Multidimensional Array - Test if all values are equivilant in nested array

Question

So, I have an array that looks like this:

[ ["","",""], ["","",""], ["","",""] ]

How would I test it in javascript to see if, perhaps, the 1st index was filled with X's?

EX: [ ["X","X","X"], ["","",""], ["","",""] ]

I keep thinking I should do something like this, but it feels like there would be a quicker way...

var counter = 0, 
    win,
    b = [ ["X","X","X"], ["","",""], ["","",""] ],
    num = b[0].length;
for(var i=0; i<num; i++){
    if(b[0][i]==="X"){ counter++; }
}
if(num===counter){ win=true; }

imcg · Accepted Answer · 2014-01-06 20:59:28Z

2

var win = b[0].join('') === 'XXX';

answered Jan 6, 2014 at 20:59

imcg

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5 Comments

Nicholas Hazel Over a year ago

I understand, I but don't want people to just write code for me. I'd prefer to know a foundation and expand off of it :-). Thank you for the help. The simplest answers are often the finest.

Nicholas Hazel Over a year ago

Just add a loop for var 0, tests every index in the rows. Then add some modified diagonal math, etc etc. I got the answer I wanted here. Ty.

Nicholas Hazel Over a year ago

Decided to write a few functions and create a switch statement to determine the computer move. Thanks again!

Scott Sauyet Over a year ago

Or just do var win = b[0].join('') === new Array(b[0].length + 1).join('X'); if you really like this technique.

Edgar Villegas Alvarado Over a year ago

@Scott, that's my answer xD

Trevor Dixon · Accepted Answer · 2014-01-06 21:07:44Z

1

Use Array.prototype.every. https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/every

function isAnX(c) { return c === 'X'; }
var win = b[0].every(isAnX);

edited Jan 6, 2014 at 21:07

answered Jan 6, 2014 at 21:01

Trevor Dixon

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1 Comment

Nicholas Hazel Over a year ago

Totally interesting concept. I'll likely expand on this instead of the upvoted post above. Well done and thank you.

Barmar · Accepted Answer · 2014-01-06 20:59:53Z

1

win = true;
for (var i = 0; i < num; i++) {
    if (b[0][i] != "X") {
        win = false;
        break;
    }
}

answered Jan 6, 2014 at 20:59

Barmar

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1 Comment

Nicholas Hazel Over a year ago

Hahaha, shoulda thought to reverse it this way. Nice answer :-)

georg · Accepted Answer · 2014-01-06 21:10:26Z

1

Here's how to test if an array contains identical elements:

allTheSame = someArray.every(function(x, _, ary) { return x == ary[0] })

see every

or, for pre-ES5 engines:

allTheSame = function(ary) {
    for(var i = 1; i < ary.length; i++)
       if(ary[i] != ary[0]) return false;
    return true;
}

edited Jan 6, 2014 at 21:10

answered Jan 6, 2014 at 21:02

georg

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3 Comments

Nicholas Hazel Over a year ago

What does the underscore represent? Just a blank filler to dig deeper?

georg Over a year ago

@NicholasHazel: an unused parameter (index).

Nicholas Hazel Over a year ago

Ahhh, figured as much. Thanks for your answer :-)

PA. · Accepted Answer · 2014-01-06 21:09:10Z

0

use Array.prototype.reduce

 var win=b[0].reduce(function (v, c) { return v && (c=="X") }, true);

answered Jan 6, 2014 at 21:09

PA.

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2 Comments

Trevor Dixon Over a year ago

every is slightly preferable, because it will quite once it finds a false.

PA. Over a year ago

I mentioned .reduce because is more flexible. But the performance hit you pointed out is a very good reason to prefer .every thanks.

Edgar Villegas Alvarado · Accepted Answer · 2014-01-06 21:11:16Z

0

Another one, just to give one more option:

var win = new RegExp("^X{"+b[0].length+"}$").test(b[0].join(''));

Cheers

answered Jan 6, 2014 at 21:11

Edgar Villegas Alvarado

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1 Comment

Nicholas Hazel Over a year ago

I've been told that regex expressions are a slow way to process information. I use them for basic properties typically (converting timestamps, etc), but for this case I don't think this is the correct solution. Awesome try, and something I use regularly though :-)

Scott Sauyet · Accepted Answer · 2014-01-06 21:18:40Z

0

Similar to other answers, but with something of a twist:

var all = function(val, arr) {
    var fn = function(arr) {return arr.every(function(item) {return item === val;});};
    return arguments.length > 1 ? fn(arr) : fn;
};

var win = all('X', b[0]); //=> true
// or
var allXs = all('X');
var win = allXs(b[0]); //=> true

Or if you just want to know if they're all the same:

var allAlike = function(arr) {
    return (arr.length == 0) || all(arr[0], arr);
}

var win = allAlike(b[0]);

This would be easier if you had an appropriate curry-like function available.

answered Jan 6, 2014 at 21:18

Scott Sauyet

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2 Comments

Nicholas Hazel Over a year ago

So you're breaking the check into other functions. Perhaps would suit my purposes. Thank you :-)

Scott Sauyet Over a year ago

@NicholasHazel: Yes, breaking it into reusable functions. This really only makes sense if you might actually have additional uses for these functions, though. Most of the other solutions would be better if you have no further use for all or allAlike.

Edgar Villegas Alvarado · Accepted Answer · 2014-01-06 21:19:12Z

0

Hahaha, this is cool:

var win = Array(b[0].length+1).join('X') == b[0].join('')

Cheers!

answered Jan 6, 2014 at 21:19

Edgar Villegas Alvarado

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1 Comment

Nicholas Hazel Over a year ago

Much better for my purposes of the site I'm building it on than your previous post.

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