[Tutor] something relevant to array

[lina]

On Fri, Dec 23, 2011 at 8:19 PM, Andreas Perstinger
<andreas.perstinger at gmx.net> wrote:
> On 2011-12-23 12:02, lina wrote:
>>
>>     for i in range(len(result)):
>>         for j in range(len(result[i])):
>>             print(result[i][j])
>>
>> still have a little problem about print out,
>>
>> I wish to get like
>> a a
>> b b
>> c c
>> which will show in the same line,
>>
>> not as
>> a
>> b
>> c
>> a
>> b
>> c
>
>
> So you wish to print all the first elements from your sublists on the first
> line, all the second elements on the second line, and so on, right?
yes.
>
> Python 3.2 (r32:88445, Mar 25 2011, 19:28:28)
> [GCC 4.5.2] on linux2
> Type "help", "copyright", "credits" or "license" for more information.
>>>> result = [["a1", "b1", "c1"],["a2", "b2", "c2"],["a3", "b3", "c3"]]
>>>> for i in zip(*result):
> ...   print(" ".join(i))
> ...
> a1 a2 a3
> b1 b2 b3
> c1 c2 c3
>
> Explanation:
> "zip()" takes an arbitrary number of iterables and "returns an iterator of
> tuples, where the i-th tuple contains the i-th element from each of the
> argument sequences or iterables." (see the docs on
> http://docs.python.org/py3k/library/functions.html#zip):
>
>>>> print(list(zip([1, 2, 3], [4, 5, 6])))
> [(1, 4), (2, 5), (3, 6)]
>
> In your case, you would have to call "zip()" with
>
> zip(result[0], result[1], result[2], ... result[x])
>
> depending on how many files you process.
>
> But using the *-operator you can unpack "result" (which is a list of
> sublists), so that "zip" will see all the sublists as separate arguments.
> See also
> http://docs.python.org/py3k/tutorial/controlflow.html#unpacking-argument-lists
>
> For printing you just join all the elements of one tuple to a string.
>
> HTH, Andreas

Thanks for all. really good suggestions.

Best regards and have a weekend.
>
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