Two dimensional function not returning array of values?









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I'm trying to plot a 2-dimensional function (specifically, a 2-d Laplace solution). I defined my function and it returns the right value when I put in specific numbers, but when I try running through an array of values (x,y below), it still returns only one number. I tried with a random function of x and y (e.g., f(x,y) = x^2 + y^2) and it gives me an array of values.



def V_func(x,y):
a = 5
b = 4
Vo = 4
n = np.arange(1,100,2)
sum_list =

for indx in range(len(n)):
sum_term = (1/n[indx])*(np.cosh(n[indx]*np.pi*x/a))/(np.cosh(n[indx]*np.pi*b/a))*np.sin(n[indx]*np.pi*y/a)
sum_list = np.append(sum_list,sum_term)

summation = np.sum(sum_list)
V = 4*Vo/np.pi * summation

return V

x = np.linspace(-4,4,50)
y = np.linspace(0,5,50)
V_func(x,y)


Out: 53.633709914177224










share|improve this question























  • sum_list starts as a list . sum_term looks like it would produce an array the same size as x and y. Then you append this to sum_list using np.append (why not sum_list.append?). So sum_list ends up a 1d array (read the np.append docs). Then you np.sum that reducing it to one number (read its docs). It isn't clear where the 2d is supposed to come from? From x, y, n or some outer product?
    – hpaulj
    Nov 9 at 5:04










  • From x and y, the summation is just a part of the function. Basically, I want to get a single number as an outcome but when I input an array, I'd like for it to return an array. For example, if the function z = x2 + y2 was given the same values for x and y as above, it returns an array.
    – user10476896
    Nov 9 at 9:28










  • does that simpler function produce a 1d or 2d array?
    – hpaulj
    Nov 9 at 15:11










  • 1d, then I used the meshgrid function to get a 2d array.
    – user10476896
    Nov 9 at 17:42














up vote
0
down vote

favorite












I'm trying to plot a 2-dimensional function (specifically, a 2-d Laplace solution). I defined my function and it returns the right value when I put in specific numbers, but when I try running through an array of values (x,y below), it still returns only one number. I tried with a random function of x and y (e.g., f(x,y) = x^2 + y^2) and it gives me an array of values.



def V_func(x,y):
a = 5
b = 4
Vo = 4
n = np.arange(1,100,2)
sum_list =

for indx in range(len(n)):
sum_term = (1/n[indx])*(np.cosh(n[indx]*np.pi*x/a))/(np.cosh(n[indx]*np.pi*b/a))*np.sin(n[indx]*np.pi*y/a)
sum_list = np.append(sum_list,sum_term)

summation = np.sum(sum_list)
V = 4*Vo/np.pi * summation

return V

x = np.linspace(-4,4,50)
y = np.linspace(0,5,50)
V_func(x,y)


Out: 53.633709914177224










share|improve this question























  • sum_list starts as a list . sum_term looks like it would produce an array the same size as x and y. Then you append this to sum_list using np.append (why not sum_list.append?). So sum_list ends up a 1d array (read the np.append docs). Then you np.sum that reducing it to one number (read its docs). It isn't clear where the 2d is supposed to come from? From x, y, n or some outer product?
    – hpaulj
    Nov 9 at 5:04










  • From x and y, the summation is just a part of the function. Basically, I want to get a single number as an outcome but when I input an array, I'd like for it to return an array. For example, if the function z = x2 + y2 was given the same values for x and y as above, it returns an array.
    – user10476896
    Nov 9 at 9:28










  • does that simpler function produce a 1d or 2d array?
    – hpaulj
    Nov 9 at 15:11










  • 1d, then I used the meshgrid function to get a 2d array.
    – user10476896
    Nov 9 at 17:42












up vote
0
down vote

favorite









up vote
0
down vote

favorite











I'm trying to plot a 2-dimensional function (specifically, a 2-d Laplace solution). I defined my function and it returns the right value when I put in specific numbers, but when I try running through an array of values (x,y below), it still returns only one number. I tried with a random function of x and y (e.g., f(x,y) = x^2 + y^2) and it gives me an array of values.



