Numpy array function returns inconsistent shapes
I'm trying to define a simple function ddf() that outputs the Hessian matrix of a particular mathematical function, given a 2D vector, x as the input :
import numpy as np
def ddf(x):
dd11 = 2*x[1]+8
dd12 = 2*x[0]-8*x[1]-8
dd21 = 2*x[0]-8*x[1]-8
dd22 = -8*x[0]+2
return np.array([[dd11, dd12], [dd21, dd22]])
x0 = np.zeros((2,1))
G = ddf(x0)
print(G)
I expect the output to be a 2x2 square array/matrix, however it yields what appears to be a 4x1 column instead.
Stranger still, using
G.shape
yields (2L, 2L, 1L), not (2L,2L) as expected.
My objective is to obtain G in 2x2 form. Can anyone assist? Thanks
python numpy
edited Nov 24 '18 at 15:12
Deepak Saini
1,599815
asked Nov 24 '18 at 12:50
ruphzruphz
111
add a comment |
I'm trying to define a simple function ddf() that outputs the Hessian matrix of a particular mathematical function, given a 2D vector, x as the input :
import numpy as np
def ddf(x):
dd11 = 2*x[1]+8
dd12 = 2*x[0]-8*x[1]-8
dd21 = 2*x[0]-8*x[1]-8
dd22 = -8*x[0]+2
return np.array([[dd11, dd12], [dd21, dd22]])
x0 = np.zeros((2,1))
G = ddf(x0)
print(G)
I expect the output to be a 2x2 square array/matrix, however it yields what appears to be a 4x1 column instead.
Stranger still, using
G.shape
yields (2L, 2L, 1L), not (2L,2L) as expected.
My objective is to obtain G in 2x2 form. Can anyone assist? Thanks
python numpy
edited Nov 24 '18 at 15:12
Deepak Saini
1,599815
asked Nov 24 '18 at 12:50
ruphzruphz
111
add a comment |
I'm trying to define a simple function ddf() that outputs the Hessian matrix of a particular mathematical function, given a 2D vector, x as the input :
import numpy as np
def ddf(x):
dd11 = 2*x[1]+8
dd12 = 2*x[0]-8*x[1]-8
dd21 = 2*x[0]-8*x[1]-8
dd22 = -8*x[0]+2
return np.array([[dd11, dd12], [dd21, dd22]])
x0 = np.zeros((2,1))
G = ddf(x0)
print(G)
I expect the output to be a 2x2 square array/matrix, however it yields what appears to be a 4x1 column instead.
Stranger still, using
G.shape
yields (2L, 2L, 1L), not (2L,2L) as expected.
My objective is to obtain G in 2x2 form. Can anyone assist? Thanks
python numpy
edited Nov 24 '18 at 15:12
Deepak Saini
1,599815
asked Nov 24 '18 at 12:50
ruphzruphz
111
I'm trying to define a simple function ddf() that outputs the Hessian matrix of a particular mathematical function, given a 2D vector, x as the input :
import numpy as np
def ddf(x):
dd11 = 2*x[1]+8
dd12 = 2*x[0]-8*x[1]-8
dd21 = 2*x[0]-8*x[1]-8
dd22 = -8*x[0]+2
return np.array([[dd11, dd12], [dd21, dd22]])
x0 = np.zeros((2,1))
G = ddf(x0)
print(G)
I expect the output to be a 2x2 square array/matrix, however it yields what appears to be a 4x1 column instead.
Stranger still, using
G.shape
yields (2L, 2L, 1L), not (2L,2L) as expected.
My objective is to obtain G in 2x2 form. Can anyone assist? Thanks
python numpy
python numpy
edited Nov 24 '18 at 15:12
Deepak Saini
1,599815
asked Nov 24 '18 at 12:50
ruphzruphz
111
edited Nov 24 '18 at 15:12
Deepak Saini
1,599815
asked Nov 24 '18 at 12:50
ruphzruphz
111
edited Nov 24 '18 at 15:12
Deepak Saini
1,599815
edited Nov 24 '18 at 15:12
Deepak Saini
1,599815
edited Nov 24 '18 at 15:12
Deepak Saini
1,599815
1,599815
asked Nov 24 '18 at 12:50
ruphzruphz
111
asked Nov 24 '18 at 12:50
ruphzruphz
111
asked Nov 24 '18 at 12:50
ruphzruphz
111
111
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
Your input to the function ddf() is a 2x1 matrix, meaning all of x[0] and x[1] are vectors not scalers(floats or ints). So each element of your output matrix are 1-sized vectors, as all operations in numpy are applied elements wise if arrays are passed to the functions.
Couple of things, you can do :
- It seems that you expect x[0], x[1] to be scalars, so change the input to shape (2,) in
x0 = np.zeros((2,)). - Or reshape the output as
G.reshape((2,2))to remove the extra dimension.
answered Nov 24 '18 at 13:08
Deepak SainiDeepak Saini
1,599815
add a comment |
I'm very new to python, but I think that will work:
...
