I have a array of identifiers that have been grouped into threes. For each group, I would like to randomly assign them to one of three sets and to have those assignments stored in another array. So, for a given array of grouped identifiers (I presort them):
groupings = array([1,1,1,2,2,2,3,3,3])
A possible output would be
assignments = array([0,1,2,1,0,2,2,0,1])
Ultimately, I would like to be able to generate many of these assignment lists and to do so efficiently. My current method is just to create an zeroes array and set each consecutive subarray of length 3 to a random permutation of 3.
assignment = numpy.zeros((12,10),dtype=int)
for i in range(0,12,3):
for j in range(10):
assignment[i:i+3,j] = numpy.random.permutation(3)
Is there a better/faster way?
We can do it by simply using the built-in permutation function in itertools library. It is the shortest technique to find the permutation.
To get the subarray we can use slicing to get the subarray. Step 1: Run a loop till length+1 of the given list. Step 2: Run another loop from 0 to i. Step 3: Slice the subarray from j to i.
A permutation enumeration is to list all permutations in S n . For example, there are 24 permutations of [4]: 1234, 1243, 1324, 1342, 1423, 1432, 2134, 2143, 2314, 2341, 2413, 2431, 3124, 3142, 3214, 3241, 3412, 3421, 4123, 4132, 4213, 4231, 4312, 4321.
Two things I can think about:
instead of visiting the 2D array 3 row * 1 column
in your inner loop, try to visit it 1*3
. Accessing 2D array horizontally first is usually faster than vertically first, since it gives you better spatial locality, which is good for caching.
instead of running numpy.random.permutation(3)
each time, if 3
is fixed and is a small number, try to generate the arrays of permutations beforehand and save them into a constant array of array like: (array([0,1,2]), array([0,2,1]), array([1,0,2])...)
. You just need to randomly pick one array from it each time.
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