How to fit a specific exponential function with numpy
1
I'm trying to fit a series of data to a exponential equation, I've found some great answer here: How to do exponential and logarithmic curve fitting in Python? I found only polynomial fitting But it didn't contain the step forward that I need for this question.
I'm trying to fit y and x against a equation: y = -AeBx + A. The final A has proven to be a big trouble and I don't know how to transform the equation like log(y) = log(A) + Bx as if the final A was not there.
Any help is appreciated.
python numpy curve-fitting
asked Nov 13 '18 at 2:44
Rocky LiRocky Li
2,8731316
add a comment |
1
I'm trying to fit a series of data to a exponential equation, I've found some great answer here: How to do exponential and logarithmic curve fitting in Python? I found only polynomial fitting But it didn't contain the step forward that I need for this question.
I'm trying to fit y and x against a equation: y = -AeBx + A. The final A has proven to be a big trouble and I don't know how to transform the equation like log(y) = log(A) + Bx as if the final A was not there.
Any help is appreciated.
python numpy curve-fitting
asked Nov 13 '18 at 2:44
Rocky LiRocky Li
2,8731316
Are you sure you want A twice in that equation?
– Mad Physicist
Nov 13 '18 at 2:55
Also, absorb the second A into the log on the left hand side.
– Mad Physicist
Nov 13 '18 at 2:57
@MadPhysicist Its unknown, hence can’t be calculated
– Rocky Li
Nov 13 '18 at 3:03
Did you mean y=-Ae^Bx + C? Or is C == A, and that's not a typo?
– Mad Physicist
Nov 13 '18 at 3:06
1
@MadPhysicist Yes that is not a typo, the original equation is y/A = -e^(Bx) + 1
– Rocky Li
Nov 13 '18 at 3:09
add a comment |
1
1
1
1
I'm trying to fit a series of data to a exponential equation, I've found some great answer here: How to do exponential and logarithmic curve fitting in Python? I found only polynomial fitting But it didn't contain the step forward that I need for this question.
I'm trying to fit y and x against a equation: y = -AeBx + A. The final A has proven to be a big trouble and I don't know how to transform the equation like log(y) = log(A) + Bx as if the final A was not there.
Any help is appreciated.
python numpy curve-fitting
asked Nov 13 '18 at 2:44
Rocky LiRocky Li
2,8731316
I'm trying to fit a series of data to a exponential equation, I've found some great answer here: How to do exponential and logarithmic curve fitting in Python? I found only polynomial fitting But it didn't contain the step forward that I need for this question.
I'm trying to fit y and x against a equation: y = -AeBx + A. The final A has proven to be a big trouble and I don't know how to transform the equation like log(y) = log(A) + Bx as if the final A was not there.
Any help is appreciated.
python numpy curve-fitting
python numpy curve-fitting
asked Nov 13 '18 at 2:44
Rocky LiRocky Li
2,8731316
asked Nov 13 '18 at 2:44
Rocky LiRocky Li
2,8731316
edited Nov 13 '18 at 3:10
Rocky Li
asked Nov 13 '18 at 2:44
Rocky LiRocky Li
2,8731316
asked Nov 13 '18 at 2:44
Rocky LiRocky Li
2,8731316
asked Nov 13 '18 at 2:44
Rocky LiRocky Li
2,8731316
2,8731316
Are you sure you want A twice in that equation?
– Mad Physicist
Nov 13 '18 at 2:55
Also, absorb the second A into the log on the left hand side.
– Mad Physicist
Nov 13 '18 at 2:57
@MadPhysicist Its unknown, hence can’t be calculated
– Rocky Li
Nov 13 '18 at 3:03
Did you mean y=-Ae^Bx + C? Or is C == A, and that's not a typo?
– Mad Physicist
Nov 13 '18 at 3:06
1
@MadPhysicist Yes that is not a typo, the original equation is y/A = -e^(Bx) + 1
– Rocky Li
Nov 13 '18 at 3:09
add a comment |
Are you sure you want A twice in that equation?
– Mad Physicist
Nov 13 '18 at 2:55
Also, absorb the second A into the log on the left hand side.
– Mad Physicist
Nov 13 '18 at 2:57
@MadPhysicist Its unknown, hence can’t be calculated
– Rocky Li
Nov 13 '18 at 3:03
Did you mean y=-Ae^Bx + C? Or is C == A, and that's not a typo?
– Mad Physicist
Nov 13 '18 at 3:06
1
@MadPhysicist Yes that is not a typo, the original equation is y/A = -e^(Bx) + 1
– Rocky Li
Nov 13 '18 at 3:09
Are you sure you want A twice in that equation?
