I find a lot of reference about removing duplicates in ruby but I cannot find how to create duplicate.
If I have an array like [1,2,3] how can I map it to an array with dubbed items? [1,1,2,2,3,3]
Is there a method?
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We'd like to see what you've tried. Without evidence of your effort it looks like you didn't do research or tried things, which are required. Please read "How much research effort is expected of Stack Overflow users?".the Tin Man– the Tin Man2017-03-21 22:42:03 +00:00Commented Mar 21, 2017 at 22:42
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4 Answers
Here's yet another way, creating the array directly with Array#new :
array = [1, 2, 3]
repetitions = 2
p Array.new(array.size * repetitions) { |i| array[i / repetitions] }
# [1, 1, 2, 2, 3, 3]
According to fruity, @ursus's answer, @ilya's first two answers and mine have comparable performance. transpose.flatten is slower than any of the others.
Comments
@Ursus answer is the most clean, there are possible solutions:
a = [1, 2, 3]
a.zip(a).flatten
#=> [1, 1, 2, 2, 3, 3]
Or
a.inject([]) {|a, e| a << e << e} # a.inject([]) {|a, e| n.times {a << e}; a}
=> [1, 1, 2, 2, 3, 3]
Or
[a, a].transpose.flatten # ([a] * n).transpose.flatten
=> [1, 1, 2, 2, 3, 3]
4 Comments
Sagar Pandya
A slight variation
a.each_with_object([]) { |e,o| 2.times {o.push e} }
Ilya
@sagarpandya82, thanks for the variation. There are too many possible variants, I prefer mine, it seems more readable :)
Eric Duminil
How would you modify the methods to accept an arbitrary number of repetitions?
Ilya
@Eric, I added some comments. For the first example it's possible (
zip n times) but looks ugly and slow, let me know if you are still need this :)Try this:
[1, 2, 3] * 2
=> [1, 2, 3, 1, 2, 3]
You might want it sorted:
([1, 2, 3] * 2).sort
=> [1, 1, 2, 2, 3, 3]
5 Comments
Ruslan
This is by far the most clean version
mu is too short
You're making some assumptions here. What happens if the array is
[11, 6, 42, 23]?Eric Duminil
@Ruslan: Nope. This answer might look clean, but it's wrong. What about
[2,1] or [1,'2'] ?Ruslan
One way around the sorting:
([1, '2'] * 2).sort {|x, y| x.to_s <=> y.to_s}
seph
@muistooshort - Maybe you are the one making assumptions. Yes, there is a dearth of information in the question. But making stuff up and saying that is what they have in mind is definitely wrong. Continuing the conversation with Roberto would be good.