Commatizing numbers
You are encouraged to solve this task according to the task description, using any language you may know.
Commatizing numbers (as used here, a handy expedient made-up word) is the act of adding commas to a number (or string), or the numeric part of a larger string.
- Task
Write a function that takes a string as an argument with optional arguments or parameters (the format of parameters/options is left to the programmer) that in general, adds commas (or some other characters, including blanks or tabs) to the first numeric part of a string (if it's suitable for commatizing as per the rules below), and returns that newly commatized string.
Some of the commatizing rules (specified below) are arbitrary, but they'll be a part of this task requirements, if only to make the results consistent amongst national preferences and other disciplines.
The number may be part of a larger (non-numeric) string such as:
- «US$1744 millions» ──or──
- ±25000 motes.
The string may possibly not have a number suitable for commatizing, so it should be untouched and no error generated.
If any argument (option) is invalid, nothing is changed and no error need be generated (quiet execution, no fail execution). Error message generation is optional.
The exponent part of a number is never commatized. The following string isn't suitable for commatizing: 9.7e+12000
Leading zeroes are never commatized. The string 0000000005714.882 after commatization is: 0000000005,714.882
Any period (.) in a number is assumed to be a decimal point.
The original string is never changed except by the addition of commas [or whatever character(s) is/are used for insertion], if at all.
To wit, the following should be preserved:
- leading signs (+, -) ── even superfluous signs
- leading/trailing/embedded blanks, tabs, and other whitespace
- the case (upper/lower) of the exponent indicator, e.g.: 4.8903d-002
Any exponent character(s) should be supported:
- 1247e12
- 57256.1D-4
- 4444^60
- 7500∙10**35
- 8500x10**35
- +55000↑3
- 1000**100
- 2048²
- 409632
- 10000pow(pi)
Numbers may be terminated with any non-digit character, including subscripts and/or superscript: 41421356243 or 7320509076(base 24).
The character(s) to be used for the comma can be specified, and may contain blanks, tabs, and other whitespace characters, as well as multiple characters. The default is the comma (,) character.
The period length can be specified (sometimes referred to as "thousands" or "thousands separators"). The period length can be defined as the length (or number) of the decimal digits between commas. The default period length is 3.
- E.G.: in this example, the period length is five: 56789,12340,14148
The location of where to start the scanning for the target field (the numeric part) should be able to be specified. The default is 1.
The character strings below may be placed in a file (and read) or stored as simple strings within the program.
- Strings to be used as a minimum
The value of pi (expressed in base 10) should be separated with blanks every 5 places past the decimal point,
the Zimbabwe dollar amount should use a decimal point for the "comma" separator:
- pi=3.14159265358979323846264338327950288419716939937510582097494459231
- The author has two Z$100000000000000 Zimbabwe notes (100 trillion).
- "-in Aus$+1411.8millions"
- ===US$0017440 millions=== (in 2000 dollars)
- 123.e8000 is pretty big.
- The land area of the earth is 57268900(29% of the surface) square miles.
- Ain't no numbers in this here words, nohow, no way, Jose.
- James was never known as 0000000007
- Arthur Eddington wrote: I believe there are 15747724136275002577605653961181555468044717914527116709366231425076185631031296 protons in the universe.
- ␢␢␢$-140000±100 millions.
- 6/9/1946 was a good year for some.
where the penultimate string has three leading blanks (real blanks are to be used).
- Also see
- The Wiki entry: Arthur Eddington's number of protons in the universe.
ALGOL 68
<lang algol68># returns text commatized according to the rules of the task and the #
- period, location and separator paramters #
PROC commatize = ( STRING text, INT location, INT period, STRING separator )STRING:
IF STRING str := text[ AT 1 ];
# handle the options #
INT start position := IF location = 0 THEN 1 ELSE location FI;
INT period length := IF period = 0 THEN 3 ELSE period FI;
STRING separator string := IF separator = "" THEN "," ELSE separator FI;
period length < 1 OR start position < 1 OR start position > UPB str
THEN
# invalid parameters - return the text unchanged #
text
ELIF # attempt to find a non-zero digit #
INT number pos := start position;
WHILE IF number pos > UPB str
THEN FALSE
ELSE str[ number pos ] < "1" OR str[ number pos ] > "9"
FI
DO
number pos +:= 1
OD;
number pos > UPB str
THEN # no digits in the string - return the text unchanged #
text
ELSE # have at least one digit #
STRING result := str[ 1 : number pos - 1 ];
# find the final digit #
INT number end := number pos;
WHILE IF number end >= UPB str
THEN FALSE
ELSE str[ number end + 1 ] >= "0" AND str[ number end + 1 ] <= "9"
FI
DO
number end +:= 1
OD;
# copy the digits commatizing as required #
INT digit count := ( number end - number pos ) + 1;
WHILE digit count > 1 DO
result +:= str[ number pos ];
number pos +:= 1;
digit count -:= 1;
IF digit count MOD period length = 0 THEN
# need a comma after this digit #
result +:= separator string
FI
OD;
# final digit and the rest of the string #
result +:= str[ number pos : ];
result
FI # commatize # ;
- modes and operators to allow us to specify optional parameters to the #
- commatizing procedure #
MODE COMMATIZINGOPTIONS = STRUCT( STRING text, INT location, INT period, STRING separator ); PRIO LOCATION = 9; OP LOCATION = ( STRING text, INT location )COMMATIZINGOPTIONS: COMMATIZINGOPTIONS( text, location, 0, "" ); PRIO PERIOD = 9; OP PERIOD = ( STRING text, INT period )COMMATIZINGOPTIONS: COMMATIZINGOPTIONS( text, 0, period, "" ); PRIO SEPARATOR = 9; OP SEPARATOR = ( STRING text, CHAR separator )COMMATIZINGOPTIONS: COMMATIZINGOPTIONS( text, 0, 0, separator ); OP SEPARATOR = ( STRING text, STRING separator )COMMATIZINGOPTIONS: COMMATIZINGOPTIONS( text, 0, 0, separator ); OP LOCATION = ( COMMATIZINGOPTIONS opts, INT location )COMMATIZINGOPTIONS:
