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Martin Krause
Martin Krause

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1 line of code: How to get every odd item of an Array

const oddItems = arr => arr.filter((_, i) => i & 1 === 1);
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Returns an array which contains every odd (second) item of the original array.


The repository & npm package

You can find the all the utility functions from this series at github.com/martinkr/onelinecode
The library is also published to npm as @onelinecode for your convenience.

The code and the npm package will be updated every time I publish a new article.


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Oldest comments (12)

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jonrandy profile image
Jon Randy πŸŽ–οΈ

Not looked for faster methods yet, but the current one seems overly verbose. Also, what is which for?

const oddItems = arr => arr.filter((_, index) => index % 2)
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martinkr profile image
Martin Krause

Thank you for your contribution. The "which" is a copy paste error :D
If you have any performance improvements, please share them and I'll happily adjust the code and the article.

Cheers,

Martin

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lexlohr profile image
Alex Lohr

Binary filters are even faster than the modulo:

const oddItems = arr => arr.filter((_, i) => i & 1 === 1);
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martinkr profile image
Martin Krause

Hi Alex,

I checked the performance on hasty and indeed much faster. I updated the code and the article.

Thank you!

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jonrandy profile image
Jon Randy πŸŽ–οΈ • Edited

Faster still if you omit the unnecesary === 1 - it's about even on Firefox, but consistently quicker on Chrome (10-15%) - link

const oddItems = arr => arr.filter((_, i) => i & 1)
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jonrandy profile image
Jon Randy πŸŽ–οΈ • Edited

Much faster again (on all tested browsers):

const oddItems = (arr, odds=[], i=0)=>{ for (i=0;i<arr.length;i++) (i & 1) && odds.push(arr[i]); return odds}
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martinkr profile image
Martin Krause

Hi Jon Randy,

I updated the benchmark on hasty - impressive improvement!
I updated the code and the article.

Thank you for the improved code.

Cheers!

Thank you!

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jonrandy profile image
Jon Randy πŸŽ–οΈ • Edited

You might want to make the benchmark call the functions as well as define them! :P

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giulio profile image
Giulio "Joshi"

wouldn't jumping 2 steps ensure only odd positional, but not values?
(I just assume an unsorted array of random values)

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jonrandy profile image
Jon Randy πŸŽ–οΈ

Odd positional is what we're after

 
lexlohr profile image
Alex Lohr

If you are already using a for loop, you can also jump 2 steps on every iteration:

const oddItems = (arr, odds=[], i) => { for (i = 1; i < arr.length; i = i + 2) odds.push(arr[i]); return odds}
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jonrandy profile image
Jon Randy πŸŽ–οΈ

Haha... yeah - oops

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