numpy 从二维数组中减去/添加一维数组

Question

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

Answer 1

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]])

Answer 2

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

Answer 3

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)