Numpy dot() function equivalent
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-2
down vote
favorite
This question is just out of pure curiosity. Suppose I have 2 matrices a and b.
a=np.array([[1, 2],
[2, 3],
[4, 5]])
b=np.array([[1, 2, 3, 4],
[2, 3, 4, 5]])
To find their dot product, I might use np.dot(a,b). But is there any other way to do this? I am not asking for any other alias functions. But maybe another way to do this like np.sum(a*b, axis=1) (I know that doesn't work, it is just an example). And what if I have a 3-D matrix? Is there any other way to compute their dot product as well (without using any functions)?
Thanks in advance!
python numpy matrix matrix-multiplication
asked Nov 11 at 19:50
ѕняєє ѕιиgнι
30014
|
show 2 more comments
up vote
-2
down vote
favorite
This question is just out of pure curiosity. Suppose I have 2 matrices a and b.
a=np.array([[1, 2],
[2, 3],
[4, 5]])
b=np.array([[1, 2, 3, 4],
[2, 3, 4, 5]])
To find their dot product, I might use np.dot(a,b). But is there any other way to do this? I am not asking for any other alias functions. But maybe another way to do this like np.sum(a*b, axis=1) (I know that doesn't work, it is just an example). And what if I have a 3-D matrix? Is there any other way to compute their dot product as well (without using any functions)?
Thanks in advance!
python numpy matrix matrix-multiplication
asked Nov 11 at 19:50
ѕняєє ѕιиgнι
30014
You can use the@operator:a@b.
– Willem Van Onsem
Nov 11 at 19:51
Furthermorenp.multiplydoes not produces a matrix multiplication.
– Willem Van Onsem
Nov 11 at 19:52
@Willem Van Onsem I was asking for a mathematical equivalent, not another operator or function
– ѕняєє ѕιиgнι
Nov 11 at 19:57
It's easy with an added bit of broadcasting. Think about the dimensions you need to multiply and sum
– hpaulj
Nov 11 at 19:57
@hpaulj Could you please be a bit more specific and post your answer?
– ѕняєє ѕιиgнι
Nov 11 at 19:58
|
show 2 more comments
up vote
-2
down vote
favorite
up vote
-2
down vote
favorite
This question is just out of pure curiosity. Suppose I have 2 matrices a and b.
a=np.array([[1, 2],
[2, 3],
[4, 5]])
b=np.array([[1, 2, 3, 4],
[2, 3, 4, 5]])
To find their dot product, I might use np.dot(a,b). But is there any other way to do this? I am not asking for any other alias functions. But maybe another way to do this like np.sum(a*b, axis=1) (I know that doesn't work, it is just an example). And what if I have a 3-D matrix? Is there any other way to compute their dot product as well (without using any functions)?
Thanks in advance!
python numpy matrix matrix-multiplication
asked Nov 11 at 19:50
ѕняєє ѕιиgнι
30014
This question is just out of pure curiosity. Suppose I have 2 matrices a and b.
a=np.array([[1, 2],
[2, 3],
[4, 5]])
b=np.array([[1, 2, 3, 4],
[2, 3, 4, 5]])
To find their dot product, I might use np.dot(a,b). But is there any other way to do this? I am not asking for any other alias functions. But maybe another way to do this like np.sum(a*b, axis=1) (I know that doesn't work, it is just an example). And what if I have a 3-D matrix? Is there any other way to compute their dot product as well (without using any functions)?
Thanks in advance!
python numpy matrix matrix-multiplication
python numpy matrix matrix-multiplication
asked Nov 11 at 19:50
ѕняєє ѕιиgнι
30014
asked Nov 11 at 19:50
ѕняєє ѕιиgнι
30014
edited Nov 11 at 19:55
asked Nov 11 at 19:50
ѕняєє ѕιиgнι
30014
asked Nov 11 at 19:50
ѕняєє ѕιиgнι
30014
asked Nov 11 at 19:50
ѕняєє ѕιиgнι
30014
30014
You can use the@operator:a@b.
– Willem Van Onsem
Nov 11 at 19:51
Furthermorenp.multiplydoes not produces a matrix multiplication.
– Willem Van Onsem
Nov 11 at 19:52
@Willem Van Onsem I was asking for a mathematical equivalent, not another operator or function
– ѕняєє ѕιиgнι
Nov 11 at 19:57
It's easy with an added bit of broadcasting. Think about the dimensions you need to multiply and sum
– hpaulj
Nov 11 at 19:57
@hpaulj Could you please be a bit more specific and post your answer?