def V_func(x,y):
a = 5
b = 4
Vo = 4
n = np.arange(1,100,2)
sum_list =

for indx in range(len(n)):
sum_term = (1/n[indx])*(np.cosh(n[indx]*np.pi*x/a))/(np.cosh(n[indx]*np.pi*b/a))*np.sin(n[indx]*np.pi*y/a)
sum_list = np.append(sum_list,sum_term)

summation = np.sum(sum_list)
V = 4*Vo/np.pi * summation

return V

x = np.linspace(-4,4,50)
y = np.linspace(0,5,50)
V_func(x,y)


Out: 53.633709914177224










share|improve this question















I'm trying to plot a 2-dimensional function (specifically, a 2-d Laplace solution). I defined my function and it returns the right value when I put in specific numbers, but when I try running through an array of values (x,y below), it still returns only one number. I tried with a random function of x and y (e.g., f(x,y) = x^2 + y^2) and it gives me an array of values.



def V_func(x,y):
a = 5
b = 4
Vo = 4
n = np.arange(1,100,2)
sum_list =

for indx in range(len(n)):
sum_term = (1/n[indx])*(np.cosh(n[indx]*np.pi*x/a))/(np.cosh(n[indx]*np.pi*b/a))*np.sin(n[indx]*np.pi*y/a)
sum_list = np.append(sum_list,sum_term)

summation = np.sum(sum_list)
V = 4*Vo/np.pi * summation

return V

x = np.linspace(-4,4,50)
y = np.linspace(0,5,50)
V_func(x,y)


Out: 53.633709914177224







python arrays numpy






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 9 at 2:45









mjhm

13.6k93555




13.6k93555










asked Nov 9 at 2:19









user10476896

11




11











  • sum_list starts as a list . sum_term looks like it would produce an array the same size as x and y. Then you append this to sum_list using np.append (why not sum_list.append?). So sum_list ends up a 1d array (read the np.append docs). Then you np.sum that reducing it to one number (read its docs). It isn't clear where the 2d is supposed to come from? From x, y, n or some outer product?
    – hpaulj
    Nov 9 at 5:04










  • From x and y, the summation is just a part of the function. Basically, I want to get a single number as an outcome but when I input an array, I'd like for it to return an array. For example, if the function z = x2 + y2 was given the same values for x and y as above, it returns an array.
    – user10476896
    Nov 9 at 9:28










  • does that simpler function produce a 1d or 2d array?
    – hpaulj
    Nov 9 at 15:11










  • 1d, then I used the meshgrid function to get a 2d array.
    – user10476896
    Nov 9 at 17:42
















  • sum_list starts as a list . sum_term looks like it would produce an array the same size as x and y. Then you append this to sum_list using np.append (why not sum_list.append?). So sum_list ends up a 1d array (read the np.append docs). Then you np.sum that reducing it to one number (read its docs). It isn't clear where the 2d is supposed to come from? From x, y, n or some outer product?
    – hpaulj
    Nov 9 at 5:04










  • From x and y, the summation is just a part of the function. Basically, I want to get a single number as an outcome but when I input an array, I'd like for it to return an array. For example, if the function z = x2 + y2 was given the same values for x and y as above, it returns an array.
    – user10476896
    Nov 9 at 9:28










  • does that simpler function produce a 1d or 2d array?
    – hpaulj
    Nov 9 at 15:11










  • 1d, then I used the meshgrid function to get a 2d array.
    – user10476896
    Nov 9 at 17:42















sum_list starts as a list . sum_term looks like it would produce an array the same size as x and y. Then you append this to sum_list using np.append (why not sum_list.append?). So sum_list ends up a 1d array (read the np.append docs). Then you np.sum that reducing it to one number (read its docs). It isn't clear where the 2d is supposed to come from? From x, y, n or some outer product?
– hpaulj
Nov 9 at 5:04




sum_list starts as a list . sum_term looks like it would produce an array the same size as x and y. Then you append this to sum_list using np.append (why not sum_list.append?). So sum_list ends up a 1d array (read the np.append docs). Then you np.sum that reducing it to one number (read its docs). It isn't clear where the 2d is supposed to come from? From x, y, n or some outer product?
– hpaulj
Nov 9 at 5:04












