Suppose we have 2 Arrays say :
A [] => 1 2 3 4 5
B [] => 1 2 7 4 5
is there any method in jQuery which will give the unmatched values of 2 arrays in this case :
Result [] => 3 7
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1A loop is so easy and fast, why are you looking for a jquery method ? Don't add useless layers.Denys Séguret– Denys Séguret2012-05-26 10:57:02 +00:00Commented May 26, 2012 at 10:57
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If both lists contain unique value, you could merge them both. Then, sort the array, loop and find duplicate where element i === i+1 and then remove both of these elements. You'll be left with the array [3,7]frenchie– frenchie2012-05-26 11:03:40 +00:00Commented May 26, 2012 at 11:03
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6 Answers
hiya working demo here: http://jsfiddle.net/mbKfT/
good read http://api.jquery.com/jQuery.inArray/
This uses inArray to check it the element is there if not add it to intersect array.
rest demo will sue out any doubts :)
code
var a1 = [1,2,3,4,5];
var a2 = [1,2,7,4,5];
var intersect = [];
$.each(a1, function(i, a1val) {
if ($.inArray(a1val, a2) === -1) {
intersect.push(a1val);
}
});
$.each(a2, function(i, a1val) {
if ($.inArray(a1val, a1) === -1) {
intersect.push(a1val);
}
});
$("div").text(intersect);
alert(intersect + " -- " + matches);
Comments
Answer : no.
Solution : use a standard javascript loop.
var nomatches = [];
for (var i=Math.min(A.length, B.length); i-->0;) {
if (A[i]!=B[i]) {
nomatches.push(A[i]);
nomatches.push(B[i]);
}
}
// do what you want with remaining items if A.length != B.length
If, as supposed by a Rory, you don't want to match arrays but logical sets, you can do this :
var nomatches = [];
var setA = {};
var setB = {};
for (var i=A.length; i-->0;) setA[A[i]]=1;
for (var i=B.length; i-->0;) setB[B[i]]=1;
for (var i=A.length; i-->0;) {
if (!setB[A[i]]) nomatches.push(A[i]);
}
for (var i=B.length; i-->0;) {
if (!setA[V[i]]) nomatches.push(B[i]);
}
3 Comments
Rory McCrossan
+1 as it answers the OPs question directly, but what about cases where the arrays are in different orders. Consider
5,4,3,2,1 and 1,2,3,4,5.
Denys Séguret
It depends on the requirement.
chucktator
This is not correct. Think of
A={1,2,3,4,5}; B={1,3,4,5}. With your approach 3,4 and 5 would be in nomatches.var nomatch = [], Bcopy = B.slice(0);
for (var i = 0, j; i < A.length; i++) {
j = Bcopy.indexOf(A[i]);
if (j === -1) nomatch.push(A[i]);
else Bcopy.splice(j, 1);
}
nomatch.push.apply(nomatch, Bcopy);
Note:
- This code supposes that items in
AandBare unique. indexOffor arrays must be emulated in IE8 and previous versions.
Comments
jQuery.inArray() will do some help:
var a = [1,2,3,4,5], b=[1,2,7,4,5];
var ret =
a.filter(function(el) {
return $.inArray(el, b) === -1;
}).concat(
b.filter(function(el) {
return $.inArray(el, a) === -1;
})
);
console.log(ret);
The demo.
PS: Or you could just use b.indexOf(el) === -1, then you don't need jQuery anymore.
1 Comment
MaxArt
Why using
inArray when you can use a.indexOf? It's good if dystroy already uses jQuery, but what if he doesn't?function getUnique(A, B){
var res = [];
$.grep(A, function(element) {
if($.inArray(element, B) == -1) res.push(element)
});
$.grep(B, function(element) {
if($.inArray(element, A) == -1) res.push(element);
});
return res;
}
Use:
var A = [1,2,3,4,5],
B = [1,2,3,5,7];
getUnique(A, B);
Comments
Here is another solution for modern browsers (one-liners, yes!):
var a = [1, 2, 3, 4, 5];
var b = [1, 2, 7, 4, 5];
var result = a.concat(b).filter(function(el, i) {
return (i < a.length ? b : a).indexOf(el) == -1;
});
DEMO: http://jsfiddle.net/6Na36/
If you wish to preserve index checking also, you can use this variant:
var a = [1, 2, 3, 4, 5];
var b = [1, 2, 7, 4, 5];
var result = a.concat(b).filter(function(el, i, c) {
return el != c[i < a.length ? i + a.length : i - a.length];
});
DEMO: http://jsfiddle.net/6Na36/1/
Note, both variants successfully work with arrays of different size.