Reputation: 133
I would like to find the number of places where numpy.where has evaluated as true. The following solution works, but is pretty ugly.
b = np.where(a < 5)
num = (b[0]).shape[0]
I'm coming from a language where I need to check if num > 0 before proceeding to do something with the resulting array. Is there a more elegant way of getting num, or a more Pythonic solution than finding num?
(For those familiar with IDL, I'm trying to replicate its simple b = where(a lt 5, num)
.)
Upvotes: 5
Views: 2578
Reputation: 80456
In [1]: import numpy as np
In [2]: arr = np.arange(10)
In [3]: np.count_nonzero(arr < 5)
Out[3]: 5
or
In [4]: np.sum(arr < 5)
Out[4]: 5
If you have to define b = np.where(arr < 5)[0]
anyway, use len(b)
or b.size
( len()
seems to be a tiny bit faster, but they are pretty much the same in terms of performance).
Upvotes: 7
Reputation: 10417
Another way
In [14]: a = np.arange(100)
In [15]: np.where(a > 1000)[0].size > 0
Out[15]: False
In [16]: np.where(a > 10)[0].size > 0
Out[16]: True
Upvotes: 0
Reputation: 1095
This only works if the condition of the where statement implies that the result is not evaluated to False. But then it's easy:
import numpy as np
a = np.random.randint(1,11,100)
a_5 = a[np.where(a>5)]
a_10 = a[np.where(a>10)]
a_5.any() #True
a_10.any() #False
So if you wish you can try this before assigning, of course:
a[np.where(a>5)].any() #True
a[np.where(a>10)].any() #False
Upvotes: 0