I get it! >>> def f (*a): print a print zip (a) # My mistake print zip (*a) # Gerard's solution.
>>> f (l1, l2, l3) ([1, 2, 3], [4, 5, 6], [7, 5, 34]) # Argument: tuple of lists [([1, 2, 3],), ([4, 5, 6],), ([7, 5, 34],)] # My mistake [(1, 4, 7), (2, 5, 5), (3, 6, 34)] # That's what I want Thank you all Frederic ----- Original Message ----- From: "Gerard Flanagan" <[EMAIL PROTECTED]> Newsgroups: comp.lang.python To: <python-list@python.org> Sent: Sunday, August 27, 2006 2:59 PM Subject: Re: unpaking sequences of unknown length > > Anthra Norell wrote: > > Hi, > > > > I keep working around a little problem with unpacking in cases in which > > I don't know how many elements I get. Consider this: > > > > def tabulate_lists (*arbitray_number_of_lists): > > table = zip (arbitray_number_of_lists) > > for record in table: > > # etc ... > > > > This does not work, because the zip function also has an *arg parameter, > > which expects an arbitrary length enumeration of arguments > > maybe I don't understand the problem properly, but you can use '*args' > as 'args' or as '*args', if you see what I mean!, ie. > > def tabulate_lists (*arbitray_number_of_lists): > table = zip (*arbitray_number_of_lists) > for record in table: > # etc ... > > for example: > > def sum_columns(*rows): > for col in zip(*rows): > yield sum(col) > > for i, s in enumerate( sum_columns( [1,2], [3,2], [5,1] ) ): > print 'Column %s: SUM=%s' % (i,s) > > Column 0: SUM=9 > Column 1: SUM=5 > > ----------------------------------------------------- > > alternatively: > > import itertools as it > > def sum_columns2( iterable ): > for col in it.izip( *iterable ): > yield sum(col) > > def iter_rows(): > yield [1,2] > yield [3,2] > yield [5,1] > > print list( sum_columns2( iter_rows() ) ) > > #(izip isn't necessary here, zip would do.) > > ----------------------------------- > > Gerard > > -- > http://mail.python.org/mailman/listinfo/python-list -- http://mail.python.org/mailman/listinfo/python-list
