How to transform negative elements to zero without a loop?

Question

If I have an array like

a = np.array([2, 3, -1, -4, 3])

I want to set all the negative elements to zero: [2, 3, 0, 0, 3]. How to do it with numpy without an explicit for? I need to use the modified a in a computation, for example

c = a * b

where b is another array with the same length of the original a

Conclusion

import numpy as np
from time import time

a = np.random.uniform(-1, 1, 20000000)
t = time(); b = np.where(a>0, a, 0); print "1. ", time() - t
a = np.random.uniform(-1, 1, 20000000)
t = time(); b = a.clip(min=0); print "2. ", time() - t
a = np.random.uniform(-1, 1, 20000000)
t = time(); a[a < 0] = 0; print "3. ", time() - t
a = np.random.uniform(-1, 1, 20000000)
t = time(); a[np.where(a<0)] = 0; print "4. ", time() - t
a = np.random.uniform(-1, 1, 20000000)
t = time(); b = [max(x, 0) for x in a]; print "5. ", time() - t

1. 1.38629984856
2. 0.516846179962 <- faster a.clip(min=0);
3. 0.615426063538
4. 0.944557905197
5. 51.7364809513

Answer 1

Does map do what you need it to?

b = map(lambda x: max(x, 0), a) # b == [2, 3, 0, 0, 3]

You can also get there with this list comprehension:

b = [max(x, 0) for x in a]

Answer 2

a = a.clip(min=0)

Answer 3

I would do this:

a[a < 0] = 0

If you want to keep the original a and only set the negative elements to zero in a copy, you can copy the array first:

c = a.copy()
c[c < 0] = 0

Answer 4

Use where

a[numpy.where(a<0)] = 0