COMMATIZINGOPTIONS( text OF opts, location, period OF opts, separator OF opts );
OP PERIOD = ( COMMATIZINGOPTIONS opts, INT period )COMMATIZINGOPTIONS:
COMMATIZINGOPTIONS( text OF opts, location OF opts, period, separator OF opts );
OP SEPARATOR = ( COMMATIZINGOPTIONS opts, CHAR separator )COMMATIZINGOPTIONS:
COMMATIZINGOPTIONS( text OF opts, location OF opts, period OF opts, separator );
OP SEPARATOR = ( COMMATIZINGOPTIONS opts, STRING separator )COMMATIZINGOPTIONS:
COMMATIZINGOPTIONS( text OF opts, location OF opts, period OF opts, separator );
OP COMMATIZE = ( STRING text )STRING: commatize( text, 0, 0, "" ); OP COMMATIZE = ( COMMATIZINGOPTIONS opts )STRING:
commatize( text OF opts, location OF opts, period OF opts, separator OF opts );
- test the commatization procedure and operators #
print( ( COMMATIZE( "pi=3.14159265358979323846264338327950288419716939937510582097494459231" PERIOD 5 SEPARATOR " " LOCATION 6 ),
newline ) );
print( ( COMMATIZE( "The author has two Z$100000000000000 Zimbabwe notes (100 trillion)." SEPARATOR "." ), newline ) ); print( ( COMMATIZE """-in Aus$+1411.8millions""", newline ) ); print( ( COMMATIZE "===US$0017440 millions=== (in 2000 dollars)", newline ) ); print( ( COMMATIZE "123.e8000 is pretty big.", newline ) ); print( ( COMMATIZE "The land area of the earth is 57268900(29% of the surface) square miles.", newline ) ); print( ( COMMATIZE "Ain't no numbers in this here words, nohow, no way, Jose.", newline ) ); print( ( COMMATIZE "James was never known as 0000000007", newline ) ); print( ( COMMATIZE "Arthur Eddington wrote: I believe there are 15747724136275002577605653961181555468044717914527116709366231425076185631031296 protons in the universe.",
newline ) );
print( ( COMMATIZE " $-140000±100 millions.", newline ) ); print( ( COMMATIZE "6/9/1946 was a good year for some.", newline ) )</lang>
- Output:
pi=3.14159 26535 89793 23846 26433 83279 50288 41971 69399 37510 58209 74944 59231 The author has two Z$100.000.000.000.000 Zimbabwe notes (100 trillion). "-in Aus$+1,411.8millions" ===US$0017,440 millions=== (in 2000 dollars) 123.e8000 is pretty big. The land area of the earth is 57,268,900(29% of the surface) square miles. Ain't no numbers in this here words, nohow, no way, Jose. James was never known as 0000000007 Arthur Eddington wrote: I believe there are 15,747,724,136,275,002,577,605,653,961,181,555,468,044,717,914,527,116,709,366,231,425,076,185,631,031,296 protons in the universe. $-140,000±100 millions. 6/9/1946 was a good year for some.
C#
<lang csharp> static string[] inputs = { "pi=3.14159265358979323846264338327950288419716939937510582097494459231", "The author has two Z$100000000000000 Zimbabwe notes (100 trillion).", "\"-in Aus$+1411.8millions\"", "===US$0017440 millions=== (in 2000 dollars)" };
void Main() { inputs.Select(s => Commatize(s, 0, 3, ","))
.ToList()
.ForEach(Console.WriteLine);
}
string Commatize(string text, int startPosition, int interval, string separator) { var matches = Regex.Matches(text.Substring(startPosition), "[0-9]*"); var x = matches.Cast<Match>().Select(match => Commatize(match, interval, separator, text)).ToList(); return string.Join("", x); }
string Commatize(Match match, int interval, string separator, string original)
{
if (match.Length <= interval)
return original.Substring(match.Index,
match.Index == original.Length ? 0 : Math.Max(match.Length, 1));
return string.Join(separator, match.Value.Split(interval)); }
public static class Extension { public static string[] Split(this string source, int interval) { return SplitImpl(source, interval).ToArray(); }
static IEnumerable<string>SplitImpl(string source, int interval) { for (int i = 1; i < source.Length; i++) { if (i % interval != 0) continue;
yield return source.Substring(i - interval, interval); } } } </lang>
D
Better to have more tests than more features. <lang d>import std.stdio, std.regex, std.range;
auto commatize(in char[] txt, in uint start=0, in uint step=3,
in string ins=",") @safe
in {
assert(step > 0);
} body {
if (start > txt.length || step > txt.length)
return txt;
// First number may begin with digit or decimal point. Exponents ignored.
enum decFloField = ctRegex!("[0-9]*\\.[0-9]+|[0-9]+");
auto matchDec = matchFirst(txt[start .. $], decFloField);
if (!matchDec)
return txt;
// Within a decimal float field:
// A decimal integer field to commatize is positive and not after a point.
enum decIntField = ctRegex!("(?<=\\.)|[1-9][0-9]*");
// A decimal fractional field is preceded by a point, and is only digits.
enum decFracField = ctRegex!("(?<=\\.)[0-9]+");
return txt[0 .. start] ~ matchDec.pre ~ matchDec.hit
.replace!(m => m.hit.retro.chunks(step).join(ins).retro)(decIntField)
.replace!(m => m.hit.chunks(step).join(ins))(decFracField)
~ matchDec.post;
}
unittest {
// An attempted solution may have one or more of the following errors:
// ignoring a number that has only zero before its decimal point
assert("0.0123456".commatize == "0.012,345,6");
// commatizing numbers other than the first
assert("1000 2.3000".commatize == "1,000 2.3000");
// only commatizing in one direction from the decimal point
assert("0001123.456789".commatize == "0001,123.456,789");
// detecting prefixes such as "Z$" requires detecting other prefixes
assert(" NZ$300000".commatize == " NZ$300,000");
// detecting a decimal field that isn't attached to the first number
assert(" 2600 and .0125".commatize == " 2,600 and .0125");
// ignoring the start value, or confusing base 0 (used here) with base 1
assert("1 77000".commatize(1) == "1 77,000");
// ignoring a number that begins with a point, or treating it as integer
assert(" .0104004".commatize == " .010,400,4");
}
void main() {
"pi=3.14159265358979323846264338327950288419716939937510582097494459231"
.commatize(0, 5, " ").writeln;
"The author has two Z$100000000000000 Zimbabwe notes (100 trillion)."