From x and y, the summation is just a part of the function. Basically, I want to get a single number as an outcome but when I input an array, I'd like for it to return an array. For example, if the function z = x2 + y2 was given the same values for x and y as above, it returns an array.
– user10476896
Nov 9 at 9:28




From x and y, the summation is just a part of the function. Basically, I want to get a single number as an outcome but when I input an array, I'd like for it to return an array. For example, if the function z = x2 + y2 was given the same values for x and y as above, it returns an array.
– user10476896
Nov 9 at 9:28












does that simpler function produce a 1d or 2d array?
– hpaulj
Nov 9 at 15:11




does that simpler function produce a 1d or 2d array?
– hpaulj
Nov 9 at 15:11












1d, then I used the meshgrid function to get a 2d array.
– user10476896
Nov 9 at 17:42




1d, then I used the meshgrid function to get a 2d array.
– user10476896
Nov 9 at 17:42












2 Answers
2






active

oldest

votes

















up vote
0
down vote













Try this:



def V_func(x,y):
a = 5
b = 4
Vo = 4
n = np.arange(1,100,2)
# sum_list =
sum_list = np.zeros(50)

for indx in range(len(n)):
sum_term = (1/n[indx])*(np.cosh(n[indx]*np.pi*x/a))/(np.cosh(n[indx]*np.pi*b/a))*np.sin(n[indx]*np.pi*y/a)
# sum_list = np.append(sum_list,sum_term)
sum_list += sum_term

# summation = np.sum(sum_list)
# V = 4*Vo/np.pi * summation
V = 4*Vo/np.pi * sum_list

return V





share|improve this answer






















  • When I do that, I end up with an array of 2500 instead of 50. I'm having trouble incorporating the summation of terms for n=1,3,5... in the function.
    – user10476896
    Nov 9 at 2:54










  • Ah! I see, edited. Is that better?
    – mjhm
    Nov 9 at 3:07











  • Wow, I see now! Thanks so much, I guess np.sum was messing up the output.
    – user10476896
    Nov 9 at 20:05

















up vote
0
down vote













Define a pair of arrays:



In [6]: x = np.arange(3); y = np.arange(10,13)
In [7]: x,y
Out[7]: (array([0, 1, 2]), array([10, 11, 12]))


Try a simple function of the 2



In [8]: x + y
Out[8]: array([10, 12, 14])


Since they have the same size, they can be summed (or otherwise combined) elementwise. The result has the same shape as the 2 inputs.



Now try 'broadcasting'. x[:,None] has shape (3,1)



In [9]: x[:,None] + y
Out[9]:
array([[10, 11, 12],
[11, 12, 13],
[12, 13, 14]])


The result is (3,3), the first 3 from the reshaped x, the second from y.



I can generate the pair of arrays with meshgrid:



In [10]: I,J = np.meshgrid(x,y,sparse=True, indexing='ij')
In [11]: I
Out[11]:
array([[0],
[1],
[2]])
In [12]: J
Out[12]: array([[10, 11, 12]])
In [13]: I + J
Out[13]:
array([[10, 11, 12],
[11, 12, 13],
[12, 13, 14]])


Note the added parameters in meshgrid. So that's how we go about generating 2d values from a pair of 1d arrays.



Now look at what sum does. As you use it in the function:



In [14]: np.sum(I + J)
Out[14]: 108


the result is a scalar. See the docs. If I specify an axis I get an array.



In [15]: np.sum(I + J, axis=0)
Out[15]: array([33, 36, 39])


If you gave V_func the right x and y, sum_list could be a 3d array. That axis-less sum reduces it to a scalar.



