When only a single argument is supplied to numpy's where
function it returns the indices of the input array (the condition
) that evaluate as true (same behaviour as numpy.nonzero
). This can be used to extract the indices of an array that satisfy a given condition.
import numpy as np
a = np.arange(20).reshape(2,10)
# a = array([[ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9],
# [10, 11, 12, 13, 14, 15, 16, 17, 18, 19]])
# Generate boolean array indicating which values in a are both greater than 7 and less than 13
condition = np.bitwise_and(a>7, a<13)
# condition = array([[False, False, False, False, False, False, False, False, True, True],
# [True, True, True, False, False, False, False, False, False, False]], dtype=bool)
# Get the indices of a where the condition is True
ind = np.where(condition)
# ind = (array([0, 0, 1, 1, 1]), array([8, 9, 0, 1, 2]))
keep = a[ind]
# keep = [ 8 9 10 11 12]
If you do not need the indices, this can be achieved in one step using extract
, where you agian specify the condition
as the first argument, but give the array
to return the values from where the condition is true as the second argument.
# np.extract(condition, array)
keep = np.extract(condition, a)
# keep = [ 8 9 10 11 12]
Two further arguments x
and y
can be supplied to where
, in which case the output will contain the values of x
where the condition is True
and the values of y
where the condition is False
.
# Set elements of a which are NOT greater than 7 and less than 13 to zero, np.where(condition, x, y)
a = np.where(condition, a, a*0)
print(a)
# Out: array([[ 0, 0, 0, 0, 0, 0, 0, 0, 8, 9],
# [10, 11, 12, 0, 0, 0, 0, 0, 0, 0]])