This question has been asked a couple of times already: Using numpy to build an array of all combinations of two arrays itertools product speed up The first link has a working numpy solution, that is claimed to be several times faster than itertools, though no benchmarks are provided. This code was written by a … Read more
iter.next() was removed in python 3. Use next(iter) instead. So in your example change itertools.cycle().next() to next(itertools.cycle()) There is a good example here along with various other porting to python 3 tips. It also compares various other next() idioms in python 2.x vs python 3.x
itertools.product generates its results lazily, but this is not true for the arguments. They are evaluated eagerly. Each iterable argument is first converted to a tuple: The evaluation of the arguments (not the production of results) is very similar to the Python implementation shown in the docs: … pools = [tuple(pool) for pool in args] … Read more
Since your data seem to guarantee 3 unique crops per country (“I am compiling a table of top-3 crops by county.”), it suffices to sort the values and assign back. import numpy as np cols = [‘Crop1’, ‘Crop2’, ‘Crop3’] df1[cols] = np.sort(df1[cols].to_numpy(), axis=1) County Crop1 Crop2 Crop3 Total_pop 0 Harney apples grain melons 2000 1 … Read more
here’s a solution based on pypy’s library code (thanks to agf’s suggestion in the comments). the state is available via the .state attribute and can be reset via .goto(state) where state is an index into the sequence (starting at 0). there’s a demo at the end (you need to scroll down, i’m afraid). this is … Read more
itertools.product() takes a repeat keyword argument; set it to k: product(range(2), repeat=k) Demo: >>> from itertools import product >>> for k in range(2, 5): … print list(product(range(2), repeat=k)) … [(0, 0), (0, 1), (1, 0), (1, 1)] [(0, 0, 0), (0, 0, 1), (0, 1, 0), (0, 1, 1), (1, 0, 0), (1, 0, 1), … Read more
Try to use the iterator generated by the permutations instead of recreating a list with it : perm_iterator = itertools.permutations(list(graph.Nodes)) for item in perm_iterator: do_the_stuff(item) by doing this, python will keep in memory only the currently used permutation, not all the permutations (in term of memory usage, it is really better 😉 ) On the … Read more
The simplest way is to use itertools.product: a = [“foo”, “melon”] b = [True, False] c = list(itertools.product(a, b)) >> [(“foo”, True), (“foo”, False), (“melon”, True), (“melon”, False)]
itertools.groupby collects together contiguous items with the same key. If you want all items with the same key, you have to sort self.data first. for mid, group in itertools.groupby( sorted(self.data,key=operator.itemgetter(1)), key=operator.itemgetter(1)):