Selectively replace multiple instances of a character within a string: Difference between revisions

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(Added XPL0 example.)
(Added Go)
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{{Template:Strings}}
{{Template:Strings}}
<br>
<br>

=={{header|Go}}==
{{trans|Wren}}
<lang go>package main

import (
"fmt"
"strings"
)

func main() {
s := "abracadabra"
ss := []byte(s)
var ixs []int
for ix, c := range s {
if c == 'a' {
ixs = append(ixs, ix)
}
}
repl := "ABaCD"
for i := 0; i < 5; i++ {
ss[ixs[i]] = repl[i]
}
s = string(ss)
s = strings.Replace(s, "b", "E", 1)
s = strings.Replace(s, "r", "F", 2)
s = strings.Replace(s, "F", "r", 1)
fmt.Println(s)
}</lang>

{{out}}
<pre>
AErBcadCbFD
</pre>


=={{header|Raku}}==
=={{header|Raku}}==

Revision as of 14:26, 31 May 2022

Selectively replace multiple instances of a character within a string is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.
Task

This is admittedly a trivial task but I thought it would be interesting to see how succinctly (or otherwise) different languages can handle it.

Given the string: "abracadabra", replace programatically:

  • the first 'a' with 'A'
  • the second 'a' with 'B'
  • the fourth 'a' with 'C'
  • the fifth 'a' with 'D'
  • the first 'b' with 'E'
  • the second 'r' with 'F'


Note that there is no replacement for the third 'a', second 'b' or first 'r'.

The answer should, of course, be : "AErBcadCbFD".

Other tasks related to string operations:
Metrics
Counting
Remove/replace
Anagrams/Derangements/shuffling
Find/Search/Determine
Formatting
Song lyrics/poems/Mad Libs/phrases
Tokenize
Sequences


Go

Translation of: Wren

<lang go>package main

import ( 

   "fmt"
   "strings"

)

func main() { 

   s := "abracadabra"
   ss := []byte(s)
   var ixs []int
   for ix, c := range s {
       if c == 'a' {
           ixs = append(ixs, ix)
       }
   }
   repl := "ABaCD"
   for i := 0; i < 5; i++ {
       ss[ixs[i]] = repl[i]
   }
   s = string(ss)
   s = strings.Replace(s, "b", "E", 1)
   s = strings.Replace(s, "r", "F", 2)
   s = strings.Replace(s, "F", "r", 1)
   fmt.Println(s)

}</lang>

Output:
AErBcadCbFD

Raku

Set up to not particularly rely on absolute structure of the word. Demonstrate with both the original 'abracadabra' and with a random shuffled instance.

<lang perl6>sub mangle ($str is copy) { 

   $str.match(:ex, 'a')».from.map: { $str.substr-rw($_, 1) = 'ABaCD'.comb[$++] };
   $str.=subst('b', 'E');
   $str.substr-rw($_, 1) = 'F' given $str.match(:ex, 'r')».from[1];
   $str

}

say $_, ' -> ', .&mangle given 'abracadabra';

say $_, ' -> ', .&mangle given 'abracadabra'.comb.pick(*).join;</lang>

Output:
abracadabra -> AErBcadCbFD
caarabadrab -> cABraECdFDb

Wren

Library: Wren-seq
Library: Wren-str
Library: Wren-regex

Not particularly succinct but, thanks to a recently added library method, better than it would have been :) <lang ecmascript>import "./seq" for Lst import "./str" for Str

var s = "abracadabra" var sl = s.toList var ixs = Lst.indicesOf(sl, "a")[2] var repl = "ABaCD" for (i in 0..4) sl[ixs[i]] = repl[i] s = sl.join() s = Str.replace(s, "b", "E", 1) s = Str.replace(s, "r", "F", 2, 1) System.print(s)</lang>

Output:
AErBcadCbFD

Alternatively, using regular expressions (embedded script) producing output as before. <lang ecmascript>import "./regex" for Regex

var s = "abracadabra" var split = Regex.compile("a").split(s) var repl = "ABaCD" var res = "" for (i in 0...split.count-1) res = res + split[i] + repl[i] s = res + split[-1] s = Regex.compile("b").replace(s, "E") s = Regex.compile("r").replaceAll(s, "F", 2, 1) System.print(s)</lang>

XPL0

<lang XPL0>string 0; proc Mangle(S); char S, A, B, R; [A:= "ABaCD"; B:= "Eb"; R:= "rF"; while S(0) do 

   [case S(0) of
     ^a: [S(0):= A(0);  A:= A+1];
     ^b: [S(0):= B(0);  B:= B+1];
     ^r: [S(0):= R(0);  R:= R+1]
   other [];
   S:= S+1;
   ];

];

char S; [S:= "abracadabra"; Text(0, S); Text(0, " -> "); Mangle(S); Text(0, S); CrLf(0); S:= "caarabadrab"; Text(0, S); Text(0, " -> "); Mangle(S); Text(0, S); CrLf(0); ]</lang>

Output:
abracadabra -> AErBcadCbFD
caarabadrab -> cABraECdFDb
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