.commatize(0, 3, ".").writeln;
foreach (const line; "commatizing_numbers_using_defaults.txt".File.byLine)
line.commatize.writeln;
}</lang>
- Output:
pi=3.14159 26535 89793 23846 26433 83279 50288 41971 69399 37510 58209 74944 59231 The author has two Z$100.000.000.000.000 Zimbabwe notes (100 trillion). "-in Aus$+1,411.8millions" ===US$0017,440 millions=== (in 2000 dollars) 123.e8000 is pretty big. The land area of the earth is 57,268,900(29% of the surface) square miles. Ain't no numbers in this here words, nohow, no way, Jose. James was never known as 0000000007 Arthur Eddington wrote: I believe there are 15,747,724,136,275,002,577,605,653,961,181,555,468,044,717,914,527,116,709,366,231,425,076,185,631,031,296 protons in the universe. $-140,000±100 millions. 6/9/1946 was a good year for some.
Go
<lang go>package main
import (
"fmt" "regexp" "strings"
)
var reg = regexp.MustCompile(`(\.[0-9]+|[1-9]([0-9]+)?(\.[0-9]+)?)`)
func reverse(s string) string {
r := []rune(s)
for i, j := 0, len(r)-1; i < len(r)/2; i, j = i+1, j-1 {
r[i], r[j] = r[j], r[i]
}
return string(r)
}
func commatize(s string, startIndex, period int, sep string) string {
if startIndex < 0 || startIndex >= len(s) || period < 1 || sep == "" {
return s
}
m := reg.FindString(s[startIndex:]) // this can only contain ASCII characters
if m == "" {
return s
}
splits := strings.Split(m, ".")
ip := splits[0]
if len(ip) > period {
pi := reverse(ip)
for i := (len(ip) - 1) / period * period; i >= period; i -= period {
pi = pi[:i] + sep + pi[i:]
}
ip = reverse(pi)
}
if strings.Contains(m, ".") {
dp := splits[1]
if len(dp) > period {
for i := (len(dp) - 1) / period * period; i >= period; i -= period {
dp = dp[:i] + sep + dp[i:]
}
}
ip += "." + dp
}
return s[:startIndex] + strings.Replace(s[startIndex:], m, ip, 1)
}
func main() {
tests := [...]string{
"123456789.123456789",
".123456789",
"57256.1D-4",
"pi=3.14159265358979323846264338327950288419716939937510582097494459231",
"The author has two Z$100000000000000 Zimbabwe notes (100 trillion).",
"-in Aus$+1411.8millions",
"===US$0017440 millions=== (in 2000 dollars)",
"123.e8000 is pretty big.",
"The land area of the earth is 57268900(29% of the surface) square miles.",
"Ain't no numbers in this here words, nohow, no way, Jose.",
"James was never known as 0000000007",
"Arthur Eddington wrote: I believe there are " +
"15747724136275002577605653961181555468044717914527116709366231425076185631031296" +
" protons in the universe.",
" $-140000±100 millions.",
"6/9/1946 was a good year for some.",
}
fmt.Println(commatize(tests[0], 0, 2, "*"))
fmt.Println(commatize(tests[1], 0, 3, "-"))
fmt.Println(commatize(tests[2], 0, 4, "__"))
fmt.Println(commatize(tests[3], 0, 5, " "))
fmt.Println(commatize(tests[4], 0, 3, "."))
for _, test := range tests[5:] {
fmt.Println(commatize(test, 0, 3, ","))
}
}</lang>
- Output:
1*23*45*67*89.12*34*56*78*9 .123-456-789 5__7256.1D-4 pi=3.14159 26535 89793 23846 26433 83279 50288 41971 69399 37510 58209 74944 59231 The author has two Z$100.000.000.000.000 Zimbabwe notes (100 trillion). -in Aus$+1,411.8millions ===US$0017,440 millions=== (in 2000 dollars) 123.e8000 is pretty big. The land area of the earth is 57,268,900(29% of the surface) square miles. Ain't no numbers in this here words, nohow, no way, Jose. James was never known as 0000000007 Arthur Eddington wrote: I believe there are 15,747,724,136,275,002,577,605,653,961,181,555,468,044,717,914,527,116,709,366,231,425,076,185,631,031,296 protons in the universe. $-140,000±100 millions. 6/9/1946 was a good year for some.
J
These rules are relatively baroque, which demands long names and minimally complex statements, thus:
<lang J>require'regex' commatize=:3 :0"1 L:1 0
(i.0) commatize y
NB. deal with all those rules about options
opts=. boxopen x
char=. (#~ ' '&=@{.@(0&#)@>) opts
num=. ;opts-.char
delim=. 0 {:: char,<','
'begin period'=. _1 0+2{.num,(#num)}.1 3
NB. initialize
prefix=. begin {.y
text=. begin }. y
NB. process
'start len'=. ,'[1-9][0-9]*' rxmatch text
if.0=len do. y return. end.
number=. (start,:len) [;.0 text
numb=. (>:period|<:#number){.number
fixed=. numb,;delim&,each (-period)<\ (#numb)}.number
prefix,(start{.text),fixed,(start+len)}.text
)</lang>
In use, this might look like:
<lang J> (5;5;' ') commatize 'pi=3.14159265358979323846264338327950288419716939937510582097494459231' pi=3.14159 26535 89793 23846 26433 83279 50288 41971 69399 37510 58209 74944 59231
'.' commatize 'The author has two Z$100000000000000 Zimbabwe notes (100 trillion).'
The author has two Z$100.000.000.000.000 Zimbabwe notes (100 trillion).
commatize '-in Aus$+1411.8millions'
-in Aus$+1,411.8millions
commatize '===US$0017440 millions=== (in 2000 dollars)'
===US$0017,440 millions=== (in 2000 dollars)
commatize '123.e8000 is pretty big.'
123.e8000 is pretty big.
commatize 'The land area of the earth is 57268900(29% of the surface) square miles.'
The land area of the earth is 57,268,900(29% of the surface) square miles.
commatize 'Aint no numbers in this here words, nohow, no way, Jose.'
Ain't no numbers in this here words, nohow, no way, Jose.
commatize 'James was never known as 0000000007'
James was never known as 0000000007
commatize 'Arthur Eddington wrote: I believe there are 15747724136275002577605653961181555468044717914527116709366231425076185631031296 protons in the universe.'
