What is the fastest way of counting the number of permutations? I have the following problem: First I have this: ncombos = itertools.combinations_with_replacement(['a1', 'a2', 'a3'], years*n) ('a1', 'a1', 'a1') ('a1', 'a1', 'a2') ('a1', 'a1', 'a3') ('a1', 'a2', 'a2') .... etc..... ('a3', 'a3', 'a3') The aim is to go through each one and calculate the number of permutations that each one has and construct an array with these values. I implemented this using: nodes = np.ones(len(leafs)); i=0 #This will store the number of permutations for j in ncombos: nodes =len(list(set(itertools.permutations(np.asanyarray(j), n)))) i = i+1 np.asanyarray(j) converts the ('a1','a1','a1') into formal ['a1','a1', 'a1'] which is need for permutations() to work. set erases the permutations which are identical. list makes a list of this. len calculates how many permutations can i make with a1, a1, a1. So basically all I want is to count the number of permutations... However my code is extremely!!! slow ! Thank you! Continue reading...
