itertools

The following never prints anything in Python 3.6 from itertools import product, count for f in product(count(), [1,2]): print(f) Instead, it just sits there and burns CPU. The issue seems to be that product never returns an iterator if it's over an infinite space because it evaluates the full product first. This is surprising given that the product is supposed to be a generator. I would have expected this to start counting up (to infinity), something like the behavior of this generator (taken directly from the docs ): for tup in ((x,y) for x in count() for y in [1,2]): print(tup) But whereas

I'm trying to use itertools.combinations to return unique combinations. I've searched through several similar questions but have not been able to find an answer. An example: >>> import itertools >>> e = ['r','g','b','g'] >>> list(itertools.combinations(e,3)) [('r', 'g', 'b'), ('r', 'g', 'g'), ('r', 'b', 'g'), ('g', 'b', 'g')] For my purposes, (r,g,b) is identical to (r,b,g) and so I would want to return only (rgb),(rgg) and (gbg). This is just an illustrative example and I would want to ignore all such 'duplicates'. The list e could contain up to 5 elements. Each individual element would be

I wish to execute the following code: temp = [] temp.append([1,2]) temp.append([3,4]) temp.append([5,6]) print list(itertools.product(temp[0],temp[1],temp[2])) However, I would like to execute it for temp with arbitrary length. I.e. something more like: print list(itertools.product(temp)) How do I format the input correctly for itertools.product to produce the same result in the first segment of code without explicitly knowing how many entries there are in temp? print list(itertools.product(*temp)) Use * to unpack the argument iterable as separate positional arguments. Or can do that: Combine

As answer to my question Find the 1 based position to which two lists are the same I got the hint to use the C-library itertools to speed up things. To verify I coded the following test using cProfile: from itertools import takewhile, izip def match_iter(self, other): return sum(1 for x in takewhile(lambda x: x[0] == x[1], izip(self, other))) def match_loop(self, other): element = -1 for element in range(min(len(self), len(other))): if self[element] != other[element]: element -= 1 break return element +1 def test(): a = [0, 1, 2, 3, 4] b = [0, 1, 2, 3, 4, 0] print("match_loop a=%s, b=%s,

I need to get only the permutations that have letters and numbers (The permutation can not be. "A, B, C, D" I need it like this: "A, B, C, 1") In short, the permutations can not contain only letters, not just numbers. Must be a combination of both. My code: import itertools print list(itertools.combinations([0,1,2,3,4,'a','b','c','d'], 4)) Then I get: [(0, 1, 2, 3), (0, 1, 2, 4), (0, 1, 2, 'a'), (0, 1, 2, 'b'), (0, 1, 2, 'c'), (0, 1, 2, 'd'), (0, 1, 3, 4), (0, 1, 3, 'a'), (0, 1, 3, 'b'), (0, 1, 3, 'c'), (0, 1, 3, 'd'), (0, 1, 4, 'a'), (0, 1, 4, 'b'), (0, 1, 4, 'c'), (0, 1, 4, 'd'), (0, 1, 'a',

This question already has an answer here: difference between dict(groupby) and groupby [duplicate] 4 answers I've executed the following script: from itertools import groupby from pprint import pprint as prnt dt = [('23271800', 0.00066790780636275307), ('23271812', 0.0010018617095441298), ('26112103', 0.00066790780636275307), ('27111616', 0.0056772163540834012), # ... many lines deleted ... ('40161500', 0.00040074468381765189) ] agg = groupby(dt, lambda x: x[0]) lst = list(agg) lst1 = map(lambda x: (x[0], list(x[1])), lst) prnt(lst1) For the item '23271800' it should report [('23271800', 0

I use python pandas to perform grouping and aggregation across data frames, but I would like to now perform specific pairwise aggregation of rows (n choose 2, statistical combination). Here is the example data, where I would like to look at all pairs of genes in [mygenes]: import pandas import itertools mygenes=['ABC1', 'ABC2', 'ABC3', 'ABC4'] df = pandas.DataFrame({'Gene' : ['ABC1', 'ABC2', 'ABC3', 'ABC4','ABC5'], 'case1' : [0,1,1,0,0], 'case2' : [1,1,1,0,1], 'control1':[0,0,1,1,1], 'control2':[1,0,0,1,0] }) >>> df Gene case1 case2 control1 control2 0 ABC1 0 1 0 1 1 ABC2 1 1 0 0 2 ABC3 1 1 1