Multivariate interpolation?
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I have this kind of data. Speed in x axis and power in y axis. This gives one plot. But, there are a number of let's say C values that give other plots also on the speed vs power diagram.
The data is:
C = 12
speed:[127.1, 132.3, 154.3, 171.1, 190.7, 195.3]
power:[2800, 3400.23, 5000.1, 6880.7, 9711.1, 10011.2 ]
C = 14
speed:[113.1, 125.3, 133.3, 155.1, 187.7, 197.3]
power:[2420, 3320, 4129.91, 6287.17, 10800.34, 13076.5 ]
Now, I want to be able to interpolate at [[12.2, 122.1], [12.4, 137.3], [12.5, 154.9], [12.6, 171.4], [12.7, 192.6], [12.8, 198.5]]
for example.
I have read this answer. I am not sure if this is the way to do it.
I tried:
data = np.array([[12, 127.1, 2800], [12, 132.3, 3400.23], [12, 154.3, 5000.1], [12, 171.1, 6880.7],
[12, 190.7, 9711.1], [12, 195.3, 10011.2],
[14, 113.1, 2420], [14, 125.3, 3320], [14, 133.3, 4129.91], [14, 155.1, 6287.17],
[14, 187.7, 10800.34], [14, 197.3, 13076.5]])
coords = np.array([[12.2, 122.1], [12.4, 137.3], [12.5, 154.9], [12.6, 171.4], [12.7, 192.6], [12.8, 198.5]])
z = ndimage.map_coordinates(data, coords.T, order=2, mode='nearest')
but, I am receiving:
array([13076.5, 13076.5, 13076.5, 13076.5, 13076.5, 13076.5])
I am not sure how to deal with this kind of interpolation.
python numpy scipy interpolation
add a comment |
up vote
1
down vote
favorite
I have this kind of data. Speed in x axis and power in y axis. This gives one plot. But, there are a number of let's say C values that give other plots also on the speed vs power diagram.
The data is:
C = 12
speed:[127.1, 132.3, 154.3, 171.1, 190.7, 195.3]
power:[2800, 3400.23, 5000.1, 6880.7, 9711.1, 10011.2 ]
C = 14
speed:[113.1, 125.3, 133.3, 155.1, 187.7, 197.3]
power:[2420, 3320, 4129.91, 6287.17, 10800.34, 13076.5 ]
Now, I want to be able to interpolate at [[12.2, 122.1], [12.4, 137.3], [12.5, 154.9], [12.6, 171.4], [12.7, 192.6], [12.8, 198.5]]
for example.
I have read this answer. I am not sure if this is the way to do it.
I tried:
data = np.array([[12, 127.1, 2800], [12, 132.3, 3400.23], [12, 154.3, 5000.1], [12, 171.1, 6880.7],
[12, 190.7, 9711.1], [12, 195.3, 10011.2],
[14, 113.1, 2420], [14, 125.3, 3320], [14, 133.3, 4129.91], [14, 155.1, 6287.17],
[14, 187.7, 10800.34], [14, 197.3, 13076.5]])
coords = np.array([[12.2, 122.1], [12.4, 137.3], [12.5, 154.9], [12.6, 171.4], [12.7, 192.6], [12.8, 198.5]])
z = ndimage.map_coordinates(data, coords.T, order=2, mode='nearest')
but, I am receiving:
array([13076.5, 13076.5, 13076.5, 13076.5, 13076.5, 13076.5])
I am not sure how to deal with this kind of interpolation.
python numpy scipy interpolation
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I have this kind of data. Speed in x axis and power in y axis. This gives one plot. But, there are a number of let's say C values that give other plots also on the speed vs power diagram.
The data is:
C = 12
speed:[127.1, 132.3, 154.3, 171.1, 190.7, 195.3]
power:[2800, 3400.23, 5000.1, 6880.7, 9711.1, 10011.2 ]
C = 14
speed:[113.1, 125.3, 133.3, 155.1, 187.7, 197.3]
power:[2420, 3320, 4129.91, 6287.17, 10800.34, 13076.5 ]
Now, I want to be able to interpolate at [[12.2, 122.1], [12.4, 137.3], [12.5, 154.9], [12.6, 171.4], [12.7, 192.6], [12.8, 198.5]]
for example.
I have read this answer. I am not sure if this is the way to do it.
I tried:
data = np.array([[12, 127.1, 2800], [12, 132.3, 3400.23], [12, 154.3, 5000.1], [12, 171.1, 6880.7],
[12, 190.7, 9711.1], [12, 195.3, 10011.2],
[14, 113.1, 2420], [14, 125.3, 3320], [14, 133.3, 4129.91], [14, 155.1, 6287.17],
[14, 187.7, 10800.34], [14, 197.3, 13076.5]])
coords = np.array([[12.2, 122.1], [12.4, 137.3], [12.5, 154.9], [12.6, 171.4], [12.7, 192.6], [12.8, 198.5]])
z = ndimage.map_coordinates(data, coords.T, order=2, mode='nearest')
but, I am receiving:
array([13076.5, 13076.5, 13076.5, 13076.5, 13076.5, 13076.5])
I am not sure how to deal with this kind of interpolation.
python numpy scipy interpolation
I have this kind of data. Speed in x axis and power in y axis. This gives one plot. But, there are a number of let's say C values that give other plots also on the speed vs power diagram.
