Javascript：两个数组相乘和求和

Question

I have two arrays of equal length, and I need to multiply the corresponding (by index) values in each, and sum the results.

For example

var arr1 = [2,3,4,5];
var arr2 = [4,3,3,1];

would result in 34 (4*2+3*3+4*3+5*1).

What's the simplest to read way to write this?

Answer 1

var arr1 = [2,3,4,5];
var arr2 = [4,3,3,1];
console.log(arr1.reduce(function(r,a,i){return r+a*arr2[i]},0));
34

This shows the "functional" approach rather than the "imperative" approach for calculating the dot product of two vectors. Functional approach (which tends to be more concise) is preferred in such a simple function implementation as requested by the OP.

Answer 2

var sum = 0;
for(var i=0; i< arr1.length; i++) {
    sum += arr1[i]*arr2[i];
}

Answer 3

var a = [1,2,3,4,5];
var b = [5,4,3,2,1];

a.map(function(x, index){ //here x = a[index]
 return b[index] + x 
});

=>[6,6,6,6,6]

//if you want to add the elements of an array:

a.reduce(function(x, y){
 return x + y
});

=>15

You can read about Array.map here . and Array.reduce here

Answer 4

Other answers are almost certainly more efficient, but just to give you a recursive viewpoint (it's nicer in some other languages). It does assume the two arrays are of equal length as you didn't specify what to do if they're not.

function sumProducts(array1, array2) {
    if(array1.length) 
        return array1.pop() * array2.pop() + sumProducts(array1, array2);

    return 0;
}

Edit:

katspaugh suggested flipping the returns which is ever so slightly more efficient (don't have to ! the length).

Answer 5

var arr1 = [2,3,4,5];
var arr2 = [4,3,3,1];


var result = 0;
for (var i=0; i < arr1.length; i++) {
  result += (arr1[i] * arr2[i]);
}

alert(result);

Try it here: http://jsfiddle.net/VQKPt/

Answer 6

var i, result = 0;
for(i = 0; i < arr1.length; i++)
    result += arr1[i]*arr2[i];
alert(result);

Not that this will cause an error if arr2 is shorter than arr1, but you said they're equal length, so I didn't bother checking for it.

Answer 7

My vote for simplest-to-read way to write this goes to the humble for loop:

var ii, sumOfProds = 0;
for (ii = 0; ii < arr1.length && ii < arr2.length; ii++) {
    sumOfProds += arr1[ii] * arr2[ii];
}

Answer 8

function mul (arr1, arr2) {
    var n_array = (arr1,arr2).map(x => x * x)
    return n_array
    }
var a = [1,2,3]
var b = [1,2,3]
console.log(mul(a,b))

Answer 9

Something like this:

var sum = 0;
for (var i=0, len = arr1.length; i < len; i++) {     // optimized looping
   sum += arr1[i] * arr2[i];
}

Answer 10

This seems pretty straight forward to me

var result=0;
for (var i=0; i<arr1.length;i++){
    result+=arr1[i]*arr2[i];   
}

Answer 11

Declare functions that does the operations you want.

var sum      = function(a, b){ return a + b; };
var subtract = function(a, b){ return a - b; };
var multiply = function(a, b){ return a * b; };
var divide   = function(a, b){ return a / b; };

Then you have a very readable way to perform any operation on two arrays like this:

var array1 = [1,2,3];
var array2 = [2,4,8];

operateArrays(array1, array2, sum);      //[3, 6, 11]
operateArrays(array1, array2, subtract); //[-1, -2, -5]
operateArrays(array1, array2, multiply); //[2, 8, 24]
operateArrays(array1, array2, divide);   //[0.5, 0.5, 0.375]

Using this function

/**
* Divide each number of an array of numbers by another array of numbers
* @param  {Array}    arrayA  The array of numbers
* @param  {Array}    arrayB  The array of numbers
* @param  {Function} fn      Function that performs an operation
* @return {Array}            The resulting array
* @author Victor N. www.vitim.us
*/
function operateArrays(arrayA, arrayB, fn){
    if(arrayA.length!==arrayB.length) throw new Error("Cannot operate arrays of different lengths");
    return arrayB.map(function(b, i){
        return fn(arrayA[i], b);
    });
}

Answer 12

var arr = [1,2,3,4];
var arr2 = [1,1,1,2];

You can use:

var squares = arr.concat(arr2).reduce((t,n)=>t+n);

or:

var squares = arr.map((a, i) => a + arr2[i]).reduce((t,n) => t+n);

console.log(squares);

Answer 13

使用 ES6 .reduce() 的单行解决方案：

const sum_products = arr1.reduce((sum, val, i) => sum + (val * arr2[i]), 0)

Answer 14

Here are 3 functions that all accomplish the same thing. They are listed in order of increasingly modern JavaScript syntax.

The first function uses basic language features that have been around since the beginning. This will work on really old browsers such as IE6.

The second function uses features from ECMASScript 5 which was introduced around 2009, and will work on IE9+.

The third function uses ECMASScript 2015 arrow functions and will not work on any version of IE, not even IE 11 unless you are using some type of build-step in your code to transpile down to ES5 like Babel or Typescript

/**
 * Multiplies the corresponding values of two arrays and then sums the product.
 * Supported in all legacy browsers
 * @param {number[]} arr1 The first array
 * @param {number[]} arr2 The second array
 **/
function sumOfProductsECMAScript4(arr1, arr2) {
    var total = 0;
    for(var i = 0; i < arr1.length; i += 1) {
        total += arr1[i] * arr2[i];
    }
}

/**
 * Multiplies the corresponding values of two arrays and then sums the product.
 * Supported in all mainstream browsers except IE 8 and older.
 * @param {number[]} arr1 The first array
 * @param {number[]} arr2 The second array
 **/
function sumOfProductsECMASScript5(arr1, arr2) {
    // The map function creates a new array of the product of arr1 and arr2 values
    // The reduce function then takes this array of products and adds up all the values
    return arr1
        .map(function (value, index) { return value * arr2[index]; })
        .reduce(function (sum, current) { return sum + current; }, 0);
}

/**
 * Multiplies the corresponding values of two arrays and then sums the product.
 * Supported in all mainstream browsers except IE.
 * @param {number[]} arr1 The first array
 * @param {number[]} arr2 The second array
 **/
function sumOfProductsECMASScript2015(arr1, arr2) {
    // The map function creates a new array of the product of arr1 and arr2 values
    // The reduce function then takes this array of products and adds up all the values
    return arr1
        .map((v, i) => v * arr2[i])
        .reduce((sum, v) => sum + v, 0);
}



// Define your arrays
let arr1 = [2,3,4,5];
let arr2 = [4,3,3,1];

// Usage
let answer = sumOfProductsECMASScript4(arr1, arr2);
let answer2 = sumOfProductsECMASScript5(arr1, arr2);
let answer3 = sumOfProductsECMASScript2015(arr1, arr2);

Answer 15

if you were doing much of this kind of thing, you would probably use a library like mathjs :

var arr1 = [2,3,4,5];
var arr2 = [4,3,3,1];
result=math.dot(arr1,arr2);

Answer 16

make it map and reduce

 var arr1 = [2,3,4,5]; var arr2 = [4,3,3,1]; var result = arr1.map((v,i) => v * arr2[i]).reduce((x, y) => x + y, 0) console.log(result)