Cumsum reset at NaN

working4coins picture working4coins · Aug 12, 2013 · Viewed 7.8k times · Source

If I have a pandas.core.series.Series named ts of either 1's or NaN's like this:

3382   NaN
3381   NaN
...
3369   NaN
3368   NaN
...
15     1
10   NaN
11     1
12     1
13     1
9    NaN
8    NaN
7    NaN
6    NaN
3    NaN
4      1
5      1
2    NaN
1    NaN
0    NaN

I would like to calculate cumsum of this serie but it should be reset (set to zero) at the location of the NaNs like below:

3382   0
3381   0
...
3369   0
3368   0
...
15     1
10     0
11     1
12     2
13     3
9      0
8      0
7      0
6      0
3      0
4      1
5      2
2      0
1      0
0      0

Ideally I would like to have a vectorized solution !

I ever see a similar question with Matlab : Matlab cumsum reset at NaN?

but I don't know how to translate this line d = diff([0 c(n)]);

Answer

emprice picture emprice · Aug 12, 2013

A simple Numpy translation of your Matlab code is this:

import numpy as np

v = np.array([1., 1., 1., np.nan, 1., 1., 1., 1., np.nan, 1.])
n = np.isnan(v)
a = ~n
c = np.cumsum(a)
d = np.diff(np.concatenate(([0.], c[n])))
v[n] = -d
np.cumsum(v)

Executing this code returns the result array([ 1., 2., 3., 0., 1., 2., 3., 4., 0., 1.]). This solution will only be as valid as the original one, but maybe it will help you come up with something better if it isn't sufficient for your purposes.