我需要获得一个阵列的所有可能子集，其中至少包含2个项目且未知最大值。有人可以帮我一点吗？
说我有这个......
[1,2,3]
......我怎么得到这个？
[
[1,2]
, [1,3]
, [2,3]
, [1,2,3]
]
答案 0 :(得分：56)
窃取this JavaScript组合生成器后，我添加了一个参数来提供最小长度，
var combine = function(a, min) {
var fn = function(n, src, got, all) {
if (n == 0) {
if (got.length > 0) {
all[all.length] = got;
}
return;
}
for (var j = 0; j < src.length; j++) {
fn(n - 1, src.slice(j + 1), got.concat([src[j]]), all);
}
return;
}
var all = [];
for (var i = min; i < a.length; i++) {
fn(i, a, [], all);
}
all.push(a);
return all;
}
使用，提供一个数组，以及所需的最小子集长度，
var subsets = combine([1, 2, 3], 2);
输出是，
[[1, 2], [1, 3], [2, 3], [1, 2, 3]]
答案 1 :(得分：12)
通过这个question的小调整，我希望我的解决方案更有效，因为它使用位运算符来生成所有子集。
var sets = (function(input, size) {
var results = [], result, mask, i, total = Math.pow(2, input.length);
for (mask = size; mask < total; mask++) {
result = [];
i = input.length - 1;
do {
if ((mask & (1 << i)) !== 0) {
result.push(input[i]);
}
} while (i--);
if (result.length >= size) {
results.push(result);
}
}
return results;
})(['a','b','c','d','e','f'], 2);
console.log(sets);
答案 2 :(得分：8)
以下是使用ECMAScript 2015 generator function查找所有组合的方法：
function* generateCombinations(arr) {
function* doGenerateCombinations(offset, combo) {
yield combo;
for (let i = offset; i < arr.length; i++) {
yield* doGenerateCombinations(i + 1, combo.concat(arr[i]));
}
}
yield* doGenerateCombinations(0, []);
}
for (let combo of generateCombinations([1, 2, 3, 4, 5])) {
console.log(JSON.stringify(combo));
}
要根据问题的要求限制到最小尺寸，只需确保组合的长度，然后再产生它：
function* generateCombinations(arr, minSize) {
function* doGenerateCombinations(offset, combo) {
if (combo.length >= minSize) {
yield combo;
}
for (let i = offset; i < arr.length; i++) {
yield* doGenerateCombinations(i + 1, combo.concat(arr[i]));
}
}
yield* doGenerateCombinations(0, []);
}
for (let combo of generateCombinations([1, 2, 3, 4, 5], 2)) {
console.log(JSON.stringify(combo));
}
在yield点限制允许以可读方式使此功能适应其他常见用例，例如，选择所有精确大小的组合：
function* generateCombinations(arr, size) {
function* doGenerateCombinations(offset, combo) {
if (combo.length == size) {
yield combo;
} else {
for (let i = offset; i < arr.length; i++) {
yield* doGenerateCombinations(i + 1, combo.concat(arr[i]));
}
}
}
yield* doGenerateCombinations(0, []);
}
for (let combo of generateCombinations([1, 2, 3, 4, 5], 2)) {
console.log(JSON.stringify(combo));
}
答案 3 :(得分：6)
这个算法要求递归...这就是我要做的事情
var arr = [1,2,3,4,5];
function getSubArrays(arr){
if (arr.length === 1) return [arr];
else {
subarr = getSubArrays(arr.slice(1));
return subarr.concat(subarr.map(e => e.concat(arr[0])), [[arr[0]]]);
}
}
console.log(JSON.stringify(getSubArrays(arr)));
上述算法的另一个奇特版本;
var arr = [1,2,3,4,5],
sas = ([n,...ns],sa) => !ns.length ? [[n]]
: (sa = sas(ns),
sa.concat(sa.map(e => e.concat(n)),[[n]]));
为了了解最新情况，请一步一步走
getSubArrays函数。因此，[1,2,3,4,5]的尾部为[2,3,4,5]。[5]），我们就会将[[5]]返回到之前的getSubArrays函数调用。getSubArrays函数中arr [4,5]而subarr被分配到[[5]]。[[5]].concat([[5]].map(e => e.concat(4), [[4]])实际上[[5], [5,4], [4]]返回到之前的getSubArrays函数调用。getSubArrays函数中arr [3,4,5]而subarr被分配到[[5], [5,4], [4]]。答案 4 :(得分：5)
组合，短篇：
def test_filter_graded(self):
eval1 = mock.MagicMock()
eval2 = mock.MagicMock
eval1.is_completely_graded.return_value = True
eval2.is_completely_graded.return_value = False
filter_graded([eval1, eval2])
self.assertEqual(len(result), 1)
并致电
function combinations(array) {
return new Array(1 << array.length).fill().map(
(e1,i) => array.filter((e2, j) => i & 1 << j));
}
答案 5 :(得分：2)
使用二进制数
// eg. [2,4,5] ==> {[],[2],[4],[5],[2,4],[4,5],[2,5], [2,4,5]}
var a = [2, 4, 5], res = [];
for (var i = 0; i < Math.pow(2, a.length); i++) {
var bin = (i).toString(2), set = [];
bin = new Array((a.length-bin.length)+1).join("0")+bin;
console.log(bin);
for (var j = 0; j < bin.length; j++) {
if (bin[j] === "1") {
set.push(a[j]);
}
}
res.push(set);
}
console.table(res);
答案 6 :(得分：1)
function combinations(array) {
return new Array(1 << array.length).fill().map(
(e1, i) => array.filter((e2, j) => i & 1 << j));
}
console.log(combinations([1, 2, 3]).filter(a => a.length >= 2))你们能解释一下吗？
答案 7 :(得分：0)