The data is:
C = 12
speed:[127.1, 132.3, 154.3, 171.1, 190.7, 195.3]
power:[2800, 3400.23, 5000.1, 6880.7, 9711.1, 10011.2 ]
C = 14
speed:[113.1, 125.3, 133.3, 155.1, 187.7, 197.3]
power:[2420, 3320, 4129.91, 6287.17, 10800.34, 13076.5 ]
Now, I want to be able to interpolate at [[12.2, 122.1], [12.4, 137.3], [12.5, 154.9], [12.6, 171.4], [12.7, 192.6], [12.8, 198.5]]
for example.
I have read this answer. I am not sure if this is the way to do it.
I tried:
data = np.array([[12, 127.1, 2800], [12, 132.3, 3400.23], [12, 154.3, 5000.1], [12, 171.1, 6880.7],
[12, 190.7, 9711.1], [12, 195.3, 10011.2],
[14, 113.1, 2420], [14, 125.3, 3320], [14, 133.3, 4129.91], [14, 155.1, 6287.17],
[14, 187.7, 10800.34], [14, 197.3, 13076.5]])
coords = np.array([[12.2, 122.1], [12.4, 137.3], [12.5, 154.9], [12.6, 171.4], [12.7, 192.6], [12.8, 198.5]])
z = ndimage.map_coordinates(data, coords.T, order=2, mode='nearest')
but, I am receiving:
array([13076.5, 13076.5, 13076.5, 13076.5, 13076.5, 13076.5])
I am not sure how to deal with this kind of interpolation.
python numpy scipy interpolation
python numpy scipy interpolation
edited Nov 21 at 20:04
asked Nov 21 at 19:45
George
2,763749109
2,763749109
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3 Answers
3
active
oldest
votes
up vote
2
down vote
accepted
map_coordinates
assumes you have items at each integer index, kind of like you would in an image. I.e. (0, 0), (0, 1)..., (0, 100), (1, 0), (1, 1), ..., (100, 0), (100, 1), ..., (100, 100) are all coordinates which are well defined if you have a 100x100 image. This is not your case. You have data at coordinates (12, 127.1), (12, 132.3), etc.
You can use griddata
instead. Depending on how you want to interpolate, you'll get different results:
In [24]: data = np.array([[12, 127.1, 2800], [12, 132.3, 3400.23], [12, 154.3, 5000.1], [12, 171.1, 6880.7],
...: [12, 190.7, 9711.1], [12, 195.3, 10011.2],
...: [14, 113.1, 2420], [14, 125.3, 3320], [14, 133.3, 4129.91], [14, 155.1, 6287.17],
...: [14, 187.7, 10800.34], [14, 197.3, 13076.5]])
In [25]: from scipy.interpolate import griddata
In [28]: coords = np.array([[12.2, 122.1], [12.4, 137.3], [12.5, 154.9], [12.6, 171.4], [12.7, 192.6], [12.8, 198.5]])
In [29]: griddata(data[:, 0:2], data[:, -1], coords)
Out[29]:
array([ nan, 3895.22854545, 5366.64369048, 7408.68906748,
10791.779 , nan])
In [31]: griddata(data[:, 0:2], data[:, -1], coords, method='nearest')
Out[31]: array([ 3320. , 4129.91, 5000.1 , 6880.7 , 9711.1 , 13076.5 ])
In [32]: griddata(data[:, 0:2], data[:, -1], coords, method='cubic')
Out[32]:
array([ nan, 3998.75479082, 5357.54672326, 7297.94115979,
10647.04183455, nan])
method='cubic'
probably has the highest fidelity for "random" data, but only you can decide which method is appropriate for your data and what you're trying to do (default is method='linear'
, used in [29]
above).
Note that some of the answers are nan
. This is because you've given input that isn't inside of the "bounding polygon" that your points form in 2D space.
Here's a visualization to show you what I mean:
In [49]: x = plt.scatter(x=np.append(data[:, 0], [12.2, 12.8]), y=np.append(data[:, 1], [122.1, 198.5]), c=['green']*len(data[:, 0]) + ['red']*2)
In [50]: plt.show()
I didn't connect the points in green, but you can see the two points in red are outside of the polygon that would be formed if I had connected those dots. You can't interpolate outside of that range, so you get nan
. To see why, consider the 1D case. If I ask you what's the value at index 2.5 of [0,1,2,3]
, a reasonable response would be 2.5
. However, if I ask what's at the value of index 100...a priori we have no idea what's at 100, it's way too far outside the range of what you can see. So we can't really give an answer. Saying it's 100
is wrong for this functionality, since that would be extrapolation, not interpolation.
HTH.
