Replace the value at a specific index (array)

Question

Good evening everyone,

I have an array which contains 24 characters

"hours": [
        "1",
        "1",
        "1",
        "1",
        "0",
        "0",
        "0",
        "0",
        "0",
        "0",
        "0",
        "0",
        "0",
        "0",
        "0",
        "0",
        "0",
        "0",
        "0",
        "0",
        "0",
        "0",
        "0",
        "0"
    ]

I would like to search through another array of indexes; for example [2, 5, 9, 10]

Then, go to the positions(of the first array) (2, 5, 9 and 10) of the first array and change the character to this position by a "1", and move the rest of the array to "0"

Can someone guide me?

Thank you !

Please provide a better example of the second array, and how do you get 2, 5, 9 and 10 from this second array? — Martin Geldart
– Martin Geldart, Commented Aug 19, 2018 at 19:32
Yes for sure Martin, I tried for (var i = 0; i < hours.length; i++) { if(i === positions[i]) { hours[i] = "1" } else { hours[i] = "0" } } the problem is that after, indexArray went to undefined:/ — Miguelit0
– Miguelit0, Commented Aug 19, 2018 at 20:44

Martin Geldart · Accepted Answer · 2018-08-19 20:28:10Z

2

A solution:

var hours = ["1", "1", "1", "1", "0", "0", "0", "0", "0", "0", "0", "0", "0", "0", "0", "0", "0", "0", "0", "0", "0", "0", "0", "0"],
    positions = [2, 5, 9, 10];

for (var i = 0; i < hours.length; i++) {
  hours[i] = positions.includes(i) ? "1" : "0";   
}

console.log(hours);

// ["0", "0", "1", "0", "0", "1", "0", "0", "0", "1", "1", "0", "0", "0", "0", "0", "0", "0", "0", "0", "0", "0", "0", "0"]

Demo: https://jsfiddle.net/gzynud0c/1/

As pointed out in the comments, the above solution works in all major browsers, including IE9+, without using a transpiler like Babel or using a polyfill.

The following works in all browsers.

for (var i = 0; i < hours.length; i++) {
  hours[i] = ~positions.indexOf(i) ? "1" : "0";   
}

Demo: https://jsfiddle.net/s59jt870/3/

edited Aug 19, 2018 at 20:28

answered Aug 19, 2018 at 19:53

Martin Geldart

4172 silver badges8 bronze badges

Sign up to request clarification or add additional context in comments.

5 Comments

TheCarver Over a year ago

+1 because unless the OP is using a transpiler like Babel or polyfills for ES6 features, this is the only answer that will work in IE (9+).

Martin Geldart Over a year ago

I'm guessing by the experience level of the question, the chances of the OP using ES6 are slim. I've updated my answer to include a cross-browser solution, too.

Koushik Chatterjee Over a year ago

Remember, searching each index in the lookup array will be performed for each index of the actual array and for those where the indexes will be missing it will search the entire lookup array.

Martin Geldart Over a year ago

Totally agreed... but the "entire lookup array" will only ever be 24 long. If it were unlimited, I'd be more concerned.

Koushik Chatterjee Over a year ago

lookup array length is 5 and actual array lengthis 24, so approx 24*5, assume a 50*20 and not bigger one. so 1000 isn't it? where it can be achieved by only 50+10 i.ie 60 operations

Koushik Chatterjee · Accepted Answer · 2018-08-19 21:23:53Z

2

I feel the most simple approach here will be to set all the values of hours array (mutate or not depends on requirement) to "0" first, and then visit only those locations we have to set "1" and set it there!

indices.reduce((h,v)=>(h[v]="1") && h, hours.map(()=>"0"))

let hours = ["1", "1", "1", "1", "0", "0", "0", "0", "0", "0", "0", "0", "0", "0", "0", "0", "0", "0", "0", "0", "0", "0", "0", "0"],
    indices = [2, 5, 9, 10];

let res = indices.reduce((h,v)=>(h[v]="1") && h, hours.map(()=>"0"));

console.log(res);

.as-console-wrapper { max-height: 100% !important; top: 0; }

edited Aug 19, 2018 at 21:23

answered Aug 19, 2018 at 20:16

Koushik Chatterjee

4,1933 gold badges20 silver badges32 bronze badges

5 Comments

Martin Geldart Over a year ago

Nice! I'd say shortest not simplest :)

Koushik Chatterjee Over a year ago

this will just perform minimal number of operations as well!

