如何有效地替换一维数组中相同的连续值

Question

Surprisingly, after a fair bit of research, I did not find any post sparking a good idea to solve this simple problem.

I have a 1D numpy array of shape (n, ) which is mostly zeros with a few other positive values.

D = np.zeros(10)
D[2] = 1
D[5] = 4
D[7] = 3

Out: array([0., 0., 1., 0., 0., 4., 0., 3., 0., 0.])

However, sometimes, one of the values (which is a trigger at the given sample) might be repeated during the next (or few next samples).

D = np.zeros(10)
D[2] = 1
D[3] = 1
D[5] = 4
D[7] = 3

Out: array([0., 0., 1., 1., 0., 4., 0., 3., 0., 0.])

In this case, I want to process the array to replace the identical consecutive values by 0; in this case, to replace the second 1 by 0 .

In: array([0., 0., 1., 1., 0., 4., 0., 3., 0., 0.])
Out: array([0., 0., 1., 0., 0., 4., 0., 3., 0., 0.])

For one additional complexity level, I would like to define a tolerance, and if every element right after a value x>0 is close enough to x , then they get replaced by 0.

In: array([0., 0., 1., 0.98, 1.01, 4., 0., 3., 3.1, 0.]), tolerance = 0.05
Out: array([0., 0., 1., 0., 0., 4., 0., 3., 3.1, 0.])

This is fairly easy to do by looping on the array with either a while or a for loop. However, as this is for an online application, I was looking for a vectorized solution in numpy. If anyone has any suggestion, any function I could look at to do this as efficiently as possible, please comment!

Answer 1

Try using np.diff() like so:

D[np.where(np.diff(D)==0)] = 0

Edit: Including OP's solution to include tolerance and prepend:

D[np.where(np.diff(D, prepend=[0])<=tolerance)] = 0

Edit 2: @obchardon's excellent suggestion to remove the unnecessary np.where :

D[abs(np.diff(D, prepend=[0]))<=tolerance] = 0