Python：查找一列中的最新日期，而另一列中没有匹配的日期

Question

我有两个日期列，代表了客户设施的条目和退出。

ID	ENTRY_DATE	EXIT_DATE	ORICANT_ENTRYDATE
003246	2022-03-22	NAN	2012-10-01
003246	2015-07-24	2022-03-22	2012-10-10-01
003246		​	2012-10-01
003246	2001-02-02	2010-04-05	2001-02-02

对于桌子中的所有ID实例输入日期代表该ID在设施之间移动但不离开护理的时间跨度的开始，并将其返回，并将其返回原始_entrydate。

在示例中，前三行的原始_entrydate值是2012-10-01，因为该entry_date与exit_date不匹配，这表明与护理的分离，日期显示了这持续了两年零几个月。如果还有该ID的其他记录，则该过程将重置并找到与该分离之前的任何记录，直到下一个分隔的任何记录。

Answer 1

I solved my problem in the most clunky way imaginable--by creating nested if-else statements:

res_phys_levels['ORIGINAL_AdmDt']=''

for i in range(0, len(res_phys_levels)):
    start_ID = res_phys_levels.iloc[i]['Individual_ID']
    aDate = res_phys_levels.iloc[i]['Admit_Date']
    id_count = res_phys_levels.Individual_ID.value_counts()[start_ID]
    if id_count == 1:     #if there's only one instance of Individual ID in table, then ORIGINAL_AdmDt = Admit_Date
        res_phys_levels.at[i, 'ORIGINAL_AdmDt'] = aDate
    else:                 #if there's more than one instance of Individual_ID, then--
        j = i+1        
        next_ID = res_phys_levels.iloc[j]['Individual_ID']
        if start_ID != next_ID:
            res_phys_levels.at[i, 'ORIGINAL_AdmDt'] = aDate
        else: 
            sDate = res_phys_levels.iloc[j]['SEPARATION_DATE']
            if aDate != sDate: 
                res_phys_levels.at[i, 'ORIGINAL_AdmDt'] = aDate
            else: 
                aDate = res_phys_levels.iloc[j]['Admit_Date']
                k = j+1
                next_ID = res_phys_levels.iloc[k]['Individual_ID']
                if start_ID != next_ID:
                    res_phys_levels.at[i, 'ORIGINAL_AdmDt'] = aDate
                else: 
                    sDate = res_phys_levels.iloc[k]['SEPARATION_DATE']
                    if aDate != sDate: 
                        res_phys_levels.at[i, 'ORIGINAL_AdmDt'] = aDate
                    else: 
                        aDate = res_phys_levels.iloc[k]['Admit_Date']
                        m = k+1
                        next_ID = res_phys_levels.iloc[m]['Individual_ID']
                        if start_ID != next_ID:
                            res_phys_levels.at[i, 'ORIGINAL_AdmDt'] = aDate
                        else:
                            sDate = res_phys_levels.iloc[m]['SEPARATION_DATE']
                            if aDate != sDate:
                                res_phys_levels.at[i, 'ORIGINAL_AdmDt'] = aDate
                            else:
                                aDate = res_phys_levels.iloc[k]['Admit_Date']
                                n = m+1
                                next_ID = res_phys_levels.iloc[n]['Individual_ID']
                                if start_ID != next_ID:
                                    res_phys_levels.at[i, 'ORIGINAL_AdmDt'] = aDate

This worked for two reasons: the dataframe was sorted by 'Individual_ID' ascending and 'Admit_Date' descending--and the nested if-else statements allowed为了比较索引[i]的“私有_id”与随后的行进行比较，直到所有可能性都用尽。

我也知道每个ID最多有4行。

但是 - 请向我展示一种更好，更多的Pythonic方法！