Arthur Eddington wrote: I believe there are 15,747,724,136,275,002,577,605,653,961,181,555,468,044,717,914,527,116,709,366,231,425,076,185,631,031,296 protons in the universe.
commatize ' $-140000±100 millions.' $-140,000±100 millions. commatize '6/9/1946 was a good year for some.'
6/9/1946 was a good year for some.</lang>
Java
<lang java>import java.io.File; import java.util.*; import java.util.regex.*;
public class CommatizingNumbers {
public static void main(String[] args) throws Exception {
commatize("pi=3.14159265358979323846264338327950288419716939937510582"
+ "097494459231", 6, 5, " ");
commatize("The author has two Z$100000000000000 Zimbabwe notes (100 "
+ "trillion).", 0, 3, ".");
try (Scanner sc = new Scanner(new File("input.txt"))) {
while(sc.hasNext())
commatize(sc.nextLine());
}
}
static void commatize(String s) {
commatize(s, 0, 3, ",");
}
static void commatize(String s, int start, int step, String ins) {
if (start < 0 || start > s.length() || step < 1 || step > s.length())
return;
Matcher m = Pattern.compile("([1-9][0-9]*)").matcher(s.substring(start));
StringBuffer result = new StringBuffer(s.substring(0, start));
if (m.find()) {
StringBuilder sb = new StringBuilder(m.group(1)).reverse();
for (int i = step; i < sb.length(); i += step)
sb.insert(i++, ins);
m.appendReplacement(result, sb.reverse().toString());
}
System.out.println(m.appendTail(result)); }
}</lang>
pi=3.14159 26535 89793 23846 26433 83279 50288 41971 69399 37510 58209 74944 59231 The author has two Z$100.000.000.000.000 Zimbabwe notes (100 trillion). "-in Aus$+1,411.8millions" ===US$0017,440 millions=== (in 2000 dollars) 123.e8000 is pretty big. The land area of the earth is 57,268,900(29% of the surface) square miles. Ain't no numbers in this here words, nohow, no way, Jose. James was never known as 0000000007 Arthur Eddington wrote: I believe there are 15,747,724,136,275,002,577,605,653,961,181,555,468,044,717,914,527,116,709,366,231,425,076,185,631,031,296 protons in the universe. $-140,000±100 millions. 6/9/1946 was a good year for some.
Julia
<lang julia>input = [
["pi=3.14159265358979323846264338327950288419716939937510582097494459231", " ", 5], [raw"The author has two Z$100000000000000 Zimbabwe notes (100 trillion).", "."], [raw"-in Aus$+1411.8millions"], [raw"===US$0017440 millions=== (in 2000 dollars)"], ["123.e8000 is pretty big."], ["The land area of the earth is 57268900(29% of the surface) square miles."], ["Ain\'t no numbers in this here words, nohow, no way, Jose."], ["James was never known as 0000000007"], ["Arthur Eddington wrote: I believe there are 15747724136275002577605653961181555468044717914527116709366231425076185631031296 protons in the universe."], [raw" $-140000±100 millions."], ["6/9/1946 was a good year for some."]]
function commatize(tst)
grouping = (length(tst) == 3) ? tst[3] : 3
sep = (length(tst) > 1) ? tst[2] : ","
rmend(s) = replace(s, Regex("$sep\\Z") =>"")
greg = Regex(".{$grouping}")
cins(str) = reverse(rmend(replace(reverse(str), greg => s -> s * sep)))
mat = match(Regex("(?<![eE\\/])([1-9]\\d{$grouping,})"), tst[1])
if mat != nothing
return replace(tst[1], mat.match => cins)
end
return tst[1]
end
for tst in input
println(commatize(tst))
end
</lang>
- Output:
pi=3.14159 26535 89793 23846 26433 83279 50288 41971 69399 37510 58209 74944 59231 The author has two Z$100.000.000.000.000 Zimbabwe notes (100 trillion). -in Aus$+1,411.8millions ===US$0017,440 millions=== (in 2000 dollars) 123.e8000 is pretty big. The land area of the earth is 57,268,900(29% of the surface) square miles. Ain't no numbers in this here words, nohow, no way, Jose. James was never known as 0000000007 Arthur Eddington wrote: I believe there are 15,747,724,136,275,002,577,605,653,961,181,555,468,044,717,914,527,116,709,366,231,425,076,185,631,031,296 protons in the universe.
$-140,000±100 millions.6/9/1946 was a good year for some.
Kotlin
<lang scala>// version 1.1.4-3
val r = Regex("""(\.[0-9]+|[1-9]([0-9]+)?(\.[0-9]+)?)""")
fun String.commatize(startIndex: Int = 0, period: Int = 3, sep: String = ","): String {
if ((startIndex !in 0 until this.length) || period < 1 || sep == "") return this
val m = r.find(this, startIndex)
if (m == null) return this
val splits = m.value.split('.')
var ip = splits[0]
if (ip.length > period) {
val sb = StringBuilder(ip.reversed())
for (i in (ip.length - 1) / period * period downTo period step period) {
sb.insert(i, sep)
}
ip = sb.toString().reversed()
}
if ('.' in m.value) {
var dp = splits[1]
if (dp.length > period) {
val sb2 = StringBuilder(dp)
for (i in (dp.length - 1) / period * period downTo period step period) {
sb2.insert(i, sep)
}
dp = sb2.toString()
}
ip += "." + dp
}
return this.take(startIndex) + this.drop(startIndex).replaceFirst(m.value, ip)
}
fun main(args: Array<String>) {
val tests = arrayOf(
"123456789.123456789",
".123456789",
"57256.1D-4",
"pi=3.14159265358979323846264338327950288419716939937510582097494459231",
"The author has two Z$100000000000000 Zimbabwe notes (100 trillion).",
"-in Aus$+1411.8millions",
"===US$0017440 millions=== (in 2000 dollars)",
"123.e8000 is pretty big.",
"The land area of the earth is 57268900(29% of the surface) square miles.",
"Ain't no numbers in this here words, nohow, no way, Jose.",
"James was never known as 0000000007",
"Arthur Eddington wrote: I believe there are " +
"15747724136275002577605653961181555468044717914527116709366231425076185631031296" +
" protons in the universe.",
" $-140000±100 millions.",
"6/9/1946 was a good year for some."
