Matrix multiplication with multiple numpy arrays

What is the quickest way to multiply a matrix against a numpy array of vectors? I need to multiply a matrix A by every single vector in a list of 1000 vectors. Using a for loop is taking too long, so I was wondering if there's a way to multiply them all at once?

Example:

arr = [[1,1,1], [1,1,1],[1,1,1]]



A=

[2 2 2]

[2 2 2]

So I need to multiply Av for each v in arr. The result:

arr = [[6,6], [6,6], [6,6]]

Is there a faster way than:

new_arr = 

for v in arr:

    sol = np.matmul(A, v)

    new_arr.append(sol)

edited Nov 15 '18 at 23:03

Matthieu Brucher

16.8k32244

asked Nov 15 '18 at 22:28

AP730

183

1

What kind of multiplication? Elementwise across rows?

– CJR
Nov 15 '18 at 22:30

1

Please provide sample data with expected output.

– Alexander
Nov 15 '18 at 22:32

Your terminology is a little vague. What's the shape of A. Is the other thing a list or array? If array what's the dtype? What's the shape of the 'vectors'?

– hpaulj
Nov 15 '18 at 22:54

The basic rule for matmul (and dot) is last dimension of A pairs with the 2nd to the last of B (or the only one of v). Have you tried matmul(A, arr.T)?

– hpaulj
Nov 15 '18 at 23:19

If A is (2,3) and arr is (4,3), it should be clearer that you want to pair the 3's, and get a (2,4) or (4,2) result. To get that a transpose of either A` or arr is required.

– hpaulj
Nov 15 '18 at 23:38

add a comment |

Example:

arr = [[1,1,1], [1,1,1],[1,1,1]]



A=

[2 2 2]

[2 2 2]

So I need to multiply Av for each v in arr. The result:

arr = [[6,6], [6,6], [6,6]]

Is there a faster way than:

new_arr = 

for v in arr:

    sol = np.matmul(A, v)

    new_arr.append(sol)

edited Nov 15 '18 at 23:03

Matthieu Brucher

16.8k32244

asked Nov 15 '18 at 22:28

AP730

183

1

What kind of multiplication? Elementwise across rows?

– CJR
Nov 15 '18 at 22:30

1

Please provide sample data with expected output.

– Alexander
Nov 15 '18 at 22:32

Your terminology is a little vague. What's the shape of A. Is the other thing a list or array? If array what's the dtype? What's the shape of the 'vectors'?

– hpaulj
Nov 15 '18 at 22:54

The basic rule for matmul (and dot) is last dimension of A pairs with the 2nd to the last of B (or the only one of v). Have you tried matmul(A, arr.T)?

– hpaulj
Nov 15 '18 at 23:19

If A is (2,3) and arr is (4,3), it should be clearer that you want to pair the 3's, and get a (2,4) or (4,2) result. To get that a transpose of either A` or arr is required.

– hpaulj
Nov 15 '18 at 23:38

add a comment |

Example:

arr = [[1,1,1], [1,1,1],[1,1,1]]



A=

[2 2 2]

[2 2 2]

So I need to multiply Av for each v in arr. The result:

arr = [[6,6], [6,6], [6,6]]

Is there a faster way than:

new_arr = 

for v in arr:

    sol = np.matmul(A, v)

    new_arr.append(sol)

edited Nov 15 '18 at 23:03

Matthieu Brucher

16.8k32244

asked Nov 15 '18 at 22:28

AP730

183

Example:

arr = [[1,1,1], [1,1,1],[1,1,1]]



A=

[2 2 2]

[2 2 2]

So I need to multiply Av for each v in arr. The result:

arr = [[6,6], [6,6], [6,6]]

Is there a faster way than:

new_arr = 

for v in arr:

    sol = np.matmul(A, v)

    new_arr.append(sol)

python python-3.x numpy scipy

edited Nov 15 '18 at 23:03

Matthieu Brucher

16.8k32244

asked Nov 15 '18 at 22:28

AP730

183

edited Nov 15 '18 at 23:03

Matthieu Brucher

16.8k32244

asked Nov 15 '18 at 22:28

AP730

183

edited Nov 15 '18 at 23:03

Matthieu Brucher

16.8k32244

edited Nov 15 '18 at 23:03

Matthieu Brucher

16.8k32244

edited Nov 15 '18 at 23:03

Matthieu Brucher

16.8k32244

asked Nov 15 '18 at 22:28

AP730

183

asked Nov 15 '18 at 22:28

AP730

183

asked Nov 15 '18 at 22:28

AP730

183

1

What kind of multiplication? Elementwise across rows?

– CJR
Nov 15 '18 at 22:30

1

Please provide sample data with expected output.

– Alexander
Nov 15 '18 at 22:32

Your terminology is a little vague. What's the shape of A. Is the other thing a list or array? If array what's the dtype? What's the shape of the 'vectors'?

– hpaulj
Nov 15 '18 at 22:54

The basic rule for matmul (and dot) is last dimension of A pairs with the 2nd to the last of B (or the only one of v). Have you tried matmul(A, arr.T)?

