##### Counting Number of Element in a smart way
 Counting Number of Element in a smart way quest Not Blown Up Yet Posts: 50 Threads: 25 Joined: Nov 2020 Reputation: Nov-09-2020, 09:26 PM Hello, I have an array I want to count the elements in the array and I can count the number of elements with this code ```a0 = sum(np.all((results-np.array([0,0,0]))==0, axis=1)) x1 = sum(np.all((results-np.array([0,0,1]))==0, axis=1)) x2 = sum(np.all((results-np.array([0,1,0]))==0, axis=1)) x3 = sum(np.all((results-np.array([0,1,1]))==0, axis=1)) x4 = sum(np.all((results-np.array([1,0,0]))==0, axis=1)) x5 = sum(np.all((results-np.array([1,0,1]))==0, axis=1)) x6 = sum(np.all((results-np.array([1,1,0]))==0, axis=1)) x7 = sum(np.all((results-np.array([1,1,1]))==0, axis=1))```But I want to do it in a shorter way. I don't want to specify my all triplets. This time I have 8 different elements so I could write it by hand but if I have 100 different triplets, I couldn't use this code Is there any nicer way to do it? Bests Reply bowlofred Da Bishop Posts: 1,190 Threads: 3 Joined: Mar 2020 Reputation: Nov-09-2020, 09:58 PM If the only objection is writing out the triplets, you could do the following. It runs the same commands, just does it in a loop. ```from itertool import product triplets = tuple(product((0, 1), repeat = 3)) sums = {} for triple in triplets: sums[triple] = sum(np.all((results-np.array(triple))==0, axis=1))```The sums will be in a dict `sums` rather than as separate variables. But you can pull them out, or you could change how the keys are named. However, this doesn't change the fundamental calls made. Perhaps someone else can see if there's a better way to have numpy do the sum more directly. quest likes this post Reply quest Not Blown Up Yet Posts: 50 Threads: 25 Joined: Nov 2020 Reputation: Nov-09-2020, 10:24 PM Thank you very much ! Reply

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