In code like this you need to keep track of array shapes. Include test prints if needed; don't just assume anything; test it. Pay attention to how dimensions grow and shrink as they pass through various operations.






share|improve this answer




















  • Thanks so much, I figured it out.
    – user10476896
    Nov 9 at 20:05










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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
0
down vote













Try this:



def V_func(x,y):
a = 5
b = 4
Vo = 4
n = np.arange(1,100,2)
# sum_list =
sum_list = np.zeros(50)

for indx in range(len(n)):
sum_term = (1/n[indx])*(np.cosh(n[indx]*np.pi*x/a))/(np.cosh(n[indx]*np.pi*b/a))*np.sin(n[indx]*np.pi*y/a)
# sum_list = np.append(sum_list,sum_term)
sum_list += sum_term

# summation = np.sum(sum_list)
# V = 4*Vo/np.pi * summation
V = 4*Vo/np.pi * sum_list

return V





share|improve this answer






















  • When I do that, I end up with an array of 2500 instead of 50. I'm having trouble incorporating the summation of terms for n=1,3,5... in the function.
    – user10476896
    Nov 9 at 2:54










  • Ah! I see, edited. Is that better?
    – mjhm
    Nov 9 at 3:07











  • Wow, I see now! Thanks so much, I guess np.sum was messing up the output.
    – user10476896
    Nov 9 at 20:05














up vote
0
down vote













Try this:



def V_func(x,y):
a = 5
b = 4
Vo = 4
n = np.arange(1,100,2)
# sum_list =
sum_list = np.zeros(50)

for indx in range(len(n)):
sum_term = (1/n[indx])*(np.cosh(n[indx]*np.pi*x/a))/(np.cosh(n[indx]*np.pi*b/a))*np.sin(n[indx]*np.pi*y/a)
# sum_list = np.append(sum_list,sum_term)
sum_list += sum_term

# summation = np.sum(sum_list)
# V = 4*Vo/np.pi * summation
V = 4*Vo/np.pi * sum_list

return V





share|improve this answer






















  • When I do that, I end up with an array of 2500 instead of 50. I'm having trouble incorporating the summation of terms for n=1,3,5... in the function.
    – user10476896
    Nov 9 at 2:54










  • Ah! I see, edited. Is that better?
    – mjhm
    Nov 9 at 3:07











  • Wow, I see now! Thanks so much, I guess np.sum was messing up the output.
    – user10476896
    Nov 9 at 20:05












up vote
0
down vote










up vote
0
down vote









Try this:



def V_func(x,y):
a = 5
b = 4
Vo = 4
n = np.arange(1,100,2)
# sum_list =
sum_list = np.zeros(50)

for indx in range(len(n)):
sum_term = (1/n[indx])*(np.cosh(n[indx]*np.pi*x/a))/(np.cosh(n[indx]*np.pi*b/a))*np.sin(n[indx]*np.pi*y/a)
# sum_list = np.append(sum_list,sum_term)
sum_list += sum_term

# summation = np.sum(sum_list)
# V = 4*Vo/np.pi * summation
V = 4*Vo/np.pi * sum_list

return V





share|improve this answer














Try this:



def V_func(x,y):
a = 5
b = 4
Vo = 4
n = np.arange(1,100,2)
# sum_list =
sum_list = np.zeros(50)

for indx in range(len(n)):
sum_term = (1/n[indx])*(np.cosh(n[indx]*np.pi*x/a))/(np.cosh(n[indx]*np.pi*b/a))*np.sin(n[indx]*np.pi*y/a)
# sum_list = np.append(sum_list,sum_term)
sum_list += sum_term

# summation = np.sum(sum_list)
# V = 4*Vo/np.pi * summation
V = 4*Vo/np.pi * sum_list

return V






share|improve this answer














share|improve this answer



share|improve this answer








edited Nov 9 at 3:06

























answered Nov 9 at 2:43









mjhm

13.6k93555




13.6k93555











  • When I do that, I end up with an array of 2500 instead of 50. I'm having trouble incorporating the summation of terms for n=1,3,5... in the function.
    – user10476896
    Nov 9 at 2:54










  • Ah! I see, edited. Is that better?
    – mjhm
    Nov 9 at 3:07











  • Wow, I see now! Thanks so much, I guess np.sum was messing up the output.
    – user10476896
    Nov 9 at 20:05
















  • When I do that, I end up with an array of 2500 instead of 50. I'm having trouble incorporating the summation of terms for n=1,3,5... in the function.
    – user10476896
    Nov 9 at 2:54










  • Ah! I see, edited. Is that better?
    – mjhm
    Nov 9 at 3:07











  • Wow, I see now! Thanks so much, I guess np.sum was messing up the output.
    – user10476896
    Nov 9 at 20:05















