使用掩码和factorize：

# identify 1s
m = df['C'].eq(1)
# group consecutive values
g = m.ne(m.shift()).cumsum()
# identify stretches of 5+ 1s
m2 = m & df.groupby(g)['C'].transform('size').ge(5)

out = (df.loc[m2]
         .assign(group=pd.factorize(g[m2])[0]+1)
       )

输出量：

A       B  C  group
0       10    john  1      1
1       12    paul  1      1
2       23  kishan  1      1
3       12    teja  1      1
4       12    zebo  1      1
17    3434     coo  1      2
18     545     xoo  1      2
19    6565     zoo  1      2
20  232321     qoo  1      2
21   34454     woo  1      2
22  546556     eoo  1      2

你可以groupby列C的差，得到cumsum，其中差不等于零，并转换每个组的大小，这样你就只能保留大于或等于5的组

df[df['C'].eq(1) & df.groupby(df['C'].diff().ne(0).cumsum()).transform('size').gt(4)]

         A       B  C
0       10    john  1
1       12    paul  1
2       23  kishan  1
3       12    teja  1
4       12    zebo  1
17    3434     coo  1
18     545     xoo  1
19    6565     zoo  1
20  232321     qoo  1
21   34454     woo  1
22  546556     eoo  1

如果您想要一个组列，

# create the groups by calculating the diff and getting the cumsum
df['group'] = df['C'].diff().ne(0).cumsum()
# boolean indexing to keep values where C == 1 AND the size of each group is greater than 4
df[df['C'].eq(1) & df.groupby('group').transform('size').gt(4)]

         A       B  C  group
0       10    john  1      1
1       12    paul  1      1
2       23  kishan  1      1
3       12    teja  1      1
4       12    zebo  1      1
17    3434     coo  1      7
18     545     xoo  1      7
19    6565     zoo  1      7
20  232321     qoo  1      7
21   34454     woo  1      7
22  546556     eoo  1      7