I want to index a numpy array based on specific numbers, e.g., I have the following np array,
b = np.array([[1, 2, 3, 2, 1, 3],[2, 1, 3, 5, 4, 1]])
And I want to change to zero the values different from 1 and 2. I tried,
b[b not in [1, 2]] = 0
but did work. Any ideas on how to do it?
You can use numpy.isin():
import numpy as np
b = np.array([[1, 2, 3, 2, 1, 3],[2, 1, 3, 5, 4, 1]])
v = np.array([1,2])
m = np.isin(b, v)
b[~m] = 0
print(b)
Output:
[[1 2 0 2 1 0]
[2 1 0 0 0 1]]
In [266]: b = np.array([[1, 2, 3, 2, 1, 3],[2, 1, 3, 5, 4, 1]])
Since all values you want to change are greater than 2:
In [267]: b>2
Out[267]:
array([[False, False, True, False, False, True],
[False, False, True, True, True, False]])
In [268]: np.where(b>2,0,b)
Out[268]:
array([[1, 2, 0, 2, 1, 0],
[2, 1, 0, 0, 0, 1]])
or pairing two tests:
In [271]: (b==1)|(b==2)
Out[271]:
array([[ True, True, False, True, True, False],
[ True, True, False, False, False, True]])
In [272]: np.where((b==1)|(b==2),b,0)
Out[272]:
array([[1, 2, 0, 2, 1, 0],
[2, 1, 0, 0, 0, 1]])
We could change b in place, but I chose where so I can test various ideas without changing b.
isin uses a similar idea if one set is significantly smaller than the other, which is true in this case.