1
Thanks!Nice solution! Thanks for the info(upv)
– George
Nov 21 at 21:23
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up vote
1
down vote
Assuming that your function is of the form power = F(C, speed), you can use scipy.interpolate.interp2d
:
import scipy.interpolate as sci
speed = [127.1, 132.3, 154.3, 171.1, 190.7, 195.3]
C = [12]*len(speed)
power = [2800, 3400.23, 5000.1, 6880.7, 9711.1, 10011.2 ]
speed += [113.1, 125.3, 133.3, 155.1, 187.7, 197.3]
C += [14]*(len(speed) - len(C))
power += [2420, 3320, 4129.91, 6287.17, 10800.34, 13076.5 ]
f = sci.interp2d(C, speed, power)
coords = np.array([[12.2, 122.1], [12.4, 137.3], [12.5, 154.9], [12.6, 171.4], [12.7, 192.6], [12.8, 198.5]])
power_interp = np.concatenate([f(*coord) for coord in coords])
with np.printoptions(precision=1, suppress=True, linewidth=9999):
print(power_interp)
This outputs:
[1632.4 2659.5 3293.4 4060.2 5074.8 4506.6]
which seems a little low. The reason for this is that interp2d
by default uses a linear spline fit, and your data is definitely nonlinear. You can get better results by directly accessing the spline fitting routines via LSQBivariateSpline
:
xknots = (min(C), max(C))
yknots = (min(speed), max(speed))
f = sci.LSQBivariateSpline(C, speed, power, tx=xknots, ty=yknots, kx=2, ky=3)
power_interp = f(*coords.T, grid=False)
with np.printoptions(precision=1, suppress=True, linewidth=9999):
print(power_interp)
This outputs:
[ 2753.2 3780.8 5464.5 7505.2 10705.9 11819.6]
which seems more reasonable.
Thanks for the solution!The other answer seemed for me a little more straightforward and gave better results based on what I am expecting.(upv)
– George
Nov 21 at 21:24
@George I added a bit in my answer about using the the underlying spline fitting functionality inscipy.interpolate
to get a better fit to your data. Given the sparsity of your data, the results you get fromgriddata
may still be more accurate. On the other hand, a spline fit will still give reasonable results for interpolated data outside of the "bounding polygon" that @MattMessersmith mentions (instead of justNaN
)
– tel
Nov 21 at 22:35
:Thanks for the info!I wanted to ask.kx=2
means degree 2 forC
? Andky=3
degree 3 forspeed
?I expect a parabolic plot, so useky=2
?Am I right?
– George
Nov 22 at 7:47
@George If you have that kind of prior knowledge about your system, by all means you should include it in your models. If you know thatpower
is quadratic (ie parabola-shaped) in terms of bothC
andspeed
, then usekx, ky = 2, 2
. Ifpower
is quadratic inspeed
but linear inC
, usekx, ky = 1, 2
instead.
– tel
Nov 22 at 8:15
:Ok, thanks for the tips!
– George
Nov 22 at 8:22
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up vote
0
down vote
Seems to me that all you have here is:
speed = F(C) and power = G(C)
So you don't need any multivariate interpolation, just interp1d
to create one function for the speed, and another for the power...
Hi.In order to compute the power, I need power, speed and C points.So, what I have is :f(C, speed)
– George
Nov 21 at 20:56
then @tel's answer
– Silmathoron
Nov 21 at 21:02
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
map_coordinates
assumes you have items at each integer index, kind of like you would in an image. I.e. (0, 0), (0, 1)..., (0, 100), (1, 0), (1, 1), ..., (100, 0), (100, 1), ..., (100, 100) are all coordinates which are well defined if you have a 100x100 image. This is not your case. You have data at coordinates (12, 127.1), (12, 132.3), etc.
You can use griddata
instead. Depending on how you want to interpolate, you'll get different results:
In [24]: data = np.array([[12, 127.1, 2800], [12, 132.3, 3400.23], [12, 154.3, 5000.1], [12, 171.1, 6880.7],
...: [12, 190.7, 9711.1], [12, 195.3, 10011.2],
...: [14, 113.1, 2420], [14, 125.3, 3320], [14, 133.3, 4129.91], [14, 155.1, 6287.17],
...: [14, 187.7, 10800.34], [14, 197.3, 13076.5]])
In [25]: from scipy.interpolate import griddata
In [28]: coords = np.array([[12.2, 122.1], [12.4, 137.3], [12.5, 154.9], [12.6, 171.4], [12.7, 192.6], [12.8, 198.5]])
In [29]: griddata(data[:, 0:2], data[:, -1], coords)
Out[29]:
array([ nan, 3895.22854545, 5366.64369048, 7408.68906748,
10791.779 , nan])
In [31]: griddata(data[:, 0:2], data[:, -1], coords, method='nearest')
Out[31]: array([ 3320. , 4129.91, 5000.1 , 6880.7 , 9711.1 , 13076.5 ])
In [32]: griddata(data[:, 0:2], data[:, -1], coords, method='cubic')
Out[32]:
array([ nan, 3998.75479082, 5357.54672326, 7297.94115979,
10647.04183455, nan])
method='cubic'
probably has the highest fidelity for "random" data, but only you can decide which method is appropriate for your data and what you're trying to do (default is method='linear'
, used in [29]
above).
Note that some of the answers are nan
. This is because you've given input that isn't inside of the "bounding polygon" that your points form in 2D space.
