What's the fastest/one-liner way to convert an array like this:
[1, 1, 1, 1, 2, 2, 3, 5, 5, 5, 8, 13, 21, 21, 21]
...into an array of objects like this:
[{1 => 4}, {2 => 2}, {3 => 1}, {5 => 3}, {8 => 1}, {13 => 1}, {21 => 3}]
views:
210answers:
4
+4
A:
super basic, finally understand inject:
array.inject({}) { |h,v| h[v] ||= 0; h[v] += 1; h }
Not totally there, but almost
viatropos
2009-10-01 08:10:28
why do you ask if you know the answer?
Peter Kofler
2009-10-01 08:13:15
It'scertainly not illegal to answer your own question. The SO FAQ even says it's okay, as long as it's in a question format.
Twisol
2009-10-01 08:14:53
to help other people find good information. it doesn't matter who answers it. don't worry, I don't have the desire to go around writing questions to things I have answers lined up for :).
viatropos
2009-10-01 09:30:09
There is a badge for that even called self-learner: Answered your own question with at least 3 up votes, so nothing illegal here.
khelll
2009-10-01 09:40:41
+3
A:
To achieve your desired format, you could append a call to map to your solution:
array.inject({}) { |h,v| h[v] ||= 0; h[v] += 1; h }.map {|k, v| {k => v}}
Although it still is a one-liner, it starts to get messy.
jgre
2009-10-01 08:28:00
You can clean that up a bit by using a Hash with a default value: `array.inject(Hash.new(0)) { |h,v| h[v] += 1; h }.map {|k, v| {k => v}}`
rampion
2009-10-01 12:05:54
And to get it sorted (like in the question example), add `.sort_by { |o| o.keys[0] }` to it. *Now* it's messy. :)
andre-r
2009-10-01 14:39:01
+2
A:
array.inject(Hash.new(0)) {|h,v| h[v] += 1; h }
Not a huge difference but I prefer to be explicit with what I am doing. Saying Hash.new(0) replaces {} and h[v] ||= 0
Ryan Neufeld
2009-10-01 16:41:41
+2
A:
Requires 1.8.7
a = [1, 1, 1, 1, 2, 2, 3, 5, 5, 5, 8, 13, 21, 21, 21]
h = {}
a.group_by {|e| e}.each {|k,v| h[k] = v.length}
p h # => {5=>3, 1=>4, 2=>2, 13=>1, 8=>1, 3=>1, 21=>3}
glenn jackman
2009-10-02 02:59:07