将多个数组中的每个项平均化为一个数组

Question

我的大脑正在融化。。。我正在努力实现以下目标：

我知道有多少个数组，每个数组有多少个元素。这些数字是动态的，但假设有：3个数组，每个数组中有18个元素。

例子：

["106","142","112","77","115","127","87","127","156","118","91","93","107","151","110","79","40","186"]

["117","139","127","108","172","113","79","128","121","104","105","117","139","109","137","109","82","137"]

["111","85","110","112","108","109","107","89","104","108","123","93","125","174","129","113","162","159"]

现在我想得到所有三个数组的元素1的平均值，以及所有三个数组的元素2的平均值，以此类推。

最终结果应该是所有18个元素的平均值的一个数组。

比如：

var result_array = [];
for (i = 0; i < 3; i++) {  
  result_array.push(arrayone[i] + arraytwo[i] + arraythree[i]) / 3
}

如果3是固定的，这将起作用，但数组的数量是动态的。

希望这有意义...

Answer 1

这个解决方案是用Java编写的，但是你可以在逻辑方面得到帮助。希望有帮助。

ArrayList<Integer> averageArrayList = new ArrayList<>();
int arrayListLength = arrayList1.length(); //assuming previous arraylists have same size.


   for(int i=0; i<arrayListLength; i++){
        int averageValue = (arrayList1.get(i) + arrayList2.get(i) + arrayList3.get(i)) / 3;
        //adds average value to current index.
        averageArrayList.add(i, averageValue);
    }
//averageArrayList is ready with your values..

Answer 2

您确实用underscore.js标记了这个问题，所以您可以使用_. zip方法。这将数组中的所有第一个元素放在一起，依此类推。然后您可以对每个数组进行平均。

见代码笔。

js prettyprint-override">var arr1 = ["106","142","112","77","115","127","87","127","156","118","91","93","107","151","110","79","40","186"]
var arr2 = ["117","139","127","108","172","113","79","128","121","104","105","117","139","109","137","109","82","137"]
var arr3 = ["111","85","110","112","108","109","107","89","104","108","123","93","125","174","129","113","162","159"]
// ... as many more arrays as you want

var avgEmAll = function (arrays) {
  // zip with array of arrays https://stackoverflow.com/a/10394791/327074
  return _.zip.apply(null, arrays).map(avg)
}

// average an array https://stackoverflow.com/a/10624256/327074
var avg = function (x) {
  return x.reduce(function (y, z) {return Number(y) + Number(z)}) / x.length
}
console.log(avgEmAll([arr1, arr2, arr3]))

使用ES6箭头功能（代码笔）：

const avgEmAll = arrays => _.zip.apply(null, arrays).map(avg)
const sum = (y, z) => Number(y) + Number(z)
const avg = x => x.reduce(sum) / x.length

Answer 3

var arrays = [
    [106,142,112,77,115,127,87,127,156,118,91,93,107,151,110,79,40,186],
    [117,139,127,108,172,113,79,128,121,104,105,117,139,109,137,109,82,137],
    [111,85,110,112,108,109,107,89,104,108,123,93,125,174,129,113,162,159],
    [104,153,110,112,108,109,107,89,104,108,123,93,125,174,129,113,162,159]
    /* Can be any amount of arrays */
    ],
    result = [];

//Rounding to nearest whole number.
for(var i = 0; i < arrays[0].length; i++){
  var num = 0;
  //still assuming all arrays have the same amount of numbers
  for(var i2 = 0; i2 < arrays.length; i2++){ 
    num += arrays[i2][i];
  }
  result.push(Math.round(num / arrays.length));
}

alert(result);