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
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
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
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
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
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)]