– hpaulj
Nov 15 '18 at 23:19

If A is (2,3) and arr is (4,3), it should be clearer that you want to pair the 3's, and get a (2,4) or (4,2) result. To get that a transpose of either A` or arr is required.

– hpaulj
Nov 15 '18 at 23:38

add a comment |

1

What kind of multiplication? Elementwise across rows?

– CJR
Nov 15 '18 at 22:30

1

Please provide sample data with expected output.

– Alexander
Nov 15 '18 at 22:32

Your terminology is a little vague. What's the shape of A. Is the other thing a list or array? If array what's the dtype? What's the shape of the 'vectors'?

– hpaulj
Nov 15 '18 at 22:54

The basic rule for matmul (and dot) is last dimension of A pairs with the 2nd to the last of B (or the only one of v). Have you tried matmul(A, arr.T)?

– hpaulj
Nov 15 '18 at 23:19

If A is (2,3) and arr is (4,3), it should be clearer that you want to pair the 3's, and get a (2,4) or (4,2) result. To get that a transpose of either A` or arr is required.

– hpaulj
Nov 15 '18 at 23:38

What kind of multiplication? Elementwise across rows?

– CJR
Nov 15 '18 at 22:30

Please provide sample data with expected output.

– Alexander
Nov 15 '18 at 22:32

Your terminology is a little vague. What's the shape of A. Is the other thing a list or array? If array what's the dtype? What's the shape of the 'vectors'?

– hpaulj
Nov 15 '18 at 22:54

The basic rule for matmul (and dot) is last dimension of A pairs with the 2nd to the last of B (or the only one of v). Have you tried matmul(A, arr.T)?

– hpaulj
Nov 15 '18 at 23:19

If A is (2,3) and arr is (4,3)

, it should be clearer that you want to pair the 3's, and get a (2,4) or (4,2) result.  To get that a transpose of either

A` or arr is required.

– hpaulj
Nov 15 '18 at 23:38

If A is (2,3) and arr is (4,3)

, it should be clearer that you want to pair the 3's, and get a (2,4) or (4,2) result.  To get that a transpose of either

A` or arr is required.

– hpaulj
Nov 15 '18 at 23:38

add a comment |

1 Answer
1

active

oldest

votes

Seems like you want a dot product:

new_arr = np.dot(arr, A.T)

where arr and A are numpy arrays:

arr = np.array([[1,1,1], [1,1,1],[1,1,1]])

A = np.array([[2,2, 2],[2,2,2]])

Result:

array([[6, 6],

       [6, 6],

       [6, 6]])

According to your edit, the dot product you want may be:

new_arr = np.dot(A, arr).T

Both return the same, but it's not the same computation.

edited Nov 15 '18 at 23:11

answered Nov 15 '18 at 23:02

Matthieu Brucher

16.8k32244

What is A.T in this case?

– AP730
Nov 15 '18 at 23:06

The transpose of A using numpy.

– Matthieu Brucher
Nov 15 '18 at 23:07

Depending on the content of the two matrices, what you want may also be np.dot(A, arr).T. Might be this you want according to your edit.

– Matthieu Brucher
Nov 15 '18 at 23:10

Oh okay. That makes sense. Basically I have a matrix of dimension 50 x 30000. And I need to multiply this matrix by every vector in a numpy array. There are 500 vectors of dim=30000. Would this still be correct?

– AP730
Nov 15 '18 at 23:13

1

Oh shoot. Thank you so much! I completely got my two matrices backwards. It works now :)

– AP730
Nov 15 '18 at 23:29

|
show 7 more comments

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

Seems like you want a dot product:

new_arr = np.dot(arr, A.T)

where arr and A are numpy arrays:

arr = np.array([[1,1,1], [1,1,1],[1,1,1]])

A = np.array([[2,2, 2],[2,2,2]])

Result:

array([[6, 6],

       [6, 6],

       [6, 6]])

According to your edit, the dot product you want may be:

new_arr = np.dot(A, arr).T

Both return the same, but it's not the same computation.

edited Nov 15 '18 at 23:11

answered Nov 15 '18 at 23:02

Matthieu Brucher

16.8k32244

What is A.T in this case?

– AP730
Nov 15 '18 at 23:06

The transpose of A using numpy.

– Matthieu Brucher
Nov 15 '18 at 23:07

Depending on the content of the two matrices, what you want may also be np.dot(A, arr).T. Might be this you want according to your edit.

– Matthieu Brucher
Nov 15 '18 at 23:10

Oh okay. That makes sense. Basically I have a matrix of dimension 50 x 30000. And I need to multiply this matrix by every vector in a numpy array. There are 500 vectors of dim=30000. Would this still be correct?

