Square array from linear array python
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I would like to get a square matrix B from a linear vector A such that B = A * transpose(A). A is a numpy array and np.shape(A) returns (10,). I would like B to be a (10,10) array. I tried B = np.matmut(A, A[np.newaxis]) but I get an error :
shapes (10,) and (1,10) not aligned: 10 (dim 0) != 1 (dim 0)
python arrays numpy matrix vector
edited Nov 23 '18 at 19:16
MarianD
4,46761432
asked Nov 23 '18 at 14:08
Mathieu LecoqMathieu Lecoq
83
add a comment |
I would like to get a square matrix B from a linear vector A such that B = A * transpose(A). A is a numpy array and np.shape(A) returns (10,). I would like B to be a (10,10) array. I tried B = np.matmut(A, A[np.newaxis]) but I get an error :
shapes (10,) and (1,10) not aligned: 10 (dim 0) != 1 (dim 0)
python arrays numpy matrix vector
edited Nov 23 '18 at 19:16
MarianD
4,46761432
asked Nov 23 '18 at 14:08
Mathieu LecoqMathieu Lecoq
83
What I want is the equivalent of "B=A*ctranspose(A)" in matlab
– Mathieu Lecoq
Nov 23 '18 at 14:12
add a comment |
I would like to get a square matrix B from a linear vector A such that B = A * transpose(A). A is a numpy array and np.shape(A) returns (10,). I would like B to be a (10,10) array. I tried B = np.matmut(A, A[np.newaxis]) but I get an error :
shapes (10,) and (1,10) not aligned: 10 (dim 0) != 1 (dim 0)
python arrays numpy matrix vector
edited Nov 23 '18 at 19:16
MarianD
4,46761432
asked Nov 23 '18 at 14:08
Mathieu LecoqMathieu Lecoq
83
I would like to get a square matrix B from a linear vector A such that B = A * transpose(A). A is a numpy array and np.shape(A) returns (10,). I would like B to be a (10,10) array. I tried B = np.matmut(A, A[np.newaxis]) but I get an error :
shapes (10,) and (1,10) not aligned: 10 (dim 0) != 1 (dim 0)
python arrays numpy matrix vector
python arrays numpy matrix vector
edited Nov 23 '18 at 19:16
MarianD
4,46761432
asked Nov 23 '18 at 14:08
Mathieu LecoqMathieu Lecoq
83
edited Nov 23 '18 at 19:16
MarianD
4,46761432
asked Nov 23 '18 at 14:08
Mathieu LecoqMathieu Lecoq
83
edited Nov 23 '18 at 19:16
MarianD
4,46761432
edited Nov 23 '18 at 19:16
MarianD
4,46761432
edited Nov 23 '18 at 19:16
MarianD
4,46761432
4,46761432
asked Nov 23 '18 at 14:08
Mathieu LecoqMathieu Lecoq
83
asked Nov 23 '18 at 14:08
Mathieu LecoqMathieu Lecoq
83
asked Nov 23 '18 at 14:08
Mathieu LecoqMathieu Lecoq
83
83
What I want is the equivalent of "B=A*ctranspose(A)" in matlab
– Mathieu Lecoq
Nov 23 '18 at 14:12
add a comment |
What I want is the equivalent of "B=A*ctranspose(A)" in matlab
– Mathieu Lecoq
Nov 23 '18 at 14:12
What I want is the equivalent of "B=A*ctranspose(A)" in matlab
– Mathieu Lecoq
Nov 23 '18 at 14:12
What I want is the equivalent of "B=A*ctranspose(A)" in matlab
– Mathieu Lecoq
Nov 23 '18 at 14:12
add a comment |
3 Answers
3
active
oldest
votes
you can do this using outer:
import numpy as np
vector = np.arange(10)
np.outer(vector, vector)
answered Nov 23 '18 at 14:20
jeremycgjeremycg
19.1k44257
add a comment |
The solution is a little ugly, but it does what you need.
import numpy as np
vector = np.array([1,2,3,4,5,6,7,8,9,10],)
matrix = np.dot(vector[:,None],vector[None,:])
print(matrix)
You can also do the following:
import numpy as np
vector = np.array([1,2,3,4,5,6,7,8,9,10],)
matrix = vector*vector[:,None]
print(matrix)
The issue comes from the fact that transposing a one dimensional array does not have the effect you might expect.
