将组分配给 Pandas 列中的连续 1

Question

I have a column in pandas having values 0 and 1. I want to assign group number where there are more than 9 consecutive 1.

Example: Say my column values are: [1,1,1,1,0,0,0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,0,0,0,1,1,1,1,1,1,1,1,1,1,1,1,1,0,0,0,1,1,1,1,1,1,1,1,1,1,1]

I want a new column or change the same column to: [0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,0,0,0,2,2,2,2,2,2,2,2,2,2,2,2,2,0,0,0,3,3,3,3,3,3,3,3,3,3,3]

I got upto a point where I can replace all the consecutive 1s (count greater than 9) by another number say 2. Here is the code:

def f(col, threshold=9):
    mask = col.groupby((col != col.shift()).cumsum()).transform('count').gt(threshold)
    mask &= col.eq(1)
    #print (mask)
    col.update(col.loc[mask].replace(1,2))
    return col

Answer 1

Find consecutive groups of 1s and determine the size of those groups. Use where to mask any groups of 0s, or groups of 1s that are too small, then ngroup will allow you to label them properly. NaN rows get labeled -1 and you want the counting to start at 1, so adding 1 fixes both of these simultaneously.

import pandas as pd
s = pd.Series([1,1,1,1,0,0,0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,1,
               1,1,1,1,1,1,1,0,0,0,0,1,1,1,1,1,1,1,1,1,1,1,1,
               1,0,0,0,1,1,1,1,1,1,1,1,1,1,1])

u = s.ne(s.shift()).cumsum().where(s.eq(1))  # Label consecutive groups of 1s, NaN 0s
u = u.groupby(u).transform('size').gt(9)     # True only if 1s and size > 9.

# Any smaller groups or 0s get NaN'd by `where` which are labeled -1 by `ngroup`
result = u.groupby(u.ne(u.shift()).cumsum().where(u)).ngroup()+1

print(results.tolist())
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 
 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 
 2, 2, 2, 0, 0, 0, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3]

Answer 2

My approach:

s = pd.Series([1,1,1,1,0,0,0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,1,
               1,1,1,1,1,1,1,0,0,0,0,1,1,1,1,1,1,1,1,1,1,1,1,
               1,0,0,0,1,1,1,1,1,1,1,1,1,1,1])

# groupby and filter those with >=9 ones
u = s.groupby(s.ne(1).cumsum()).transform('sum').ge(9) & s

# count the groups of True:
(~u.shift(fill_value=False) & u).cumsum().mul(u)

Output:

[0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 2 2 2
 2 2 2 2 2 2 2 2 2 2 0 0 0 3 3 3 3 3 3 3 3 3 3 3]