用于具有行间和定义函数的分组数据的 Python For 循环

Question

我正在尝试计算一个指示列，以便如果给定客户在给定服务结束日期后 60 天内返回另一项服务，则它返回值 1。下面是数据表和所需的指标列。

我已按客户和服务对数据进行了分类和分组。现在，我正在尝试整理一个公式，该公式查看每一行的结束日期，并检查该给定客户（组）的任何未来服务开始日期是否在 60 天内。

到目前为止，我已经对数据进行了分组，并尝试应用 for 循环......但我不断收到错误消息。我是 Python 新手，所以我不太确定如何设置代码和语法以使其正常工作。任何帮助将不胜感激 - 谢谢！

df = pd.DataFrame({'Customer':['John','John','Deb','Sara','Sara','Sara','Sara','Sara','Mike','Mike','Mike'],
               'Service':['A','B','C','A','B','C','D','E','E','F','G'],
               'Start':['1/2/2020','3/1/2020','2/5/2020','2/6/2020','3/6/2020','8/2/2020','9/15/2020','10/2/2020','5/4/2020','6/8/2020','7/1/2020'],
               'End':['1/3/2020','3/11/2020','2/7/2020','2/9/2020','3/9/2020','8/12/2020','9/19/2020','10/12/2020','5/8/2020','6/18/2020','7/8/2020']})


df['Start'] = pd.to_datetime(df['Start'], infer_datetime_format=True, errors='coerce')
df['End'] = pd.to_datetime(df['End'], infer_datetime_format=True, errors='coerce')
df.dtypes

grp=df.groupby(['Customer','Service'])
grp.head(15)

def serv_days(Start,End):
    for row in df.iterrows():
        if (row[1].Start == row.End):
            continue
        if ((row[1].Start-row.End).days < 60):
            return 1
        else:
            return 0

for grpnm,each_grp in grp:
    for row in each_grp.iterrows():
        print(serv_days(each_grp,row[1].Start,row.loc('End')))

Answer 1

正如您所说，我们可以将当前行与下一行进行比较，以查看当前行客户是否等于下一行客户，并且当前行结束日期与下一行开始日期的差异小于 60 天。
随心所欲，我们不需要groupby，我们需要：

• 使用df[1:] 移动数据框，这意味着删除第一行并保留其他行
• 我们需要reset_index 才能将index 与原始df 合并，所以现在我们有一个数据框将当前行和下一行合并到同一行中
• 只需比较customer 和customer_next（_next 后缀由merge 函数suffixes 参数添加）和差异End Start_next

df = pd.DataFrame({'Customer':['John','John','Deb','Sara','Sara','Sara','Sara','Sara','Mike','Mike','Mike'],
               'Service':['A','B','C','A','B','C','D','E','E','F','G'],
               'Start':['1/2/2020','3/1/2020','2/5/2020','2/6/2020','3/6/2020','8/2/2020','9/15/2020','10/2/2020','5/4/2020','6/8/2020','7/1/2020'],
               'End':['1/3/2020','3/11/2020','2/7/2020','2/9/2020','3/9/2020','8/12/2020','9/19/2020','10/12/2020','5/8/2020','6/18/2020','7/8/2020']})

df['Start'] = pd.to_datetime(df['Start'], infer_datetime_format=True, errors='coerce')
df['End'] = pd.to_datetime(df['End'], infer_datetime_format=True, errors='coerce')

new_df = pd.merge(df,df[1:].reset_index(drop=True),suffixes=['','_next'],left_index=True,right_index=True,how='left')
new_df['indicator'] = (new_df['Customer'] == new_df['Customer_next']) & ((new_df['Start_next']-new_df['End'])<pd.Timedelta('60 days'))
new_df = new_df[['Customer','Service','Start','End','indicator']]

    Customer    Service   Start         End         indicator
0   John        A         2020-01-02    2020-01-03  TRUE
1   John        B         2020-03-01    2020-03-11  FALSE
2   Deb         C         2020-02-05    2020-02-07  FALSE
3   Sara        A         2020-02-06    2020-02-09  TRUE
4   Sara        B         2020-03-06    2020-03-09  FALSE
5   Sara        C         2020-08-02    2020-08-12  TRUE
6   Sara        D         2020-09-15    2020-09-19  TRUE
7   Sara        E         2020-10-02    2020-10-12  FALSE
8   Mike        E         2020-05-04    2020-05-08  TRUE
9   Mike        F         2020-06-08    2020-06-18  TRUE
10  Mike        G         2020-07-01    2020-07-08  FALSE