Javascript 检查一个数组是否是另一个数组的子集
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Check if an array is subset of another array
提问by yangmei
Let's say I have two arrays,
假设我有两个数组，
var PlayerOne = ['B', 'C', 'A', 'D'];
var PlayerTwo = ['D', 'C'];
What is the best way to check if arrayTwo is subset of arrayOne using javascript?
使用javascript检查arrayTwo是否是arrayOne的子集的最佳方法是什么？
The reason: I was trying to sort out the basic logic for a game Tic tac toe, and got stuck in the middle. Here's my code anyway... Thanks heaps!
原因：我试图理清一个游戏井字游戏的基本逻辑，结果卡在了中间。无论如何，这是我的代码......谢谢！
var TicTacToe = {
PlayerOne: ['D','A', 'B', 'C'],
PlayerTwo: [],
WinOptions: {
WinOne: ['A', 'B', 'C'],
WinTwo: ['A', 'D', 'G'],
WinThree: ['G', 'H', 'I'],
WinFour: ['C', 'F', 'I'],
WinFive: ['B', 'E', 'H'],
WinSix: ['D', 'E', 'F'],
WinSeven: ['A', 'E', 'I'],
WinEight: ['C', 'E', 'G']
},
WinTicTacToe: function(){
var WinOptions = this.WinOptions;
var PlayerOne = this.PlayerOne;
var PlayerTwo = this.PlayerTwo;
var Win = [];
for (var key in WinOptions) {
var EachWinOptions = WinOptions[key];
for (var i = 0; i < EachWinOptions.length; i++) {
if (PlayerOne.includes(EachWinOptions[i])) {
(got stuck here...)
}
}
// if (PlayerOne.length < WinOptions[key]) {
// return false;
// }
// if (PlayerTwo.length < WinOptions[key]) {
// return false;
// }
//
// if (PlayerOne === WinOptions[key].sort().join()) {
// console.log("PlayerOne has Won!");
// }
// if (PlayerTwo === WinOptions[key].sort().join()) {
// console.log("PlayerTwo has Won!");
// } (tried this method but it turned out to be the wrong logic.)
}
},
};
TicTacToe.WinTicTacToe();
回答by simhumileco
Here is the correct solution:
这是正确的解决方案：
Using ES7(ECMAScript 2016):
使用 ES7（ECMAScript 2016）：
const result = PlayerTwo.every(val => PlayerOne.includes(val));
Snippet:
片段：
const PlayerOne = ['B', 'C', 'A', 'D'];
const PlayerTwo = ['D', 'C'];
const result = PlayerTwo.every(val => PlayerOne.includes(val));
console.log(result);Using ES5(ECMAScript 2009):
使用 ES5( ECMAScript 2009)：
var result = PlayerTwo.every(function(val) {
return PlayerOne.indexOf(val) >= 0;
});
Snippet:
片段：
var PlayerOne = ['B', 'C', 'A', 'D'];
var PlayerTwo = ['D', 'C'];
var result = PlayerTwo.every(function(val) {
return PlayerOne.indexOf(val) >= 0;
});
console.log(result);回答by tothemario
If you are using ES6:
如果您使用的是 ES6：
!PlayerTwo.some(val => PlayerOne.indexOf(val) === -1);
If you have to use ES5, use a polyfill for the somefunction the Mozilla documentation, then use regular function syntax:
如果您必须使用 ES5，请为Mozilla 文档中的some函数使用 polyfill ，然后使用常规函数语法：
!PlayerTwo.some(function(val) { return PlayerOne.indexOf(val) === -1 });
回答by SuperNova
You can use this simple piece of code.
您可以使用这段简单的代码。
PlayerOne.every(function(val) { return PlayerTwo.indexOf(val) >= 0; })
回答by SuperNova
If PlayerTwo is subset of PlayerOne, then length of set(PlayerOne + PlayerTwo) must be equal to length of set(PlayerOne).
如果 PlayerTwo 是 PlayerOne 的子集，则 set(PlayerOne + PlayerTwo) 的长度必须等于 set(PlayerOne) 的长度。
var PlayerOne = ['B', 'C', 'A', 'D'];
var PlayerTwo = ['D', 'C'];
// Length of set(PlayerOne + PlayerTwo) == Length of set(PlayerTwo)
Array.from(new Set(PlayerOne) ).length == Array.from(new Set(PlayerOne.concat(PlayerTwo)) ).length
回答by Armen Zakaryan
function isSubsetOf(set, subset) {
return Array.from(new Set([...set, ...subset])).length === set.length;
}
回答by Sanford Staab
This seems most clear to me:
这对我来说似乎最清楚：
function isSubsetOf(set, subset) {
for (let i = 0; i < set.length; i++) {
if (subset.indexOf(set[i]) == -1) {
return false;
}
}
return true;
}
It also has the advantage of breaking out as soon as a non-member is found.
它还具有在发现非会员后立即爆发的优势。
回答by dmmfll
Here is a solution that exploits the set data type and its hasfunction.
这是一个利用集合数据类型及其has功能的解决方案。
let PlayerOne = ['B', 'C', 'A', 'D', ],
PlayerTwo = ['D', 'C', ],
[one, two] = [PlayerOne, PlayerTwo, ]
.map( e => new Set(e) ),
matches = Array.from(two)
.filter( e => one.has(e) ),
isOrisNot = matches.length ? '' : ' not',
message = `${PlayerTwo} is${isOrisNot} a subset of ${PlayerOne}`;
console.log(message)
Out: D,C is a subset of B,C,A,D
