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 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
--
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