JavaScript - 从具有m个元素的n个数组生成组合

时间:2013-03-08 16:37:29

标签: javascript permutation combinations

我在编写代码时遇到麻烦,需要在JavaScript中生成包含m个元素的n个数组的组合。我已经在其他语言中看到了类似的问题,但答案中包含语法或库魔法,我不确定如何翻译。

考虑这些数据:

[[0,1], [0,1,2,3], [0,1,2]]

3个数组,其中包含不同数量的元素。我想要做的是通过组合每个数组中的项来获得所有组合。

例如:

0,0,0 // item 0 from array 0, item 0 from array 1, item 0 from array 2
0,0,1
0,0,2
0,1,0
0,1,1
0,1,2
0,2,0
0,2,1
0,2,2

等等。

如果数组的数量是固定的,那么很容易进行硬编码实现。但阵列的数量可能会有所不同:

[[0,1], [0,1]]
[[0,1,3,4], [0,1], [0], [0,1]]

非常感谢任何帮助。

10 个答案:

答案 0 :(得分:81)

这是一个非常简单和简单的使用递归辅助函数:

function cartesian() {
    var r = [], arg = arguments, max = arg.length-1;
    function helper(arr, i) {
        for (var j=0, l=arg[i].length; j<l; j++) {
            var a = arr.slice(0); // clone arr
            a.push(arg[i][j]);
            if (i==max)
                r.push(a);
            else
                helper(a, i+1);
        }
    }
    helper([], 0);
    return r;
}

用法:

cartesian([0,1], [0,1,2,3], [0,1,2]);

要使该函数采用数组数组,只需将签名更改为function cartesian(arg),以便arg是参数而不是all arguments

答案 1 :(得分:9)

您可以通过构建子数组来采用迭代方法。

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var parts = [[0, 1], [0, 1, 2, 3], [0, 1, 2]],
    result = parts.reduce((a, b) => a.reduce((r, v) => r.concat(b.map(w => [].concat(v, w))), []));

console.log(result.map(a => a.join(', ')));
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.as-console-wrapper { max-height: 100% !important; top: 0; }
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答案 2 :(得分:6)

我建议使用简单的递归generator function

// Generate all combinations of array elements:
function* cartesian(head, ...tail) {
  let remainder = tail.length ? cartesian(...tail) : [[]];
  for (let r of remainder) for (let h of head) yield [h, ...r];
}


// Example:
for (let c of cartesian([0,1], [0,1,2,3], [0,1,2])) {
  console.log(...c);
}

答案 3 :(得分:3)

在做了一点研究后,我发现了一个先前的相关问题: Finding All Combinations of JavaScript array values

我已经从那里调整了一些代码,以便它返回一个包含所有排列的数组数组:

function(arraysToCombine) {
    var divisors = [];
    for (var i = arraysToCombine.length - 1; i >= 0; i--) {
       divisors[i] = divisors[i + 1] ? divisors[i + 1] * arraysToCombine[i + 1].length : 1;
    }

    function getPermutation(n, arraysToCombine) {
       var result = [], 
           curArray;    
       for (var i = 0; i < arraysToCombine.length; i++) {
          curArray = arraysToCombine[i];
          result.push(curArray[Math.floor(n / divisors[i]) % curArray.length]);
       }    
       return result;
    }

    var numPerms = arraysToCombine[0].length;
    for(var i = 1; i < arraysToCombine.length; i++) {
        numPerms *= arraysToCombine[i].length;
    }

    var combinations = [];
    for(var i = 0; i < numPerms; i++) {
        combinations.push(getPermutation(i, arraysToCombine));
    }
    return combinations;
}

我在http://jsfiddle.net/7EakX/放了一个工作副本,它带有你之前给出的数组([[0,1],[0,1,2,3],[0,1,2]])并将结果输出到浏览器控制台。

答案 4 :(得分:3)

只是为了好玩,这是我的第一个答案中解决方案的更多功能变体:

function cartesian() {
    var r = [], args = Array.from(arguments);
    args.reduceRight(function(cont, factor, i) {
        return function(arr) {
            for (var j=0, l=factor.length; j<l; j++) {
                var a = arr.slice(); // clone arr
                a[i] = factor[j];
                cont(a);
            }
        };
    }, Array.prototype.push.bind(r))(new Array(args.length));
    return r;
}

替代方案,为了全速运行,我们可以动态编译自己的循环:

function cartesian() {
    return (cartesian.cache[arguments.length] || cartesian.compile(arguments.length)).apply(null, arguments);
}
cartesian.cache = [];
cartesian.compile = function compile(n) {
    var args = [],
        indent = "",
        up = "",
        down = "";
    for (var i=0; i<n; i++) {
        var arr = "$"+String.fromCharCode(97+i),
            ind = String.fromCharCode(105+i);
        args.push(arr);
        up += indent+"for (var "+ind+"=0, l"+arr+"="+arr+".length; "+ind+"<l"+arr+"; "+ind+"++) {\n";
        down = indent+"}\n"+down;
        indent += "  ";
        up += indent+"arr["+i+"] = "+arr+"["+ind+"];\n";
    }
    var body = "var res=[],\n    arr=[];\n"+up+indent+"res.push(arr.slice());\n"+down+"return res;";
    return cartesian.cache[n] = new Function(args, body);
}

