Hi folks,
I came across something I thought should have an obvious answer and
came up blank. I'm hoping folks could give me a nice solution to this
problem.
Given the output below:
a = [1,2,3,4].zip([5,6,7,8])
=> [[1, 5], [2, 6], [3, 7], [4, 8]]
I'd like to recover the original two arrays, something like
unzip(a) #=> [[1,2,3,4],[5,6,7,8]]
The best ordinary solution I could come up with was this:
b = a.each_with_object([,]) do |pair, memo|
memo[0] << pair[0]
memo[1] << pair[1]
end
Shorter, but perhaps weirder, is:
require "matrix"
b = Matrix[*a].column_vectors.map(&:to_a)
Although either of these solutions are likely good enough for my
needs, neither feel natural to me. Anyone have something better?
-greg
*facepalm* Array#transpose is what I needed, I can't believe I forgot
that. Thanks goes to David Black via IM.
-greg
···
On Tue, Jul 14, 2009 at 9:27 AM, Gregory Brown<gregory.t.brown@gmail.com> wrote:
require "matrix"
b = Matrix[*a].column_vectors.map(&:to_a)
Gregory Brown wrote:
require "matrix"
b = Matrix[*a].column_vectors.map(&:to_a)
*facepalm* Array#transpose is what I needed, I can't believe I forgot
that. Thanks goes to David Black via IM.
That's funny - I was just trying to do more or less the same thing the
other day. Given an array of [k,v] pairs, I needed to turn it into an
array of k's and an array of v's.
I searched for a built-in method to do this, couldn't find it, so went
for the obvious solution:
ks = src.collect { |k,v| k }
vs = src.collect { |k,v| v }
But now I know a better way:
ks, vs = src.transpose
Thanks for sharing!
···
On Tue, Jul 14, 2009 at 9:27 AM, Gregory > Brown<gregory.t.brown@gmail.com> wrote:
--
Posted via http://www.ruby-forum.com/\.