When I do that, I end up with an array of 2500 instead of 50. I'm having trouble incorporating the summation of terms for n=1,3,5... in the function.
– user10476896
Nov 9 at 2:54




When I do that, I end up with an array of 2500 instead of 50. I'm having trouble incorporating the summation of terms for n=1,3,5... in the function.
– user10476896
Nov 9 at 2:54












Ah! I see, edited. Is that better?
– mjhm
Nov 9 at 3:07





Ah! I see, edited. Is that better?
– mjhm
Nov 9 at 3:07













Wow, I see now! Thanks so much, I guess np.sum was messing up the output.
– user10476896
Nov 9 at 20:05




Wow, I see now! Thanks so much, I guess np.sum was messing up the output.
– user10476896
Nov 9 at 20:05












up vote
0
down vote













Define a pair of arrays:



In [6]: x = np.arange(3); y = np.arange(10,13)
In [7]: x,y
Out[7]: (array([0, 1, 2]), array([10, 11, 12]))


Try a simple function of the 2



In [8]: x + y
Out[8]: array([10, 12, 14])


Since they have the same size, they can be summed (or otherwise combined) elementwise. The result has the same shape as the 2 inputs.



Now try 'broadcasting'. x[:,None] has shape (3,1)



In [9]: x[:,None] + y
Out[9]:
array([[10, 11, 12],
[11, 12, 13],
[12, 13, 14]])


The result is (3,3), the first 3 from the reshaped x, the second from y.



I can generate the pair of arrays with meshgrid:



In [10]: I,J = np.meshgrid(x,y,sparse=True, indexing='ij')
In [11]: I
Out[11]:
array([[0],
[1],
[2]])
In [12]: J
Out[12]: array([[10, 11, 12]])
In [13]: I + J
Out[13]:
array([[10, 11, 12],
[11, 12, 13],
[12, 13, 14]])


Note the added parameters in meshgrid. So that's how we go about generating 2d values from a pair of 1d arrays.



Now look at what sum does. As you use it in the function:



In [14]: np.sum(I + J)
Out[14]: 108


the result is a scalar. See the docs. If I specify an axis I get an array.



In [15]: np.sum(I + J, axis=0)
Out[15]: array([33, 36, 39])


If you gave V_func the right x and y, sum_list could be a 3d array. That axis-less sum reduces it to a scalar.



In code like this you need to keep track of array shapes. Include test prints if needed; don't just assume anything; test it. Pay attention to how dimensions grow and shrink as they pass through various operations.






share|improve this answer




















  • Thanks so much, I figured it out.
    – user10476896
    Nov 9 at 20:05














up vote
0
down vote













Define a pair of arrays:



In [6]: x = np.arange(3); y = np.arange(10,13)
In [7]: x,y
Out[7]: (array([0, 1, 2]), array([10, 11, 12]))


Try a simple function of the 2



In [8]: x + y
Out[8]: array([10, 12, 14])


Since they have the same size, they can be summed (or otherwise combined) elementwise. The result has the same shape as the 2 inputs.



Now try 'broadcasting'. x[:,None] has shape (3,1)



In [9]: x[:,None] + y
Out[9]:
array([[10, 11, 12],
[11, 12, 13],
[12, 13, 14]])


The result is (3,3), the first 3 from the reshaped x, the second from y.



I can generate the pair of arrays with meshgrid:



In [10]: I,J = np.meshgrid(x,y,sparse=True, indexing='ij')
In [11]: I
Out[11]:
array([[0],
[1],
[2]])
In [12]: J
Out[12]: array([[10, 11, 12]])
In [13]: I + J
Out[13]:
array([[10, 11, 12],
[11, 12, 13],
[12, 13, 14]])


Note the added parameters in meshgrid. So that's how we go about generating 2d values from a pair of 1d arrays.



Now look at what sum does. As you use it in the function:



In [14]: np.sum(I + J)
Out[14]: 108


the result is a scalar. See the docs. If I specify an axis I get an array.



In [15]: np.sum(I + J, axis=0)
Out[15]: array([33, 36, 39])


If you gave V_func the right x and y, sum_list could be a 3d array. That axis-less sum reduces it to a scalar.



