How do I replace repetitive elements in an array matching another array?

Question

I have two arrays:

array_main = [23432, 3434, 312, 32432] 
array_second = [23432, 312]

I want to replace the elements in array_main with 0 matching the elements of array_second, so the output should look like:

array_main = [0, 3434, 0, 32432]

How do I do it?

Boris Stitnicky · Accepted Answer · 2013-06-06 21:59:03Z

5

array_main.map { |e| array_second.include?( e ) ? 0 : e }

And if you drop that requirement about replacing with 0, you can simply write

array_main - array_second

answered Jun 6, 2013 at 21:59

Boris Stitnicky

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5 Comments

Zippie Over a year ago

i am not sure if the questioner wanted to replace it in place ( in the array_main), but if so, then (s)he should use the destructive method map!, instead of map

Boris Stitnicky Over a year ago

@Zippie: Since which verion of Ruby is map! available, that I managed to miss it?

Boris Stitnicky Over a year ago

@Zippie: That map! is a funny name anyway, who has ever heard of mappings destroying their domain? It's real fun to try e = array_main.map! and then manually call e.next several times, and see how array_main is destroyed. Is this useful, or what?

the Tin Man Over a year ago

Why is map! funny? It's an in-place mutation of the original array, which is useful if you don't want to create a new variable. It's potentially faster too, because no extra variable allocation should be happening. And, it's been part of Ruby for a long time.

Boris Stitnicky Over a year ago

@theTinMan: No, no, I don't mean that map! itself is funny, it is an amazing piece of functionality, that I regret is not available also for Hash. Just the name is funny, map with bang!!! Modifies the functional domain!!! That sounds like a math joke, it makes me laugh. Functional programming flowing in reverse!!! a = 1, 2, 3; e = a.map!; e.each { 42 }. (I'm drunk again.)

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