– ѕняєє ѕιиgнι
Nov 11 at 19:58
|
show 2 more comments
You can use the@operator:a@b.
– Willem Van Onsem
Nov 11 at 19:51
Furthermorenp.multiplydoes not produces a matrix multiplication.
– Willem Van Onsem
Nov 11 at 19:52
@Willem Van Onsem I was asking for a mathematical equivalent, not another operator or function
– ѕняєє ѕιиgнι
Nov 11 at 19:57
It's easy with an added bit of broadcasting. Think about the dimensions you need to multiply and sum
– hpaulj
Nov 11 at 19:57
@hpaulj Could you please be a bit more specific and post your answer?
– ѕняєє ѕιиgнι
Nov 11 at 19:58
You can use the
@ operator: a@b.– Willem Van Onsem
Nov 11 at 19:51
You can use the
@ operator: a@b.– Willem Van Onsem
Nov 11 at 19:51
Furthermore
np.multiply does not produces a matrix multiplication.– Willem Van Onsem
Nov 11 at 19:52
Furthermore
np.multiply does not produces a matrix multiplication.– Willem Van Onsem
Nov 11 at 19:52
@Willem Van Onsem I was asking for a mathematical equivalent, not another operator or function
– ѕняєє ѕιиgнι
Nov 11 at 19:57
@Willem Van Onsem I was asking for a mathematical equivalent, not another operator or function
– ѕняєє ѕιиgнι
Nov 11 at 19:57
It's easy with an added bit of broadcasting. Think about the dimensions you need to multiply and sum
– hpaulj
Nov 11 at 19:57
It's easy with an added bit of broadcasting. Think about the dimensions you need to multiply and sum
– hpaulj
Nov 11 at 19:57
@hpaulj Could you please be a bit more specific and post your answer?
– ѕняєє ѕιиgнι
Nov 11 at 19:58
@hpaulj Could you please be a bit more specific and post your answer?
– ѕняєє ѕιиgнι
Nov 11 at 19:58
|
show 2 more comments
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
In [66]: a=np.array([[1, 2],
...: [2, 3],
...: [4, 5]])
...:
...: b=np.array([[1, 2, 3, 4],
...: [2, 3, 4, 5]])
...:
...:
In [67]: np.dot(a,b)
Out[67]:
array([[ 5, 8, 11, 14],
[ 8, 13, 18, 23],
[14, 23, 32, 41]])
In [68]: a@b
Out[68]:
array([[ 5, 8, 11, 14],
[ 8, 13, 18, 23],
[14, 23, 32, 41]])
In [69]: np.einsum('ij,jk',a,b)
Out[69]:
array([[ 5, 8, 11, 14],
[ 8, 13, 18, 23],
[14, 23, 32, 41]])
Broadcasted multiply and sum:
In [71]: (a[:,:,None]*b[None,:,:]).sum(axis=1)
Out[71]:
array([[ 5, 8, 11, 14],
[ 8, 13, 18, 23],
[14, 23, 32, 41]])
In [72]: (a[:,:,None]*b[None,:,:]).shape
Out[72]: (3, 2, 4)
answered Nov 12 at 3:38
hpaulj
109k674140
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
In [66]: a=np.array([[1, 2],
...: [2, 3],
...: [4, 5]])
...:
...: b=np.array([[1, 2, 3, 4],
...: [2, 3, 4, 5]])
...:
...:
In [67]: np.dot(a,b)
Out[67]:
array([[ 5, 8, 11, 14],
[ 8, 13, 18, 23],
[14, 23, 32, 41]])
In [68]: a@b
Out[68]:
array([[ 5, 8, 11, 14],
[ 8, 13, 18, 23],
[14, 23, 32, 41]])
In [69]: np.einsum('ij,jk',a,b)
Out[69]:
array([[ 5, 8, 11, 14],
[ 8, 13, 18, 23],
[14, 23, 32, 41]])
Broadcasted multiply and sum:
In [71]: (a[:,:,None]*b[None,:,:]).sum(axis=1)
Out[71]:
array([[ 5, 8, 11, 14],
