Identifying consecutive occurrences of a value in a column of a pandas DataFrame

Not sure if this is optimized, but you can give it a try:

from itertools import groupby
import pandas as pd

l = []
for k, g in groupby(df.Count):
    size = sum(1 for _ in g)
    if k == 1 and size >= 2:
        l = l + [1]*size
    else:
        l = l + [0]*size

df['new_Value'] = pd.Series(l)

df

Count   new_Value
0   1   0
1   0   0
2   1   1
3   1   1
4   0   0
5   0   0
6   1   1
7   1   1
8   1   1
9   0   0

You could:

df['consecutive'] = df.Count.groupby((df.Count != df.Count.shift()).cumsum()).transform('size') * df.Count

to get:

   Count  consecutive
0      1            1
1      0            0
2      1            2
3      1            2
4      0            0
5      0            0
6      1            3
7      1            3
8      1            3
9      0            0

From here you can, for any threshold:

threshold = 2
df['consecutive'] = (df.consecutive > threshold).astype(int)

to get:

   Count  consecutive
0      1            0
1      0            0
2      1            1
3      1            1
4      0            0
5      0            0
6      1            1
7      1            1
8      1            1
9      0            0

or, in a single step:

(df.Count.groupby((df.Count != df.Count.shift()).cumsum()).transform('size') * df.Count >= threshold).astype(int)

In terms of efficiency, using pandas methods provides a significant speedup when the size of the problem grows:

 df = pd.concat([df for _ in range(1000)])

%timeit (df.Count.groupby((df.Count != df.Count.shift()).cumsum()).transform('size') * df.Count >= threshold).astype(int)
1000 loops, best of 3: 1.47 ms per loop

compared to:

%%timeit
l = []
for k, g in groupby(df.Count):
    size = sum(1 for _ in g)
    if k == 1 and size >= 2:
        l = l + [1]*size
    else:
        l = l + [0]*size    
pd.Series(l)

10 loops, best of 3: 76.7 ms per loop