2

I have a list of lists of tuples 

A= [ [(1,2,3),(4,5,6)], [(7,8,9),(8,7,6),(5,4,3)],[(2,1,0),(1,3,5)] ]

The outer list can have any number of inner lists, the inner lists can have any number of tuples, a tuple always has 3 integers.

I want to generate all combination of tuples, one from each list:

[(1,2,3),(7,8,9),(2,1,0)] [(1,2,3),(7,8,9),(1,3,5)] [(1,2,3),(8,7,6),(2,1,0)] ... [(4,5,6),(5,4,3),(1,3,5)]

A simple way to do it is to use a function similar to itertools.poduct() but it must be called like this 

itertools.product([(1,2,3),(4,5,6)], [(7,8,9),(8,7,6),(5,4,3)],[(2,1,0),(1,3,5)])

i.e the outer list is removed. And I don't know how to do that. Is there a better way to generate all combinations of tuples?

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  • How deep can the nested levels of lists be. Just 2, as in your example? Commented Nov 6, 2009 at 16:34
  • Duplicate: stackoverflow.com/search?q=%5Bpython%5D+flatten, stackoverflow.com/questions/120886/…, stackoverflow.com/questions/406121/… Commented Nov 6, 2009 at 16:38
  • @S Lott, related but not duplicates. (I thought so too, initially, the "flatten" keyword is misleading I think. Thought of editing title but didn't find a better expression; maybe "enumerate combinations"... Commented Nov 6, 2009 at 16:45
  • Maybe wrong word, but thought it was about "flatten" when I wrote the question. No, it is not homework. I got my degree '83, been working sice then. But new with python. trying to solve a quiz in a swedish newspaper lead to all my questions so far. Commented Nov 6, 2009 at 16:59
  • 2
    "Cartesian product" is maybe the word you're looking for. Commented Nov 8, 2009 at 20:27

7 Answers 7

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10 
itertools.product(*A)

For more details check the python tutorial

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2

This works for your example, if there is only one level of nested lists (no lists of lists of lists):

itertools.product(*A)
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2

you can probably call itertools.product like so:

itertools.product(*A) # where A is your list of lists of tuples

This way it expands your list's elements into arguments for the function you are calling.

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2

Late to the party but ...

I'm new to python and come from a lisp background. This is what I came up with (check out the var names for lulz):

def flatten(lst): if lst: car,*cdr=lst if isinstance(car,(list)): if cdr: return flatten(car) + flatten(cdr) return flatten(car) if cdr: return [car] + flatten(cdr) return [car]

Seems to work. Test:

A = [ [(1,2,3),(4,5,6)], [(7,8,9),(8,7,6),(5,4,3)],[(2,1,0),(1,3,5)] ] flatten(A)

Result:

[(1, 2, 3), (4, 5, 6), (7, 8, 9), (8, 7, 6), (5, 4, 3), (2, 1, 0), (1, 3, 5)]

Note: the line car,*cdr=lst only works in Python 3.0

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3 Comments

-1 The statement car,*cdr=lst isn't even legal Python. There is no direct syntax in Python to split a list in its head and tail (sorry 'bout that).
@ThomasH: It is valid in Python 3. See python.org/dev/peps/pep-3132. +1 to counteract needless downvote.
@StevenRumbalski Thanks for the pointer. I thought I was checking the Python 3 docs too, but obviously missed the *target.
1

This is not exactly one step, but this would do what you want if for some reason you don't want to use the itertools solution:

def crossprod(listoflists): if len(listoflists) == 1: return listoflists else: result = [] remaining_product = prod(listoflists[1:]) for outertupe in listoflists[0]: for innercombo in remaining_product[0]: newcombo = [outertupe] newcombo.append(innercombo) result.append(newcombo) return result
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def flatten(A) answer = [] for i in A: if type(i) == list: ans.extend(i) else: ans.append(i) return ans
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you need to fix your indentation.
Terribly sorry for indentation errors. Turns out that copying code from my local IDE does not preserve proper indentation when I paste here
-1

This may also be achieved using list comprehension.

In [62]: A = [ [(1,2,3),(4,5,6)], [(7,8,9),(8,7,6),(5,4,3)],[(2,1,0),(1,3,5)] ] In [63]: improved_list = [num for elem in A for num in elem] In [64]: improved_list Out[64]: [(1, 2, 3), (4, 5, 6), (7, 8, 9), (8, 7, 6), (5, 4, 3), (2, 1, 0), (1, 3, 5)]
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