– Mad Physicist
Nov 13 '18 at 2:55
Are you sure you want A twice in that equation?
– Mad Physicist
Nov 13 '18 at 2:55
Also, absorb the second A into the log on the left hand side.
– Mad Physicist
Nov 13 '18 at 2:57
Also, absorb the second A into the log on the left hand side.
– Mad Physicist
Nov 13 '18 at 2:57
@MadPhysicist Its unknown, hence can’t be calculated
– Rocky Li
Nov 13 '18 at 3:03
@MadPhysicist Its unknown, hence can’t be calculated
– Rocky Li
Nov 13 '18 at 3:03
Did you mean y=-Ae^Bx + C? Or is C == A, and that's not a typo?
– Mad Physicist
Nov 13 '18 at 3:06
Did you mean y=-Ae^Bx + C? Or is C == A, and that's not a typo?
– Mad Physicist
Nov 13 '18 at 3:06
1
1
@MadPhysicist Yes that is not a typo, the original equation is y/A = -e^(Bx) + 1
– Rocky Li
Nov 13 '18 at 3:09
@MadPhysicist Yes that is not a typo, the original equation is y/A = -e^(Bx) + 1
– Rocky Li
Nov 13 '18 at 3:09
add a comment |
1 Answer
1
active
oldest
votes
3
You can always just use scipy.optimize.curve_fit as long as your equation isn't too crazy:
import matplotlib.pyplot as plt
import numpy as np
import scipy.optimize as sio
def f(x, A, B):
return -A*np.exp(B*x) + A
A = 2
B = 1
x = np.linspace(0,1)
y = f(x, A, B)
scale = (max(y) - min(y))*.10
noise = np.random.normal(size=x.size)*scale
y += noise
fit = sio.curve_fit(f, x, y)
plt.scatter(x, y)
plt.plot(x, f(x, *fit[0]))
plt.show()
This produces:
answered Nov 13 '18 at 3:12
teltel
7,30121431
Interesting and nice explanation, I'll try it out.
– Rocky Li
Nov 13 '18 at 3:14
Worked! I just needed to flip B to negative because it doesn't deal with negative numbers apparently.
– Rocky Li
Nov 13 '18 at 3:21
1
@RockyLi Great, glad to hear it. And if you're ever bored and looking for some math to do, you may be able to derive an analytical expression for the fit using the least-squares approach. There's a well-known closed form equation for the fit of the standard exponential formy = A*e**(B*x)
– tel
Nov 13 '18 at 3:33
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
3
You can always just use scipy.optimize.curve_fit as long as your equation isn't too crazy:
import matplotlib.pyplot as plt
import numpy as np
import scipy.optimize as sio
def f(x, A, B):
return -A*np.exp(B*x) + A
A = 2
B = 1
x = np.linspace(0,1)
y = f(x, A, B)
scale = (max(y) - min(y))*.10
noise = np.random.normal(size=x.size)*scale
y += noise
fit = sio.curve_fit(f, x, y)
plt.scatter(x, y)
plt.plot(x, f(x, *fit[0]))
plt.show()
This produces:
answered Nov 13 '18 at 3:12
teltel
7,30121431
Interesting and nice explanation, I'll try it out.
– Rocky Li
Nov 13 '18 at 3:14
Worked! I just needed to flip B to negative because it doesn't deal with negative numbers apparently.
– Rocky Li
Nov 13 '18 at 3:21
1
@RockyLi Great, glad to hear it. And if you're ever bored and looking for some math to do, you may be able to derive an analytical expression for the fit using the least-squares approach. There's a well-known closed form equation for the fit of the standard exponential formy = A*e**(B*x)
– tel
Nov 13 '18 at 3:33
add a comment |
3
You can always just use scipy.optimize.curve_fit as long as your equation isn't too crazy:
import matplotlib.pyplot as plt
import numpy as np
import scipy.optimize as sio
def f(x, A, B):
return -A*np.exp(B*x) + A
A = 2
B = 1
x = np.linspace(0,1)
y = f(x, A, B)
scale = (max(y) - min(y))*.10
noise = np.random.normal(size=x.size)*scale
y += noise
fit = sio.curve_fit(f, x, y)
plt.scatter(x, y)
plt.plot(x, f(x, *fit[0]))
plt.show()
This produces:
answered Nov 13 '18 at 3:12
teltel
7,30121431
Interesting and nice explanation, I'll try it out.
– Rocky Li
Nov 13 '18 at 3:14
Worked! I just needed to flip B to negative because it doesn't deal with negative numbers apparently.