– AP730
Nov 15 '18 at 23:13

1

Oh shoot. Thank you so much! I completely got my two matrices backwards. It works now :)

– AP730
Nov 15 '18 at 23:29

|
show 7 more comments

Seems like you want a dot product:

new_arr = np.dot(arr, A.T)

where arr and A are numpy arrays:

arr = np.array([[1,1,1], [1,1,1],[1,1,1]])

A = np.array([[2,2, 2],[2,2,2]])

Result:

array([[6, 6],

       [6, 6],

       [6, 6]])

According to your edit, the dot product you want may be:

new_arr = np.dot(A, arr).T

Both return the same, but it's not the same computation.

edited Nov 15 '18 at 23:11

answered Nov 15 '18 at 23:02

Matthieu Brucher

16.8k32244

What is A.T in this case?

– AP730
Nov 15 '18 at 23:06

The transpose of A using numpy.

– Matthieu Brucher
Nov 15 '18 at 23:07

Depending on the content of the two matrices, what you want may also be np.dot(A, arr).T. Might be this you want according to your edit.

– Matthieu Brucher
Nov 15 '18 at 23:10

Oh okay. That makes sense. Basically I have a matrix of dimension 50 x 30000. And I need to multiply this matrix by every vector in a numpy array. There are 500 vectors of dim=30000. Would this still be correct?

– AP730
Nov 15 '18 at 23:13

1

Oh shoot. Thank you so much! I completely got my two matrices backwards. It works now :)

– AP730
Nov 15 '18 at 23:29

|
show 7 more comments

Seems like you want a dot product:

new_arr = np.dot(arr, A.T)

where arr and A are numpy arrays:

arr = np.array([[1,1,1], [1,1,1],[1,1,1]])

A = np.array([[2,2, 2],[2,2,2]])

Result:

array([[6, 6],

       [6, 6],

       [6, 6]])

According to your edit, the dot product you want may be:

new_arr = np.dot(A, arr).T

Both return the same, but it's not the same computation.

edited Nov 15 '18 at 23:11

answered Nov 15 '18 at 23:02

Matthieu Brucher

16.8k32244

Seems like you want a dot product:

new_arr = np.dot(arr, A.T)

where arr and A are numpy arrays:

arr = np.array([[1,1,1], [1,1,1],[1,1,1]])

A = np.array([[2,2, 2],[2,2,2]])

Result:

array([[6, 6],

       [6, 6],

       [6, 6]])

According to your edit, the dot product you want may be:

new_arr = np.dot(A, arr).T

Both return the same, but it's not the same computation.

edited Nov 15 '18 at 23:11

answered Nov 15 '18 at 23:02

Matthieu Brucher

16.8k32244

edited Nov 15 '18 at 23:11

answered Nov 15 '18 at 23:02

Matthieu Brucher

16.8k32244

answered Nov 15 '18 at 23:02

Matthieu Brucher

16.8k32244

answered Nov 15 '18 at 23:02

Matthieu Brucher

16.8k32244

What is A.T in this case?

– AP730
Nov 15 '18 at 23:06

The transpose of A using numpy.

– Matthieu Brucher
Nov 15 '18 at 23:07

Depending on the content of the two matrices, what you want may also be np.dot(A, arr).T. Might be this you want according to your edit.

– Matthieu Brucher
Nov 15 '18 at 23:10

Oh okay. That makes sense. Basically I have a matrix of dimension 50 x 30000. And I need to multiply this matrix by every vector in a numpy array. There are 500 vectors of dim=30000. Would this still be correct?

– AP730
Nov 15 '18 at 23:13

1

Oh shoot. Thank you so much! I completely got my two matrices backwards. It works now :)

– AP730
Nov 15 '18 at 23:29

|
show 7 more comments

What is A.T in this case?

– AP730
Nov 15 '18 at 23:06

The transpose of A using numpy.

– Matthieu Brucher
Nov 15 '18 at 23:07

Depending on the content of the two matrices, what you want may also be np.dot(A, arr).T. Might be this you want according to your edit.

– Matthieu Brucher
Nov 15 '18 at 23:10

Oh okay. That makes sense. Basically I have a matrix of dimension 50 x 30000. And I need to multiply this matrix by every vector in a numpy array. There are 500 vectors of dim=30000. Would this still be correct?

– AP730
Nov 15 '18 at 23:13

1

Oh shoot. Thank you so much! I completely got my two matrices backwards. It works now :)

– AP730
Nov 15 '18 at 23:29

What is A.T in this case?

– AP730
Nov 15 '18 at 23:06

The transpose of A using numpy.

– Matthieu Brucher
Nov 15 '18 at 23:07

Depending on the content of the two matrices, what you want may also be np.dot(A, arr).T. Might be this you want according to your edit.

– Matthieu Brucher
Nov 15 '18 at 23:10

Oh okay. That makes sense. Basically I have a matrix of dimension 50 x 30000. And I need to multiply this matrix by every vector in a numpy array. There are 500 vectors of dim=30000. Would this still be correct?

– AP730
Nov 15 '18 at 23:13

Oh shoot. Thank you so much! I completely got my two matrices backwards. It works now :)

– AP730
Nov 15 '18 at 23:29

|
show 7 more comments

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