Re: complementary lists?
Kay Schluehr kay.schlu...@gmx.net writes: On 29 Apr., 05:41, Ross ross.j...@gmail.com wrote: If I have a list x = [1,2,3,4,5,6,7,8,9] and another list that is a subset of x: y = [1,4,7] , is there a quick way that I could return the complementary subset to y z=[2,3,5,6,8,9] ? The reason I ask is because I have a generator function that generates a list of tuples and I would like to divide this list into complementary lists. z = [u for u in x if u not in y] or z = [u for u in x if u not in set(y)] The above will evaluate set(y) for each element in x. s = set(y) z = [u for u in x if u not in s] -- Arnaud -- http://mail.python.org/mailman/listinfo/python-list
Re: complementary lists?
On Apr 28, 11:16 pm, Arnaud Delobelle arno...@googlemail.com wrote: Kay Schluehr kay.schlu...@gmx.net writes: On 29 Apr., 05:41, Ross ross.j...@gmail.com wrote: If I have a list x = [1,2,3,4,5,6,7,8,9] and another list that is a subset of x: y = [1,4,7] , is there a quick way that I could return the complementary subset to y z=[2,3,5,6,8,9] ? The reason I ask is because I have a generator function that generates a list of tuples and I would like to divide this list into complementary lists. z = [u for u in x if u not in y] or z = [u for u in x if u not in set(y)] The above will evaluate set(y) for each element in x. s = set(y) z = [u for u in x if u not in s] -- Arnaud y = [1, 2, 3, 4, 5] s = set(y) s.complimentary() ['you are handsome', 'you are smart', 'nice jacket'] -- http://mail.python.org/mailman/listinfo/python-list
Re: complementary lists?
On Apr 28, 11:16 pm, Arnaud Delobelle arno...@googlemail.com wrote: Kay Schluehr kay.schlu...@gmx.net writes: On 29 Apr., 05:41, Ross ross.j...@gmail.com wrote: If I have a list x = [1,2,3,4,5,6,7,8,9] and another list that is a subset of x: y = [1,4,7] , is there a quick way that I could return the complementary subset to y z=[2,3,5,6,8,9] ? The reason I ask is because I have a generator function that generates a list of tuples and I would like to divide this list into complementary lists. z = [u for u in x if u not in y] or z = [u for u in x if u not in set(y)] The above will evaluate set(y) for each element in x. s = set(y) z = [u for u in x if u not in s] -- Arnaud ls = [1, 2, 3, 4, 5] s = set(ls) s.complimentary() ['your are handsome', 'your are smart', 'nice jacket' -- http://mail.python.org/mailman/listinfo/python-list
complementary lists?
If I have a list x = [1,2,3,4,5,6,7,8,9] and another list that is a subset of x: y = [1,4,7] , is there a quick way that I could return the complementary subset to y z=[2,3,5,6,8,9] ? The reason I ask is because I have a generator function that generates a list of tuples and I would like to divide this list into complementary lists. -- http://mail.python.org/mailman/listinfo/python-list
Re: complementary lists?
On 29 Apr., 05:41, Ross ross.j...@gmail.com wrote: If I have a list x = [1,2,3,4,5,6,7,8,9] and another list that is a subset of x: y = [1,4,7] , is there a quick way that I could return the complementary subset to y z=[2,3,5,6,8,9] ? The reason I ask is because I have a generator function that generates a list of tuples and I would like to divide this list into complementary lists. z = [u for u in x if u not in y] or z = [u for u in x if u not in set(y)] Since you are dealing with tuples and a natural order might not be that relevant you can also set objects in the first place: z = x - y -- http://mail.python.org/mailman/listinfo/python-list
Re: complementary lists?
On Apr 28, 10:41�pm, Ross ross.j...@gmail.com wrote: If I have a list x = [1,2,3,4,5,6,7,8,9] and another list that is a subset of x: �y = [1,4,7] , is there a quick way that I could return the complementary subset to y z=[2,3,5,6,8,9] ? The reason I ask is because I have a generator function that generates a list of tuples and I would like to divide this list into complementary lists. s = set([1,2,3,4,5,6,7,8,9]) s.difference(set([1,4,7])) set([2, 3, 5, 6, 8, 9]) -- http://mail.python.org/mailman/listinfo/python-list
Re: complementary lists?
Ross ross.j...@gmail.com writes: If I have a list x = [1,2,3,4,5,6,7,8,9] and another list that is a subset of x: y = [1,4,7] , is there a quick way that I could return the complementary subset to y z=[2,3,5,6,8,9] ? x = [1,2,3,4,5,6,7,8,9] y = [1,4,7] print sorted(set(x)-set(y)) [2, 3, 5, 6, 8, 9] -- http://mail.python.org/mailman/listinfo/python-list
