Say I have 2 arrays like this:
# base set
a = [1, 2, 3, 4, 5, 6, 7, 8, 9]
# sub set
b = [5, 1, 8, 3]
What's the optimal way to sort b to the same order as a?
a.sort_like(b) #=> [1, 3, 5, 8]
What is this operation called?
-
1Your receiver and the argument is strange. It would be more natural if it were the other way around.sawa– sawa2011-04-19 23:58:03 +00:00Commented Apr 19, 2011 at 23:58
-
either way works for me.Lance Pollard– Lance Pollard2011-04-19 23:59:12 +00:00Commented Apr 19, 2011 at 23:59
-
using this in the code right now, love it.Lance Pollard– Lance Pollard2011-04-20 00:05:24 +00:00Commented Apr 20, 2011 at 0:05
-
possible duplicate of How to sort an array in Ruby to a particular order?Andrew Grimm– Andrew Grimm2011-04-20 02:35:40 +00:00Commented Apr 20, 2011 at 2:35
Add a comment
|
3 Answers
I think this is what you want:
a & b
1 Comment
sawa
The operation is called "set intersection", and does not sort, but it just removes from (the
dup of) the receiver what's not in the argument. You can go with randomguy's naming.This will do it, I'm not sure about the most efficient way.
def sort_like(other)
items = []
other.each do |find|
each do |item|
items.append item if item == find
end
end
items
end
Comments
If b is a subset of a, as you have defined, then the result is going to always be the sorted version of b. In which case, the you just want b.sort (it won't mutate b, it will return a new array with the sorted contents of b).