– Rocky Li
Nov 13 '18 at 3:21
1
@RockyLi Great, glad to hear it. And if you're ever bored and looking for some math to do, you may be able to derive an analytical expression for the fit using the least-squares approach. There's a well-known closed form equation for the fit of the standard exponential formy = A*e**(B*x)
– tel
Nov 13 '18 at 3:33
add a comment |
3
3
3
You can always just use scipy.optimize.curve_fit as long as your equation isn't too crazy:
import matplotlib.pyplot as plt
import numpy as np
import scipy.optimize as sio
def f(x, A, B):
return -A*np.exp(B*x) + A
A = 2
B = 1
x = np.linspace(0,1)
y = f(x, A, B)
scale = (max(y) - min(y))*.10
noise = np.random.normal(size=x.size)*scale
y += noise
fit = sio.curve_fit(f, x, y)
plt.scatter(x, y)
plt.plot(x, f(x, *fit[0]))
plt.show()
This produces:
answered Nov 13 '18 at 3:12
teltel
7,30121431
You can always just use scipy.optimize.curve_fit as long as your equation isn't too crazy:
import matplotlib.pyplot as plt
import numpy as np
import scipy.optimize as sio
def f(x, A, B):
return -A*np.exp(B*x) + A
A = 2
B = 1
x = np.linspace(0,1)
y = f(x, A, B)
scale = (max(y) - min(y))*.10
noise = np.random.normal(size=x.size)*scale
y += noise
fit = sio.curve_fit(f, x, y)
plt.scatter(x, y)
plt.plot(x, f(x, *fit[0]))
plt.show()
This produces:
answered Nov 13 '18 at 3:12
teltel
7,30121431
answered Nov 13 '18 at 3:12
teltel
7,30121431
answered Nov 13 '18 at 3:12
teltel
7,30121431
answered Nov 13 '18 at 3:12
teltel
7,30121431
7,30121431
Interesting and nice explanation, I'll try it out.
– Rocky Li
Nov 13 '18 at 3:14
Worked! I just needed to flip B to negative because it doesn't deal with negative numbers apparently.
– Rocky Li
Nov 13 '18 at 3:21
1
@RockyLi Great, glad to hear it. And if you're ever bored and looking for some math to do, you may be able to derive an analytical expression for the fit using the least-squares approach. There's a well-known closed form equation for the fit of the standard exponential formy = A*e**(B*x)
– tel
Nov 13 '18 at 3:33
add a comment |
Interesting and nice explanation, I'll try it out.
– Rocky Li
Nov 13 '18 at 3:14
Worked! I just needed to flip B to negative because it doesn't deal with negative numbers apparently.
– Rocky Li
Nov 13 '18 at 3:21
1
@RockyLi Great, glad to hear it. And if you're ever bored and looking for some math to do, you may be able to derive an analytical expression for the fit using the least-squares approach. There's a well-known closed form equation for the fit of the standard exponential formy = A*e**(B*x)
– tel
Nov 13 '18 at 3:33
Interesting and nice explanation, I'll try it out.
– Rocky Li
Nov 13 '18 at 3:14
Interesting and nice explanation, I'll try it out.
– Rocky Li
Nov 13 '18 at 3:14
Worked! I just needed to flip B to negative because it doesn't deal with negative numbers apparently.
– Rocky Li
Nov 13 '18 at 3:21
Worked! I just needed to flip B to negative because it doesn't deal with negative numbers apparently.
– Rocky Li
Nov 13 '18 at 3:21
1
1
@RockyLi Great, glad to hear it. And if you're ever bored and looking for some math to do, you may be able to derive an analytical expression for the fit using the least-squares approach. There's a well-known closed form equation for the fit of the standard exponential form
y = A*e**(B*x)– tel
Nov 13 '18 at 3:33
@RockyLi Great, glad to hear it. And if you're ever bored and looking for some math to do, you may be able to derive an analytical expression for the fit using the least-squares approach. There's a well-known closed form equation for the fit of the standard exponential form
y = A*e**(B*x)– tel
Nov 13 '18 at 3:33
add a comment |
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Are you sure you want A twice in that equation?
– Mad Physicist
Nov 13 '18 at 2:55
Also, absorb the second A into the log on the left hand side.
– Mad Physicist
Nov 13 '18 at 2:57
@MadPhysicist Its unknown, hence can’t be calculated
– Rocky Li
Nov 13 '18 at 3:03
Did you mean y=-Ae^Bx + C? Or is C == A, and that's not a typo?
– Mad Physicist
Nov 13 '18 at 3:06
1
@MadPhysicist Yes that is not a typo, the original equation is y/A = -e^(Bx) + 1
– Rocky Li
Nov 13 '18 at 3:09