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To: python-list@python.org
From: Terry Reedy <tjreedy@udel.edu>
Subject: Re: Converting a list of lists to a single list
Date: Wed, 24 Jul 2013 11:56:30 -0400
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On 7/23/2013 7:02 PM, Terry Reedy wrote:
> On 7/23/2013 5:52 PM, steve@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 flattening. The omission is the 
difficulty.

> A recursive generator suffices.

But see below for how to use itertools.product.

>> I want to convert an arbitrary number of lists with an arbitrary
>> number of elements in each list into a single list as follows.

While others answered the Python2-oriented question ("How do I produce a 
list from a list of lists"), I answered the Python-3 oriented question 
of how to produce an iterator from an iterable of iterables. This scales 
better to an input with lots of long sequences. There is usually no need 
to manifest the output as a list, as the typical use of the list will be 
to iterate it.

> def crossflat(lofl):
>      if lofl:
>          first = lofl.pop(0)
>          for o in first:
>             yield o
>             yield from crossflat(lofl.copy())
>
> A0, A1, A2 = 100, 101, 102
> B0, B1, B2 = 10, 11, 12
> C0, C1, C2 = 0, 1, 2
> LL = [[A0, A1, A2], [B0, B1, B2], [C0, C1, C2]]
> cfLL = list(crossflat(LL))
> print(cfLL)
> assert cfLL == [
>     A0, B0, C0, C1, C2, B1, C0, C1, C2, B2, C0, C1, C2,
>     A1, B0, C0, C1, C2, B1, C0, C1, C2, B2, C0, C1, C2,
>     A2, B0, C0, C1, C2, B1, C0, C1, C2, B2, C0, C1, C2]
>
> passes

Here is filtered flattened product version. I think it clumsier than 
directly producing the items wanted, but it is good to know of this 
approach as a backup.

from itertools import product

def flatprod(iofi):  # iterable of iterables
     lofi = list(iofi)
     now = [object()] * len(lofi)
     for new in product(*lofi):
         i = 0
         while now[i] == new[i]:
             i += 1
         yield from new[i:]
         now = new

cfLL = list(flatprod(LL))

Same assert as before passes.

-- 
Terry Jan Reedy