Ruby two dimensional array sort based on first element
ruby, ruby-on-rails, sorting
Solution
a.sort_by {|i| i.first.split('/').map(&:to_i).reverse }
#[
# ["3/2013", 15.0, 0.0, 0.0],
# ["4/2013", 15.0, 0.0, 0.0],
# ["5/2013", 20.0, 6.0, 6.0],
# ["7/2013", 73.0, 66.0, 66.0],
# ["50/2013", 11530.0, 12625.27, 12087.75],
# ["2/2014", 5926.0, 6058.0, 5876.4]
#]
Problem
I have a two dimensional array which has following structure ``` a = [["5/2013", 20.0, 6.0, 6.0], ["7/2013", 73.0, 66.0, 66.0], ["50/2013", 11530.0, 12625.27, 12087.75], ["2/2014", 5926.0, 6058.0, 5876.4], ["3/2013", 15.0, 0.0, 0.0], ["4/2013", 15.0, 0.0, 0.0]] ``` I want to sort the array based on first element, first element of each array presents the week no in year(i.e. "2/2014" means 2nd week in 2014, which is greater than "50/2013") the result will be like this ``` ["3/2013", 15.0, 0.0, 0.0], ["4/2013", 15.0, 0.0, 0.0], ["5/2013", 20.0, 6.0, 6.0], ["7/2013", 73.0, 66.0, 66.0], ["50/2013", 11530.0, 12625.27, 12087.75], ["2/2014", 5926.0, 6058.0, 5876.4], ] ``` I tried with this one ``` a.sort{|a,b| a[0].split('/')[1].to_i <=> b[0].split('/')[1].to_i && a[0].split('/') [0].to_i <=> b[0].split('/')[0].to_i} ``` but it does not help. I am new in ruby and rails. Can anyone please help me to solve my problem.