Javascript：跨多个数组查找共享值

Question

What this should be able to do is take in a 2D array filled with one letter values and return an array of all the shared values. This is what I have so far:

var res = array[0].filter(function(x){
        return array.every(function(y){
             return y.indexOf(x) >= 0
        })
    });
return res;  

This is in some form of working state but only under certain condition which makes it very hit and miss. Working as intended:

var array = [["x","x"],
             ["x","x","x"]];

This returns the expected array of ["x","x"] but when like this:

var array = [["x","x","x"],
             ["x","x"]];

It returns ["x","x","x"]

As you can see the two arrays only share 2 common x's but the code doesn't reflect that in different situations. Also it should be able to handle arrays with other letters like so:

var array = [["x","x","z","y","y"],
             ["x,"x","x","y"],
             ["x","x","z","y"]];

With something like this it should return ["x","x","y"] as all arrays share 2 common x's and 1 common y

Answer 1

Use a combination of .every and .filter , use .indexOf to check whether a element exists in an array or not.

 var array = [ ["x", "x", "z", "y", "y"], ["x", "x", "x", "y"], ["x", "x", "z", "y"] ]; var res = array[0].filter(function(x) { return array.every(function(y) { if (y.indexOf(x) != -1) { y[y.indexOf(x)] = Infinity; return true; } return false; }) }) alert(res) 

Answer 2

Here's another way, using Array methods only availble from IE9 etc.

 function compareValues() { var arrs = [].slice.call(arguments).sort(function(a,b) { return a.length > b.length; // always iterate shortest array }); return arrs.shift().filter(function(x, i) { // filter the first array return arrs.every(function(arr) { // if all other arrays return arr[i] === x; // have the same value at the same index }) }); } var result = compareValues(["x","x","x","y"], ["x","x","z","y"], ["x","x","z","y","y"]); alert(result); 

Answer 3

This is a proposal with Array.prototype.reduce() , Array.prototype.filter() and Array.prototype.indexOf() for nondestructive search.

 var array = [ ["x", "x", "z", "y", "y"], ["x", "x", "x", "y"], ["x", "x", "z", "y"] ], result = array.reduce(function (r, a) { var last = {}; return r.filter(function (b) { var p = a.indexOf(b, last[b] || 0); if (~p) { last[b] = p + 1; return true; } }); }); document.write('<pre>' + JSON.stringify(result, 0, 4) + '</pre>');