Cartesian Product of two lists (itertools)

[Thorsten Kampe]

Hi,

is there a way to make itertools.product generate triples instead of
pairs from two lists?

For example:

list1 = [1, 2]; list2 = [4, 5]; list3 = [7, 8]
from itertools import product
list(product(list1, list2, list3))

[(1, 4, 7), (1, 4, 8), (1, 5, 7), (1, 5, 8), (2, 4, 7), (2, 4, 8), (2,
5, 7), (2, 5, 8)]

so far so good... Now...[((1, 4), 7), ((1, 4), 8), ((1, 5), 7), ((1, 5), 8), ((2, 4), 7), ((2,
4), 8), ((2, 5), 7), ((2, 5), 8)]

Oops, pairs of pairs instead triples. Not what I wanted.

What's the best way to pre-process the arguments to "itertools.product"
or to post-process the result of "itertools.product" to get what I
want?!

I have an older utility which I would like to replace with
itertools.product. The old one uses a rather clumsy way to indicate that
a triple was wanted:

def cartes(seq0, seq1, modus = 'pair'):
""" return the Cartesian Product of two sequences """
if modus == 'pair':
return [[item0, item1] for item0 in seq0 for item1 in seq1]
elif modus == 'triple':
return [item0 + [item1] for item0 in seq0 for item1 in seq1]


Thorsten

 

[Mensanator]

Hi,

is there a way to make itertools.product generate triples instead of
pairs from two lists?

For example:>>> list1 = [1, 2]; list2 = [4, 5]; list3 = [7, 8]
[(1, 4, 7), (1, 4, 8), (1, 5, 7), (1, 5, 8), (2, 4, 7), (2, 4, 8), (2,
5, 7), (2, 5, 8)]

so far so good... Now...>>> list(product(product(list1, list2), list3))

[((1, 4), 7), ((1, 4), 8), ((1, 5), 7), ((1, 5), 8), ((2, 4), 7), ((2,
4), 8), ((2, 5), 7), ((2, 5), 8)]

Oops, pairs of pairs instead triples. Not what I wanted.

What's the best way to pre-process the arguments to "itertools.product"
or to post-process the result of "itertools.product" to get what I
want?!

I have an older utility which I would like to replace with
itertools.product. The old one uses a rather clumsy way to indicate that
a triple was wanted:

def cartes(seq0, seq1, modus = 'pair'):
� � """ return the Cartesian Product of two sequences """
� � if � modus == 'pair':
� � � � return [[item0, item1] for item0 in seq0 for item1 in seq1]
� � elif modus == 'triple':
� � � � return [item0 + [item1] for item0 in seq0 for item1 in seq1]

Thorsten

Will this work for you?

list4 = [(i,) for i in list3]
list4

[(7,), (8,)]
[((1, 4), (7,)), ((1, 4), (8,)), ((1, 5), (7,)), ((1, 5), (8,)), ((2,
4), (7,)), ((2, 4), (8,)), ((2, 5), (7,)), ((2, 5), (8,))]
return tuple(itertools.chain.from_iterable(listOfLists))

list5 = [flatten(i) for i in a]
list5

[(1, 4, 7), (1, 4, 8), (1, 5, 7), (1, 5, 8), (2, 4, 7), (2, 4, 8), (2,
5, 7), (2, 5, 8)]

 

[Terry Reedy]

Hi,

is there a way to make itertools.product generate triples instead of
pairs from two lists?

For example:

list1 = [1, 2]; list2 = [4, 5]; list3 = [7, 8]
from itertools import product
list(product(list1, list2, list3))

[(1, 4, 7), (1, 4, 8), (1, 5, 7), (1, 5, 8), (2, 4, 7), (2, 4, 8), (2,
5, 7), (2, 5, 8)]

so far so good... Now...[((1, 4), 7), ((1, 4), 8), ((1, 5), 7), ((1, 5), 8), ((2, 4), 7), ((2,
4), 8), ((2, 5), 7), ((2, 5), 8)]

Oops, pairs of pairs instead triples. Not what I wanted.

What's the best way to pre-process the arguments to "itertools.product"
or to post-process the result of "itertools.product" to get what I
want?!

I have an older utility which I would like to replace with
itertools.product. The old one uses a rather clumsy way to indicate that
a triple was wanted:

A pair of function, cart_pair, cart_trip, would have been better.
Or auto recognition of the number of sequences passed in.

def cartes(seq0, seq1, modus = 'pair'):
""" return the Cartesian Product of two sequences """
if modus == 'pair':
return [[item0, item1] for item0 in seq0 for item1 in seq1]
elif modus == 'triple':
return [item0 + [item1] for item0 in seq0 for item1 in seq1]

The second branch only produces a triple if seq0 is a sequence of pairs.
This must be called with something like
res = cartes(cartes(list1,list2),list3,'triple')
Just replace that with your first only-once itertools call
list(product(list1, list2, list3))

 

[Mark Wooding]

[((1, 4), 7), ((1, 4), 8), ((1, 5), 7), ((1, 5), 8), ((2, 4), 7), ((2,
4), 8), ((2, 5), 7), ((2, 5), 8)]
[...]

What's the best way to pre-process the arguments to "itertools.product"
or to post-process the result of "itertools.product" to get what I
want?!

Python has powerful destructuring capabilities:

[(x, y, z) for ((x, y), z) in lopsided_list]

This seems simpler and is probably faster than a generalized flatten for
this size of problem, but I wouldn't use it to rearrange tuples with
more than four or five elements.

-- [mdw]