Reputation: 8067
This relates to this earlier post: Numpy random choice of tuples
I have a 2D numpy array and want to choose from it using a 2D probability array. The only way I could think to do this was to flatten and then use the modulo and remainder to convert the result back to a 2D index
import numpy as np
# dummy data
x=np.arange(100).reshape(10,10)
# dummy probability array
p=np.zeros([10,10])
p[4:7,1:4]=1.0/9
xy=np.random.choice(x.flatten(),1,p=p.flatten())
index=[int(xy/10),(xy%10)[0]] # convert back to index
print(index)
which gives
[5, 2]
but is there a cleaner way that avoids flattening and the modulo? i.e. I could pass a list of coordinate tuples as x, but how can I then handle the weights?
Upvotes: 2
Views: 1901
Reputation: 88226
I don't think it's possible to directly specify a 2D shaped array of probabilities. So raveling should be fine. However to get the corresponding 2D shaped indices from the flat index you can use np.unravel_index
index= np.unravel_index(xy.item(), x.shape)
# (4, 2)
For multiple indices, you can just stack the result:
xy=np.random.choice(x.flatten(),3,p=p.flatten())
indices = np.unravel_index(xy, x.shape)
# (array([4, 4, 5], dtype=int64), array([1, 2, 3], dtype=int64))
np.c_[indices]
array([[4, 1],
[4, 2],
[5, 3]], dtype=int64)
where np.c_
stacks along the right hand axis and gives the same result as
np.column_stack(indices)
Upvotes: 3
Reputation: 1992
You could use numpy.random.randint
to generate an index, for example:
# assumes p is a square array
ij = np.random.randint(p.shape[0], size=p.ndim) # size p.ndim = 2 generates 2 coords
# need to convert to tuple to index correctly
p[tuple(i for i in ij))]
>>> 0.0
You can also index multiple random values at once:
ij = np.random.randint(p.shape[0], size=(p.ndim, 5)) # get 5 values
p[tuple(i for i in ij))]
>>> array([0. , 0. , 0. , 0.11111111, 0. ])
Upvotes: 0