Here's a visualization to show you what I mean:
In [49]: x = plt.scatter(x=np.append(data[:, 0], [12.2, 12.8]), y=np.append(data[:, 1], [122.1, 198.5]), c=['green']*len(data[:, 0]) + ['red']*2)
In [50]: plt.show()
I didn't connect the points in green, but you can see the two points in red are outside of the polygon that would be formed if I had connected those dots. You can't interpolate outside of that range, so you get nan
. To see why, consider the 1D case. If I ask you what's the value at index 2.5 of [0,1,2,3]
, a reasonable response would be 2.5
. However, if I ask what's at the value of index 100...a priori we have no idea what's at 100, it's way too far outside the range of what you can see. So we can't really give an answer. Saying it's 100
is wrong for this functionality, since that would be extrapolation, not interpolation.
HTH.
1
Thanks!Nice solution! Thanks for the info(upv)
– George
Nov 21 at 21:23
add a comment |
up vote
2
down vote
accepted
map_coordinates
assumes you have items at each integer index, kind of like you would in an image. I.e. (0, 0), (0, 1)..., (0, 100), (1, 0), (1, 1), ..., (100, 0), (100, 1), ..., (100, 100) are all coordinates which are well defined if you have a 100x100 image. This is not your case. You have data at coordinates (12, 127.1), (12, 132.3), etc.
You can use griddata
instead. Depending on how you want to interpolate, you'll get different results:
In [24]: data = np.array([[12, 127.1, 2800], [12, 132.3, 3400.23], [12, 154.3, 5000.1], [12, 171.1, 6880.7],
...: [12, 190.7, 9711.1], [12, 195.3, 10011.2],
...: [14, 113.1, 2420], [14, 125.3, 3320], [14, 133.3, 4129.91], [14, 155.1, 6287.17],
...: [14, 187.7, 10800.34], [14, 197.3, 13076.5]])
In [25]: from scipy.interpolate import griddata
In [28]: coords = np.array([[12.2, 122.1], [12.4, 137.3], [12.5, 154.9], [12.6, 171.4], [12.7, 192.6], [12.8, 198.5]])
In [29]: griddata(data[:, 0:2], data[:, -1], coords)
Out[29]:
array([ nan, 3895.22854545, 5366.64369048, 7408.68906748,
10791.779 , nan])
In [31]: griddata(data[:, 0:2], data[:, -1], coords, method='nearest')
Out[31]: array([ 3320. , 4129.91, 5000.1 , 6880.7 , 9711.1 , 13076.5 ])
In [32]: griddata(data[:, 0:2], data[:, -1], coords, method='cubic')
Out[32]:
array([ nan, 3998.75479082, 5357.54672326, 7297.94115979,
10647.04183455, nan])
method='cubic'
probably has the highest fidelity for "random" data, but only you can decide which method is appropriate for your data and what you're trying to do (default is method='linear'
, used in [29]
above).
Note that some of the answers are nan
. This is because you've given input that isn't inside of the "bounding polygon" that your points form in 2D space.
Here's a visualization to show you what I mean:
In [49]: x = plt.scatter(x=np.append(data[:, 0], [12.2, 12.8]), y=np.append(data[:, 1], [122.1, 198.5]), c=['green']*len(data[:, 0]) + ['red']*2)
In [50]: plt.show()
I didn't connect the points in green, but you can see the two points in red are outside of the polygon that would be formed if I had connected those dots. You can't interpolate outside of that range, so you get nan
. To see why, consider the 1D case. If I ask you what's the value at index 2.5 of [0,1,2,3]
, a reasonable response would be 2.5
. However, if I ask what's at the value of index 100...a priori we have no idea what's at 100, it's way too far outside the range of what you can see. So we can't really give an answer. Saying it's 100
is wrong for this functionality, since that would be extrapolation, not interpolation.
HTH.
1
Thanks!Nice solution! Thanks for the info(upv)
– George
Nov 21 at 21:23
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
map_coordinates
assumes you have items at each integer index, kind of like you would in an image. I.e. (0, 0), (0, 1)..., (0, 100), (1, 0), (1, 1), ..., (100, 0), (100, 1), ..., (100, 100) are all coordinates which are well defined if you have a 100x100 image. This is not your case. You have data at coordinates (12, 127.1), (12, 132.3), etc.
You can use griddata
instead. Depending on how you want to interpolate, you'll get different results:
In [24]: data = np.array([[12, 127.1, 2800], [12, 132.3, 3400.23], [12, 154.3, 5000.1], [12, 171.1, 6880.7],
...: [12, 190.7, 9711.1], [12, 195.3, 10011.2],
...: [14, 113.1, 2420], [14, 125.3, 3320], [14, 133.3, 4129.91], [14, 155.1, 6287.17],
...: [14, 187.7, 10800.34], [14, 197.3, 13076.5]])
In [25]: from scipy.interpolate import griddata
In [28]: coords = np.array([[12.2, 122.1], [12.4, 137.3], [12.5, 154.9], [12.6, 171.4], [12.7, 192.6], [12.8, 198.5]])
In [29]: griddata(data[:, 0:2], data[:, -1], coords)
Out[29]:
array([ nan, 3895.22854545, 5366.64369048, 7408.68906748,
10791.779 , nan])
In [31]: griddata(data[:, 0:2], data[:, -1], coords, method='nearest')
Out[31]: array([ 3320. , 4129.91, 5000.1 , 6880.7 , 9711.1 , 13076.5 ])
In [32]: griddata(data[:, 0:2], data[:, -1], coords, method='cubic')
Out[32]:
array([ nan, 3998.75479082, 5357.54672326, 7297.94115979,
10647.04183455, nan])
method='cubic'
probably has the highest fidelity for "random" data, but only you can decide which method is appropriate for your data and what you're trying to do (default is method='linear'
, used in [29]
above).
