我说我有
foo
如何将其转换为
template <typename ... Args>
foo(Args...) -> foo<sizeof...(Args)>;
?
我的做法是:
f8
只是想知道还有没有更短的方法?
答案 0 :(得分:2)
// What question mentions
const item = {
"Key1": "Value1",
"Key2": ["Value2", "Value3"]
}
let res = item.Key2.map(i => [item.Key1, i])
console.log(res)
// What the world wants!
const item = {
"Key1": ["ValueA", "ValueB"],
"Key2": ["Value1", "Value2"]
}
res = item.Key1.reduce((acc, i1) => acc.concat(item.Key2.map(i2 => [i1, i2])), [])
console.log(res)
答案 1 :(得分:2)
您可以构建笛卡尔积。
const
item = { Key1: "Value1", Key2: ["Value2", "Value3"] },
values = Object.values(item).map(v => [].concat(v)),
result = values.reduce((a, b) => a.reduce((r, v) => r.concat(b.map(w => [].concat(v, w))), []));
console.log(result);
答案 2 :(得分:0)
const item = { Key1: "Value1", Key2: ["Value2", "Value3"]};
let target = item.Key2.reduce((i, _) => {
i.push([item.Key1, _]);
return i;
}, []);
console.log(target)