itertools

I have four lists: LISTA = ['A1', 'A2'] LISTB = ['B1_C', 'B2_D'] LISTC = ['C1', 'C2'] LISTD = ['D1', 'D2'] I'd like to get the Cartesian product of LISTA and LISTB , and then depending on the value of B, I'd like to add either the product of C, or the product of D. (A1 B1_C C1) (A1 B1_C C2) (A2 B1_C C1) (A2 B1_C C2) (A1 B2_D D1) (A1 B2_D D2) (A2 B2_D D1) (A2 B2_D D2) I can get the first part with itertools.product(LISTA, LISTB) , but I've been looking through itertools for how to achieve the second part and I'm not sure the best way to go. Suggestions? Here is an interactive demonstration of a

I want to make a combination of 1's and 0's in a 2d array like the following: [[ 1, 1, 1, 1, 0, 0, 0, 0 ], [ 1, 1, 1, 0, 1, 0, 0, 0 ], [ 1, 1, 1, 0, 0, 1, 0, 0 ], [ 1, 1, 1, 0, 0, 0, 1, 0 ], [ 1, 1, 1, 0, 0, 0, 0, 1 ], . . . ] That means a combination of four 1's and four 0's. I have looked at the itertools module's permutations() and combinations() , but couldn't find any suitable functions to perform this combination. You can also use combinations to generate unique combinations directly: n = 8 n1 = 4 for x in itertools.combinations( xrange(n), n1 ) : print [ 1 if i in x else 0 for i in

I have a pandas.DataFrame with integer column names, which has zeroes and ones. An example of the input: 12 13 14 15 1 0 0 1 0 2 0 0 1 1 3 1 0 0 1 4 1 1 0 1 5 1 1 1 0 6 0 0 1 0 7 0 0 1 1 8 1 1 0 1 9 0 0 1 1 10 0 0 1 1 11 1 1 0 1 12 1 1 1 1 13 1 1 1 1 14 1 0 1 1 15 0 0 1 1 I need to count all consecutive ones which has a length/sum which is >=2, iterating through columns and returning also indices where an array of the consecutive ones occurs (start, end). The preferred output would be a 3D DataFrame, where subcolumns "count" and "indices" refer to integer column names from the input. An

I have often used itertools module in Python but it feels like cheating if I don't know the logic behind it. Here is the code to find combinations of string when order is not important. def combinations(iterable, r): # combinations('ABCD', 2) --> AB AC AD BC BD CD # combinations(range(4), 3) --> 012 013 023 123 pool = tuple(iterable) n = len(pool) if r > n: return indices = list(range(r)) yield tuple(pool[i] for i in indices) while True: for i in reversed(range(r)): if indices[i] != i + n - r: break else: return indices[i] += 1 for j in range(i+1, r): indices[j] = indices[j-1] + 1 yield tuple

For some context, I'm trying to enumerate the number of unique situations that can occur when calculating the Banzhaf power indices for four players, when there is no dictator and there are either four or five winning coalitions. I am using the following code to generate a set of lists that I want to iterate over. from itertools import chain, combinations def powerset(iterable): s = list(iterable) return chain.from_iterable(map(list, combinations(s, r)) for r in range(2, len(s)+1)) def superpowerset(iterable): s = powerset(iterable) return chain.from_iterable(map(list, combinations(s, r)) for

I have a list with numbers: numbers = [1, 2, 3, 4] . I would like to have a list where they repeat n times like so (for n = 3 ): [1, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4] . The problem is that I would like to only use itertools for this, since I am very constrained in performance. I tried to use this expression: list(itertools.chain.from_iterable(itertools.repeat(numbers, 3))) But it gives me this kind of result: [1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4] which is obviously not what I need. Is there a way to do this with itertools only, without using sorting, loops and list comprehensions? The closest I