我有一个JavaScript数组,如:
[["$6"], ["$12"], ["$25"], ["$25"], ["$18"], ["$22"], ["$10"]]
如何将单独的内部数组合并为一个,例如:
["$6", "$12", "$25", ...]
我有一个JavaScript数组,如:
[["$6"], ["$12"], ["$25"], ["$25"], ["$18"], ["$22"], ["$10"]]
如何将单独的内部数组合并为一个,例如:
["$6", "$12", "$25", ...]
当前回答
您可以使用Undercore:
var x = [[1], [2], [3, 4]];
_.flatten(x); // => [1, 2, 3, 4]
其他回答
使用此方法展开两个阵列
arr1.concat(...arr2);
const flatten = array => array.reduce((a, b) => a.concat(Array.isArray(b) ? flatten(b) : b), []);
根据请求,分解一行基本上就是这样。
function flatten(array) {
// reduce traverses the array and we return the result
return array.reduce(function(acc, b) {
// if is an array we use recursion to perform the same operations over the array we found
// else we just concat the element to the accumulator
return acc.concat( Array.isArray(b) ? flatten(b) : b);
}, []); // we initialize the accumulator on an empty array to collect all the elements
}
这是一个基于artif3x答案的Typescript版本,为Scala粉丝提供了flatMap的额外实现。
function flatten<T>(items: T[][]): T[] {
return items.reduce((prev, next) => prev.concat(next), []);
}
function flatMap<T, U>(items: T[], f: (t: T) => U[]): U[] {
return items.reduce((prev, next) => prev.concat(f(next)), new Array<U>());
}
let arr = [["$6"], ["$12"], ["$25"], ["$25"], ["$18"], ["$22"], ["$10"], ["$0"], ["$15"],["$3"], ["$75"], ["$5"], ["$100"], ["$7"], ["$3"], ["$75"], ["$5"]];
arr = arr.reduce((a, b) => a.concat(b)); // flattened
let arr = [1, [2, 3, [4, 5, [6, 7], [8, 9, 10, 11, 12]]]];
function flattenList(nestedArr) {
let newFlattenList = [];
const handleFlat = (array) => {
let count = 0;
while (count < array.length) {
let item = array[count];
if (Array.isArray(item)) {
handleFlat(item);
} else {
newFlattenList.push(item);
}
count++;
}
};
handleFlat(nestedArr);
return newFlattenList;
}`enter code here`
console.log(flattenList(arr));
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