Python Pandas Count Interval Of Occurrence

I have implemented solution with iterating rows but it takes too long because of size of dataframe. The problem is this: I have a dataframe like this (ignore the first 3 columns):

Solution 1:

We can do that by writing a function that iterate the list (D). We go through the list, initialize a counter by 1, whenever we find one we increment, whenever we find 0, we affect the value and re-do the same process.

import pandas as pd
import copy

df = pd.DataFrame([1,1,1,0,1,1,0,0,0,1,1,1,0])

df.columns = ['D']
d= copy.copy(df.D)

deftransform(l):
  count=1for index,x inenumerate(l): 
    if x==0:
      l[index]=count
      count=1else:
      l[index]=0
      count+=1return l

df['intervales']=transform(t)
df['D']=d

print df

The Output:

     D  intervales
0   1           0
1   1           0
2   1           0
3   0           4
4   1           0
5   1           0
6   0           3
7   0           1
8   0           1
9   1           0
10  1           0
11  1           0
12  0           4

I tried to do that using itertools, but it leads to treat many cases.

# import itertools# l= [list(g) for k,g in itertools.groupby(df.D,lambda x:x in [0]) ]