JavaScript数组在奇数索引处拆分

Question

I have an array of products this contains a product name and a manufacturer, eg "product","manufacturer","product","manufacturer" etc..

How can I spilt the array into a separate array so that I can have the products and manufacturers in separate arrays? I want to spilt it at every odd index so that I can take the manufacture out of the array and add this into a new array?

Thanks.

Answer 1

/**
 * @param candid Array of results
 * @return Returns an array where index 0 = array of even ones, and index 1 = array of odd ones
*/
function splitArray(candid) {
    var oddOnes = [],
        evenOnes = [];
    for(var i=0; i<candid.length; i++)
        (i % 2 == 0 ? evenOnes : oddOnes).push(candid[i]);
    return [evenOnes, oddOnes];
}

Answer 2

Try with something like this:

var arr = ["product","manufacturer","product","manufacturer"];
var products = [];
var manufacturers = [];
for (var i = 0; i < arr.length; i += 2) {
    products.push(arr[i]);
    arr[i+1] && manufacturers.push(arr[i + 1]);
}

Answer 3

Underscore.js groupBy does the work pretty good.

_.groupBy(["product", "manufacturer","product", "manufacturer"], function(val, index){ return index % 2; });
=> {1: ["product", "product"], 2: ["manufacturer", "manufacturer"]}

Answer 4

This should work:

function separate(array) {
    var products = [],
        manufacturers = [];
    for (var i = 0, length = array.length; i < length; i++) {
        if (i % 2 === 0) {
            products.push(array[i]);    
        } else {
            manufacturers.push(array[i]);  
        }        
    }
    return {
        products: products,
        manufacturers: manufacturers 
    };
}

jsFiddle Example

Answer 5

Edit : my response was a little bit late to the show. I'd strongly prefer if you picked one of the other folks who did the same thing.

If your array looks like this:

var result = ["Product #1", "ABC Company", "Product #2", "DEF Company"];

You can extract the products and manufacturers using a for loop

var products = [];
var manufacturers = [];
for (var i = 0; i < results.length; ++i) {
    if (i % 2 === 0) {
        products.push(products[i]);
    } else {
        manufacturers.push(products[i]);
    }
}

Answer 6

Use math and the mod operator.

Get thru your array with an index (i, perhaps) and check the value of i

if (i % 2 == 0){
goes_to_even_array();
]else{
goes_odd_array();
}

Answer 7

Ok, here's another entry:

var arr = ['p0','m0','p1','m1','p2'];
var products = [];
var manufacturers = [];
var i = 0, iLen = arr.length;

while (i < iLen) {
  manufacturers.push(arr[i++]);
  products.push(arr[i++]);
}

and if order doesn't matter then:

var i = arr.length;
while (i) {
  i && manufacturers.push(arr[--i]);
  i && products.push(arr[--i]);
}

and if you absolutely trust length to be even:

var i = arr.length;
while (i) {
  manufacturers.push(arr[--i]);
  products.push(arr[--i]);
}

but that will go into an endless loop if length is odd.

Answer 8

Another method using recursion :)

var split = function(arr) {
    var splitHelper = function(arr, odd, even) {
        if (arr.length == 0) {
            return {odd: odd, even: even};
        }
        odd.push(arr.splice(0, 1)[0]);
        arr[0] && even.push(arr.splice(0, 1)[0]);
        return splitHelper(arr, odd, even);
    };
    return splitHelper(arr, [], []);
};

var arr = ["product", "manufacturer", "product", "manufacturer"];
var arrays = split(arr);
// Test
alert("odd: " + arrays.odd);
alert("even: " + arrays.even);