Note that some of the answers are nan
. This is because you've given input that isn't inside of the "bounding polygon" that your points form in 2D space.
Here's a visualization to show you what I mean:
In [49]: x = plt.scatter(x=np.append(data[:, 0], [12.2, 12.8]), y=np.append(data[:, 1], [122.1, 198.5]), c=['green']*len(data[:, 0]) + ['red']*2)
In [50]: plt.show()
I didn't connect the points in green, but you can see the two points in red are outside of the polygon that would be formed if I had connected those dots. You can't interpolate outside of that range, so you get nan
. To see why, consider the 1D case. If I ask you what's the value at index 2.5 of [0,1,2,3]
, a reasonable response would be 2.5
. However, if I ask what's at the value of index 100...a priori we have no idea what's at 100, it's way too far outside the range of what you can see. So we can't really give an answer. Saying it's 100
is wrong for this functionality, since that would be extrapolation, not interpolation.
HTH.
map_coordinates
assumes you have items at each integer index, kind of like you would in an image. I.e. (0, 0), (0, 1)..., (0, 100), (1, 0), (1, 1), ..., (100, 0), (100, 1), ..., (100, 100) are all coordinates which are well defined if you have a 100x100 image. This is not your case. You have data at coordinates (12, 127.1), (12, 132.3), etc.
You can use griddata
instead. Depending on how you want to interpolate, you'll get different results:
In [24]: data = np.array([[12, 127.1, 2800], [12, 132.3, 3400.23], [12, 154.3, 5000.1], [12, 171.1, 6880.7],
...: [12, 190.7, 9711.1], [12, 195.3, 10011.2],
...: [14, 113.1, 2420], [14, 125.3, 3320], [14, 133.3, 4129.91], [14, 155.1, 6287.17],
...: [14, 187.7, 10800.34], [14, 197.3, 13076.5]])
In [25]: from scipy.interpolate import griddata
In [28]: coords = np.array([[12.2, 122.1], [12.4, 137.3], [12.5, 154.9], [12.6, 171.4], [12.7, 192.6], [12.8, 198.5]])
In [29]: griddata(data[:, 0:2], data[:, -1], coords)
Out[29]:
array([ nan, 3895.22854545, 5366.64369048, 7408.68906748,
10791.779 , nan])
In [31]: griddata(data[:, 0:2], data[:, -1], coords, method='nearest')
Out[31]: array([ 3320. , 4129.91, 5000.1 , 6880.7 , 9711.1 , 13076.5 ])
In [32]: griddata(data[:, 0:2], data[:, -1], coords, method='cubic')
Out[32]:
array([ nan, 3998.75479082, 5357.54672326, 7297.94115979,
10647.04183455, nan])
method='cubic'
probably has the highest fidelity for "random" data, but only you can decide which method is appropriate for your data and what you're trying to do (default is method='linear'
, used in [29]
above).
Note that some of the answers are nan
. This is because you've given input that isn't inside of the "bounding polygon" that your points form in 2D space.
Here's a visualization to show you what I mean:
In [49]: x = plt.scatter(x=np.append(data[:, 0], [12.2, 12.8]), y=np.append(data[:, 1], [122.1, 198.5]), c=['green']*len(data[:, 0]) + ['red']*2)
In [50]: plt.show()
I didn't connect the points in green, but you can see the two points in red are outside of the polygon that would be formed if I had connected those dots. You can't interpolate outside of that range, so you get nan
. To see why, consider the 1D case. If I ask you what's the value at index 2.5 of [0,1,2,3]
, a reasonable response would be 2.5
. However, if I ask what's at the value of index 100...a priori we have no idea what's at 100, it's way too far outside the range of what you can see. So we can't really give an answer. Saying it's 100
is wrong for this functionality, since that would be extrapolation, not interpolation.
HTH.
