您可以用途：

# Convert to datetime64 if needed
df['DATE'] = pd.to_datetime(df['DATE'])

# Pandas part: reshape your dataframe
out = df.sort_values('DATE').pivot_table(index='CaseID', columns='EVENT', values='DATE', aggfunc='last')
m = out.notna().all(axis=1)

# Numpy part: compute business day
arr = out[m].values.astype('datetime64[D]')
out.loc[m, 'bdays'] = np.busday_count(arr[:, 0], arr[:, 1])

输出：

>>> out
EVENT           A          B  bdays
CaseID                             
1      2023-01-20 2023-02-01    8.0
2      2023-03-04 2023-04-30   40.0
6      2023-02-14        NaT    NaN
8      2023-01-02        NaT    NaN
33     2023-02-26        NaT    NaN
100    2023-03-01 2023-03-02    1.0

您的输入数据采用管道分隔的CSV文件的形式。它非常简单，因此不需要任何模块导入来处理它。
构建一个以CaseID为键的字典。每个关联的值都应该是一个字典，可以同时包含“A”和“B”键。日期应该在与这些键关联的列表中。
您需要解析日期以找到最大值（最近的），然后做一些算术运算。
给定输入文件foo.csv，其中包含以下内容：

|CaseID|EVENT|DATE|
|1|A|1/1/23|
|1|A|1/3/23|
|1|A|1/20/23|
|1|B|2/1/23|
|8|A|1/2/23|
|100|A|3/1/23|
|100|B|3/2/23|
|6|A|2/14/23|
|33|A|2/26/23|
|2|A|3/4/23|
|2|B|4/30/23|

代码可能看起来像这样：

from datetime import datetime

def parse(d):
    return datetime.strptime(d, '%m/%d/%y')

mydict = dict()

with open('foo.csv') as data:
    next(data) # skip column headers
    for line in data:
        _, case, event, date, *_ = line.split('|')
        _date = parse(date)
        if case in mydict:
            mydict[case].setdefault(event, []).append(_date)
        else:
            mydict[case] = {event: [_date]}

for k, v in mydict.items():
    if 'A' in v and 'B' in v:
        maxa = max(v['A'])
        maxb = max(v['B'])
        print('CaseID', k, abs((maxa-maxb).days))

输出：

CaseID 1 12
CaseID 100 1
CaseID 2 57

假设数据如所示（即caseID值和DATE排序-否则您可以在进一步处理之前排序），然后首先将CSV数据读入pandas DataFrame;然后用途：

#convert date strings to datetime format
df['DATE'] = pd.to_datetime(df['DATE'], dayfirst = False)
#drop duplicated A rows
df = df.drop_duplicates(subset = ['CaseID', 'EVENT'], keep = 'last')
#calculate difference of days
df['days'] = df.groupby('CaseID')['DATE'].diff().dt.days
#and finally lose the case without an A and a B event
df = df.dropna()

这导致：

CaseID EVENT       DATE  days
3        1     B 2023-02-01  12.0
6      100     B 2023-03-02   1.0
10       2     B 2023-04-30  57.0

另一种方法是按事件和ID分组，获得最大日期，再次按ID分组，并使用apply获得日期之间的差异。

from io import StringIO

import pandas as pd

table = """
CaseID,EVENT,DATE
1,A,1/1/23
1,A,1/3/23
1,A,1/20/23
1,B,2/1/23
8,A,1/2/23
100,A,3/1/23
100,B,3/2/23
6,A,2/14/23
33,A,2/26/23
2,A,3/4/23
2,B,4/30/23
"""

df = pd.read_csv(StringIO(table), sep=',')
df['DATE'] = pd.to_datetime(df['DATE'])
d = df.groupby(['CaseID', 'EVENT']).max().reset_index().groupby('CaseID').aggregate({'DATE': lambda x: x.diff().to_numpy()[-1]})

输出：

DATE
CaseID        
1      12 days
2      57 days
6          NaT
8          NaT
33         NaT
100     1 days

import pandas as pd

def get_time_lapsed(df, case_id, event, date):
    df = df.sort_values(by=[case_id, date])
    df["time_lapsed"] = df.groupby(case_id)[date].diff()
    return df

def test_get_time_lapsed():
    df = pd.DataFrame(
        {
            "CaseID": [1, 1, 1, 1, 8, 100, 100, 35, 35, 6, 33, 2, 2],
            "EVENT": ["A", "A", "A", "B", "A", "A", "B", "A", "B", "A", "A", "A", "B"],
            "DATE": [
                "1/1/23",
                "1/3/23",
                "1/20/23",
                "2/1/23",
                "1/2/23",
                "3/1/23",
                "3/2/23",
                "2/13/23",
                "2/27/23",
                "2/14/23",
                "2/26/23",
                "3/4/23",
                "4/30/23",
            ],
        }
    )
    df["DATE"] = pd.to_datetime(df["DATE"])
    df = get_time_lapsed(df, "CaseID", "EVENT", "DATE")
    print(df)

CaseID EVENT       DATE time_lapsed
0        1     A 2023-01-01         NaT
1        1     A 2023-01-03      2 days
2        1     A 2023-01-20     17 days
3        1     B 2023-02-01     12 days
11       2     A 2023-03-04         NaT
12       2     B 2023-04-30     57 days
9        6     A 2023-02-14         NaT
4        8     A 2023-01-02         NaT
10      33     A 2023-02-26         NaT
7       35     A 2023-02-13         NaT
8       35     B 2023-02-27     14 days
5      100     A 2023-03-01         NaT
6      100     B 2023-03-02      1 days