CODEWITHSUNDEEP
pythonarraysnumpy
Ehsan
Ehsan

Reputation: 12407

Randomly relocate the selected elements of a numpy array according to a distribution?(expand the dimensions of a 2D array to 3D by random numbers)

I have a 2D numpy array of 0s and 1s. 

a = np.array([[1, 0, 0, 1], [0, 1, 0, 1]])

What I need is to create a new array a_new according to this:
For each 1 in location [l, k] of array a, pick a random number according to a desired distribution (e.g. shift = np.int64(np.ceil(np.random.gamma(1, 3)))) and put it at a_new[l, k, shift] if shift is smaller than N, otherwise ignore that 1.

Here is a loop implantation of it. Is there a faster (maybe array operation) solution to this. The matrix a's size is large. 

import numpy as np

N = 5
a = np.array([[1, 0, 0, 1], [0, 1, 0, 1]])
a_new = np.zeros((a.shape[0], a.shape[1], N))
for k in np.arange(a.shape[1]):
    for l in np.arange(a.shape[0]):
        if a[l, k]:
            shift = np.int64(np.ceil(np.random.gamma(1, 3)))
            if (shift < N):
                a_new[l, k, shift] = 1

output sample: 

a 
[[1 0 0 1]
 [0 1 0 1]]
a_new 
[[[0. 0. 1. 0. 0.]
  [0. 0. 0. 0. 0.]
  [0. 0. 0. 0. 0.]
  [0. 0. 1. 0. 0.]]

 [[0. 0. 0. 0. 0.]
  [0. 0. 0. 1. 0.]
  [0. 0. 0. 0. 0.]
  [0. 0. 0. 0. 0.]]]

Upvotes: 2

Views: 220

Answers (3)

Ehsan
Ehsan

Reputation: 12407

I found the solution with advanced indexing in numpy that is significantly faster. Basically we can use non zero elements of a as indices to first two dimensions of a_new and the random values of shift as the indices to the third dimension of a_new with some filtering beforehand (remove out of bound random numbers, and reshaping shift array to have same shape as nonzero sub-array of a. Here is the working code:
import numpy as np

N = 5
a = np.array([[1, 0, 0, 1], [0, 1, 0, 1]])
a_new = np.zeros((a.shape[0], a.shape[1], N))
shift = np.int64(np.ceil(np.random.gamma(1, 3, a.shape)))
a[shift > N-1] = 0
shift[shift > N-1] = 0
shift = (shift * a).reshape(1, -1)
shift = shift[shift > 0]
non_zero_a = np.nonzero(a)
a_new[non_zero_a[0], non_zero_a[1], shift] = 1

Upvotes: 0

Paul Panzer
Paul Panzer

Reputation: 53029

Here is a trick using np.bincount:

a = np.array([[1, 0, 0, 1], [0, 1, 0, 1]])
N = 5
X,Y = a.nonzero()
Z = np.ceil(np.random.gamma(1,3,X.shape)).astype(int)
Z
# array([ 2,  1, 15,  2])
flat = np.ravel_multi_index((X[Z<N],Y[Z<N],Z[Z<N]),a.shape+(N,))
np.bincount(flat,None,a.size*N).reshape(*a.shape,N)
# array([[[0, 0, 1, 0, 0],
#         [0, 0, 0, 0, 0],
#         [0, 0, 0, 0, 0],
#         [0, 1, 0, 0, 0]],
#
#        [[0, 0, 0, 0, 0],
#         [0, 0, 0, 0, 0],
#         [0, 0, 0, 0, 0],
#         [0, 0, 1, 0, 0]]])

UPDATE: With multiplicities:

a = np.array([[1, 0, 0, 3], [0, 10, 0, 1]])

N = 5
X,Y = a.nonzero()
times = a[X,Y]
X = X.repeat(times)
Y = Y.repeat(times)
Z = np.ceil(np.random.gamma(1,3,X.shape)).astype(int)
flat = np.ravel_multi_index((X[Z<N],Y[Z<N],Z[Z<N]),a.shape+(N,))
np.bincount(flat,None,a.size*N).reshape(*a.shape,N)
# array([[[0, 1, 0, 0, 0],
#        [0, 0, 0, 0, 0],
#        [0, 0, 0, 0, 0],
#        [0, 1, 0, 0, 0]],
#
#       [[0, 0, 0, 0, 0],
#        [0, 4, 2, 0, 3],
#        [0, 0, 0, 0, 0],
#        [0, 0, 0, 0, 0]]])

Upvotes: 1

Nik P
Nik P

Reputation: 224

You could also try the following which may help,

import numpy as np
N = 5
a = np.array([[1, 0, 0, 1], [0, 1, 0, 1]])
row, col = np.where(a==1)
a_new = np.zeros([a.shape[0], a.shape[1], N])
a_new[row,col] = (np.int64(np.ceil(np.random.gamma(1,3,[a.shape[1],N])))<N)

Upvotes: 0

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