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
CoffeeScript版本:
diff = (val for val in array1 when val not in array2)
其他回答
只是修整字符串以确保....空格不会影响差异
function arr_diff(a1, a2) {
var a=[], diff=[];
for(var i=0;i<a1.length;i++)
a[a1[i]]=true;
for(var i=0;i<a2.length;i++)
if(a[a2[i].trim()]) delete a[a2[i].trim()];
else a[a2[i].trim()]=true;
for(var k in a)
diff.push(k);
return diff;
}
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' ]
// }
function diffArray(newArr, oldArr) {
var newSet = new Set(newArr)
var diff = []
oldArr.forEach((a) => {
if(!newSet.delete(a))diff.push(a)
})
return diff.concat(Array.from(newSet))
}
ES6中一个更简洁的方法是以下解决方案。
var a1 = ['a', 'b'];
var a2 = ['a', 'b', 'c', 'd'];
区别
a2.filter(d => !a1.includes(d)) // gives ["c", "d"]
十字路口
a2.filter(d => a1.includes(d)) // gives ["a", "b"]
析取联合(对称差分)
[ ...a2.filter(d => !a1.includes(d)),
...a1.filter(d => !a2.includes(d)) ]
贡献一个jQuery解决方案,我目前正在使用:
if (!Array.prototype.diff) {
Array.prototype.diff = function (a) {
return $.grep(this, function (i) { return $.inArray(i, a) === -1; });
};
}