How to replace characters in string with values from array?

Question

I have two arrays.

var a = ['one', 'two', 'three'];
var b = ['two', 'three', 'four'];
var string = 'The only one and two and three';

I tried to use for-loop.

for ( var i = 0; i < string.length; i++) {
    string = string.replace(a[0], b[0]);
    string = string.replace(a[1], b[1]);
    string = string.replace(a[2], b[2]);
}

But the problem is that after first iteration replaced value replaces again! I want to replace one with two, two with three and three with four.

Expected result: The only two and three and four

I get: The only four and four and four

It's a tricky question, but why do you iterate till i < string.length? I think, It should be a.length or b.length. — Selvakumar Arumugam
– Selvakumar Arumugam, Commented Dec 9, 2013 at 18:12

raina77ow · Accepted Answer · 2013-12-09 18:14:01Z

5

One possible approach:

var dict = {};
a.forEach(function(el, i) {
    dict[el] = b[i];
});

var patt = a.join('|');
var res = string.replace(new RegExp(patt, 'g'), function(word) {
    return dict[word];
});
console.log(res); // The only two and three and four

Demo. It's really simple actually: first you create a dictionary (where keys are words to be replaced, and values are, well, words to replace them), second, you create an 'alternation' regex (with | symbol - you'll need to quote metacharacters if there are any). Finally, you go through the string with a single replace using this created pattern - and a replacement function, that looks for a particular 'correction word' in the dictionary.

answered Dec 9, 2013 at 18:14

raina77ow

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3 Comments

raina77ow Over a year ago

An alternative approach would be to reorder the replacements so that the ones in the 'corrections' list will go first. But that way is shaky - cross-references will break it right away.

raina77ow Over a year ago

As for a.join - actually I heard that creating an alternation from a list of words to look for is considered an optimization technique. )

Selvakumar Arumugam Over a year ago

Just posted an alternate approach stackoverflow.com/a/20478025/297641 not sure if I missed anything :)

bfavaretto · Accepted Answer · 2013-12-09 18:13:24Z

3

You don't need a loop, just replace backwards:

var a = ['one', 'two', 'three'];
var b = ['two', 'three', 'four'];
var string = 'The only one and two and three';

string = string.replace(a[2], b[2]);
string = string.replace(a[1], b[1]);
string = string.replace(a[0], b[0]);

Note: this works for this example, it's not generic.

answered Dec 9, 2013 at 18:13

bfavaretto

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3 Comments

Selvakumar Arumugam Over a year ago

Not a generic answer, it relies on data. I think what he posted here is an example of his real issue.

bfavaretto Over a year ago

@Vega The answer states that explicitly.

Selvakumar Arumugam Over a year ago

+1 Not sure how I missed it, but at least you had it as a note.

Selvakumar Arumugam · Accepted Answer · 2013-12-09 18:35:19Z

3

Just posting an alternate approach which splits the original string and replace it from dict object.

The dict object is built before the replace as it is essential to know what is being replaced.

var a = ['one', 'two', 'three'];
var b = ['two', 'three', 'four'];
var string = 'The only one and two and three';

var dict = {};
for (var i = 0; i < a.length; i++) {
    dict[a[i]] = b[i];
}

var stringtokens = string.split(' ');
for (var i = 0; i < stringtokens.length; i++) {
    if (dict.hasOwnProperty(stringtokens[i])){
        stringtokens[i] = dict[stringtokens[i]];
    }
}

console.log(stringtokens.join(' '));

Working fiddle

answered Dec 9, 2013 at 18:35

Selvakumar Arumugam

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3 Comments

omar-ali Over a year ago

OH, I'm sorry! You did exactly what I did and answered, I'll just delete it.

omar-ali Over a year ago

This should be marked as the correct answer!It's the best way, no need to count string lengths or worry about keeping track of what's changed and what's not..

Vlad Holubiev Over a year ago

But what if there are no spaces in input string? I pasted in question simplified code, but in my script I need to replace characters that aren't splited by space. Sure, this is interesting solution, but not in my case.

letiagoalves · Accepted Answer · 2013-12-09 18:21:54Z

2

Do it backwards:

var a = ['one', 'two', 'three'];
var b = ['two', 'three', 'four'];
var string = 'The only one and two and three';

for (var i = string.length-1; i >= 0; i--) {
    string = string.replace(a[i], b[i]);
}

Working demo

edited Dec 9, 2013 at 18:21

answered Dec 9, 2013 at 18:15

letiagoalves

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1 Comment

Blue Skies Over a year ago

Seems like your initial i should be string.length - 1 to keep it in bounds. No harm here, but technically should be changed.

Bohdan Yurov · Accepted Answer · 2013-12-09 18:13:24Z

1

You may reverse every array to achieve this. Another way is regular pattern.

Also, your code makes no sense. Why you iterating through string, if you need to iterate through arrays?

This would work:

for ( var i = 0; i < a.length; i++) {
    string = string.replace(a[a.length - 1 - i], b[b.length - 1 - i]);
}

Also, look at this universal way: http://phpjs.org/functions/str_replace/

You may do so:

str_replace(['{name}', 'l'], ['hello', 'm'], '{name}, lars');

This function can get arrays as parameters.

answered Dec 9, 2013 at 18:13

Bohdan Yurov

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1 Comment

Blue Skies Over a year ago

I think it should be a[a.length - 1 - i], b[b.length - 1 - i]

Red Alert · Accepted Answer · 2013-12-09 18:13:37Z

0

Just do it the other way around. The problem is that after you replace one with two, you replace all two with three, and then you do the same thing with three and four, leaving you will all four at the end. If you reverse the order of your replaces, that won't happen.

answered Dec 9, 2013 at 18:13

Red Alert

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