comparing two dimensional array js

Question

I am coding something in JS and I have to test code - I have to check if elements in 2 arrays are the same. So I've got an array: boreholes = [[66000, 457000],[1111,2222]....]; and I want to check if this array contain element for eg. [66000,457000] so I did: boreholes.indexOf([66000,457000]) but it returns -1, so I iterate trough array by:

for (var i = 0; i< boreholes.length; i++){
 if (boreholes[i] == [66000, 457000]){
  console.log('ok');
  break;
 }
};

but still I've got nothing. Can someone explain me what am I doing wrong?

Esailija · Accepted Answer · 2012-09-06 09:52:01Z

3

You are comparing distinct objects. When comparing objects, the comparison only evaluates to true when the 2 objects being compared are the same object. I.E

var a = [1,2,3];
var b = a;
a === b //true
b = [1,2,3];
a === b //false, b is not the same object

To compare arrays like this, you need to compare all of their elements separately:

for (var i = 0; i < boreholes.length; i++) {
    if (boreholes[i][0] == 66000 && boreholes[i][1] == 457000) {
        console.log('ok');
        break;
    }
}

answered Sep 6, 2012 at 9:52

Esailija

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2 Comments

Krystian Over a year ago

are you sure this is only way to do this? can you explain me why i have to do this this way?

Esailija Over a year ago

@Krystian If you could do this your way, the language would just implicitly do the same individual comparisons I am doing explicitly. It's just how it works with non-scalar values.

Vishal · Accepted Answer · 2012-09-06 10:20:51Z

3

You cannot compare arrays like array1 == array2 in javascript like you're trying to do here.

Here is a kludge method to compare two arrays:

function isEqual(array1, array2){
  return (array1.join('-') == array2.join('-'));
}

You can now use this method in your code like:

for (var i = 0; i< boreholes.length; i++){
 if (isEqual(boreholes[i], [66000, 457000]){
  console.log('ok');
  break;
 }
};

edited Sep 6, 2012 at 10:20

answered Sep 6, 2012 at 9:55

Vishal

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1 Comment

Alex K Over a year ago

beware, the condition will be true if the array is [660004, 57000], perhaps you should add a space into the join

webdeb · Accepted Answer · 2015-06-26 18:42:29Z

2

Currently I had the same problem, did it with the toString() method

var array1 = [1,2,3,[1,2,3]]
var array2 = [1,2,3,[1,2,3]]

array1 == array2 // false
array1.toString() == array2.toString() // true

var array3 = [1,2,3,[1,3,2]]

// Take attention
array1.toString() == array3.toString() // false

edited Jun 26, 2015 at 18:42

answered Jun 26, 2015 at 13:58

webdeb

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2 Comments

Krystian Over a year ago

great fresh solution, I've resolved that problem 3 years ago by accepted answer, but your solution seems to be very nice for me, thanks. :)

webdeb Over a year ago

hehe, sure, maybe some else will find it nice too ;)

Felix Gertz · Accepted Answer · 2012-09-06 10:01:48Z

0

You could also doing it with the Underscore.js-framework for functional programming.

function containsElements(elements) {
  _.find(boreholes, function(ele){ return _.isEqual(ele, elements); });
}

if(containsElements([66000, 457000])) {
  console.log('ok');
}

answered Sep 6, 2012 at 10:01

Felix Gertz

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Comments

Alex K · Accepted Answer · 2012-09-06 10:16:05Z

0

The question isn't quite clear if there can be more than 2 elements in an array, so this might work

var boreholes = [[66000, 457000],[1111,2222]]; 
var it = [66000, 457000];

function hasIt(boreholes, check) {
  var len = boreholes.length;
  for (var a = 0; a < len; a++) {
    if (boreholes[a][0] == check[0] && boreholes[a][1] == check[1]) {
       // ok
       return true;
    }
  }
  return false;
}

if (hasIt(boreholes, it)) {
  // ok, it has it
}

answered Sep 6, 2012 at 10:16

Alex K

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3 Comments

Alex K Over a year ago

can there be 3 or more elements in the element that needs to be checked against bareholes?

Krystian Over a year ago

ohh sorry.. all elements are xy cooridantes, so there can not be more than 2 elements in one element.

Alex K Over a year ago

oh ok, then you can easily use this implementation

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