answered Nov 21 at 21:05
Matt Messersmith
5,81021729
5,81021729
1
Thanks!Nice solution! Thanks for the info(upv)
– George
Nov 21 at 21:23
add a comment |
1
Thanks!Nice solution! Thanks for the info(upv)
– George
Nov 21 at 21:23
1
1
Thanks!Nice solution! Thanks for the info(upv)
– George
Nov 21 at 21:23
Thanks!Nice solution! Thanks for the info(upv)
– George
Nov 21 at 21:23
add a comment |
up vote
1
down vote
Assuming that your function is of the form power = F(C, speed), you can use scipy.interpolate.interp2d
:
import scipy.interpolate as sci
speed = [127.1, 132.3, 154.3, 171.1, 190.7, 195.3]
C = [12]*len(speed)
power = [2800, 3400.23, 5000.1, 6880.7, 9711.1, 10011.2 ]
speed += [113.1, 125.3, 133.3, 155.1, 187.7, 197.3]
C += [14]*(len(speed) - len(C))
power += [2420, 3320, 4129.91, 6287.17, 10800.34, 13076.5 ]
f = sci.interp2d(C, speed, power)
coords = np.array([[12.2, 122.1], [12.4, 137.3], [12.5, 154.9], [12.6, 171.4], [12.7, 192.6], [12.8, 198.5]])
power_interp = np.concatenate([f(*coord) for coord in coords])
with np.printoptions(precision=1, suppress=True, linewidth=9999):
print(power_interp)
This outputs:
[1632.4 2659.5 3293.4 4060.2 5074.8 4506.6]
which seems a little low. The reason for this is that interp2d
by default uses a linear spline fit, and your data is definitely nonlinear. You can get better results by directly accessing the spline fitting routines via LSQBivariateSpline
:
xknots = (min(C), max(C))
yknots = (min(speed), max(speed))
f = sci.LSQBivariateSpline(C, speed, power, tx=xknots, ty=yknots, kx=2, ky=3)
power_interp = f(*coords.T, grid=False)
with np.printoptions(precision=1, suppress=True, linewidth=9999):
print(power_interp)
This outputs:
[ 2753.2 3780.8 5464.5 7505.2 10705.9 11819.6]
which seems more reasonable.
Thanks for the solution!The other answer seemed for me a little more straightforward and gave better results based on what I am expecting.(upv)
– George
Nov 21 at 21:24
@George I added a bit in my answer about using the the underlying spline fitting functionality inscipy.interpolate
to get a better fit to your data. Given the sparsity of your data, the results you get fromgriddata
may still be more accurate. On the other hand, a spline fit will still give reasonable results for interpolated data outside of the "bounding polygon" that @MattMessersmith mentions (instead of justNaN
)
– tel
Nov 21 at 22:35
:Thanks for the info!I wanted to ask.kx=2
means degree 2 forC
? Andky=3
degree 3 forspeed
?I expect a parabolic plot, so useky=2
?Am I right?
– George
Nov 22 at 7:47
@George If you have that kind of prior knowledge about your system, by all means you should include it in your models. If you know thatpower
is quadratic (ie parabola-shaped) in terms of bothC
andspeed
, then usekx, ky = 2, 2
. Ifpower
is quadratic inspeed
but linear inC
, usekx, ky = 1, 2
instead.
– tel
Nov 22 at 8:15
:Ok, thanks for the tips!
– George
Nov 22 at 8:22
add a comment |
up vote
1
down vote
Assuming that your function is of the form power = F(C, speed), you can use scipy.interpolate.interp2d
:
import scipy.interpolate as sci
speed = [127.1, 132.3, 154.3, 171.1, 190.7, 195.3]
C = [12]*len(speed)
power = [2800, 3400.23, 5000.1, 6880.7, 9711.1, 10011.2 ]
speed += [113.1, 125.3, 133.3, 155.1, 187.7, 197.3]
C += [14]*(len(speed) - len(C))
power += [2420, 3320, 4129.91, 6287.17, 10800.34, 13076.5 ]
f = sci.interp2d(C, speed, power)
coords = np.array([[12.2, 122.1], [12.4, 137.3], [12.5, 154.9], [12.6, 171.4], [12.7, 192.6], [12.8, 198.5]])
power_interp = np.concatenate([f(*coord) for coord in coords])
with np.printoptions(precision=1, suppress=True, linewidth=9999):
print(power_interp)
This outputs:
[1632.4 2659.5 3293.4 4060.2 5074.8 4506.6]
which seems a little low. The reason for this is that interp2d
by default uses a linear spline fit, and your data is definitely nonlinear. You can get better results by directly accessing the spline fitting routines via LSQBivariateSpline
:
xknots = (min(C), max(C))
yknots = (min(speed), max(speed))
f = sci.LSQBivariateSpline(C, speed, power, tx=xknots, ty=yknots, kx=2, ky=3)
power_interp = f(*coords.T, grid=False)
with np.printoptions(precision=1, suppress=True, linewidth=9999):
print(power_interp)
This outputs:
[ 2753.2 3780.8 5464.5 7505.2 10705.9 11819.6]
which seems more reasonable.
Thanks for the solution!The other answer seemed for me a little more straightforward and gave better results based on what I am expecting.(upv)
– George
Nov 21 at 21:24
@George I added a bit in my answer about using the the underlying spline fitting functionality inscipy.interpolate
to get a better fit to your data. Given the sparsity of your data, the results you get fromgriddata
may still be more accurate. On the other hand, a spline fit will still give reasonable results for interpolated data outside of the "bounding polygon" that @MattMessersmith mentions (instead of justNaN
)
– tel
Nov 21 at 22:35
:Thanks for the info!I wanted to ask.kx=2
means degree 2 forC
? Andky=3
degree 3 forspeed
?I expect a parabolic plot, so useky=2
?Am I right?
– George
Nov 22 at 7:47
@George If you have that kind of prior knowledge about your system, by all means you should include it in your models. If you know thatpower
is quadratic (ie parabola-shaped) in terms of bothC
andspeed
, then usekx, ky = 2, 2
. Ifpower
is quadratic inspeed
but linear inC
, usekx, ky = 1, 2
instead.