In code like this you need to keep track of array shapes. Include test prints if needed; don't just assume anything; test it. Pay attention to how dimensions grow and shrink as they pass through various operations.






share|improve this answer




















  • Thanks so much, I figured it out.
    – user10476896
    Nov 9 at 20:05












up vote
0
down vote










up vote
0
down vote









Define a pair of arrays:



In [6]: x = np.arange(3); y = np.arange(10,13)
In [7]: x,y
Out[7]: (array([0, 1, 2]), array([10, 11, 12]))


Try a simple function of the 2



In [8]: x + y
Out[8]: array([10, 12, 14])


Since they have the same size, they can be summed (or otherwise combined) elementwise. The result has the same shape as the 2 inputs.



Now try 'broadcasting'. x[:,None] has shape (3,1)



In [9]: x[:,None] + y
Out[9]:
array([[10, 11, 12],
[11, 12, 13],
[12, 13, 14]])


The result is (3,3), the first 3 from the reshaped x, the second from y.



I can generate the pair of arrays with meshgrid:



In [10]: I,J = np.meshgrid(x,y,sparse=True, indexing='ij')
In [11]: I
Out[11]:
array([[0],
[1],
[2]])
In [12]: J
Out[12]: array([[10, 11, 12]])
In [13]: I + J
Out[13]:
array([[10, 11, 12],
[11, 12, 13],
[12, 13, 14]])


Note the added parameters in meshgrid. So that's how we go about generating 2d values from a pair of 1d arrays.



Now look at what sum does. As you use it in the function:



In [14]: np.sum(I + J)
Out[14]: 108


the result is a scalar. See the docs. If I specify an axis I get an array.



In [15]: np.sum(I + J, axis=0)
Out[15]: array([33, 36, 39])


If you gave V_func the right x and y, sum_list could be a 3d array. That axis-less sum reduces it to a scalar.



In code like this you need to keep track of array shapes. Include test prints if needed; don't just assume anything; test it. Pay attention to how dimensions grow and shrink as they pass through various operations.






share|improve this answer












Define a pair of arrays:



In [6]: x = np.arange(3); y = np.arange(10,13)
In [7]: x,y
Out[7]: (array([0, 1, 2]), array([10, 11, 12]))


Try a simple function of the 2



In [8]: x + y
Out[8]: array([10, 12, 14])


Since they have the same size, they can be summed (or otherwise combined) elementwise. The result has the same shape as the 2 inputs.



Now try 'broadcasting'. x[:,None] has shape (3,1)



In [9]: x[:,None] + y
Out[9]:
array([[10, 11, 12],
[11, 12, 13],
[12, 13, 14]])


The result is (3,3), the first 3 from the reshaped x, the second from y.



I can generate the pair of arrays with meshgrid:



In [10]: I,J = np.meshgrid(x,y,sparse=True, indexing='ij')
In [11]: I
Out[11]:
array([[0],
[1],
[2]])
In [12]: J
Out[12]: array([[10, 11, 12]])
In [13]: I + J
Out[13]:
array([[10, 11, 12],
[11, 12, 13],
[12, 13, 14]])


Note the added parameters in meshgrid. So that's how we go about generating 2d values from a pair of 1d arrays.



Now look at what sum does. As you use it in the function:



In [14]: np.sum(I + J)
Out[14]: 108


the result is a scalar. See the docs. If I specify an axis I get an array.



In [15]: np.sum(I + J, axis=0)
Out[15]: array([33, 36, 39])


If you gave V_func the right x and y, sum_list could be a 3d array. That axis-less sum reduces it to a scalar.



In code like this you need to keep track of array shapes. Include test prints if needed; don't just assume anything; test it. Pay attention to how dimensions grow and shrink as they pass through various operations.







share|improve this answer












share|improve this answer



share|improve this answer










answered Nov 9 at 18:00









hpaulj

107k673137




107k673137











  • Thanks so much, I figured it out.
    – user10476896
    Nov 9 at 20:05
















  • Thanks so much, I figured it out.
    – user10476896
    Nov 9 at 20:05















Thanks so much, I figured it out.
– user10476896
Nov 9 at 20:05




Thanks so much, I figured it out.
– user10476896
Nov 9 at 20:05

















 

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