)
println(tests[0].commatize(period = 2, sep = "*")) println(tests[1].commatize(period = 3, sep = "-")) println(tests[2].commatize(period = 4, sep = "__")) println(tests[3].commatize(period = 5, sep = " ")) println(tests[4].commatize(sep = ".")) for (test in tests.drop(5)) println(test.commatize())
}</lang>
- Output:
1*23*45*67*89.12*34*56*78*9 .123-456-789 5__7256.1D-4 pi=3.14159 26535 89793 23846 26433 83279 50288 41971 69399 37510 58209 74944 59231 The author has two Z$100.000.000.000.000 Zimbabwe notes (100 trillion). -in Aus$+1,411.8millions ===US$0017,440 millions=== (in 2000 dollars) 123.e8000 is pretty big. The land area of the earth is 57,268,900(29% of the surface) square miles. Ain't no numbers in this here words, nohow, no way, Jose. James was never known as 0000000007 Arthur Eddington wrote: I believe there are 15,747,724,136,275,002,577,605,653,961,181,555,468,044,717,914,527,116,709,366,231,425,076,185,631,031,296 protons in the universe. $-140,000±100 millions. 6/9/1946 was a good year for some.
Perl
Displaying before/after only when changes applied. <lang perl>@input = (
['pi=3.14159265358979323846264338327950288419716939937510582097494459231', ' ', 5], ['The author has two Z$100000000000000 Zimbabwe notes (100 trillion).', '.'], ['-in Aus$+1411.8millions'], ['===US$0017440 millions=== (in 2000 dollars)'], ['123.e8000 is pretty big.'], ['The land area of the earth is 57268900(29% of the surface) square miles.'], ['Ain\'t no numbers in this here words, nohow, no way, Jose.'], ['James was never known as 0000000007'], ['Arthur Eddington wrote: I believe there are 15747724136275002577605653961181555468044717914527116709366231425076185631031296 protons in the universe.'], [' $-140000±100 millions.'], ['5/9/1946 was a good year for some.']
);
for $i (@input) {
$old = @$i[0];
$new = commatize(@$i);
printf("%s\n%s\n\n", $old, $new) if $old ne $new;
}
sub commatize {
my($str,$sep,$by) = @_; $sep = ',' unless $sep; $by = 3 unless $by;
$str =~ s/ # matching rules:
(?<![eE\/]) # not following these characters
([1-9]\d{$by,}) # leading non-zero digit, minimum number of digits required
/c_ins($1,$by,$sep)/ex; # substitute matched text with subroutine output
return $str;
}
sub c_ins {
my($s,$by,$sep) = @_;
($c = reverse $s) =~ s/(.{$by})/$1$sep/g;
$c =~ s/$sep$//;
return reverse $c;
}</lang>
- Output:
pi=3.14159265358979323846264338327950288419716939937510582097494459231 pi=3.14159 26535 89793 23846 26433 83279 50288 41971 69399 37510 58209 74944 59231 The author has two Z$100000000000000 Zimbabwe notes (100 trillion). The author has two Z$100.000.000.000.000 Zimbabwe notes (100 trillion). -in Aus$+1411.8millions -in Aus$+1,411.8millions ===US$0017440 millions=== (in 2000 dollars) ===US$0017,440 millions=== (in 2000 dollars) The land area of the earth is 57268900(29% of the surface) square miles. The land area of the earth is 57,268,900(29% of the surface) square miles. Arthur Eddington wrote: I believe there are 15747724136275002577605653961181555468044717914527116709366231425076185631031296 protons in the universe. Arthur Eddington wrote: I believe there are 15,747,724,136,275,002,577,605,653,961,181,555,468,044,717,914,527,116,709,366,231,425,076,185,631,031,296 protons in the universe. $-140000±100 millions. $-140,000±100 millions.
Perl 6
<lang perl6>for ('pi=3.14159265358979323846264338327950288419716939937510582097494459231', {:6at, :5by, :ins(' ')}),
('The author has two Z$100000000000000 Zimbabwe notes (100 trillion).', {:ins<.>}),
'-in Aus$+1411.8millions',
'===US$0017440 millions=== (in 2000 dollars)',
'123.e8000 is pretty big.',
'The land area of the earth is 57268900(29% of the surface) square miles.',
'Ain\'t no numbers in this here words, nohow, no way, Jose.',
'James was never known as 0000000007',
'Arthur Eddington wrote: I believe there are 15747724136275002577605653961181555468044717914527116709366231425076185631031296 protons in the universe.',
' $-140000±100 millions.',
'6/9/1946 was a good year for some.'
{
say "Before: ", .[0];
say " After: ", .[1] ?? .[0].&commatize( |.[1] ) !! .&commatize;
}
sub commatize($s, :$at = 0, :$ins = ',', :$by = 3) {
$s.subst: :continue($at), :1st, / <[1..9]> <[0..9]>* /,
*.flip.comb(/<{ ".**1..$by" }>/).join($ins).flip;
}</lang>
- Output:
Before: pi=3.14159265358979323846264338327950288419716939937510582097494459231 After: pi=3.14159 26535 89793 23846 26433 83279 50288 41971 69399 37510 58209 74944 59231 Before: The author has two Z$100000000000000 Zimbabwe notes (100 trillion). After: The author has two Z$100.000.000.000.000 Zimbabwe notes (100 trillion). Before: -in Aus$+1411.8millions After: -in Aus$+1,411.8millions Before: ===US$0017440 millions=== (in 2000 dollars) After: ===US$0017,440 millions=== (in 2000 dollars) Before: 123.e8000 is pretty big. After: 123.e8000 is pretty big. Before: The land area of the earth is 57268900(29% of the surface) square miles. After: The land area of the earth is 57,268,900(29% of the surface) square miles. Before: Ain't no numbers in this here words, nohow, no way, Jose. After: Ain't no numbers in this here words, nohow, no way, Jose. Before: James was never known as 0000000007 After: James was never known as 0000000007 Before: Arthur Eddington wrote: I believe there are 15747724136275002577605653961181555468044717914527116709366231425076185631031296 protons in the universe. After: Arthur Eddington wrote: I believe there are 15,747,724,136,275,002,577,605,653,961,181,555,468,044,717,914,527,116,709,366,231,425,076,185,631,031,296 protons in the universe. Before: $-140000±100 millions. After: $-140,000±100 millions. Before: 6/9/1946 was a good year for some. After: 6/9/1946 was a good year for some.
