How do you map-replace characters in Javascript si

问题:

I've been trying to figure out how to map a set of characters in a string to another set similar to the tr function in Perl.

I found this site that shows equivalent functions in JS and Perl, but sadly no tr equivalent.

the tr (transliteration) function in Perl maps characters one to one, so

     data =~ tr|\-_|+/|;

would map

     - => + and _ => /

How can this be done efficiently in JavaScript?

回答1:

There isn't a built-in equivalent, but you can get close to one with replace:

data = data.replace(/[\-_]/g, function (m) {
    return {
        '-': '+',
        '_': '/'
    }[m];
});

回答2:

Method:

String.prototype.mapReplace = function(map) {
    var regex = [];
    for(var key in map)
        regex.push(key.replace(/[\-\[\]\/\{\}\(\)\*\+\?\.\\\^\$\|]/g, "\\$&"));
    return this.replace(new RegExp(regex.join('|'),"g"),function(word){
        return map[word];
    });
};

A perfect example:

var s = "I think Peak rocks!"
s.mapReplace({"I think":"Actually","rocks":"sucks"})
// console: "Actually Peak sucks!"

回答3:

I can't vouch for 'efficient' but this uses a regex and a callback to provide the replacement character.

function tr( text, search, replace ) {
    // Make the search string a regex.
    var regex = RegExp( '[' + search + ']', 'g' );
    var t = text.replace( regex, 
            function( chr ) {
                // Get the position of the found character in the search string.
                var ind = search.indexOf( chr );
                // Get the corresponding character from the replace string.
                var r = replace.charAt( ind );
                return r;
            } );
    return t;
}

For long strings of search and replacement characters, it might be worth putting them in a hash and have the function return from that. ie, tr/abcd/QRST/ becomes the hash { a: Q, b: R, c: S, d: T } and the callback returns hash[ chr ].

回答4:

This will map all as to b and all y to z

var map = { a: 'b', y: 'z' };
var str = 'ayayay';

for (var i = 0; i < str.length; i++)
    str[i] = map[str[i]] || str[i];

EDIT:

Apparently you can't do that with strings. Here's an alternative:

var map = { a: 'b', y: 'z' };
var str = 'ayayay', str2 = [];

for (var i = 0; i < str.length; i++)
    str2.push( map[str[i]] || str[i] );
str2.join('');

回答5:

This functions, which are similar how it's built in Perl.

function s(a, b){ $_ = $_.replace(a, b); }
function tr(a, b){ [...a].map((c, i) => s(new RegExp(c, "g"), b[i])); }

$_ = "Εμπεδοκλης ο Ακραγαντινος";

tr("ΑΒΓΔΕΖΗΘΙΚΛΜΝΟΠΡΣΤΥΦΧΩ", "ABGDEZITIKLMNOPRSTIFHO");
tr("αβγδεζηθικλμνοπρστυφχω", "abgdezitiklmnoprstifho");
s(/Ξ/g, "X"); s(/Ψ/g, "Ps");
s(/ξ/g, "x"); s(/ψ/g, "Ps");
s(/ς/g, "s");

console.log($_);

回答6:

In Perl, one can also write

tr{-_}{+/}

as

my %trans = (
   '-' => '+',
   '_' => '/',
);

my $class = join '', map quotemeta, keys(%trans);
my $re = qr/[$class]/;

s/($re)/$trans{$1}/g;

This latter version can surely be implemented in JS without much trouble.

(My version lacks the duplication of Jonathan Lonowski's solution.)