我的方法是从 df.groupby('id').transform 调用一个辅助函数.我觉得这比它可能的更复杂和更慢，但它似乎有效.

My approach is to call a helper function from df.groupby('id').transform. I feel this is more complicated and slower than it could be, but it seems to work.

# test data

    date    id  cum_count_desired
2017-03-01  A   1
2017-03-01  B   1
2017-03-01  C   1
2017-05-01  B   2
2017-05-01  D   1
2017-07-01  A   2
2017-07-01  D   2
2017-08-01  C   1
2017-09-01  B   2
2017-09-01  B   3

# preprocessing

df['date'] = pd.to_datetime(df['date'])
df.set_index('date', inplace=True)
# Encode the ID strings to numbers to have a column
# to work with after grouping by ID
df['id_code'] = pd.factorize(df['id'])[0]

# solution

def cumcounter(x):
    y = [x.loc[d - pd.DateOffset(months=4):d].count() for d in x.index]
    gr = x.groupby('date')
    adjust = gr.rank(method='first') - gr.size() 
    y += adjust
    return y

df['cum_count'] = df.groupby('id')['id_code'].transform(cumcounter)

# output

df[['id', 'id_num', 'cum_count_desired', 'cum_count']]

           id  id_num  cum_count_desired  cum_count
date                                               
2017-03-01  A       0                  1          1
2017-03-01  B       1                  1          1
2017-03-01  C       2                  1          1
2017-05-01  B       1                  2          2
2017-05-01  D       3                  1          1
2017-07-01  A       0                  2          2
2017-07-01  D       3                  2          2
2017-08-01  C       2                  1          1
2017-09-01  B       1                  2          2
2017-09-01  B       1                  3          3

调整

如果同一 ID 在同一天出现多次，我使用的切片方法会多计算每个同一天的 ID，因为当列表推导时，基于日期的切片会立即抓取所有当天的值遇到多个 ID 出现的日期.修复:

The need for adjust

If the same ID occurs multiple times on the same day, the slicing approach that I use will overcount each of the same-day IDs, because the date-based slice immediately grabs all of the same-day values when the list comprehension encounters the date on which multiple IDs show up. Fix:

1. 按日期对当前 DataFrame 进行分组.
2. 对每个日期组中的每一行进行排名.
3. 从这些排名中减去每个日期组中的总行数.这会产生一个以日期为索引的升序负整数系列，以 0 结尾.
4. 将这些非正整数调整添加到 y.

这只影响给定测试数据中的一行——倒数第二行，因为 B 在同一天出现两次.

This only affects one row in the given test data -- the second-last row, because B appears twice on the same day.

要计算与 4 个日历月前一样旧或更新的行数，即包括 4 个月时间间隔的左端点，请保持此行不变:

To count rows as old as or newer than 4 calendar months ago, i.e., to include the left endpoint of the 4-month time interval, leave this line unchanged:

y = [x.loc[d - pd.DateOffset(months=4):d].count() for d in x.index]

要计算比 4 个日历月前新的行，即要排除 4 个月时间间隔的左端点，请改用:

To count rows strictly newer than 4 calendar months ago, i.e., to exclude the left endpoint of the 4-month time interval, use this instead:

y = [d.loc[d - pd.DateOffset(months=4, days=-1):d].count() for d in x.index]