numpy matrix multiplication each col and sum
up vote
1
down vote
favorite
I have some matrices:
A = np.array([[0.8, 0.2],
[0.4, 0.6]])
B = np.array([0.4, 0.2])
C = np.array([0.28, 0.06])
And what I want to do is:
S = A.shape[0]
what_I_need = np.zeros([S])
for s in range(S):
what_I_need[s] = np.sum(A[:, s] * C) * B[s]
print(what_I_need)
The result is
[0.0992 0.0184]
Is there an elegant or pythonic way to do this without for loop? Thanks
python numpy
asked Nov 11 at 7:26
Sheng
295
add a comment |
up vote
1
down vote
favorite
I have some matrices:
A = np.array([[0.8, 0.2],
[0.4, 0.6]])
B = np.array([0.4, 0.2])
C = np.array([0.28, 0.06])
And what I want to do is:
S = A.shape[0]
what_I_need = np.zeros([S])
for s in range(S):
what_I_need[s] = np.sum(A[:, s] * C) * B[s]
print(what_I_need)
The result is
[0.0992 0.0184]
Is there an elegant or pythonic way to do this without for loop? Thanks
python numpy
asked Nov 11 at 7:26
Sheng
295
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I have some matrices:
A = np.array([[0.8, 0.2],
[0.4, 0.6]])
B = np.array([0.4, 0.2])
C = np.array([0.28, 0.06])
And what I want to do is:
S = A.shape[0]
what_I_need = np.zeros([S])
for s in range(S):
what_I_need[s] = np.sum(A[:, s] * C) * B[s]
print(what_I_need)
The result is
[0.0992 0.0184]
Is there an elegant or pythonic way to do this without for loop? Thanks
python numpy
asked Nov 11 at 7:26
Sheng
295
I have some matrices:
A = np.array([[0.8, 0.2],
[0.4, 0.6]])
B = np.array([0.4, 0.2])
C = np.array([0.28, 0.06])
And what I want to do is:
S = A.shape[0]
what_I_need = np.zeros([S])
for s in range(S):
what_I_need[s] = np.sum(A[:, s] * C) * B[s]
print(what_I_need)
The result is
[0.0992 0.0184]
Is there an elegant or pythonic way to do this without for loop? Thanks
python numpy
python numpy
asked Nov 11 at 7:26
Sheng
295
asked Nov 11 at 7:26
Sheng
295
asked Nov 11 at 7:26
Sheng
295
asked Nov 11 at 7:26
Sheng
295
asked Nov 11 at 7:26
Sheng
295
295
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
4
down vote
accepted
With matrix-multiplication -
A.T.dot(C)*B # or C.dot(A)*B
With a single call to np.einsum -
np.einsum('ij,i,j->j',A,C,B) #use `optimize` flag for better perf.
answered Nov 11 at 7:29
Divakar
152k1478167
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
With matrix-multiplication -
A.T.dot(C)*B # or C.dot(A)*B
With a single call to np.einsum -
np.einsum('ij,i,j->j',A,C,B) #use `optimize` flag for better perf.
answered Nov 11 at 7:29
Divakar
152k1478167
add a comment |
up vote
4
down vote
accepted
With matrix-multiplication -
A.T.dot(C)*B # or C.dot(A)*B
With a single call to np.einsum -
np.einsum('ij,i,j->j',A,C,B) #use `optimize` flag for better perf.
answered Nov 11 at 7:29
Divakar
152k1478167
add a comment |
up vote
4
down vote
accepted
up vote
4
down vote
accepted
With matrix-multiplication -
A.T.dot(C)*B # or C.dot(A)*B
With a single call to np.einsum -
np.einsum('ij,i,j->j',A,C,B) #use `optimize` flag for better perf.
answered Nov 11 at 7:29
Divakar
152k1478167
With matrix-multiplication -
A.T.dot(C)*B # or C.dot(A)*B
With a single call to np.einsum -
np.einsum('ij,i,j->j',A,C,B) #use `optimize` flag for better perf.
answered Nov 11 at 7:29
Divakar
152k1478167
answered Nov 11 at 7:29
Divakar
152k1478167
answered Nov 11 at 7:29
Divakar
152k1478167
answered Nov 11 at 7:29
Divakar
152k1478167
152k1478167
add a comment |
add a comment |
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