Add element to the beginning of array

Question

I have a JS array so:

var car_type_1 = [12,16,23,31,33,34,39,45,49,54,59,62,62,62,63,63,64,66,69,71,73,75,78,80,82,85,87,89,91,94,95]; var car_type_2 = [17,20,28,35,37,38, and so on];

I created another array which stores the subtracted values between the consecutive numbers in the old array:

 function diff(arr) { return arr.slice(1).map(function(n, i) { return n - arr[i]; }); } var car_type_1_diff = diff(car_type_1); //so on

Understandly, the new array has 1 less element compared to the old array. What I want to do is add the first number of the old array to the new array, so it should be like:

 car_type_1_diff = [**12**, 4, 7, 8, 2, 1, 5, 6, 4, 5, 5, 3, 0, 0, 1, 0, 1, 2, 3, 2, 2, 2, 3, 2, 2, 3, 2, 2, 2, 3, 1]

12 should be pushed to the beginning of the new array 1. How can I do this?

Others have answered this below with words. I'm dropping in a link to devdocs.io here as it's infinitely more useful than the MDN link provided: devdocs.io/javascript/global_objects/array/unshift — vhs
– vhs, Commented Jun 4, 2017 at 9:06

vhs · Accepted Answer · 2017-06-04 11:49:09Z

3

In ES6, when constructing the new array car_type_1_diff you can also use the spread operator like so: var car_type_1_diff = [12, ...diff(car_type_1)]

edited Jun 4, 2017 at 11:49

vhs

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answered May 2, 2017 at 22:37

Tom Chen

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1 Comment

vhs Over a year ago

This is the preferred approach when compared to unshift when array mutation is not desirable and helps build more predictable applications.

Bruno Peres · Accepted Answer · 2017-05-02 22:35:31Z

Yes, you can use unshift method.

Example from the referenced docs:

var arr = [1, 2]; arr.unshift(0); // result of call is 3, the new array length // arr is [0, 1, 2] arr.unshift(-2, -1); // = 5 // arr is [-2, -1, 0, 1, 2] arr.unshift([-3]); // arr is [[-3], -2, -1, 0, 1, 2]

Note that unshift returns an integer and mutates the array, a behavior which may be undesirable if you're shooting for immutable state.

ioakeimo · Accepted Answer · 2017-05-02 22:33:28Z

0

Possibly you are looking for unshift() method.

answered May 2, 2017 at 22:33

ioakeimo

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1 Comment

vhs Over a year ago

It would be helpful to see an example usage of unshift in the answer.

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