Phix
Note that printf() has comma handling built in, fro example sprintf("%,d",1234) yields "1,234". <lang Phix>procedure commatize(string s, string sep=",", integer start=1, integer step=3) integer l = length(s)
for i=start to l do
if find(s[i],"123456789") then
for j=i+1 to l+1 do
if j>l or not find(s[j],"0123456789") then
for k=j-1-step to i by -step do
s[k+1..k] = sep
end for
exit
end if
end for
exit
end if
end for
printf(1,"%s\n",{s})
end procedure
commatize("pi=3.14159265358979323846264338327950288419716939937510582097494459231"," ",6,5) commatize("The author has two Z$100000000000000 Zimbabwe notes (100 trillion).",".") commatize("\"-in Aus$+1411.8millions\"") commatize("===US$0017440 millions=== (in 2000 dollars)") commatize("123.e8000 is pretty big.") commatize("The land area of the earth is 57268900(29% of the surface) square miles.") commatize("Ain't no numbers in this here words, nohow, no way, Jose.") commatize("James was never known as 0000000007") commatize("Arthur Eddington wrote: I believe there are 15747724136275002577605653961181555468044717914527116709366231425076185631031296 protons in the universe.") commatize(" $-140000±100 millions.") commatize("6/9/1946 was a good year for some.")</lang>
- Output:
pi=3.14159 26535 89793 23846 26433 83279 50288 41971 69399 37510 58209 74944 59231 The author has two Z$100.000.000.000.000 Zimbabwe notes (100 trillion). "-in Aus$+1,411.8millions" ===US$0017,440 millions=== (in 2000 dollars) 123.e8000 is pretty big. The land area of the earth is 57,268,900(29% of the surface) square miles. Ain't no numbers in this here words, nohow, no way, Jose. James was never known as 0000000007 Arthur Eddington wrote: I believe there are 15,747,724,136,275,002,577,605,653,961,181,555,468,044,717,914,527,116,709,366,231,425,076,185,631,031,296 protons in the universe. $-140,000±100 millions. 6/9/1946 was a good year for some.
Python
<lang Python> import re as RegEx
def Commatize( _string, _startPos=0, _periodLen=3, _separator="," ):
outString = ""
strPos = 0
matches = RegEx.findall( "[0-9]*", _string )
for match in matches[:-1]: if not match: outString += _string[ strPos ] strPos += 1 else: if len(match) > _periodLen: leadIn = match[:_startPos] periods = [ match [ i:i + _periodLen ] for i in range ( _startPos, len ( match ), _periodLen ) ] outString += leadIn + _separator.join( periods ) else: outString += match
strPos += len( match )
return outString
print ( Commatize( "pi=3.14159265358979323846264338327950288419716939937510582097494459231", 0, 5, " " ) ) print ( Commatize( "The author has two Z$100000000000000 Zimbabwe notes (100 trillion).", 0, 3, "." )) print ( Commatize( "\"-in Aus$+1411.8millions\"" )) print ( Commatize( "===US$0017440 millions=== (in 2000 dollars)" )) print ( Commatize( "123.e8000 is pretty big." )) print ( Commatize( "The land area of the earth is 57268900(29% of the surface) square miles." )) print ( Commatize( "Ain't no numbers in this here words, nohow, no way, Jose." )) print ( Commatize( "James was never known as 0000000007" )) print ( Commatize( "Arthur Eddington wrote: I believe there are 15747724136275002577605653961181555468044717914527116709366231425076185631031296 protons in the universe." )) print ( Commatize( "␢␢␢$-140000±100 millions." )) print ( Commatize( "6/9/1946 was a good year for some." )) </lang>
Racket
Note that the post-number part of the split catches the date in 6/9/1946 (I think that's
desirable) but 1946-09-06 would still be commatized as 1,946-09-06 -- it
would take date recognition to capture that case.
All tests pass (so it's as good as Perl, I guess).
<lang racket>#lang racket (require (only-in srfi/13 [string-reverse gnirts]))
- returns a string with the "comma"s inserted every step characters from the RIGHT of n.
- because of the right handedness of this, there is a lot of reversal going on
(define ((insert-commas comma step) n)
(define px (pregexp (format ".{1,~a}" step)))
(string-join (add-between (reverse (map gnirts (regexp-match* px (gnirts n)))) comma) ""))
(define (commatize s #:start (start 0) #:comma (comma ",") #:step (step 3))
(define ins-comms (insert-commas comma step)) ; specific to our comma and step
(define split-into-numbers
(match-lambda
[(regexp
#px"^([^1-9]*)([1-9][0-9.]*)(\\S*)(.*)$" ; see below for description of bits
(list _ ; the whole match
(app split-into-numbers pre) ; recur on left
num ; the number bit before any exponent or other
; interestingness
post-number ; from exponent to the first space
(app split-into-numbers post))) ; recur on right
(define skip (substring num 0 start))
(match-define
(regexp #px"^(.*?)(\\..*)?$"
(list _ ; whole match
(app ins-comms n) ; the bit that gets the commas added
(or (? string? d) ; if it matches, then the raw string is in d
(and #f (app (lambda (f) "") d))))) ; if (...)? doesn't match it returns
; #f which we thunk to an empty string
(substring num start)) ; do the match on the unskipped bit
(string-append pre skip n d post-number post)] ; stitch it back together
[else else])) ; if it doesn't match leave as is
;; kick it off
(split-into-numbers s))
(module+ test
(require tests/eli-tester)
(test
(commatize "pi=3.14159265358979323846264338327950288419716939937510582097494459231"
#:start 6 #:comma " " #:step 5)
=>"pi=3.14159 26535 89793 23846 26433 83279 50288 41971 69399 37510 58209 74944 59231"
(commatize "The author has two Z$100000000000000 Zimbabwe notes (100 trillion)." #:comma ".")
=>"The author has two Z$100.000.000.000.000 Zimbabwe notes (100 trillion)."
