Ian Witham wrote: > This looks like a job for List Comprehensions! > > >>> list = [[1, 2, 3], [4, 5, 6], [7, 8, 9]] > >>> new_list = [item[1] for item in list] > >>> new_list > [2, 5, 8] > >>>
Alternately, if you want *all* columns, you can use zip() to transpose the lists: In [1]: lst = [[1, 2, 3], [4, 5, 6], [7, 8, 9]] In [2]: zip(*lst) Out[2]: [(1, 4, 7), (2, 5, 8), (3, 6, 9)] PS. Don't use 'list' as the name of a list, it shadows the builtin 'list' which is the <type> of list. Similarly, avoid str, dict, set and file as names. Kent > looks good? > > Ian > > On 8/21/07, *Orest Kozyar* <[EMAIL PROTECTED] > <mailto:[EMAIL PROTECTED]>> wrote: > > I've got a "2D" list (essentially a list of lists where all sublists > are of > the same length). The sublists are polymorphic. One "2D" list I > commonly > work with is: > > [ [datetime object, float, int, float], > [datetime object, float, int, float], > [datetime object, float, int, float] ] > > I'd like to be able to quickly accumulate the datetime column into a > list. > Is there a simple/straightforward syntax (such as list[:][0]) to do > this, or > do we need to use a for loop? I expect what I have in mind is > similar to > the Python array type, except it would be polymorphic. > > Thanks, > Orest > > _______________________________________________ > Tutor maillist - Tutor@python.org <mailto:Tutor@python.org> > http://mail.python.org/mailman/listinfo/tutor > > > > ------------------------------------------------------------------------ > > _______________________________________________ > Tutor maillist - Tutor@python.org > http://mail.python.org/mailman/listinfo/tutor _______________________________________________ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor