On 7/23/2013 7:02 PM, Terry Reedy wrote:
On 7/23/2013 5:52 PM, st...@divillo.com wrote:
I think that itertools may be able to do what I want but I have not
been able to figure out how.
What you want is a flattened product with unchanged components of the
successive products omitted in the
Wow, thanks everyone. Very helpful indeed!
On Tuesday, July 23, 2013 2:52:21 PM UTC-7, st...@divillo.com wrote:
I think that itertools may be able to do what I want but I have not been able
to figure out how.
I want to convert an arbitrary number of lists with an arbitrary number of
I think that itertools may be able to do what I want but I have not been able
to figure out how.
I want to convert an arbitrary number of lists with an arbitrary number of
elements in each list into a single list as follows.
Say I have three lists:
[[A0,A1,A2], [B0,B1,B2] [C0,C1,C2]]
I would
El 23/07/13 23:52, st...@divillo.com escribió:
[[A0,A1,A2], [B0,B1,B2] [C0,C1,C2]]
Hi,
I think you are looking for itertools.chain, or in this case,
itertools.chain.from_iterable:
In [1]: x = [['A0','A1','A2'], ['B0','B1','B2'], ['C0','C1','C2']]
In [2]: import itertools
In [3]: [ y for y
On 23/07/2013 22:52, st...@divillo.com wrote:
I think that itertools may be able to do what I want but I have not been able
to figure out how.
I want to convert an arbitrary number of lists with an arbitrary number of
elements in each list into a single list as follows.
Say I have three
:
On 23 July 2013 17:52, st...@divillo.com wrote:
Say I have three lists:
[[A0,A1,A2], [B0,B1,B2] [C0,C1,C2]]
I would like to convert those to a single list that looks like this:
On 7/23/2013 5:52 PM, st...@divillo.com wrote:
I think that itertools may be able to do what I want but I have not
been able to figure out how.
A recursive generator suffices.
I want to convert an arbitrary number of lists with an arbitrary
number of elements in each list into a single list
On Wed, Jul 24, 2013 at 8:34 AM, Rafael Durán Castañeda
rafadurancastan...@gmail.com wrote:
In [3]: [ y for y in itertools.chain.from_iterable(x)]
Out[3]: ['A0', 'A1', 'A2', 'B0', 'B1', 'B2', 'C0', 'C1', 'C2']
Complete aside, given that this has already been pointed out as
solving a different