我已经修改了接受的解决方案，以便在min等于0时考虑空集（空集是任何给定集的子集）。
这是一个完整的示例页面，用于复制粘贴，准备好运行一些输出。
<html>
<head>
<meta http-equiv="Content-type" content="text/html;charset=UTF-8">
<title>All Subsets</title>
<script type="text/javascript">
// get all possible subsets of an array with a minimum of X (min) items and an unknown maximum
var FindAllSubsets = function(a, min) {
var fn = function(n, src, got, all) {
if (n == 0) {
if (got.length > 0) {
all[all.length] = got;
}
return;
}
for (var j = 0; j < src.length; j++) {
fn(n - 1, src.slice(j + 1), got.concat([src[j]]), all);
}
return;
}
var all = [];
// empty set is a subset of the set (only when min number of elements can be 0)
if(min == 0)
all.push([-1]); // array with single element '-1' denotes empty set
for (var i = min; i < a.length; i++) {
fn(i, a, [], all);
}
all.push(a);
return all;
}
function CreateInputList(){
var inputArr = [];
var inputArrSize = 4;
var maxInputValue = 10;
for(i=0; i < inputArrSize; i++){
var elem = Math.floor(Math.random()*maxInputValue);
// make sure to have unique elements in the array
while(inputArr.contains(elem)){ // OR - while(inputArr.indexOf(elem) > -1){
elem = Math.floor(Math.random()*maxInputValue);
}
inputArr.push(elem);
}
return inputArr;
}
Array.prototype.contains = function(obj) {
var i = this.length;
while (i--) {
if (this[i] === obj) {
return true;
}
}
return false;
}
function ArrayPrinter(arr){
var csv = 'input = [';
var i = 0;
for(i; i<arr.length - 1; i++){
csv += arr[i] + ', ';
}
csv += arr[i];
var divResult = document.getElementById('divResult');
divResult.innerHTML += csv + ']<br />';
}
// assumes inner array with single element being '-1' an empty set
function ArrayOfArraysPrinter(arr){
var csv = 'subsets = ';
var i = 0;
for(i; i<arr.length; i++){
csv += '[';
var j = 0;
var inArr = arr[i];
for(j; j<inArr.length - 1; j++){
csv += inArr[j] + ', ';
}
// array with single element '-1' denotes empty set
csv += inArr[j] == -1 ? '<E>' : inArr[j];
csv += ']';
if(i < arr.length - 1)
csv += ' ';
}
csv += ' (# of subsets =' + arr.length + ')';
var divResult = document.getElementById('divResult');
divResult.innerHTML += csv + '<br />';
}
function Main(){
// clear output
document.getElementById('divResult').innerHTML = '';
// sample run (min = 0)
document.getElementById('divResult').innerHTML += '<hr/>MIN = 0 (must include empty set)<br />';
var list = CreateInputList();
ArrayPrinter(list);
var subsets = FindAllSubsets(list, 0);
ArrayOfArraysPrinter(subsets);
document.getElementById('divResult').innerHTML += '<hr />';
// sample run (min = 1)
document.getElementById('divResult').innerHTML += 'MIN = 1<br />';
var list = CreateInputList();
ArrayPrinter(list);
var subsets = FindAllSubsets(list, 1);
ArrayOfArraysPrinter(subsets);
document.getElementById('divResult').innerHTML += '<hr />';
// sample run (min = 2)
document.getElementById('divResult').innerHTML += 'MIN = 2<br />';
var list = CreateInputList();
ArrayPrinter(list);
var subsets = FindAllSubsets(list, 2);
ArrayOfArraysPrinter(subsets);
document.getElementById('divResult').innerHTML += '<hr />';
// sample run (min = 3)
document.getElementById('divResult').innerHTML += 'MIN = 3<br />';
var list = CreateInputList();
ArrayPrinter(list);
var subsets = FindAllSubsets(list, 3);
ArrayOfArraysPrinter(subsets);
document.getElementById('divResult').innerHTML += '<hr />';
// sample run (min = 4)
document.getElementById('divResult').innerHTML += 'MIN = 4<br />';
var list = CreateInputList();
ArrayPrinter(list);
var subsets = FindAllSubsets(list, 4);
ArrayOfArraysPrinter(subsets);
document.getElementById('divResult').innerHTML += '<hr />';
}
</script>
</head>
<body>
<input type="button" value="All Subsets" onclick="Main()" />
<br />
<br />
<div id="divResult"></div>
</body>
</html>
答案 8 :(得分：0)
如果元素顺序很重要：
// same values, different order:
[1,2]
[2,1]
[1,3]
[3,1]
然后你可能还想考虑一种排列。
// ---------------------
// Permutation
// ---------------------
function permutate (src, minLen, maxLen){
minLen = minLen-1 || 0;
maxLen = maxLen || src.length+1;
var Asource = src.slice(); // copy the original so we don't apply results to the original.
