9/28/2023 0 Comments Itertools permutationsmul )) # running product > list ( accumulate ( data, max )) # running maximum # Amortize a 5% loan of 1000 with 4 annual payments of 90 > cashflows = > list ( accumulate ( cashflows, lambda bal, pmt : bal * 1.05 + pmt )) # Chaotic recurrence relation > logistic_map = lambda x, _ : r * x * ( 1 - x ) > r = 3.8 > x0 = 0.4 > inputs = repeat ( x0, 36 ) # only the initial value is used > > data = > list ( accumulate ( data, operator. R-length tuples, in sorted order, with repeated elementsĪA AB AC AD BA BB BC BD CA CB CC CD DA DB DC DD R-length tuples, in sorted order, no repeated elements R-length tuples, all possible orderings, no repeated elements Zip_longest('ABCD', 'xy', fillvalue='-') -> Ax By C- D-Ĭartesian product, equivalent to a nested for-loop It1, it2, … itn splits one iterator into n Sub-iterators grouped by value of keyfunc(v) Seq, seq, starting when pred failsĮlements of seq where pred(elem) is falseįilterfalse(lambda x: x%2, range(10)) -> 0 2 4 6 8 Iterators terminating on the shortest input sequence: IteratorĬom_iterable() -> A B C D E FĬompress('ABCDEF', ) -> A C E F Sum(map(operator.mul, vector1, vector2)).Įlem, elem, elem, … endlessly or up to n times Operator can be mapped across two vectors to form an efficient dot-product: These tools and their built-in counterparts also work well with the high-speedįunctions in the operator module. The same effect can be achieved in Pythonīy combining map() and count() to form map(f, count()). Together, they form an “iteratorĪlgebra” making it possible to construct specialized tools succinctly andįor instance, SML provides a tabulation tool: tabulate(f) which produces a The module standardizes a core set of fast, memory efficient tools that are This module implements a number of iterator building blocks inspiredīy constructs from APL, Haskell, and SML. itertools - Functions creating iterators for efficient looping ¶ We can get our final result by taking the permutations of the items in cList: result = [''.join(perm) for combo in cListįor perm in itertools.10.1. Using () instead of will make this a generator rather than a list, to delay evaluation until we really need it. Thus we can use the smaller set of combinations to filter for the length 8 that we want: cList = (combo for length in range(1, 9)įor combo in binations(newList, length) Recall that in combinations, the order of the combined items doesn't matter, while in permutations it does. Instead of using permutations, let's use combinations, of which there are far fewer ("only" 263,949). P(20, 1) = roughly 5.5 billion, which is impractical to work with. Yet here you end up with an enormous amount of permutations to filter through: P(20, 8) + P(20, 7) +. One way to resolve this is to iterate through the various lengths: pList = However, if any of your strings are longer than one character, this won't get you what you want, since it will only return permutations of exactly 8 items. If you have no empty strings in your list, you'll never need more than 8 items, so we can use 8 as the second argument: p = itertools.permutations(newList, 8) Then, you can use the second argument of itertools.permutations to set the number of items you want. The first, obvious thing to do is filter out any strings with length > 8: newList = if len(i) <= 8]
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