![]() ![]() *) Update: In my previous edit, the manual approach came out faster this was because I forgot to actually consume the generators returned by the two functions. Personally, I'd still go with using itertools.permutations and a few simple, readable filters. Of course, depending on the size of the input list and the constraints, the manual approach can present more or less of a saving, but can also be (much) more or less difficult to implement or to adapt to changing constraints. > %timeit list(manual_permutations(numbers)) all the permutations starting with a number > %timeit original_loop(numbers) This checks the two constraints while generating the permutations, so that if e.g. If n = 5 and lastlast - last >= - first: continueįor perm in manual_permutations(rest, n+1, first, last): However, it follows that: with replacement: produce all permutations n r via product without replacement: filter from the latter Permutations with replacement, n r x for x in it. If n = 4 and last - first >= - rest: continue Unobviously, Cartesian product can generate subsets of permutations. Something like this: def manual_permutations(numbers, n=0, last=0, lastlast=0): for a given value for p there can be any valid values for p and so on. If you do not even want to generate them, you will have to implement a custom recursive permutations algorithm and manually check whether e.g. You can do the same in a regular for loop with if conditions, the only difference being that you do not collect all the permutations in a list first, but iterate the generator directly: all_permutations = itertools.permutations(numbers) # no list(.)īoth approaches will still generate all the N! permutations, but most are immediately discarded and only the "right" ones are ever stored in the list. If the checks become more complicated, you do not even have to use a list comprehension. Instead of first creating a list of all the permutations and then adding some of those elements to a second list and discarding the rest (about 90% in this case), you can use a list comprehension to filter the permutations yielded by itertools as they are created. Can someone assist with a way to apply the constraints while permutations are determined so not all N! are determined? The problem with this code is that I'm wasting time finding all possible permutations just to filter them out again. # Apply calculations to current permutation # Position 3 + Position 5 3 and permutation + permutation El operador, que realiza la división de enteros, se utiliza para devolver un tipo de entero. p n / (n - r) Se puede calcular de la siguiente manera utilizando la función math.factorial (), que devuelve el factorial. Use a loop and call zip(iterables) with each permutation and the shorter list as iterables to pair their elements into a combination. I am trying to calculate all permutations with list = adhering to some constraints. El número total de permutaciones, p, se obtiene mediante la siguiente ecuación utilizando factoriales.
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