12

The array the numpy.gradient function returns depends on the number of data-points/spacing of the data-points. Is this expected behaviour? For example:

y = lambda x: x

x1 = np.arange(0,10,1)
x2 = np.arange(0,10,0.1)
x3 = np.arange(0,10,0.01)

plt.plot(x1,np.gradient(y(x1)),'r--o')
plt.plot(x2,np.gradient(y(x2)),'b--o')
plt.plot(x3,np.gradient(y(x3)),'g--o')

returns the ATTACHED plot.

Only the gradient of y(x1) returns the correct result. What is going on here? Is there a better way to compute the numerical derivative using numpy?

Cheers

jabaldonedo
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user1654183
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1 Answers1

23

In np.gradient you should tell the sample distance. To get the same results you should type:

plt.plot(x1,np.gradient(y(x1),1),'r--o')
plt.plot(x2,np.gradient(y(x2),0.1),'b--o')
plt.plot(x3,np.gradient(y(x3),0.01),'g--o')

The default sample distance is 1 and that's why it works for x1.

If the distance is not even you have to compute it manually. If you use the forward difference you can do:

d = np.diff(y(x))/np.diff(x)

If you are interested in computing central difference as np.gradient does you could do something like this:

x = np.array([1, 2, 4, 7, 11, 16], dtype=np.float)
y = lambda x: x**2

z1 = np.hstack((y(x[0]), y(x[:-1])))
z2 = np.hstack((y(x[1:]), y(x[-1])))

dx1 = np.hstack((0, np.diff(x)))
dx2 = np.hstack((np.diff(x), 0))

d = (z2-z1) / (dx2+dx1)
pabaldonedo
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    Awesome, thank you pabaldonedo. I saw that in the manual but 'sample distance' was unfamiliar terminology to me. One last question - what if the sample distance is not even? There is nothing in the manual. – user1654183 Aug 28 '13 at 23:54
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    @user1654183 I have edited my answer to include a possible solution for the case when the sample distance is not even. – pabaldonedo Aug 29 '13 at 10:15
  • Very thorough answer, you completely solved my problem. Thank you. – user1654183 Aug 29 '13 at 15:41