[Tutor] Is this a job for zip(), or some other way?

[Gregor Lingl]

Bob Gailer schrieb:

> ....
> >>> apply(zip, [ ('bob', 24, 457), ('mike', 20, 4567), ('steve', 30, 
> 576)])
> [('bob', 'mike', 'steve'), (24, 20, 30), (457, 4567, 576)]
>
>
This works also:

 >>> zip(('bob', 24, 457), ('mike', 20, 4567), ('steve', 30, 576))
[('bob', 'mike', 'steve'), (24, 20, 30), (457, 4567, 576)]

So in your case

 >>> zip(*[ ('bob', 24, 457), ('mike', 20, 4567), ('steve', 30, 576)])
[('bob', 'mike', 'steve'), (24, 20, 30), (457, 4567, 576)]


would also be appropriate.
Gregor

P.S.:

*zip*( 	seq1, ...)

    This function returns a list of tuples, where the i-th tuple
    contains the i-th element from each of the argument sequences. At
    least one sequence is required, otherwise a TypeError is raised. The
    returned list is truncated in length to the length of the shortest
    argument sequence. When there are multiple argument sequences which
    are all of the same length, zip() is similar to map() with an
    initial argument of |None|. With a single sequence argument, it
    returns a list of 1-tuples. New in version 2.0.