answered Nov 23 '18 at 14:19
Esteban QuirosEsteban Quiros
1015
add a comment |
Variation on outer product:
a = A.reshape(-1, 1) # make sure it's a column vector
B = a @ a.T
answered Nov 23 '18 at 15:32
deckarddeckard
31527
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
you can do this using outer:
import numpy as np
vector = np.arange(10)
np.outer(vector, vector)
answered Nov 23 '18 at 14:20
jeremycgjeremycg
19.1k44257
add a comment |
you can do this using outer:
import numpy as np
vector = np.arange(10)
np.outer(vector, vector)
answered Nov 23 '18 at 14:20
jeremycgjeremycg
19.1k44257
add a comment |
you can do this using outer:
import numpy as np
vector = np.arange(10)
np.outer(vector, vector)
answered Nov 23 '18 at 14:20
jeremycgjeremycg
19.1k44257
you can do this using outer:
import numpy as np
vector = np.arange(10)
np.outer(vector, vector)
answered Nov 23 '18 at 14:20
jeremycgjeremycg
19.1k44257
answered Nov 23 '18 at 14:20
jeremycgjeremycg
19.1k44257
answered Nov 23 '18 at 14:20
jeremycgjeremycg
19.1k44257
answered Nov 23 '18 at 14:20
jeremycgjeremycg
19.1k44257
19.1k44257
add a comment |
add a comment |
The solution is a little ugly, but it does what you need.
import numpy as np
vector = np.array([1,2,3,4,5,6,7,8,9,10],)
matrix = np.dot(vector[:,None],vector[None,:])
print(matrix)
You can also do the following:
import numpy as np
vector = np.array([1,2,3,4,5,6,7,8,9,10],)
matrix = vector*vector[:,None]
print(matrix)
The issue comes from the fact that transposing a one dimensional array does not have the effect you might expect.
answered Nov 23 '18 at 14:19
Esteban QuirosEsteban Quiros
1015
add a comment |
The solution is a little ugly, but it does what you need.
import numpy as np
vector = np.array([1,2,3,4,5,6,7,8,9,10],)
matrix = np.dot(vector[:,None],vector[None,:])
print(matrix)
You can also do the following:
import numpy as np
vector = np.array([1,2,3,4,5,6,7,8,9,10],)
matrix = vector*vector[:,None]
print(matrix)
The issue comes from the fact that transposing a one dimensional array does not have the effect you might expect.
answered Nov 23 '18 at 14:19
Esteban QuirosEsteban Quiros
1015
add a comment |
The solution is a little ugly, but it does what you need.
import numpy as np
vector = np.array([1,2,3,4,5,6,7,8,9,10],)
matrix = np.dot(vector[:,None],vector[None,:])
print(matrix)
You can also do the following:
import numpy as np
vector = np.array([1,2,3,4,5,6,7,8,9,10],)
matrix = vector*vector[:,None]
print(matrix)
The issue comes from the fact that transposing a one dimensional array does not have the effect you might expect.
answered Nov 23 '18 at 14:19
Esteban QuirosEsteban Quiros
1015
The solution is a little ugly, but it does what you need.
import numpy as np
vector = np.array([1,2,3,4,5,6,7,8,9,10],)
matrix = np.dot(vector[:,None],vector[None,:])
print(matrix)
You can also do the following:
import numpy as np
vector = np.array([1,2,3,4,5,6,7,8,9,10],)
matrix = vector*vector[:,None]
print(matrix)
The issue comes from the fact that transposing a one dimensional array does not have the effect you might expect.
answered Nov 23 '18 at 14:19
Esteban QuirosEsteban Quiros
1015
answered Nov 23 '18 at 14:19
Esteban QuirosEsteban Quiros
1015
answered Nov 23 '18 at 14:19
Esteban QuirosEsteban Quiros
1015
answered Nov 23 '18 at 14:19
Esteban QuirosEsteban Quiros
1015
1015
add a comment |
add a comment |
Variation on outer product:
a = A.reshape(-1, 1) # make sure it's a column vector
B = a @ a.T
answered Nov 23 '18 at 15:32
deckarddeckard
31527
add a comment |
Variation on outer product:
a = A.reshape(-1, 1) # make sure it's a column vector
B = a @ a.T
answered Nov 23 '18 at 15:32
deckarddeckard
31527
add a comment |
Variation on outer product:
a = A.reshape(-1, 1) # make sure it's a column vector
B = a @ a.T
answered Nov 23 '18 at 15:32
deckarddeckard
31527
Variation on outer product:
a = A.reshape(-1, 1) # make sure it's a column vector
B = a @ a.T
answered Nov 23 '18 at 15:32
deckarddeckard
31527
answered Nov 23 '18 at 15:32
deckarddeckard
31527
answered Nov 23 '18 at 15:32
deckarddeckard
31527
answered Nov 23 '18 at 15:32
deckarddeckard
31527
31527
add a comment |
add a comment |
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What I want is the equivalent of "B=A*ctranspose(A)" in matlab
– Mathieu Lecoq
Nov 23 '18 at 14:12