Idea是聚合每个连续组0的值，每个Id和助手Series s由Series.cumsum创建，计数True的s值，然后由于可能的多个组Id聚合max和最后一组numpy.where中的值：

N = 3
m = df['completed'].ne(0)
#if need check in completed OR scheduled column
#m = df[['completed','scheduled']].ne(0).any(axis=1)

df = ((~m).groupby([df['Id'], m.cumsum().mask(m, -1)])
          .sum()
          .groupby(level=0)
          .max()
          .reset_index(name='status')
          .assign(status = lambda x: np.where(x['status'].ge(N), 
                                            'defaulter','non_defaulter')))
       
print (df)
  Id         status
0  A      defaulter
1  B  non_defaulter
2  C      defaulter

您可以使用双groupby来计算completed中连续的1，然后确保至少有一个大于或等于N=3的延伸：

N = 3

# is the row a zero?
m = df['completed'].eq(0)

# count the consecutive zeros
(m.groupby([df['Id'], (~m).cumsum()])
   .sum().ge(N)
   # check if there is at least one stretch of value >= N
   .groupby(level=0).any()
   # convert the True/False into strings
   .map({False: 'non_defaulter', True: 'defaulter'})
   .reset_index(name='status')
)

输出量：

Id         status
0  A      defaulter
1  B  non_defaulter
2  C      defaulter