Cumsum reset at NaN

Question

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
         


        

Answer 1

Here's a slightly more pandas-onic way to do it:

v = Series([1, 1, 1, nan, 1, 1, 1, 1, nan, 1], dtype=float)
n = v.isnull()
a = ~n
c = a.cumsum()
index = c[n].index  # need the index for reconstruction after the np.diff
d = Series(np.diff(np.hstack(([0.], c[n]))), index=index)
v[n] = -d
result = v.cumsum()

Note that either of these requires that you're using pandas at least at 9da899b or newer. If you aren't then you can cast the bool dtype to an int64 or float64 dtype:

v = Series([1, 1, 1, nan, 1, 1, 1, 1, nan, 1], dtype=float)
n = v.isnull()
a = ~n
c = a.astype(float).cumsum()
index = c[n].index  # need the index for reconstruction after the np.diff
d = Series(np.diff(np.hstack(([0.], c[n]))), index=index)
v[n] = -d
result = v.cumsum()