Is there an efficient way of doing this. I have an array
a=[1,2,2,3,1,2]
I want to output the frequency of occurrence in an ascending order. Example
[[3,1],[1,2],[2,3]]
Here is my code in ruby.
b=a.group_by{|x| x}
out={}
b.each do |k,v|
out[k]=v.size
end
out.sort_by{|k,v| v}
a = [1,2,2,3,1,2]
a.each_with_object(Hash.new(0)){ |m,h| h[m] += 1 }.sort_by{ |k,v| v }
#=> [[3, 1], [1, 2], [2, 3]]
Something like this:
x = a.inject(Hash.new(0)) { |h, e| h[e] += 1 ; h }.to_a.sort{|a, b| a[1] <=> b[1]}
Are you trying to work out the algorithm or do you just want the job done? In the latter case, don't reinvent the wheel:
require 'facets'
[1, 2, 2, 3, 1, 2].frequency.sort_by(&:last)
# => [[3, 1], [1, 2], [2, 3]]
use hashing, create a hash, traverse the array, for each number in array, update the count in hash. it will take linear time O(n) and space complexity will be equal to storing the hash O(n).
def frequency(a)
a.group_by do |e|
e
end.map do |key, values|
[key, values.size]
end
end
a = [1, 2, 2, 3, 1, 2]
p frequency(a).sort_by(&:last)
