How to find index of minimum non zero element with numpy?
Question:
I have a 4×1 array that I want to search for the minimum non zero value and find its index. For example:
theta = array([0,1,2,3]).reshape(4,1)
It was suggested in a similar thread to use nonzero() or where(), but when I tried to use that in the way that was suggested, it creates a new array that doesn’t have the same indices as the original:
np.argmin(theta[np.nonzero(theta)])
gives an index of zero, which clearly isn’t right. I think this is because it creates a new array of non zero elements first. I am only interested in the first minimum value if there are duplicates.
Answers:
np.nonzero(theta) returns the index of the values that are non-zero. In your case, it returns,
[1,2,3]
Then, theta[np.nonzero(theta)] returns the values
[1,2,3]
When you do np.argmin(theta[np.nonzero(theta)]) on the previous output, it returns the index of the value 1 which is 0.
Hence, the correct approach would be:
i,j = np.where( theta==np.min(theta[np.nonzero(theta)])) where i,j are the indices of the minimum non zero element of the original numpy array
theta[i,j] or theta[i] gives the respective value at that index.
I think you @Emily were very close to the correct answer. You said:
np.argmin(theta[np.nonzero(theta)]) gives an index of zero, which clearly isn’t right. I think this is because it creates a new array of non zero elements first.
The last sentence is correct => the first one is wrong since it is expected to give the index in the new array.
Let’s now extract the correct index in the old (original) array:
nztheta_ind = np.nonzero(theta)
k = np.argmin(theta[nztheta_ind])
i = nztheta_ind[0][k]
j = nztheta_ind[1][k]
or:
[i[k] for i in nztheta_ind]
for arbitrary dimensionality of original array.
#!/usr/bin/env python
# Solution utilizing numpy masking of zero value in array
import numpy as np
import numpy.ma as ma
a = [0,1,2,3]
a = np.array(a)
print "your array: ",a
# the non-zero minimum value
minval = np.min(ma.masked_where(a==0, a))
print "non-zero minimum: ",minval
# the position/index of non-zero minimum value in the array
minvalpos = np.argmin(ma.masked_where(a==0, a))
print "index of non-zero minimum: ", minvalpos
ndim Solution
i = np.unravel_index(np.where(theta!=0, theta, theta.max()+1).argmin(), theta.shape)
Explaination
- Masking the zeros out creates
t0. There are other ways, see the perfplot.
- Finding the minimum location, returns the flattened (1D) index.
unravel_index fixes this problem, and hasn’t been suggested yet.
theta = np.triu(np.random.rand(4,4), 1) # example array
t0 = np.where(theta!=0, theta, np.nan) # 1
i0 = np.nanargmin(t0) # 2
i = np.unravel_index(i0, theta.shape) # 3
print(theta, i, theta[i]) #
mask: i = np.unravel_index(np.ma.masked_where(a==0, a).argmin(), a.shape)
nan: i = np.unravel_index(np.nanargmin(np.where(a!=0, a, np.nan)), a.shape)
max: i = np.unravel_index(np.where(a!=0, a, a.max()+1).argmin(), a.shape)
@ Jay M, that worked for me, just putting it here if I am doing it right
j,k = unravel_index(argmin(ma.masked_where(V==0, V)),V.shape)
V[j,k] is my 2D array.
I have a 4×1 array that I want to search for the minimum non zero value and find its index. For example:
theta = array([0,1,2,3]).reshape(4,1)
It was suggested in a similar thread to use nonzero() or where(), but when I tried to use that in the way that was suggested, it creates a new array that doesn’t have the same indices as the original:
np.argmin(theta[np.nonzero(theta)])
gives an index of zero, which clearly isn’t right. I think this is because it creates a new array of non zero elements first. I am only interested in the first minimum value if there are duplicates.
np.nonzero(theta) returns the index of the values that are non-zero. In your case, it returns,
[1,2,3]
Then, theta[np.nonzero(theta)] returns the values
[1,2,3]
When you do np.argmin(theta[np.nonzero(theta)]) on the previous output, it returns the index of the value 1 which is 0.
Hence, the correct approach would be:
i,j = np.where( theta==np.min(theta[np.nonzero(theta)])) where i,j are the indices of the minimum non zero element of the original numpy array
theta[i,j] or theta[i] gives the respective value at that index.
I think you @Emily were very close to the correct answer. You said:
np.argmin(theta[np.nonzero(theta)])gives an index of zero, which clearly isn’t right. I think this is because it creates a new array of non zero elements first.
The last sentence is correct => the first one is wrong since it is expected to give the index in the new array.
Let’s now extract the correct index in the old (original) array:
nztheta_ind = np.nonzero(theta)
k = np.argmin(theta[nztheta_ind])
i = nztheta_ind[0][k]
j = nztheta_ind[1][k]
or:
[i[k] for i in nztheta_ind]
for arbitrary dimensionality of original array.
#!/usr/bin/env python
# Solution utilizing numpy masking of zero value in array
import numpy as np
import numpy.ma as ma
a = [0,1,2,3]
a = np.array(a)
print "your array: ",a
# the non-zero minimum value
minval = np.min(ma.masked_where(a==0, a))
print "non-zero minimum: ",minval
# the position/index of non-zero minimum value in the array
minvalpos = np.argmin(ma.masked_where(a==0, a))
print "index of non-zero minimum: ", minvalpos
ndim Solution
i = np.unravel_index(np.where(theta!=0, theta, theta.max()+1).argmin(), theta.shape)
Explaination
- Masking the zeros out creates
t0. There are other ways, see the perfplot. - Finding the minimum location, returns the flattened (1D) index.
unravel_indexfixes this problem, and hasn’t been suggested yet.
theta = np.triu(np.random.rand(4,4), 1) # example array
t0 = np.where(theta!=0, theta, np.nan) # 1
i0 = np.nanargmin(t0) # 2
i = np.unravel_index(i0, theta.shape) # 3
print(theta, i, theta[i]) #
mask:i = np.unravel_index(np.ma.masked_where(a==0, a).argmin(), a.shape)
nan:i = np.unravel_index(np.nanargmin(np.where(a!=0, a, np.nan)), a.shape)
max:i = np.unravel_index(np.where(a!=0, a, a.max()+1).argmin(), a.shape)
@ Jay M, that worked for me, just putting it here if I am doing it right
j,k = unravel_index(argmin(ma.masked_where(V==0, V)),V.shape)
V[j,k] is my 2D array.