我的目标是以所有可能的方式合并2个数组,第一个数组将始终是单元格的第一部分,第二个数组将始终位于单元格的第二部分。 /> 我的最后一次尝试,最接近但不完全是我需要的:
function combineArrays(array1, array2) {
let arr = [];
for (let i = 0; i < array1.length; i++) {
arr[i] = array1[i];
}
for (let i = 0, j = 0; i < array1.length; i++) {
arr[i] += array2[j];
if (array2[j + 1]) {
j++;
}
else {
j = 0;
}
}
return arr;
}
编辑:
例如,数组A和数组B:
答:[&#34;我&#34;,&#34;你&#34;,&#34;他&#34;,&#34;她&#34; ]
B:[&#34; am&#34;,&#34;是&#34;,&#34;&#34;&#34; ]
预期结果:
[
&#34;我是&#34;,&#34;你是&#34;,&#34;他是&#34;,&#34;她是&#34;,
&#34;我是&#34;,&#34;你是&#39;,&#34;他是&#34;,&#34;她是&#34;,
&#34;我是&#34;,&#34;你是&#34;,&#34;他是&#34;,&#34;她是&#34;
]
答案 0 :(得分:1)
let a = ["I", "You", "He", "She"];
let b = ["am", "is", "are"];
let result = b.map(b => a.map(a => a + ' ' + b));
console.log(result);
如果您不想要嵌套数组:
let a = ["I", "You", "He", "She"];
let b = ["am", "is", "are"];
let result = b.reduce((arr, b) => arr.concat(a.map(a => a + ' ' + b)), []);
console.log(result);
对于n
数量的数组:
let a = ["I", "You", "He", "She"];
let b = ["am", "is", "are"];
let c = ["good", "bad", "happy"];
let crossProduct = (a, b) => b.reduce((arr, b) => arr.concat(a.map(a => a + ' ' + b)), []);
let result = [a, b, c].reduce(crossProduct);
console.log(result);
答案 1 :(得分:0)
var arr1 = ['a', 'b', 'c'];
var arr2 = ['d', 'e', 'f'];
var arr3 = arr1.concat(arr2);
// arr3 is a new array [ "a", "b", "c", "d", "e", "f" ]
但是你似乎想要对你实现的数组求和:
var array1 = [1,2,3,4];
var array2 = [5,6,7,8];
var sum = array1.map(function (num, idx) {
return num + array2[idx];
});
// [6,8,10,12]
答案 2 :(得分:0)
如果您的目标是合并两个数组,那么这是ES6方式:
const arr1 = ['a', 'b', 'c'];
const arr2 = ['d', 'e', 'f'];
const arr3 = [...arr1, ...arr2]; // Output: [ "a", "b", "c", "d", "e", "f" ]
但是,在你的情况下(编辑之后),你想要做更多的事情!
const arr1 = [ "I", "You", "He", "She" ];
const arr2 = [ "am", "is", "are" ];
arr2.map(eltB => arr1.map(eltA => `${eltA} ${eltB}`))
.reduce((newArray, arr) => newArray.concat(arr), []);