javascript中是否有类似于Python的zip函数?也就是说，给定多个相等长度的数组，创建一个由对组成的数组。
例如，如果我有三个这样的数组:
var array1 = [1, 2, 3];
var array2 = ['a','b','c'];
var array3 = [4, 5, 6];
输出数组应该是:
var outputArray = [[1,'a',4], [2,'b',5], [3,'c',6]]
当前回答
这将从Ddi基于迭代器的答案中删除一行:
function* zip(...toZip) {
const iterators = toZip.map((arg) => arg[Symbol.iterator]());
const next = () => toZip = iterators.map((iter) => iter.next());
while (next().every((item) => !item.done)) {
yield toZip.map((item) => item.value);
}
}
其他回答
惰性生成器解决方案的一个变体:
function* iter(it) { yield* it; } function* zip(...its) { its = its.map(iter); while (true) { let rs = its.map(it => it.next()); if (rs.some(r => r.done)) return; yield rs.map(r => r.value); } } for (let r of zip([1,2,3], [4,5,6,7], [8,9,0,11,22])) console.log(r.join()) // the only change for "longest" is some -> every function* zipLongest(...its) { its = its.map(iter); while (true) { let rs = its.map(it => it.next()); if (rs.every(r => r.done)) return; yield rs.map(r => r.value); } } for (let r of zipLongest([1,2,3], [4,5,6,7], [8,9,0,11,22])) console.log(r.join())
这是python经典的“n-group”习语zip(*[iter(a)]*n):
triples = [...zip(...Array(3).fill(iter(a)))]
python zip函数的生成器方法。
function* zip(...arrs){
for(let i = 0; i < arrs[0].length; i++){
a = arrs.map(e=>e[i])
if(a.indexOf(undefined) == -1 ){yield a }else{return undefined;}
}
}
// use as multiple iterators
for( let [a,b,c] of zip([1, 2, 3, 4], ['a', 'b', 'c', 'd'], ['hi', 'hello', 'howdy', 'how are you']) )
console.log(a,b,c)
// creating new array with the combined arrays
let outputArr = []
for( let arr of zip([1, 2, 3, 4], ['a', 'b', 'c', 'd'], ['hi', 'hello', 'howdy', 'how are you']) )
outputArr.push(arr)
我在纯JS中尝试了一下，想知道上面发布的插件是如何完成这项工作的。这是我的结果。首先我要说的是，我不知道这在IE和类似的软件中会有多稳定。这只是一个快速的模型。
init(); function init() { var one = [0, 1, 2, 3]; var two = [4, 5, 6, 7]; var three = [8, 9, 10, 11, 12]; var four = zip(one, two, one); //returns array //four = zip(one, two, three); //returns false since three.length !== two.length console.log(four); } function zip() { for (var i = 0; i < arguments.length; i++) { if (!arguments[i].length || !arguments.toString()) { return false; } if (i >= 1) { if (arguments[i].length !== arguments[i - 1].length) { return false; } } } var zipped = []; for (var j = 0; j < arguments[0].length; j++) { var toBeZipped = []; for (var k = 0; k < arguments.length; k++) { toBeZipped.push(arguments[k][j]); } zipped.push(toBeZipped); } return zipped; }
虽然不是无懈可击，但还是很有趣。
ES2020最短变体:
function * zip(arr1, arr2, i = 0) {
while(arr1[i] || arr2[i]) yield [arr1[i], arr2[i++]].filter(x => !!x);
}
[ ...zip(arr1, arr2) ] // result
如果你喜欢ES6:
const zip = (arr,...arrs) =>(
arr.map(
(v,i) => arrs.reduce((a,arr)=>[...a, arr[i]], [v])))
