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]]
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
Python有两个压缩序列的函数:zip和itertools.zip_longest。Javascript中相同功能的实现如下所示:
Python的zip在JS/ES6上的实现
const zip = (...arrays) => {
const length = Math.min(...arrays.map(arr => arr.length));
return Array.from({ length }, (value, index) => arrays.map((array => array[index])));
};
结果:
console.log(zip(
[1, 2, 3, 'a'],
[667, false, -378, '337'],
[111],
[11, 221]
));
[[1, 667, 111, 11]
console.log(zip(
[1, 2, 3, 'a'],
[667, false, -378, '337'],
[111, 212, 323, 433, '1111']
));
[[1、667、111],[2,假的,212年],[3、-378、323],[' a ', “337”,433]]
console.log(zip(
[1, 2, 3, 'a'],
[667, false, -378, '337'],
[111],
[]
));
[]
Python的zip_longest在JS/ES6上的实现
(https://docs.python.org/3.5/library/itertools.html?highlight=zip_longest # itertools.zip_longest)
const zipLongest = (placeholder = undefined, ...arrays) => {
const length = Math.max(...arrays.map(arr => arr.length));
return Array.from(
{ length }, (value, index) => arrays.map(
array => array.length - 1 >= index ? array[index] : placeholder
)
);
};
结果:
console.log(zipLongest(
undefined,
[1, 2, 3, 'a'],
[667, false, -378, '337'],
[111],
[]
));
[[1,667, 111, undefined], [2, false, undefined, undefined], [3, -378, undefined, undefined], ['a', '337', undefined, 未定义
console.log(zipLongest(
null,
[1, 2, 3, 'a'],
[667, false, -378, '337'],
[111],
[]
));
[[1, 667, 111, null], [2, false, null, null], [3, -378, Null, Null], ['a', '337', Null, Null]]
console.log(zipLongest(
'Is None',
[1, 2, 3, 'a'],
[667, false, -378, '337'],
[111],
[]
));
[[1,667, 111, 'Is None'], [2, false, 'Is None', 'Is None'], [3, -378,“没有”,“没有 ' ], [ ' ”、“337”、“没有”、“ 没有']]
其他回答
不是Javascript本身内置的。一些常见的Javascript框架(如Prototype)提供了一个实现,或者你也可以自己编写。
下面是一个快速有效的方法,使用iter-ops库,operator zip:
const {pipe, zip} = require('iter-ops');
const i = pipe(array1, zip(array2, array3));
console.log(...i); //=> [ 1, 'a', 4 ] [ 2, 'b', 5 ] [ 3, 'c', 6 ]
标准库将所有输入作为可迭代对象处理,因此它们只迭代一次。它可以以同样的方式处理所有类型的可迭代对象——iterable, AsyncIterable, Iterator, AsyncIterator。
附注:我是iter-ops的作者。
Python有两个压缩序列的函数:zip和itertools.zip_longest。Javascript中相同功能的实现如下所示:
Python的zip在JS/ES6上的实现
const zip = (...arrays) => {
const length = Math.min(...arrays.map(arr => arr.length));
return Array.from({ length }, (value, index) => arrays.map((array => array[index])));
};
结果:
console.log(zip(
[1, 2, 3, 'a'],
[667, false, -378, '337'],
[111],
[11, 221]
));
[[1, 667, 111, 11]
console.log(zip(
[1, 2, 3, 'a'],
[667, false, -378, '337'],
[111, 212, 323, 433, '1111']
));
[[1、667、111],[2,假的,212年],[3、-378、323],[' a ', “337”,433]]
console.log(zip(
[1, 2, 3, 'a'],
[667, false, -378, '337'],
[111],
[]
));
[]
Python的zip_longest在JS/ES6上的实现
(https://docs.python.org/3.5/library/itertools.html?highlight=zip_longest # itertools.zip_longest)
const zipLongest = (placeholder = undefined, ...arrays) => {
const length = Math.max(...arrays.map(arr => arr.length));
return Array.from(
{ length }, (value, index) => arrays.map(
array => array.length - 1 >= index ? array[index] : placeholder
)
);
};
结果:
console.log(zipLongest(
undefined,
[1, 2, 3, 'a'],
[667, false, -378, '337'],
[111],
[]
));
[[1,667, 111, undefined], [2, false, undefined, undefined], [3, -378, undefined, undefined], ['a', '337', undefined, 未定义
console.log(zipLongest(
null,
[1, 2, 3, 'a'],
[667, false, -378, '337'],
[111],
[]
));
[[1, 667, 111, null], [2, false, null, null], [3, -378, Null, Null], ['a', '337', Null, Null]]
console.log(zipLongest(
'Is None',
[1, 2, 3, 'a'],
[667, false, -378, '337'],
[111],
[]
));
[[1,667, 111, 'Is None'], [2, false, 'Is None', 'Is None'], [3, -378,“没有”,“没有 ' ], [ ' ”、“337”、“没有”、“ 没有']]
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)
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