var Aout = [];
var minMax = function(arr){
var len = arr.length;
if(len > minLen && len <= maxLen){
Aout.push(arr);
}
}
var picker = function (arr, holder, collect) {
if (holder.length) {
collect.push(holder);
}
var len = arr.length;
for (var i=0; i<len; i++) {
var arrcopy = arr.slice();
var elem = arrcopy.splice(i, 1);
var result = holder.concat(elem);
minMax(result);
if (len) {
picker(arrcopy, result, collect);
} else {
collect.push(result);
}
}
}
picker(Asource, [], []);
return Aout;
}
var combos = permutate(["a", "b", "c"], 2);
for(var i=0; i<combos.length; i++){
var item = combos[i];
console.log("combos[" + i + "]" + " = [" + item.toString() + "]");
}
警告!!! - 您的计算机无法处理包含＆gt; 10项的数组。
答案 9 :(得分：0)
首次提交答案！希望它能帮助某人。我将此用于类似的递归解决方案，其中涉及到返回数组数组，我发现 flatMap() 方法非常有用。
var arr = [1,2,3,4,5];
function getAllCombos(arr){
if(arr[0] === undefined) return [arr]
return getAllCombos(arr.slice(1)).flatMap(el => [el.concat(arr[0]), el])
}
console.log(JSON.stringify(getAllCombos(arr)));
如果您希望console.log（）输出以[1、2、3、4、5]而不是[5、4、3、2、1]开头，则可以添加sort()这样的混合方法：
var arr = [1,2,3,4,5];
function getAllCombos(arr){
if(arr[0] === undefined) return [arr]
return getAllCombos(arr.slice(1)).flatMap(el => [el.concat(arr[0]).sort(), el])
}
console.log(JSON.stringify(getAllCombos(arr)));
答案 10 :(得分：0)
这可能会也可能不会很好地进行基准测试，但这是另一种方式，而且非常简洁：
const combinations = arr => arr.reduce((acc, item) => {
return acc.concat(acc.map(x => [...x, item]));
}, [[]]);
console.log(combinations([1, 2, 3]).filter(a => a.length > 1));答案 11 :(得分：-1)
这里是C ++中的等效代码；
#include<iostream>
#include<math.h>
#include<vector>
using namespace std;
// Prints all Subset having min 2 elements
void printSubset(vector<int> subsetVect, int subsetElement) {
if( subsetElement>= 2) {
vector<int>::iterator itr;
cout<<"\n[";
for( itr = subsetVect.begin(); itr!=subsetVect.end(); itr++)
cout<<*itr<<',';
cout<<"]";
}
}
// Creates all possible subsets
void allSubsets( int arr[], int n ) {
/*set_size of power set(all possible subsets) of a set with set_size
n is ((2^n) -1)*/
int pow_set_size = pow(2, n);
int counter, j;
int subsetElement=0;
/*Run from counter 000..0 to 111..1*/
for( counter =0; counter< pow_set_size; counter++ ) {
vector<int> subsetVect;
subsetElement = 0;
for( j=0; j<n; j++) {
/* Check if jth bit in the counter is set
If set then print jth element from set */
if( counter & ( 1 << j )){
subsetVect.push_back(arr[j]);
subsetElement++;
}
}
printSubset(subsetVect, subsetElement);
}
}
int main(){
int arr[]={1,2,3}, n=3;
allSubsets(arr,n);// Ans. [1,2],[1,3],[2,3],[1,2,3]
}