System.write("\nmfkh(input2, %(k)) = ")
i = 0
for (c in s2) {
System.write((i % 2 == 0) ? c : c.bytes[0])
i = i + 1
}
var result = maxDistance - mostFreqKSimilarity.call(s1, s2)
System.print("\nSDF(input1, input2, %(k), %(maxDistance)) = %(result)\n")
}
var pairs = [
["research", "seeking"], ["night", "nacht"], ["my", "a"], ["research", "research"], ["aaaaabbbb", "ababababa"], ["significant", "capabilities"]
] for (pair in pairs) mostFreqKSDF.call(pair[0], pair[1], 2, 10)
var s1 = "LCLYTHIGRNIYYGSYLYSETWNTGIMLLLITMATAFMGYVLPWGQMSFWGATVITNLFSAIPYIGTNLV" var s2 = "EWIWGGFSVDKATLNRFFAFHFILPFTMVALAGVHLTFLHETGSNNPLGLTSDSDKIPFHPYYTIKDFLG" mostFreqKSDF.call(s1, s2, 2, 100) s1 = "abracadabra12121212121abracadabra12121212121" s2 = s1[-1..0] mostFreqKSDF.call(s1, s2, 2, 100)</lang>
- Output:
Sane as Kotlin entry.
- References
- ↑ SEKER, Sadi E.; Altun, Oguz; Ayan, Ugur; Mert, Cihan (2014), "A Novel String Distance Function based on Most Frequent K Characters", wp:International Journal of Machine Learning and Computing (IJMLC), 4, wp:International Association of Computer Science and Information Technology Press (IACSIT Press), pp. 177-183, https://arxiv.org/abs/1401.6596.