答案 5 :(得分:2)

var f = function(arr){
    if(typeof arr !== 'object'){
        return false;
    }

    arr = arr.filter(function(elem){ return (elem !== null); }); // remove empty elements - make sure length is correct
    var len = arr.length;

    var nextPerm = function(){ // increase the counter(s)
        var i = 0;

        while(i < len)
        {
            arr[i].counter++;

            if(arr[i].counter >= arr[i].length){
                arr[i].counter = 0;
                i++;
            }else{
                return false;
            }
        }

        return true;
    };

    var getPerm = function(){ // get the current permutation
        var perm_arr = [];

        for(var i = 0; i < len; i++)
        {
            perm_arr.push(arr[i][arr[i].counter]);
        }

        return perm_arr;
    };

    var new_arr = [];

    for(var i = 0; i < len; i++) // set up a counter property inside the arrays
    {
        arr[i].counter = 0;
    }

    while(true)
    {
        new_arr.push(getPerm()); // add current permutation to the new array

        if(nextPerm() === true){ // get next permutation, if returns true, we got them all
            break;
        }
    }

    return new_arr;
};

答案 6 :(得分:2)

这是另一种做法。我将所有数组的索引视为一个数字,其数字都是不同的数字(如时间和日期),使用数组的长度作为基数。

所以,使用你的第一组数据,第一个数字是基数2,第二个数字是基数4,第三个数字是基数3.计数器开始000,然后是001,002,然后是010.数字对应于数组中的索引,并且由于保留了顺序,这没有问题。

我在这里工作的小提琴:http://jsfiddle.net/Rykus0/DS9Ea/1/

这是代码:

// Arbitrary base x number class 
var BaseX = function(initRadix){
    this.radix     = initRadix ? initRadix : 1;    
    this.value     = 0;
    this.increment = function(){
        return( (this.value = (this.value + 1) % this.radix) === 0);
    }
}

function combinations(input){
    var output    = [],    // Array containing the resulting combinations
        counters  = [],    // Array of counters corresponding to our input arrays
        remainder = false, // Did adding one cause the previous digit to rollover?
        temp;              // Holds one combination to be pushed into the output array

    // Initialize the counters
    for( var i = input.length-1; i >= 0; i-- ){
        counters.unshift(new BaseX(input[i].length));
    }

    // Get all possible combinations
    // Loop through until the first counter rolls over
    while( !remainder ){
        temp      = [];   // Reset the temporary value collection array
        remainder = true; // Always increment the last array counter

        // Process each of the arrays
        for( i = input.length-1; i >= 0; i-- ){
            temp.unshift(input[i][counters[i].value]); // Add this array's value to the result

            // If the counter to the right rolled over, increment this one.
            if( remainder ){
                remainder = counters[i].increment();
            }
        }
        output.push(temp); // Collect the results.
    }

    return output;
}

// Input is an array of arrays
console.log(combinations([[0,1], [0,1,2,3], [0,1,2]]));

答案 7 :(得分:1)

使用ES6递归样式的另一种实现

Array.prototype.cartesian = function(a,...as){
  return a ? this.reduce((p,c) => (p.push(...a.cartesian(...as).map(e => as.length ? [c,...e] : [c,e])),p),[])
           : this;
};

console.log(JSON.stringify([0,1].cartesian([0,1,2,3], [[0],[1],[2]])));

答案 8 :(得分:1)

您可以使用递归函数获取所有组合

const charSet = [["A", "B"],["C", "D", "E"],["F", "G", "H", "I"]];

let loopOver = (arr, str = '', final = []) => {
  if (arr.length > 1) {
    arr[0].forEach(v => loopOver(arr.slice(1), str + v, final))
  } else {
    arr[0].forEach(v => final.push(str + v))
  }
  return final
}

console.log(loopOver(charSet))


使用三进制仍可以简化此代码,但出于可读性的考虑,我更喜欢第一个版本

const charSet = [["A", "B"],["C", "D", "E"],["F", "G", "H", "I"]];

let loopOver = (arr, str = '') => arr[0].map(v => arr.length > 1 ? loopOver(arr.slice(1), str + v) : str + v).flat()

console.log(loopOver(charSet))

答案 9 :(得分:0)

const charSet = [["A", "B"],["C", "D", "E"],["F", "G", "H", "I"]];
console.log(charSet.reduce((a,b)=>a.flatMap(x=>b.map(y=>x+y)),['']))