(commatize "-in Aus$+1411.8millions")
=>"-in Aus$+1,411.8millions"
(commatize "===US$0017440 millions=== (in 2000 dollars)")
=>"===US$0017,440 millions=== (in 2,000 dollars)"
(commatize "123.e8000 is pretty big.")
=>"123.e8000 is pretty big."
(commatize "The land area of the earth is 57268900(29% of the surface) square miles.")
=>"The land area of the earth is 57,268,900(29% of the surface) square miles."
(commatize "Ain't no numbers in this here words, nohow, no way, Jose.")
=>"Ain't no numbers in this here words, nohow, no way, Jose."
(commatize "James was never known as 0000000007")
=>"James was never known as 0000000007"
(commatize "Arthur Eddington wrote: I believe there are 15747724136275002577605653961181555468044717914527116709366231425076185631031296 protons in the universe.")
=>"Arthur Eddington wrote: I believe there are 15,747,724,136,275,002,577,605,653,961,181,555,468,044,717,914,527,116,709,366,231,425,076,185,631,031,296 protons in the universe."
(commatize " $-140000±100 millions.")
=>" $-140,000±100 millions."
(commatize "6/9/1946 was a good year for some.")
=>"6/9/1946 was a good year for some."))
</lang>
REXX
The hardest part of the comma function is to locate where a usable number starts and ends. <lang rexx>/*REXX program add commas (or other chars) to a number within a string (or a char str).*/ @. = @.1= "pi=3.14159265358979323846264338327950288419716939937510582097494459231" @.2= "The author has two Z$100000000000000 Zimbabwe notes (100 trillion)." @.3= "-in Aus$+1411.8millions" @.4= "===US$0017440 millions=== (in 2000 dollars)" @.5= "123.e8000 is pretty big." @.6= "The land area of the earth is 57268900(29% of the surface) square miles." @.7= "Ain't no numbers in this here words, nohow, no way, Jose." @.8= "James was never known as 0000000007" @.9= "Arthur Eddington wrote: I believe there are 15747724136275002577605653961181555468044717914527116709366231425076185631031296 protons in the universe." @.10= " $-140000±100 millions." @.11= "6/9/1946 was a good year for some."
do i=1 while @.i\==; if i\==1 then say /*process each string.*/
say 'before──►'@.i /*show the before str.*/
if i==1 then say ' after──►'comma(@.i, 'blank', 5, , 6) /* p=5, start=6. */
if i==2 then say ' after──►'comma(@.i, ".") /*comma=decimal point.*/
if i>2 then say ' after──►'comma(@.i) /*use the defaults. */
end /*j*/
exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ comma: procedure; parse arg _,c,p,t,s /*obtain the number & optional options.*/
arg ,cU . /*obtain an uppercase version of C. */
c=word(c ',', 1) /*obtain the commatizing character(s).*/
if cU=='BLANK' then c=' ' /*special case for a "blank" separator.*/
o=word(p 3, 1) /*obtain the optional period length. */
p=abs(o) /*obtain the positive period length. */
t=word(t 999999999, 1) /*obtain max # of "commas" to insert. */
s=word(s 1, 1) /*obtain the optional start position.*/
if \datatype(p, 'W') | \datatype(t, "W") | \datatype(s, 'W') | ,
t<1 | s<1 | p==0 | arg()>5 then return _ /*any invalid options? */
n=_'.9'; #=123456789; k=0 /*define some handy-dandy variables. */
if o<0 then do /*using a negative period length ? */
b=verify(_, ' ', , s) /*position of first blank in string. */
e=length(_) - verify( reverse(_), ' ') + 1 - p
end
else do /*using a positive period length. */
b=verify(n, #, 'M', s) /*position of first useable decimal dig*/
z=max(1, verify(n, #'0.', "M", s)) /* " " last " "*/
e=verify(n, #'0', , max(1, verify(n, #"0.", 'M', s) ) ) - p - 1
end
if e>0 & b>0 then do j=e to b by -p while k<t /*commatize the digits.*/
_=insert(c, _, j) /*comma spray ───► #.*/
k= k + 1 /*bump the commatizing.*/
end /*j*/
return _</lang>
- output when using the internal default inputs:
before──►pi=3.14159265358979323846264338327950288419716939937510582097494459231 after──►pi=3.14159 26535 89793 23846 26433 83279 50288 41971 69399 37510 58209 74944 59231 before──►The author has two Z$100000000000000 Zimbabwe notes (100 trillion). after──►The author has two Z$100.000.000.000.000 Zimbabwe notes (100 trillion). before──►-in Aus$+1411.8millions after──►-in Aus$+1,411.8millions before──►===US$0017440 millions=== (in 2000 dollars) after──►===US$0017,440 millions=== (in 2000 dollars) before──►123.e8000 is pretty big. after──►123.e8000 is pretty big. before──►The land area of the earth is 57268900(29% of the surface) square miles. after──►The land area of the earth is 57,268,900(29% of the surface) square miles. before──►Ain't no numbers in this here words, nohow, no way, Jose. after──►Ain't no numbers in this here words, nohow, no way, Jose. before──►James was never known as 0000000007 after──►James was never known as 0000000007 before──►Arthur Eddington wrote: I believe there are 15747724136275002577605653961181555468044717914527116709366231425076185631031296 protons in the universe. after──►Arthur Eddington wrote: I believe there are 15,747,724,136,275,002,577,605,653,961,181,555,468,044,717,914,527,116,709,366,231,425,076,185,631,031,296 protons in the universe. before──► $-140000±100 millions. after──► $-140,000±100 millions. before──►6/9/1946 was a good year for some. after──►6/9/1946 was a good year for some.