G = ddf(x0)
G = np.reshape(G, (2,2))
print(G)
It yields a (2,2) as you wanted.
answered Nov 24 '18 at 12:59
karsteankarstean
1114
It does indeed, thank you very much. I also found that reshaping the 2x1 vector (x) as a 1x2 vector does the job
– ruphz
Nov 24 '18 at 13:06
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Your input to the function ddf() is a 2x1 matrix, meaning all of x[0] and x[1] are vectors not scalers(floats or ints). So each element of your output matrix are 1-sized vectors, as all operations in numpy are applied elements wise if arrays are passed to the functions.
Couple of things, you can do :
- It seems that you expect x[0], x[1] to be scalars, so change the input to shape (2,) in
x0 = np.zeros((2,)). - Or reshape the output as
G.reshape((2,2))to remove the extra dimension.
answered Nov 24 '18 at 13:08
Deepak SainiDeepak Saini
1,599815
add a comment |
Your input to the function ddf() is a 2x1 matrix, meaning all of x[0] and x[1] are vectors not scalers(floats or ints). So each element of your output matrix are 1-sized vectors, as all operations in numpy are applied elements wise if arrays are passed to the functions.
Couple of things, you can do :
- It seems that you expect x[0], x[1] to be scalars, so change the input to shape (2,) in
x0 = np.zeros((2,)). - Or reshape the output as
G.reshape((2,2))to remove the extra dimension.
answered Nov 24 '18 at 13:08
Deepak SainiDeepak Saini
1,599815
add a comment |
Your input to the function ddf() is a 2x1 matrix, meaning all of x[0] and x[1] are vectors not scalers(floats or ints). So each element of your output matrix are 1-sized vectors, as all operations in numpy are applied elements wise if arrays are passed to the functions.
Couple of things, you can do :
- It seems that you expect x[0], x[1] to be scalars, so change the input to shape (2,) in
x0 = np.zeros((2,)). - Or reshape the output as
G.reshape((2,2))to remove the extra dimension.
answered Nov 24 '18 at 13:08
Deepak SainiDeepak Saini
1,599815
Your input to the function ddf() is a 2x1 matrix, meaning all of x[0] and x[1] are vectors not scalers(floats or ints). So each element of your output matrix are 1-sized vectors, as all operations in numpy are applied elements wise if arrays are passed to the functions.
Couple of things, you can do :
- It seems that you expect x[0], x[1] to be scalars, so change the input to shape (2,) in
x0 = np.zeros((2,)). - Or reshape the output as
G.reshape((2,2))to remove the extra dimension.
answered Nov 24 '18 at 13:08
Deepak SainiDeepak Saini
1,599815
answered Nov 24 '18 at 13:08
Deepak SainiDeepak Saini
1,599815
answered Nov 24 '18 at 13:08
Deepak SainiDeepak Saini
1,599815
answered Nov 24 '18 at 13:08
Deepak SainiDeepak Saini
1,599815
1,599815
add a comment |
add a comment |
I'm very new to python, but I think that will work:
...
G = ddf(x0)
G = np.reshape(G, (2,2))
print(G)
It yields a (2,2) as you wanted.
answered Nov 24 '18 at 12:59
karsteankarstean
1114
It does indeed, thank you very much. I also found that reshaping the 2x1 vector (x) as a 1x2 vector does the job
– ruphz
Nov 24 '18 at 13:06
add a comment |
I'm very new to python, but I think that will work:
...
G = ddf(x0)
G = np.reshape(G, (2,2))
print(G)
It yields a (2,2) as you wanted.
answered Nov 24 '18 at 12:59
karsteankarstean
1114
It does indeed, thank you very much. I also found that reshaping the 2x1 vector (x) as a 1x2 vector does the job
– ruphz
Nov 24 '18 at 13:06
add a comment |
I'm very new to python, but I think that will work:
...
G = ddf(x0)
G = np.reshape(G, (2,2))
print(G)
It yields a (2,2) as you wanted.
answered Nov 24 '18 at 12:59
karsteankarstean
1114
I'm very new to python, but I think that will work:
...
G = ddf(x0)
G = np.reshape(G, (2,2))
print(G)
It yields a (2,2) as you wanted.
answered Nov 24 '18 at 12:59
karsteankarstean
1114
edited Nov 24 '18 at 13:10
answered Nov 24 '18 at 12:59
karsteankarstean
1114
answered Nov 24 '18 at 12:59
karsteankarstean
1114
answered Nov 24 '18 at 12:59
karsteankarstean
1114
1114
It does indeed, thank you very much. I also found that reshaping the 2x1 vector (x) as a 1x2 vector does the job
– ruphz
Nov 24 '18 at 13:06
add a comment |
It does indeed, thank you very much. I also found that reshaping the 2x1 vector (x) as a 1x2 vector does the job
– ruphz
Nov 24 '18 at 13:06
It does indeed, thank you very much. I also found that reshaping the 2x1 vector (x) as a 1x2 vector does the job
– ruphz
Nov 24 '18 at 13:06
It does indeed, thank you very much. I also found that reshaping the 2x1 vector (x) as a 1x2 vector does the job
– ruphz
Nov 24 '18 at 13:06
add a comment |
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