你可以：
df['consecutive'] = df.Count.groupby((df.Count != df.Count.shift()).cumsum()).transform('size') * df.Count要得到：
Count consecutive0 1 11 0 02 1 23 1 24 0 05 0 06 1 37 1 38 1 39 0 0
在这里，您可以设置任何阈值：
threshold = 2df['consecutive'] = (df.consecutive > threshold).astype(int)
要得到：
Count consecutive0 1 01 0 02 1 13 1 14 0 05 0 06 1 17 1 18 1 19 0 0
或者，只需一步即可：
(df.Count.groupby((df.Count != df.Count.shift()).cumsum()).transform('size') * df.Count >= threshold).astype(int)在效率方面，
pandas当问题的规模变大时，使用方法可以显着提高速度：
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相比：
%%timeitl = []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
