What would be a shorter way to write :
if (array1[0] >= array2[0] && array1[1] >= array2[1] && ...) {
do something;
}
I tried creating a function but I was not able to make it work, I'm still quite new at this.
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Refer this link with a similar Question posted <br> stackoverflow.com/questions/3432929/…Devendra Lattu– Devendra Lattu2013-12-13 14:54:40 +00:00Commented Dec 13, 2013 at 14:54
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@DevendraLattu that question isn't at all similarJoe– Joe2013-12-13 15:08:56 +00:00Commented Dec 13, 2013 at 15:08
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3 Answers
The most elegant way would be to use .every
The every() method tests whether all elements in the array pass the test implemented by the provided function.
if (array1.every(function(e,i){ return e>=array2[i];})) {
do something;
}
answered Dec 13, 2013 at 14:58
Benjamin Gruenbaum
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3 Comments
Tibos
+1 For the most elegant solution. I deleted my answer which was basically identical, but slower :) .
Trax
I tried this one but I'm getting a syntax error "missing ; before statement".
Benjamin Gruenbaum
Yeah, I'm missing a } here, one sec... Fixed. @user3099816
This will return true if all elements of a are greater than all elements of b. It will return as early as possible rather than having to compare all of the elements.
function compare(a, b) {
for (i = 0; i < a.length; i++) {
if (a[i] < b[i]) { return false;}
}
return true
}
Comments
var isGreater = true;
for (var i = 0; i < array1.length; i++)
{
if (array1[i] < array2[i])
{
isGreater = false;
break;
}
}
if (isGreater)
{
//do something
}
You loop your first array and replace the numbers by the looping variable (i)
3 Comments
Joe
This doesn't answer the question, which is that all elements of of the first are greater than the second, not some.
Shryme
looking at his example, he was doing arr1[0] vs arr2[0], arr1[1] vs arr2[1], that's what my for loop is doing, he never asked how to check everything with everything
Joe
I disagree. Look at his logic, it's
&&ing all the conditions.