I have the following 2D-array:
a = array([[ 1, 2, 3],
[ 4, 5, 6],
[ 7, 8, 9],
[10, 11, 12],
[13, 14, 15]])
and another 1D-array:
b = array([ 1, 2, 3, 4, 5])
then I want to calculate something like
c = a - b
with the intent of getting:
c = array([[0, 1, 2],
[2, 3, 4],
[4, 5, 6],
[6, 7, 8],
[8, 9, 10]])
but instead I get the error message:
Traceback (most recent call last):
Python Shell, prompt 79, line 1
ValueError: operands could not be broadcast together with shapes (5,3) (5,)
I read the broadcasting rules but didn´t get any wiser. I could do a workaround with for-loops or similar but there should be a direct way. Thanks
You need to convert array b to a (2, 1) shape
array, use None or numpy.newaxis
in the index tuple. Here is the Indexing of Numpy array .
You can do it Like:
import numpy
a = numpy.array([[ 1, 2, 3],
[ 4, 5, 6],
[ 7, 8, 9],
[10, 11, 12],
[13, 14, 15]])
b = numpy.array([ 1, 2, 3, 4, 5])
c=a - b[:,None]
print c
Output:
Out[2]:
array([[ 0, 1, 2],
[ 2, 3, 4],
[ 4, 5, 6],
[ 6, 7, 8],
[ 8, 9, 10]])
As Divakar specified in the comments, just add a new axis to b
.
I suggest you read more about broadcasting which is very often useful to vectorize computations in numpy: interestingly enough, a.transpose() - b
wouldn't have raised an error (you'd need to transpose the result again to obtain your desired output).
In this computaion, the first array's shape is (3, 5)
, and b.shape
is (5,). So the shape of b
corresponds to the tail of the shape of a
, and broadcasting can happen. This is not the case when the shape of the first array is (5, 3)
, hence the error you obtained.
Here are some runtime tests to compare the speeds of the suggested answers, with your values for a
and b
: you can see that the differences are not really significant
In [9]: %timeit (a.T - b).T
Out[9]: 1000000 loops, best of 3: 1.32 µs per loop
In [10]: %timeit a - b[:,None]
Out[10]: 1000000 loops, best of 3: 1.25 µs per loop
In [11]: %timeit a - b[None].T
Out[11]: 1000000 loops, best of 3: 1.3 µs per loop
Sometimes is usefull to do this:
C = numpy.zeros(shape=A.shape)
for i in range(len(A)):
C[i] = A[i] - B[i]
S = C.astype(int)
print(S)
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