Explore our sites

Stack Exchange
tour help
Take the 2-minute tour ×
Code Review Stack Exchange is a question and answer site for peer programmer code reviews. It's 100% free, no registration required.
up vote 3 down vote favorite
1

I am using a long array of over 300 var & 7,000 lines of code written like this:

var a = [];
var b = [];
var c = [];
var d = [];
var e = [];

a[0] = "a";
b[0] = "b";
c[0] = "c";
d[0] = "d";
e[0] = "e";

a[1] = "1";
b[1] = "2";
c[1] = "3";
d[1] = "4";
e[1] = "5";

a[2] = "one";
b[2] = "two";
c[2] = "three";
d[2] = "four";
e[2] = "five";

Im guessing it is the same as a much cleaner and shorter - 

var a = [a,1,one];
var b = [b,2,two];
var c = [c,3,three];
var d = [d,4,four];
var e = [e,5,five];

Is there an easy or automatic way to rewrite the original array like the 2nd method?

share|improve this question
 
If you have a finite number of elements that you need to add to an array which follow a pattern, use a for loop. If you don't know how many you're going to need, a while loop will do the trick. Loop through them and use the index to assign the numbers accordingly. –  Jonny Sooter Jun 4 '13 at 16:48
add comment

3 Answers

up vote 4 down vote accepted

First of all I think you meant:

a = ["a","1","one"]; // instead of [a, 1, one]

I handle these rare kind of issues as follow:

  • Open your browser (the following code is tested against Chrome)
  • Open the JavaScript console [Control -Shift -J (Chrome:Windows/Linux)]
  • Copy/paste your own code into the console

and then paste the following code

copy('abcde'.split('').map(function(varName){
return 'var ' + varName + ' = ' + JSON.stringify(window[varName])+';';
}).join('\n'));

If your variable names are more than 1 character use this code instead: (of course the following code won't work with the file example you specified, I changed the variable names for demonstration purpose)

copy(['var1', 'myVar2', 'blablabla'].map(function(varName){
return varName + ' = ' + JSON.stringify(window[varName])+';';
}).join('\n'));
  • Hit enter

the above code will copy the following JavaScript code into your clipboard:

a = ["a","1","one"];
b = ["b","2","two"];
c = ["c","3","three"];
d = ["d","4","four"];
e = ["e","5","five"];

Paste it into your file, save it and that's all!

share|improve this answer
 
wow. Looks good, but it looks like I need to make an array of my vars 1st before I run this. Ill test it out - Thanks –  Jon Jun 5 '13 at 17:35
 
If your variables follow a specified pattern just tell me what it is and I'll see what I can do about it :) –  FGRibreau Jun 6 '13 at 7:58
 
They are product codes so dont really have a pattern. I have them in csv format so it should be easy enough. Thanks –  Jon Jun 6 '13 at 11:33
 
excellent - thanks so much. I went from 180kb file to 80kb, and I think easier to manage file. I dont have var before the variables though. It works but not best practice? –  Jon Jun 6 '13 at 14:59
 
I added 2 things to the explanation so idiots can follow - the shortkey to the JS console, and hit enter at the end! –  Jon Jun 6 '13 at 15:04
show 1 more comment
up vote 1 down vote

If you have a finite number of elements that you need to add to an array which follow a pattern, use a for loop. If you don't know how many you're going to need, a while loop will do the trick. Loop through them and use the index to assign the numbers accordingly.

for (var i = 0; i < 36; i++) {
    var a = [String.fromCharCode(65 + i).toLowerCase(), i+1, toWords(i) ];
};

console.log(a);    

var th = ['','thousand','million', 'billion','trillion'];
var dg = ['zero','one','two','three','four', 'five','six','seven','eight','nine'];
var tn = ['ten','eleven','twelve','thirteen', 'fourteen','fifteen','sixteen', 'seventeen','eighteen','nineteen'];
var tw = ['twenty','thirty','forty','fifty', 'sixty','seventy','eighty','ninety'];

function toWords(s){
    s = s.toString();
    s = s.replace(/[\, ]/g,'');
    if (s != parseFloat(s))
        return 'not a number';
    var x = s.indexOf('.'); 
    if (x == -1) 
        x = s.length; 
    if (x > 15) 
        return 'too big'; 
    var n = s.split(''); 
    var str = ''; 
    var sk = 0; 
    for (var i=0; i < x; i++) {
        if ((x-i)%3==2) {
            if (n[i] == '1') {
                str += tn[Number(n[i+1])] + ' ';
                i++;
                sk=1;
            } else if (n[i]!=0) {
                str += tw[n[i]-2] + ' ';
                sk=1;
            }
        } else if (n[i]!=0) {
            str += dg[n[i]] +' ';
            if ((x-i)%3==0)
                str += 'hundred ';
            sk=1;
        } 
        if ((x-i)%3==1) {
            if (sk)
                str += th[(x-i-1)/3] + ' ';
            sk=0;
        }
    }
    if (x != s.length) {
        var y = s.length;
        str += 'point ';
        for (var i=x+1; i<y; i++)
            str += dg[n[i]] +' ';
    } 
    return str.replace(/\s+/g,' ');
}
share|improve this answer
add comment
up vote 1 down vote

First of all:

var a = [a,1,one];
var b = [b,2,two];
var c = [c,3,three];
var d = [d,4,four];
var e = [e,5,five];

is not the same as:

var a = ["a",1,one];
var b = ["b",2,two];
var c = ["c",3,three];
var d = ["d",4,four];
var e = ["e",5,five];

Second: Technically, it would have the same effect, to declare an array step by step or in a row.

And no: there is no real way to convert your code automatically from the former to the latter. But perhaps, your question goes in another direction. Perhaps you want a solution to easily build combined arrays. For that task I suggest to take a look at underscore's zip function:

_.zip(['moe', 'larry', 'curly'], [30, 40, 50], [true, false, false]);

outcome: 

[["moe", 30, true], ["larry", 40, false], ["curly", 50, false]]

So in your case:

_.zip(["a", "b", "c", ...], ["1", "2", "3", ...], ["one", "two", "three", ...]);

would be, what you are looking for.

share|improve this answer
 
I thought if I had my arrays like the 2nd method my array js file would be a lot smaller, and easier to read etc. The filesize reduction would be my biggest goal - thanks –  Jon Jun 5 '13 at 17:30
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.