– tel
Nov 22 at 8:15
:Ok, thanks for the tips!
– George
Nov 22 at 8:22
add a comment |
up vote
1
down vote
up vote
1
down vote
Assuming that your function is of the form power = F(C, speed), you can use scipy.interpolate.interp2d
:
import scipy.interpolate as sci
speed = [127.1, 132.3, 154.3, 171.1, 190.7, 195.3]
C = [12]*len(speed)
power = [2800, 3400.23, 5000.1, 6880.7, 9711.1, 10011.2 ]
speed += [113.1, 125.3, 133.3, 155.1, 187.7, 197.3]
C += [14]*(len(speed) - len(C))
power += [2420, 3320, 4129.91, 6287.17, 10800.34, 13076.5 ]
f = sci.interp2d(C, speed, power)
coords = np.array([[12.2, 122.1], [12.4, 137.3], [12.5, 154.9], [12.6, 171.4], [12.7, 192.6], [12.8, 198.5]])
power_interp = np.concatenate([f(*coord) for coord in coords])
with np.printoptions(precision=1, suppress=True, linewidth=9999):
print(power_interp)
This outputs:
[1632.4 2659.5 3293.4 4060.2 5074.8 4506.6]
which seems a little low. The reason for this is that interp2d
by default uses a linear spline fit, and your data is definitely nonlinear. You can get better results by directly accessing the spline fitting routines via LSQBivariateSpline
:
xknots = (min(C), max(C))
yknots = (min(speed), max(speed))
f = sci.LSQBivariateSpline(C, speed, power, tx=xknots, ty=yknots, kx=2, ky=3)
power_interp = f(*coords.T, grid=False)
with np.printoptions(precision=1, suppress=True, linewidth=9999):
print(power_interp)
This outputs:
[ 2753.2 3780.8 5464.5 7505.2 10705.9 11819.6]
which seems more reasonable.
Assuming that your function is of the form power = F(C, speed), you can use scipy.interpolate.interp2d
:
import scipy.interpolate as sci
speed = [127.1, 132.3, 154.3, 171.1, 190.7, 195.3]
C = [12]*len(speed)
power = [2800, 3400.23, 5000.1, 6880.7, 9711.1, 10011.2 ]
speed += [113.1, 125.3, 133.3, 155.1, 187.7, 197.3]
C += [14]*(len(speed) - len(C))
power += [2420, 3320, 4129.91, 6287.17, 10800.34, 13076.5 ]
f = sci.interp2d(C, speed, power)
coords = np.array([[12.2, 122.1], [12.4, 137.3], [12.5, 154.9], [12.6, 171.4], [12.7, 192.6], [12.8, 198.5]])
power_interp = np.concatenate([f(*coord) for coord in coords])
with np.printoptions(precision=1, suppress=True, linewidth=9999):
print(power_interp)
This outputs:
[1632.4 2659.5 3293.4 4060.2 5074.8 4506.6]
which seems a little low. The reason for this is that interp2d
by default uses a linear spline fit, and your data is definitely nonlinear. You can get better results by directly accessing the spline fitting routines via LSQBivariateSpline
:
xknots = (min(C), max(C))
yknots = (min(speed), max(speed))
f = sci.LSQBivariateSpline(C, speed, power, tx=xknots, ty=yknots, kx=2, ky=3)
power_interp = f(*coords.T, grid=False)
with np.printoptions(precision=1, suppress=True, linewidth=9999):
print(power_interp)
This outputs:
[ 2753.2 3780.8 5464.5 7505.2 10705.9 11819.6]
which seems more reasonable.
edited Nov 21 at 22:09
answered Nov 21 at 20:57
tel
4,18911429
4,18911429
Thanks for the solution!The other answer seemed for me a little more straightforward and gave better results based on what I am expecting.(upv)
– George
Nov 21 at 21:24
@George I added a bit in my answer about using the the underlying spline fitting functionality inscipy.interpolate
to get a better fit to your data. Given the sparsity of your data, the results you get fromgriddata
may still be more accurate. On the other hand, a spline fit will still give reasonable results for interpolated data outside of the "bounding polygon" that @MattMessersmith mentions (instead of justNaN
)
– tel
Nov 21 at 22:35
:Thanks for the info!I wanted to ask.kx=2
means degree 2 forC
? Andky=3
degree 3 forspeed
?I expect a parabolic plot, so useky=2
?Am I right?
– George
Nov 22 at 7:47
@George If you have that kind of prior knowledge about your system, by all means you should include it in your models. If you know thatpower
is quadratic (ie parabola-shaped) in terms of bothC
andspeed
, then usekx, ky = 2, 2
. Ifpower
is quadratic inspeed
but linear inC
, usekx, ky = 1, 2
instead.
– tel
Nov 22 at 8:15
:Ok, thanks for the tips!
– George
Nov 22 at 8:22
add a comment |
Thanks for the solution!The other answer seemed for me a little more straightforward and gave better results based on what I am expecting.(upv)
– George
Nov 21 at 21:24
@George I added a bit in my answer about using the the underlying spline fitting functionality inscipy.interpolate
to get a better fit to your data. Given the sparsity of your data, the results you get fromgriddata
may still be more accurate. On the other hand, a spline fit will still give reasonable results for interpolated data outside of the "bounding polygon" that @MattMessersmith mentions (instead of justNaN
)
– tel
Nov 21 at 22:35
:Thanks for the info!I wanted to ask.kx=2
means degree 2 forC
? Andky=3
degree 3 forspeed
?I expect a parabolic plot, so useky=2
?Am I right?