Koushik Chatterjee Over a year ago

@Miguelit0 why not?

Miguelit0 Over a year ago

@Koushik Chatterjee Oh no no it's perfect as a solution! It's just that I'm not yet at your level! he he he he

Martin Geldart Over a year ago

This is the most efficient answer for sure. However, Miguelit0 and other users will need to be using ES6 features to use those arrow functions otherwise they won't work in any version of IE. And reduce() only works in >IE9. I'm guessing Miguelit0 is not yet using ES6. Excellent answer though +1

Nina Scholz · Accepted Answer · 2018-08-19 19:58:35Z

1

With a sorted indices array, you could take an index j for it and check the looping index i with that value and change the indices array index and return either '1' or '0'.

var hours = ["1", "1", "1", "1", "0", "0", "0", "0", "0", "0", "0", "0", "0", "0", "0", "0", "0", "0", "0", "0", "0", "0", "0", "0"],
    indices = [2, 5, 9, 10],
    replaced = hours.map((j => (v, i) => indices[j] === i && ++j ? '1' : '0')(0));

console.log(replaced.join(' '));

.as-console-wrapper { max-height: 100% !important; top: 0; }

edited Aug 19, 2018 at 19:58

answered Aug 19, 2018 at 19:53

Nina Scholz

388k26 gold badges367 silver badges417 bronze badges

Comments

Mark · Accepted Answer · 2018-08-19 21:18:57Z

It sounds like you want to look at all the items in the first item and forEach item if it's index in it in the array of index's change the value to 1 otherwise 0, right?. You can almost translate the sentence into code:

let hours= [ "1", "1", "1", "1", "0", "0", "0", "0", "0", "0", "0", "0", "0", "0", "0", "0", "0", "0", "0", "0", "0", "0", "0", "0" ]
let indexes = [2, 5, 9, 10]

hours.forEach((item, index, self) => self[index] = indexes.includes(index) ? '1' : '0' )

console.log(hours)

This mutates the original array. If you wanted a new array you could do something similar with map():

let hours= [ "1", "1", "1", "1", "0", "0", "0", "0", "0", "0", "0", "0", "0", "0", "0", "0", "0", "0", "0", "0", "0", "0", "0", "0" ]
let indexes = [2, 5, 9, 10]

// leave hours as is and create a new array
let newArray = hours.map((item, index, self) =>  indexes.includes(index) ? '1' : '0' )

console.log(newArray)

If your arrays are very large, this won't be efficient since you loop through the indexes array with every iteration. If that's the case you can convert your indexes array to something like set that allows for constant time lookups. Alternatively you could just start with an array of zeros and set the ones:

let hours= [ "1", "1", "1", "1", "0", "0", "0", "0", "0", "0", "0", "0", "0", "0", "0", "0", "0", "0", "0", "0", "0", "0", "0", "0" ]
let indexes = [2, 5, 9, 10]

let newHours = indexes.reduce((a, index) => (a[index] = '1', a), Array.from(hours).fill("0"))

console.log(newHours)

Thank you ! It gives me a good idea of how to work with the map

Collectives™ on Stack Overflow

Replace the value at a specific index (array)

4 Answers 4

5 Comments

5 Comments

Comments

1 Comment

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

4 Answers 4

5 Comments

5 Comments

Comments

1 Comment

Your Answer

Sign up or log in

Post as a guest

Related