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
ES6的纯粹魔力
const flat = A => A.reduce((A, a) => Array.isArray(a) ? [...A, ...flat(a)] : [...A, a], []);
其他回答
您可以使用“join()”和“split()”:
设arrs=[["$6"],["$12"],["$25"],["$25"],["$18"],["$22"],["$10"]];让newArr=arrs.join(“,”).split(“,“);console.log(newArr);//["$6", "$12", "$25", "$25", "$18", "$22", "$10"]
此外,还可以使用“toString()”和“split()”:
设arrs=[["$6"],["$12"],["$25"],["$25"],["$18"],["$22"],["$10"]];让newArr=arrs.toString().split(“,”);console.log(newArr);//["$6", "$12", "$25", "$25", "$18", "$22", "$10"]
然而,如果字符串包含逗号,上述两种方式都不能正常工作:
“join()”和“split()”:
设arrs=[["$,6"],["$,12"],["$2,5"],["$2,5"],[",$18"],["$22,"],["$,1,0"]];让newArr=arrs.join(“,”).split(“,“);console.log(newArr);// ["$", "6", "$", "12", "$2", "5", "$2", "5", "", "$18", "$22", "", "$", "1", "0"]
“toString()”和“split()”:
设arrs=[["$,6"],["$,12"],["$2,5"],["$2,5"],[",$18"],["$22,"],["$,1,0"]];让newArr=arrs.toString().split(“,”);console.log(newArr);// ["$", "6", "$", "12", "$2", "5", "$2", "5", "", "$18", "$22", "", "$", "1", "0"]
现在,实现这一点的最佳且简单的方法是这样连接和拆分阵列。
var multipleArrays = [["$6","$Demo"], ["$12",["Multi","Deep"]], ["$25"], ["$25"], ["$18"], ["$22"], ["$10"], ["$0"], ["$15"],["$3"], ["$75"], ["$5"], ["$100"], ["$7"], ["$3"], ["$75"], ["$5"]]
var flattened = multipleArrays.join().split(",")
此解决方案适用于多个级别,也是一个线性解决方案。
DEMO
ECMAScript 6的EDIT
由于ECMAScript 6已经标准化,您可以更改操作[].contat.apply([],arrays);对于[].contat(…数组);
var flattened = [].concat(...input);
DEMO
EDIT最高效的解决方案
解决这个问题最有效的方法是使用循环。您可以在此处比较“ops/sec”速度
var flattened=[];
for (var i=0; i<input.length; ++i) {
var current = input[i];
for (var j=0; j<current.length; ++j)
flattened.push(current[j]);
}
DEMO
希望有帮助
我知道这有点粗糙,但我所知道的扁平化字符串数组(任何深度!)(没有逗号!)的唯一简洁方法是将数组转换为字符串,然后用逗号分割字符串:
var myArray =[["$6"], ["$12"], ["$25"], ["$25"], ["$18"], ["$22"], ["$10"], ["$0"], ["$15"],["$3"], ["$75"], ["$5"], ["$100"], ["$7"], ["$3"], ["$75"], ["$5"]];
var myFlatArray = myArray.toString().split(',');
myFlatArray;
// ["$6", "$12", "$25", "$25", "$18", "$22", "$10", "$0", "$15", "$3", "$75", "$5", "$100", "$7", "$3", "$75", "$5"]
这应该适用于任何深度的仅包含字符串和数字(整数和浮点)的嵌套数组,但需要注意的是,数字将在过程中转换为字符串。这可以通过一点映射来解决:
var myArray =[[[1,2],[3,4]],[[5,6],[7,8]],[[9,0]]];
var myFlatArray = myArray.toString().split(',').map(function(e) { return parseInt(e); });
myFlatArray;
// [1, 2, 3, 4, 5, 6, 7, 8, 9, 0]
Array.prototype.flatten = Array.prototype.flatten || function() {
return [].reduce.call(this, function(flat, toFlatten) {
return flat.concat(Array.isArray(toFlatten) ? toFlatten.flatten() : toFlatten);
},[])
};
/**
* flatten an array first level
* @method flatten
* @param array {Array}
* @return {Array} flatten array
*/
function flatten(array) {
return array.reduce((acc, current) => acc.concat(current), []);
}
/**
* flatten an array recursively
* @method flattenDeep
* @param array {Array}
* @return {Array} flatten array
*/
function flattenDeep(array) {
return array.reduce((acc, current) => {
return Array.isArray(current) ? acc.concat(flattenDeep(current)) : acc.concat([current]);
}, []);
}
/**
* flatten an array recursively limited by depth
* @method flattenDepth
* @param array {Array}
* @return {Array} flatten array
*/
function flattenDepth(array, depth) {
if (depth === 0) {
return array;
}
return array.reduce((acc, current) => {
return Array.isArray(current) ? acc.concat(flattenDepth(current, --depth)) : acc.concat([current]);
}, []);
}