arr1 = ['a', 'b', 'c'] ;
arr2 = ['1', '2', '3', '4'];
我发表了
// run time
query() {
stuff1 = a;
stuff2 = 1;
}
query() {
stuff1 = a;
stuff2 = 2;
}
query() {
stuff1 = a;
stuff2 = 3;
}
query() {
stuff1 = a;
stuff2 = 4;
}
query() {
stuff1 = b;
stuff2 = 1;
}
query() {
stuff1 = b;
stuff2 = 2;
}
query() {
stuff1 = b;
stuff2 = 3;
}
query() {
stuff1 = b;
stuff2 = 4;
}
...
query() {
stuff1 = c;
stuff2 = 6;
}
如何编写代码?
_.map(arr1, (res) => { reutrn _.map(arr2, (res2, res1) => ... } blabla
我没有想法......
如果我使用zipWith,a:1 b:2 c:3 ......但我不想要它
由于
答案 0 :(得分:1)
var arr = _.chain(arr1).map((item) => {
return _.map(arr2, (item2) => {
return {
stuff1: item,
stuff2: item2
}
})
}).flatten().value();
输出: