Here’s a brute-force approach (it might be easier to understand): from itertools import chain def condense(*lists): # remember original positions positions = {} for pos, item in enumerate(chain(*lists)): if item not in positions: positions[item] = pos # condense disregarding order sets = condense_sets(map(set, lists)) # restore order result = [sorted(s, key=positions.get) for s in sets] … Read more
The reason that your first approach doesn’t work is that the the groups get “consumed” when you create that list with list(groupby(“cccccaaaaatttttsssssss”)) To quote from the groupby docs The returned group is itself an iterator that shares the underlying iterable with groupby(). Because the source is shared, when the groupby() object is advanced, the previous … Read more
The primary purpose of itertools.repeat is to supply a stream of constant values to be used with map or zip: >>> list(map(pow, range(10), repeat(2))) # list of squares [0, 1, 4, 9, 16, 25, 36, 49, 64, 81] The secondary purpose is that it gives a very fast way to loop a fixed number of … Read more
zip computes all the list at once, izip computes the elements only when requested. One important difference is that ‘zip’ returns an actual list, ‘izip’ returns an ‘izip object’, which is not a list and does not support list-specific features (such as indexing): >>> l1 = [1, 2, 3, 4, 5, 6] >>> l2 = … Read more
In Python 3 the built-in zip does the same job as itertools.izip in 2.X(returns an iterator instead of a list). The zip implementation is almost completely copy-pasted from the old izip, just with a few names changed and pickle support added. Here is a benchmark between zip in Python 2 and 3 and izip in … Read more
You can use itertools.combinations() to create the index array, and then use NumPy’s fancy indexing: import numpy as np from itertools import combinations, chain from scipy.special import comb def comb_index(n, k): count = comb(n, k, exact=True) index = np.fromiter(chain.from_iterable(combinations(range(n), k)), int, count=count*k) return index.reshape(-1, k) data = np.array([[1,2,3,4,5],[10,11,12,13,14]]) idx = comb_index(5, 3) print(data[:, idx]) output: … Read more
itertools functions return iterators, objects that produce results lazily, on demand. You could either loop over the object with a for loop, or turn the result into a list by calling list() on it: from itertools import chain, combinations def powerset(iterable): “powerset([1,2,3]) –> () (1,) (2,) (3,) (1,2) (1,3) (2,3) (1,2,3)” s = list(iterable) return … Read more
The NumPy equivalent of itertools.product() is numpy.indices(), but it will only get you the product of ranges of the form 0,…,k-1: numpy.rollaxis(numpy.indices((2, 3, 3)), 0, 4) array([[[[0, 0, 0], [0, 0, 1], [0, 0, 2]], [[0, 1, 0], [0, 1, 1], [0, 1, 2]], [[0, 2, 0], [0, 2, 1], [0, 2, 2]]], [[[1, 0, … Read more
islice() can be used to get the next n items of an iterator. Thus, list(islice(f, n)) will return a list of the next n lines of the file f. Using this inside a loop will give you the file in chunks of n lines. At the end of the file, the list might be shorter, … Read more
>>> list(itertools.product(*arrays)) [(-1, -2, -3), (-1, -2, 3), (-1, 2, -3), (-1, 2, 3), (1, -2, -3), (1, -2, 3), (1, 2, -3), (1, 2, 3)] This will feed all the pairs as separate arguments to product, which will then give you the cartesian product of them. The reason your version isn’t working is that … Read more