On Sep 26, 9:24 am, p...@informatimago.com (Pascal J. Bourguignon) wrote: > Xah Lee <xah...@gmail.com> writes: > > here's a interesting toy list processing problem. > > > I have a list of lists, where each sublist is labelled by > > a number. I need to collect together the contents of all sublists > > sharing > > the same label. So if I have the list > > > ((0 a b) (1 c d) (2 e f) (3 g h) (1 i j) (2 k l) (4 m n) (2 o p) (4 q > > r) (5 s t)) > > > where the first element of each sublist is the label, I need to > > produce: > > > output: > > ((a b) (c d i j) (e f k l o p) (g h) (m n q r) (s t)) > > > a Mathematica solution is here: > >http://xahlee.org/UnixResource_dir/writ/notations_mma.html > > > R5RS Scheme lisp solution: > >http://xahlee.org/UnixResource_dir/writ/Sourav_Mukherjee_sourav.work_... > > by Sourav Mukherjee > > > also, a Common Lisp solution can be found here: > >http://groups.google.com/group/comp.lang.lisp/browse_frm/thread/5d1de... > > It's too complex. Just write: > > (let ((list '((0 a b) (1 c d) (2 e f) (3 g h) (1 i j) (2 k l) (4 m n) > (2 o p) (4 q r) (5 s t)))) > > (mapcar (lambda (class) (reduce (function append) class :key (function > rest))) > (com.informatimago.common-lisp.list:equivalence-classes list :key > (function first))) > > ) > > --> ((S T) (Q R M N) (G H) (O P K L E F) (I J C D) (A B)) > > -- > __Pascal Bourguignon__ http://www.informatimago.com/
Ruby: [[0, 'a', 'b'], [1, 'c', 'd'], [2, 'e', 'f'], [3, 'g', 'h'], [1, 'i', 'j'], [2, 'k', 'l'], [4, 'm', 'n'], [2, 'o', 'p'], [4, 'q', 'r'], [5, 's', 't']]. group_by{|x| x.first}.values.map{|x| x.map{|y| y[1..-1]}.flatten} ==>[["s", "t"], ["a", "b"], ["c", "d", "i", "j"], ["e", "f", "k", "l", "o", "p"], ["g", "h"], ["m", "n", "q", "r"]] -- http://mail.python.org/mailman/listinfo/python-list