Javascript compare even to odd values in multidimensional array

Question

Given the following sample:

var arr = [
  [1,2,3,4,56],
  [45,5,56,67,4],
  [5,4,5,88,75],
  [2,4,5,66,7]
];

var len = arr.length; 
for(var i = 0; i < len; i++) {
  var parent = arr[i];
  for(var j = 0; j < parent.length; j++) {
    console.log(parent[j]);
  }
}

How would I determine if the value at each index in the nested arrays are lesser/greater than the next array? I tried using parent[j][0] > parent[j][1], but that returns undefined.

What's the objective? Can't you just sort the arrays and figure out from there? — elclanrs
– elclanrs, Commented Jul 22, 2013 at 22:45
The arrays are values from a table - a 4 column multi-row table. If the value in cell 1 is greater than cell 2, add specific styling. If cell 3 is greater than 4, add styling. So on and so forth for each row. The multi-dimensional array is a representation of the table. — psdpainter
– psdpainter, Commented Jul 22, 2013 at 22:50

Lee Meador · Accepted Answer · 2013-07-22 23:02:12Z

2

Try:

  var evenParent = arr[i];
  var oddParent = arr[i+1]
  for(var j = 0; j < parent.length; j++) {
    console.log(evenParent[j] + " " + oddParent[j]);
  }
  ++i;

This does test the elements in one row to the element in the same column, next row.

If you want to test the element in one column to the element in the next column, same row:

var parent = arr[i];
for (var j =0; j < parent.length; j+=2) {
    console.log(parent[j] + " " + parent[j+1]);
}

edited Jul 22, 2013 at 23:02

answered Jul 22, 2013 at 22:43

Lee Meador

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2 Comments

omma2289 Over a year ago

You would just need a check when reaching the last array if(i+1 < len)

Lee Meador Over a year ago

@koala_dev True. You should test that arr[i] and arr[i+1] both exist before using them. Or you could verify that len is even before the loop. Also, it would be better to use i+=2 in the for, I suppose, rather than ++i twice.

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