I have a NumPy array like:
a = np.array([1,2,3,4,0,0,5,6,7,8,0,0,9,10,11,12])
What is the most effective way to select all values except values (in my example is 0) at some positions?
So I need to get an array:
[1,2,3,4,5,6,7,8,9,10,11,12]
I know how to skip the one nth value with [::n] construction but is it possible to skip several values using the similar syntax?
Thank you for any help!
You probably want np.delete:
>>> np.delete(a, [4, 5, 10, 11])
array([ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12])
a[start:stop:step], which isn't your case. A copy will be made regardless, np.delete is the fastest way to produce it.I see two options:
If you want to get an index vector that you can use on multiple arrays:
import numpy as np
#your input
a = np.array([1,2,3,4,0,0,5,6,7,8,0,0,9,10,11,12])
#indices of elements that you want to remove (given)
idx = [4,5,10,11]
#get the inverted indices
idx_inv = [x for x in range(len(a)) if x not in idx]
a[idx_inv]
This output:
array([ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12])
Use np.delete:
import numpy as np
#your input
a = np.array([1,2,3,4,0,0,5,6,7,8,0,0,9,10,11,12])
#indices of elements that you want to remove (given)
idx = [4,5,10,11]
np.delete(a,idx)
This outputs:
array([ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12])
idx_inv can be computed using np.in1d much more efficiently.Using boolean or mask index array:
>>> a = np.array([1,2,3,4,0,0,5,6,7,8,0,0,9,10,11,12])
>>> a[a != 0]
array([ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12])
a[(a != 4) & (a != 0)] or a[a < 3]You can use Boolean array indexing:
import numpy as np
a = np.array([1,2,3,4,0,0,5,6,7,8,0,0,9,10,11,12])
print a[a != 0]
# Output: [ 1 2 3 4 5 6 7 8 9 10 11 12]
and you can change a != 0 to other conditions which result in a Boolean array.
