I'd like to get an NxM
matrix where numbers in each row are random samples generated from different normal distributions(same mean
but different standard deviations). The following code works:
import numpy as np
mean = 0.0 # same mean
stds = [1.0, 2.0, 3.0] # different stds
matrix = np.random.random((3,10))
for i,std in enumerate(stds):
matrix[i] = np.random.normal(mean, std, matrix.shape[1])
However, this code is not quite efficient as there is a for
loop involved. Is there a faster way to do this?
np.random.normal()
is vectorized; you can switch axes and transpose the result:
np.random.seed(444)
arr = np.random.normal(loc=0., scale=[1., 2., 3.], size=(1000, 3)).T
print(arr.mean(axis=1))
# [-0.06678394 -0.12606733 -0.04992722]
print(arr.std(axis=1))
# [0.99080274 2.03563299 3.01426507]
That is, the scale
parameter is the column-wise standard deviation, hence the need to transpose via .T
since you want row-wise inputs.
How about this?
rows = 10000
stds = [1, 5, 10]
data = np.random.normal(size=(rows, len(stds)))
scaled = data * stds
print(np.std(scaled, axis=0))
Output:
[ 0.99417905 5.00908719 10.02930637]
This exploits the fact that a two normal distributions can be interconverted by linear scaling (in this case, multiplying by standard deviation). In the output, each column (second axis) will contain a normally distributed variable corresponding to a value in stds
.
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