How can I count the number of days between a 1 and the next 0

You try this using groupby like itertool.groupby here. The extract 1st index of every group. Since you have to find difference b/w two groups there have to be same number of 1 groups and 0 groups, if it's not the case then drop the last group.

s = pd.Series(values, index = pd.to_datetime(dates))
g = s.ne(s.shift()).cumsum()
vals = s.groupby(g).apply(lambda x:x.index[0])
# vals
1    2019-08-01
2    2019-08-05
3    2019-08-08
4    2019-08-12
5    2019-08-13
6    2019-08-14
7    2019-08-16
8    2019-08-23
9    2019-08-29
dtype: object

Now we dont have same number of 1 groups and 0 groups, so ditch the group index. And make chunks for size 2 i.e now, each has 1 and 0 groups indices.

end = None if not len(vals)%2 else -1
vals = vals.iloc[:end].values.reshape((-1, 2))
# vals 
array([['2019-08-01T00:00:00.000000000', '2019-08-05T00:00:00.000000000'],
       ['2019-08-08T00:00:00.000000000', '2019-08-12T00:00:00.000000000'],
       ['2019-08-13T00:00:00.000000000', '2019-08-14T00:00:00.000000000'],
       ['2019-08-16T00:00:00.000000000', '2019-08-23T00:00:00.000000000']],
      dtype='datetime64[ns]')

Now, we have to find the difference using np.diff.

days = np.diff(vals, axis=1).squeeze()
out = pd.Series(days)
# out

0   4 days
1   4 days
2   1 days
3   7 days
dtype: timedelta64[ns]

I think something like below should work, first have a series with a date index:

ds = pd.Series(values, index = pd.to_datetime(dates))

Then you calculate the difference between consecutive values:

delta = ds - ds.shift(fill_value=ds[0]-1)

It looks like this :

pd.DataFrame({'value':ds,'delta':delta})


    value   delta
2019-08-01  1   1
2019-08-02  1   0
2019-08-05  0   -1
2019-08-06  0   0
2019-08-07  0   0
2019-08-08  1   1
2019-08-09  1   0
2019-08-12  0   -1
2019-08-13  1   1
2019-08-14  0   -1

So the start dates you need are when delta is 1, and the next zero you need is where it is -1 .So:

starts = ds.index[delta == 1]
ends = ds.index[delta == -1]
(ends - starts[:len(ends)]).days

Int64Index([4, 4, 1, 7], dtype='int64')

Note there are some cases where at the end of the data frame, you have 1s but they don't flip into 0, so I ignore those.

Start from creation of a DataFrame with date column composed of dates converted to datetime and val column composed of values:

df = pd.DataFrame({'date': pd.to_datetime(dates), 'val': values})

The idea to get the result is:

• Get dates where val == 0 (for other rows take NaT).
• Perform "backwards filling".
• Subtract date.
• From the above result (timedelta) get the days number.
• Fill outstanding NaT values (if any) with 0 (in your case this pertains to 2 last rows, which are not followed by any "0 row").
• Save the result in dist column.

The code to do it is:

df['dist'] = (df.date.where(df.val == 0).bfill(0) - df.date)\
    .dt.days.fillna(0, downcast='infer')

The result is:

         date  val  dist
0  2019-08-01    1     4
1  2019-08-02    1     3
2  2019-08-05    0     0
3  2019-08-06    0     0
4  2019-08-07    0     0
5  2019-08-08    1     4
6  2019-08-09    1     3
7  2019-08-12    0     0
8  2019-08-13    1     1
9  2019-08-14    0     0
10 2019-08-15    0     0
11 2019-08-16    1     7
12 2019-08-19    1     4
13 2019-08-20    1     3
14 2019-08-21    1     2
15 2019-08-22    1     1
16 2019-08-23    0     0
17 2019-08-26    0     0
18 2019-08-27    0     0
19 2019-08-28    0     0
20 2019-08-29    1     0
21 2019-08-30    1     0

(dist column is the distance in days).

If you need, take from the above result only rows with val != 0.