是否有一种方法可以在JavaScript中返回两个数组之间的差异?
例如:
var a1 = ['a', 'b'];
var a2 = ['a', 'b', 'c', 'd'];
// need ["c", "d"]
是否有一种方法可以在JavaScript中返回两个数组之间的差异?
例如:
var a1 = ['a', 'b'];
var a2 = ['a', 'b', 'c', 'd'];
// need ["c", "d"]
当前回答
简单地比较所有值,并返回数组与不重复的值。
var main = [9, '$', 'x', 'r', 3, 'A', '#', 0, 1];
var arr0 = ['Z', 9, 'e', '$', 'r'];
var arr1 = ['x', 'r', 3, 'A', '#'];
var arr2 = ['m', '#', 'a', 0, 'r'];
var arr3 = ['$', 1, 'n', '!', 'A'];
Array.prototype.diff = function(arrays) {
var items = [].concat.apply(this, arguments);
var diff = [].slice.call(items), i, l, x, pos;
// go through all items
for (x = 0, i = 0, l = items.length; i < l; x = 0, i++) {
// find all positions
while ((pos = diff.indexOf(items[i])) > -1) {
// remove item + increase found count
diff.splice(pos, 1) && x++;
}
// if item was found just once, put it back
if (x === 1) diff.push(items[i]);
}
// get all not duplicated items
return diff;
};
main.diff(arr0, arr1, arr2, arr3).join(''); // returns "Zeman!"
[].diff(main, arr0, arr1, arr2, arr3).join(''); // returns "Zeman!"
其他回答
使用http://phrogz.net/JS/ArraySetMath.js你可以:
var array1 = ["test1", "test2","test3", "test4"];
var array2 = ["test1", "test2","test3","test4", "test5", "test6"];
var array3 = array2.subtract( array1 );
// ["test5", "test6"]
var array4 = array1.exclusion( array2 );
// ["test5", "test6"]
类似于Ian Grainger的解决方案(但在打字稿中):
function findDiffs(arrayOne: string[], arrayTwo: string[]) {
let onlyInArrayOne = []
let onlyInArrayTwo = []
let share = []
let [arrayOneCopy, arrayTwoCopy] = [[...arrayOne], [...arrayTwo]]
arrayOneCopy.sort(); arrayTwoCopy.sort()
while (arrayOneCopy.length !== 0 && arrayTwoCopy.length !== 0) {
if (arrayOneCopy[0] == arrayTwoCopy[0]) {
share.push(arrayOneCopy[0])
arrayOneCopy.splice(0, 1)
arrayTwoCopy.splice(0, 1)
}
if (arrayOneCopy[0] < arrayTwoCopy[0]) {
onlyInArrayOne.push(arrayOneCopy[0])
arrayOneCopy.splice(0, 1)
}
if (arrayOneCopy[0] > arrayTwoCopy[0]) {
onlyInArrayTwo.push(arrayTwoCopy[0])
arrayTwoCopy.splice(0, 1)
}
}
onlyInArrayTwo = onlyInArrayTwo.concat(arrayTwoCopy)
onlyInArrayOne = onlyInArrayOne.concat(arrayOneCopy)
return {
onlyInArrayOne,
onlyInArrayTwo,
share,
diff: onlyInArrayOne.concat(onlyInArrayTwo)
}
}
// arrayOne: [ 'a', 'b', 'c', 'm', 'y' ]
// arrayTwo: [ 'c', 'b', 'f', 'h' ]
//
// Results:
// {
// onlyInArrayOne: [ 'a', 'm', 'y' ],
// onlyInArrayTwo: [ 'f', 'h' ],
// share: [ 'b', 'c' ],
// diff: [ 'a', 'm', 'y', 'f', 'h' ]
// }
困难的方法(如果你想做一些比.indexOf更奇特的东西)
var difference = function (source, target) {
return source.reduce(function (diff, current) {
if (target.indexOf(current) === -1) {
diff.push(current);
}
return diff;
}, []);
}
简单的方法
var difference = function (source, target) {
return source.filter(function (current) {
return target.indexOf(current) === -1;
});
}
const dbData = [{name:'ally'},
{name:'James'}]
const localData = [{name:'James'}]
const diff = dbData.filter(a =>!localData.some(b => { return a.name === b.name}))
我一直在寻找一个不涉及使用不同库的简单答案,我想出了我自己的答案,我想这里没有提到过。 我不知道它的效率如何,但它确实有效;
function find_diff(arr1, arr2) {
diff = [];
joined = arr1.concat(arr2);
for( i = 0; i <= joined.length; i++ ) {
current = joined[i];
if( joined.indexOf(current) == joined.lastIndexOf(current) ) {
diff.push(current);
}
}
return diff;
}
对于我的代码,我也需要删除副本,但我想这并不总是可取的。
我想主要的缺点是它可能会比较许多已经被拒绝的选择。