Scala
Java-ish version
<lang Scala>import java.io.File import java.util.Scanner import java.util.regex.Pattern
object CommatizingNumbers extends App {
def commatize(s: String): Unit = commatize(s, 0, 3, ",")
def commatize(s: String, start: Int, step: Int, ins: String): Unit = {
if (start >= 0 && start <= s.length && step >= 1 && step <= s.length) {
val m = Pattern.compile("([1-9][0-9]*)").matcher(s.substring(start))
val result = new StringBuffer(s.substring(0, start))
if (m.find) {
val sb = new StringBuilder(m.group(1)).reverse
for (i <- step until sb.length by step) sb.insert(i, ins)
m.appendReplacement(result, sb.reverse.toString)
}
println(m.appendTail(result))
}
}
commatize("pi=3.14159265358979323846264338327950288419716939937510582" + "097494459231", 6, 5, " ")
commatize("The author has two Z$100000000000000 Zimbabwe notes (100 " + "trillion).", 0, 3, ".")
val sc = new Scanner(new File("input.txt"))
while (sc.hasNext) commatize(sc.nextLine)
}</lang>
Swift
<lang Swift>import Foundation
extension String {
private static let commaReg = try! NSRegularExpression(pattern: "(\\.[0-9]+|[1-9]([0-9]+)?(\\.[0-9]+)?)")
public func commatize(start: Int = 0, period: Int = 3, separator: String = ",") -> String {
guard separator != "" else {
return self
}
let sep = Array(separator) let startIdx = index(startIndex, offsetBy: start) let matches = String.commaReg.matches(in: self, range: NSRange(startIdx..., in: self))
guard !matches.isEmpty else {
return self
}
let fullMatch = String(self[Range(matches.first!.range(at: 0), in: self)!]) let splits = fullMatch.components(separatedBy: ".") var ip = splits[0]
if ip.count > period {
var builder = Array(ip.reversed())
for i in stride(from: (ip.count - 1) / period * period, through: period, by: -period) {
builder.insert(contentsOf: sep, at: i)
}
ip = String(builder.reversed()) }
if fullMatch.contains(".") {
var dp = splits[1]
if dp.count > period {
var builder = Array(dp)
for i in stride(from: (dp.count - 1) / period * period, through: period, by: -period) {
builder.insert(contentsOf: sep, at: i)
}
dp = String(builder)
}
ip += "." + dp }
return String(prefix(start)) + String(dropFirst(start)).replacingOccurrences(of: fullMatch, with: ip) }
}
let tests = [
"123456789.123456789",
".123456789",
"57256.1D-4",
"pi=3.14159265358979323846264338327950288419716939937510582097494459231",
"The author has two Z$100000000000000 Zimbabwe notes (100 trillion).",
"-in Aus$+1411.8millions",
"===US$0017440 millions=== (in 2000 dollars)",
"123.e8000 is pretty big.",
"The land area of the earth is 57268900(29% of the surface) square miles.",
"Ain't no numbers in this here words, nohow, no way, Jose.",
"James was never known as 0000000007",
"Arthur Eddington wrote: I believe there are " +
"15747724136275002577605653961181555468044717914527116709366231425076185631031296" +
" protons in the universe.",
" $-140000±100 millions.",
"6/9/1946 was a good year for some."
]
print(tests[0].commatize(period: 2, separator: "*")) print(tests[1].commatize(period: 3, separator: "-")) print(tests[2].commatize(period: 4, separator: "__")) print(tests[3].commatize(period: 5, separator: " ")) print(tests[4].commatize(separator: "."))
for testCase in tests.dropFirst(5) {
print(testCase.commatize())
}</lang>
- Output:
1*23*45*67*89.12*34*56*78*9 .123-456-789 5__7256.1D-4 pi=3.14159 26535 89793 23846 26433 83279 50288 41971 69399 37510 58209 74944 59231 The author has two Z$100.000.000.000.000 Zimbabwe notes (100 trillion). -in Aus$+1,411.8millions ===US$0017,440 millions=== (in 2000 dollars) 123.e8000 is pretty big. The land area of the earth is 57,268,900(29% of the surface) square miles. Ain't no numbers in this here words, nohow, no way, Jose. James was never known as 0000000007 Arthur Eddington wrote: I believe there are 15,747,724,136,275,002,577,605,653,961,181,555,468,044,717,914,527,116,709,366,231,425,076,185,631,031,296 protons in the universe. $-140,000±100 millions. 6/9/1946 was a good year for some.
VBA
<lang vb>Public Sub commatize(s As String, Optional sep As String = ",", Optional start As Integer = 1, Optional step As Integer = 3)
Dim l As Integer: l = Len(s)
For i = start To l
If Asc(Mid(s, i, 1)) >= Asc("1") And Asc(Mid(s, i, 1)) <= Asc("9") Then
For j = i + 1 To l + 1
If j > l Then
For k = j - 1 - step To i Step -step
s = Mid(s, 1, k) & sep & Mid(s, k + 1, l - k + 1)
l = Len(s)
Next k
Exit For
Else
If (Asc(Mid(s, j, 1)) < Asc("0") Or Asc(Mid(s, j, 1)) > Asc("9")) Then
For k = j - 1 - step To i Step -step
s = Mid(s, 1, k) & sep & Mid(s, k + 1, l - k + 1)
l = Len(s)
Next k
Exit For
End If
End If
Next j
Exit For
End If
Next i
Debug.Print s
End Sub
Public Sub main()
commatize "pi=3.14159265358979323846264338327950288419716939937510582097494459231", " ", 6, 5 commatize "The author has two Z$100000000000000 Zimbabwe notes (100 trillion).", "." commatize """-in Aus$+1411.8millions""" commatize "===US$0017440 millions=== (in 2000 dollars)" commatize "123.e8000 is pretty big." commatize "The land area of the earth is 57268900(29% of the surface) square miles." commatize "Ain't no numbers in this here words, nohow, no way, Jose." commatize "James was never known as 0000000007" commatize "Arthur Eddington wrote: I believe there are 15747724136275002577605653961181555468044717914527116709366231425076185631031296 protons in the universe." commatize " $-140000±100 millions." commatize "6/9/1946 was a good year for some."
End Sub</lang>
- Output:
pi=3.14159 26535 89793 23846 26433 83279 50288 41971 69399 37510 58209 74944 59231 The author has two Z$100.000.000.000.000 Zimbabwe notes (100 trillion). "-in Aus$+1,411.8millions" ===US$0017,440 millions=== (in 2000 dollars) 123.e8000 is pretty big. The land area of the earth is 57,268,900(29% of the surface) square miles. Ain't no numbers in this here words, nohow, no way, Jose. James was never known as 0000000007 Arthur Eddington wrote: I believe there are 15,747,724,136,275,002,577,605,653,961,181,555,468,044,717,914,527,116,709,366,231,425,076,185,631,031,296 protons in the universe. $-140,000±100 millions. 6/9/1946 was a good year for some.