[ 8, 13, 18, 23],
[14, 23, 32, 41]])
In [72]: (a[:,:,None]*b[None,:,:]).shape
Out[72]: (3, 2, 4)
answered Nov 12 at 3:38
hpaulj
109k674140
add a comment |
up vote
1
down vote
accepted
In [66]: a=np.array([[1, 2],
...: [2, 3],
...: [4, 5]])
...:
...: b=np.array([[1, 2, 3, 4],
...: [2, 3, 4, 5]])
...:
...:
In [67]: np.dot(a,b)
Out[67]:
array([[ 5, 8, 11, 14],
[ 8, 13, 18, 23],
[14, 23, 32, 41]])
In [68]: a@b
Out[68]:
array([[ 5, 8, 11, 14],
[ 8, 13, 18, 23],
[14, 23, 32, 41]])
In [69]: np.einsum('ij,jk',a,b)
Out[69]:
array([[ 5, 8, 11, 14],
[ 8, 13, 18, 23],
[14, 23, 32, 41]])
Broadcasted multiply and sum:
In [71]: (a[:,:,None]*b[None,:,:]).sum(axis=1)
Out[71]:
array([[ 5, 8, 11, 14],
[ 8, 13, 18, 23],
[14, 23, 32, 41]])
In [72]: (a[:,:,None]*b[None,:,:]).shape
Out[72]: (3, 2, 4)
answered Nov 12 at 3:38
hpaulj
109k674140
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
In [66]: a=np.array([[1, 2],
...: [2, 3],
...: [4, 5]])
...:
...: b=np.array([[1, 2, 3, 4],
...: [2, 3, 4, 5]])
...:
...:
In [67]: np.dot(a,b)
Out[67]:
array([[ 5, 8, 11, 14],
[ 8, 13, 18, 23],
[14, 23, 32, 41]])
In [68]: a@b
Out[68]:
array([[ 5, 8, 11, 14],
[ 8, 13, 18, 23],
[14, 23, 32, 41]])
In [69]: np.einsum('ij,jk',a,b)
Out[69]:
array([[ 5, 8, 11, 14],
[ 8, 13, 18, 23],
[14, 23, 32, 41]])
Broadcasted multiply and sum:
In [71]: (a[:,:,None]*b[None,:,:]).sum(axis=1)
Out[71]:
array([[ 5, 8, 11, 14],
[ 8, 13, 18, 23],
[14, 23, 32, 41]])
In [72]: (a[:,:,None]*b[None,:,:]).shape
Out[72]: (3, 2, 4)
answered Nov 12 at 3:38
hpaulj
109k674140
In [66]: a=np.array([[1, 2],
...: [2, 3],
...: [4, 5]])
...:
...: b=np.array([[1, 2, 3, 4],
...: [2, 3, 4, 5]])
...:
...:
In [67]: np.dot(a,b)
Out[67]:
array([[ 5, 8, 11, 14],
[ 8, 13, 18, 23],
[14, 23, 32, 41]])
In [68]: a@b
Out[68]:
array([[ 5, 8, 11, 14],
[ 8, 13, 18, 23],
[14, 23, 32, 41]])
In [69]: np.einsum('ij,jk',a,b)
Out[69]:
array([[ 5, 8, 11, 14],
[ 8, 13, 18, 23],
[14, 23, 32, 41]])
Broadcasted multiply and sum:
In [71]: (a[:,:,None]*b[None,:,:]).sum(axis=1)
Out[71]:
array([[ 5, 8, 11, 14],
[ 8, 13, 18, 23],
[14, 23, 32, 41]])
In [72]: (a[:,:,None]*b[None,:,:]).shape
Out[72]: (3, 2, 4)
answered Nov 12 at 3:38
hpaulj
109k674140
answered Nov 12 at 3:38
hpaulj
109k674140
answered Nov 12 at 3:38
hpaulj
109k674140
answered Nov 12 at 3:38
hpaulj
109k674140
109k674140
add a comment |
add a comment |
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You can use the
@operator:a@b.– Willem Van Onsem
Nov 11 at 19:51
Furthermore
np.multiplydoes not produces a matrix multiplication.– Willem Van Onsem
Nov 11 at 19:52
@Willem Van Onsem I was asking for a mathematical equivalent, not another operator or function
– ѕняєє ѕιиgнι
Nov 11 at 19:57
It's easy with an added bit of broadcasting. Think about the dimensions you need to multiply and sum
– hpaulj
Nov 11 at 19:57
@hpaulj Could you please be a bit more specific and post your answer?
– ѕняєє ѕιиgнι
Nov 11 at 19:58