– George
Nov 22 at 7:47
@George If you have that kind of prior knowledge about your system, by all means you should include it in your models. If you know thatpower
is quadratic (ie parabola-shaped) in terms of bothC
andspeed
, then usekx, ky = 2, 2
. Ifpower
is quadratic inspeed
but linear inC
, usekx, ky = 1, 2
instead.
– tel
Nov 22 at 8:15
:Ok, thanks for the tips!
– George
Nov 22 at 8:22
Thanks for the solution!The other answer seemed for me a little more straightforward and gave better results based on what I am expecting.(upv)
– George
Nov 21 at 21:24
Thanks for the solution!The other answer seemed for me a little more straightforward and gave better results based on what I am expecting.(upv)
– George
Nov 21 at 21:24
@George I added a bit in my answer about using the the underlying spline fitting functionality in
scipy.interpolate
to get a better fit to your data. Given the sparsity of your data, the results you get from griddata
may still be more accurate. On the other hand, a spline fit will still give reasonable results for interpolated data outside of the "bounding polygon" that @MattMessersmith mentions (instead of just NaN
)– tel
Nov 21 at 22:35
@George I added a bit in my answer about using the the underlying spline fitting functionality in
scipy.interpolate
to get a better fit to your data. Given the sparsity of your data, the results you get from griddata
may still be more accurate. On the other hand, a spline fit will still give reasonable results for interpolated data outside of the "bounding polygon" that @MattMessersmith mentions (instead of just NaN
)– tel
Nov 21 at 22:35
:Thanks for the info!I wanted to ask.
kx=2
means degree 2 for C
? And ky=3
degree 3 for speed
?I expect a parabolic plot, so use ky=2
?Am I right?– George
Nov 22 at 7:47
:Thanks for the info!I wanted to ask.
kx=2
means degree 2 for C
? And ky=3
degree 3 for speed
?I expect a parabolic plot, so use ky=2
?Am I right?– George
Nov 22 at 7:47
@George If you have that kind of prior knowledge about your system, by all means you should include it in your models. If you know that
power
is quadratic (ie parabola-shaped) in terms of both C
and speed
, then use kx, ky = 2, 2
. If power
is quadratic in speed
but linear in C
, use kx, ky = 1, 2
instead.– tel
Nov 22 at 8:15
@George If you have that kind of prior knowledge about your system, by all means you should include it in your models. If you know that
power
is quadratic (ie parabola-shaped) in terms of both C
and speed
, then use kx, ky = 2, 2
. If power
is quadratic in speed
but linear in C
, use kx, ky = 1, 2
instead.– tel
Nov 22 at 8:15
:Ok, thanks for the tips!
– George
Nov 22 at 8:22
:Ok, thanks for the tips!
– George
Nov 22 at 8:22
add a comment |
up vote
0
down vote
Seems to me that all you have here is:
speed = F(C) and power = G(C)
So you don't need any multivariate interpolation, just interp1d
to create one function for the speed, and another for the power...
Hi.In order to compute the power, I need power, speed and C points.So, what I have is :f(C, speed)
– George
Nov 21 at 20:56
then @tel's answer
– Silmathoron
Nov 21 at 21:02
add a comment |
up vote
0
down vote
Seems to me that all you have here is:
speed = F(C) and power = G(C)
So you don't need any multivariate interpolation, just interp1d
to create one function for the speed, and another for the power...
Hi.In order to compute the power, I need power, speed and C points.So, what I have is :f(C, speed)
– George
Nov 21 at 20:56
then @tel's answer
– Silmathoron
Nov 21 at 21:02
add a comment |
up vote
0
down vote
up vote
0
down vote
Seems to me that all you have here is:
speed = F(C) and power = G(C)
So you don't need any multivariate interpolation, just interp1d
to create one function for the speed, and another for the power...
Seems to me that all you have here is:
speed = F(C) and power = G(C)
So you don't need any multivariate interpolation, just interp1d
to create one function for the speed, and another for the power...
answered Nov 21 at 20:54
Silmathoron
1,0061721
1,0061721
Hi.In order to compute the power, I need power, speed and C points.So, what I have is :f(C, speed)
– George
Nov 21 at 20:56
then @tel's answer
– Silmathoron
Nov 21 at 21:02
add a comment |
Hi.In order to compute the power, I need power, speed and C points.So, what I have is :f(C, speed)
– George
Nov 21 at 20:56
then @tel's answer
– Silmathoron
Nov 21 at 21:02
Hi.In order to compute the power, I need power, speed and C points.So, what I have is :
f(C, speed)
– George
Nov 21 at 20:56
Hi.In order to compute the power, I need power, speed and C points.So, what I have is :
f(C, speed)
– George
Nov 21 at 20:56
then @tel's answer
– Silmathoron
Nov 21 at 21:02
then @tel's answer
– Silmathoron
Nov 21 at 21:02
add a comment |
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