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

Martin Krause on November 09, 2021

const oddItems = arr => arr.filter((_, i) => i & 1 === 1); Enter fullscreen mode Exit fullscreen mode Ret...
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Jon Randy 🎖️ •

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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Alex Lohr •

Binary filters are even faster than the modulo:

const oddItems = arr => arr.filter((_, i) => i & 1 === 1);
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Jon Randy 🎖️ • • 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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Jon Randy 🎖️ • • 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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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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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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Jon Randy 🎖️ •

Haha... yeah - oops

 
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Jon Randy 🎖️ • • Edited

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

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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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Jon Randy 🎖️ •

Odd positional is what we're after

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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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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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