a = np.array([0,1,2])
b = np.array([3,4,5,6,7])
...
c = np.dot(a,b)
I want to transpose b so I can calculate the dot product of a and b.
4 Answers
You can use numpy's broadcasting for this:
import numpy as np
a = np.array([0,1,2])
b = np.array([3,4,5,6,7])
In [3]: a[:,None]*b
Out[3]:
array([[ 0, 0, 0, 0, 0],
[ 3, 4, 5, 6, 7],
[ 6, 8, 10, 12, 14]])
This has nothing to do with a dot product, though. But in the comments you said, that you want this result.
You could also use the numpy function outer:
In [4]: np.outer(a, b)
Out[4]:
array([[ 0, 0, 0, 0, 0],
[ 3, 4, 5, 6, 7],
[ 6, 8, 10, 12, 14]])
Comments
Well for this what you want is the outer product of the two arrays. The function you want to use for this is np.outer, :
a = np.array([0,1,2])
b = np.array([3,4,5,6,7])
np.outer(a,b)
array([[ 0, 0, 0, 0, 0],
[ 3, 4, 5, 6, 7],
[ 6, 8, 10, 12, 14]])
3 Comments
user8408080
What is the difference between
np.outer and np.multiply.outer?user8408080
Was just wondering, because I just put the function in my answer like two minutes before you posted your answer and I thought, yours did something different :D But now I learned about
np.add.outer which is nice :D
yatu
Ahh no the same, just that
np.ufunc.outer can be used for multiple operations. Good answer btw +1So with NumPy you could reshape swapping axes:
a = np.swapaxes([a], 1, 0)
# [[0]
# [1]
# [2]]
Then
print(a * b)
# [[ 0 0 0 0 0]
# [ 3 4 5 6 7]
# [ 6 8 10 12 14]]
Swapping b require to transpose the product, se here below.
Or usual NumPy reshape:
a = np.array([0,1,2])
b = np.array([3,4,5,6,7]).reshape(5,1)
print((a * b).T)
# [[ 0 0 0 0 0]
# [ 3 4 5 6 7]
# [ 6 8 10 12 14]]
Reshape is like b = np.array([ [bb] for bb in [3,4,5,6,7] ]) then b becomes:
# [[3]
# [4]
# [5]
# [6]
# [7]]
While reshaping
a no need to transpose:
a = np.array([0,1,2]).reshape(3,1)
b = np.array([3,4,5,6,7])
print(a * b)
# [[ 0 0 0 0 0]
# [ 3 4 5 6 7]
# [ 6 8 10 12 14]]
Just out of curiosity, good old list comprehension:
a = [0,1,2]
b = [3,4,5,6,7]
print( [ [aa * bb for bb in b] for aa in a ] )
#=> [[0, 0, 0, 0, 0], [3, 4, 5, 6, 7], [6, 8, 10, 12, 14]]
1 Comment
user8408080
OPs expected output is yours, but transposed, so you might want to add a
.T thereOthers have provided the outer and broadcasted solutions. Here's the dot one(s):
np.dot(a.reshape(3,1), b.reshape(1,5))
a[:,None].dot(b[None,:])
a[None].T.dot( b[None])
Conceptually I think it's a bit of an overkill, but due to implementation details, it actually is fastest .
1 Comment
user8408080
This is actually faster than my answer? Do you have some times? Can you explain? I'm really interested tbh
dotproduces the inner product. Whatcdo you want? I suspect you want thenp.outerproduct, not inner or matrix product.cis supposed to be.care you expecting?numpyoperation. No need to think in terms of transposition.