问题描述
我有一个 DataFrame (df)，如下所示:
I have a DataFrame (df) that looks like the following:
+----------+----+
| dd_mm_yy | id |
+----------+----+
| 01-03-17 | A |
| 01-03-17 | B |
| 01-03-17 | C |
| 01-05-17 | B |
| 01-05-17 | D |
| 01-07-17 | A |
| 01-07-17 | D |
| 01-08-17 | C |
| 01-09-17 | B |
| 01-09-17 | B |
+----------+----+
这是我想要计算的最终结果:
This the end result i would like to compute:
+----------+----+-----------+
| dd_mm_yy | id | cum_count |
+----------+----+-----------+
| 01-03-17 | A | 1 |
| 01-03-17 | B | 1 |
| 01-03-17 | C | 1 |
| 01-05-17 | B | 2 |
| 01-05-17 | D | 1 |
| 01-07-17 | A | 2 |
| 01-07-17 | D | 2 |
| 01-08-17 | C | 1 |
| 01-09-17 | B | 2 |
| 01-09-17 | B | 3 |
+----------+----+-----------+
逻辑
计算 id 中值在指定时间窗口内的累积出现次数，例如 4 个月.即每 5 个月计数器重置为 1.
Logic
To calculate the cumulative occurrences of values in id but within a specified time window, for example 4 months. i.e. every 5th month the counter resets to one.
为了获得累积出现次数，我们可以使用这个 df.groupby('id').cumcount() + 1
To get the cumulative occurences we can use this df.groupby('id').cumcount() + 1
关注 id = B 我们看到 B 的第二次出现是在 2 个月之后，所以 cum_count = 2.B 的下一次出现是在 01-09-17，回顾 4 个月我们只发现了另外一次出现，所以 cum_count = 2 等.
Focusing on id = B we see that the 2nd occurence of B is after 2 months so the cum_count = 2. The next occurence of B is at 01-09-17, looking back 4 months we only find one other occurence so cum_count = 2, etc.
推荐答案
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
的必要性如果同一 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:
- 按日期对当前 DataFrame 进行分组.
- 对每个日期组中的每一行进行排名.
- 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.
- 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]
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