Conditional operations on numpy arrays

Question

I'm new to NumPy, and I've encountered a problem with running some conditional statements on numpy arrays. Let's say I have 3 numpy arrays that look like this:

a:

[[0, 4, 4, 2],
 [1, 3, 0, 2],
 [3, 2, 4, 4]]

b:

[[6, 9, 8, 6],
 [7, 7, 9, 6],
 [8, 6, 5, 7]]

and, c:

[[0, 0, 0, 0],
 [0, 0, 0, 0],
 [0, 0, 0, 0]]

I have a conditional statement for a and b in which I would like to use the value of b (if the conditions of a and b are met) to calculate the value of c:

c[(a > 3) & (b > 8)]+=b*2

I get an error saying:

Traceback (most recent call last):
  File "<interactive input>", line 1, in <module>
ValueError: non-broadcastable output operand with shape (1,) doesn't match the broadcast shape (3,4)

Any idea how I can accomplish this?

I would like the output of c to look as follows:

[[0, 18, 0, 0],
 [0, 0, 0, 0],
 [0, 0, 0, 0]]

Thanks! Upvotes for everyone as they all worked. I accepted the answer from @Psidom for np.where as it made the most sense to me and ran in the quickest amount of time (my actual script will run these conditionals a couple million times) — bobby12345
– bobby12345, Commented Mar 1, 2017 at 20:13

akuiper · Accepted Answer · 2017-03-01 19:26:02Z

23

You can use numpy.where:

np.where((a > 3) & (b > 8), c + b*2, c)
#array([[ 0, 18,  0,  0],
#       [ 0,  0,  0,  0],
#       [ 0,  0,  0,  0]])

Or arithmetically:

c + b*2 * ((a > 3) & (b > 8))
#array([[ 0, 18,  0,  0],
#       [ 0,  0,  0,  0],
#       [ 0,  0,  0,  0]])

answered Mar 1, 2017 at 19:26

akuiper

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Comments

willeM_ Van Onsem · Accepted Answer · 2018-04-24 14:05:35Z

10

The problem is that you mask the receiving part, but do not mask the sender part. As a result:

c[(a > 3) & (b > 8)]+=b*2
# ^ 1x1 matrix        ^3x4 matrix

The dimensions are not the same. Given you want to perform element-wise addition (based on your example), you can simply add the slicing to the right part as well:

c[(a > 3) & (b > 8)]+=b[(a > 3) & (b > 8)]*2

or make it more efficient:

mask = (a > 3) & (b > 8)
c[mask] += b[mask]*2

edited Apr 24, 2018 at 14:05

answered Mar 1, 2017 at 19:25

willeM_ Van Onsem

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Comments

ZdaR · Accepted Answer · 2017-03-01 19:31:22Z

3

A slight change in the numpy expression would get the desired results:

c += ((a > 3) & (b > 8)) * b*2

Here First I create a mask matrix with boolean values, from ((a > 3) & (b > 8)), then multiply the matrix with b*2 which in turn generates a 3x4 matrix which can be easily added to c

answered Mar 1, 2017 at 19:31

ZdaR

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