Question

我有以下代码

import pandas as pd
import numpy as np
import csv


location = r'C:\Users\tmaina\Desktop\scf\output.csv'
df = pd.read_csv(location,sep='\s*,\s*',engine='python')
for i, row in df.iterrows():
    if row['COUPON_NUMBER'] == 1:
        df.OND_ORIGIN = df.DEP_FROM 
        if  df.loc[i+1,'PLDATE'] == row['PLDATE'] & row['TICKET_NUMBER'] ==df.loc[i+1,'TICKET_NUMBER'] &row['COUPON_NUMBER'] == 2:
            df.OND_DEST = df.loc[i+1,'ARR_TO']
        else:
            df.OND_DEST = df.ARR_TO
    elif row['COUPON_NUMBER'] == 2 & row['TICKET_NUMBER'] ==df.loc[i-1,'TICKET_NUMBER'] & row['PLDATE'] ==df.loc[i-1,'PLDATE']:
        df.OND_ORIGIN==df.loc[i-1,'DEP_FROM']
        df.OND_DEST = df.ARR_TO
    elif row['COUPON_NUMBER'] == 3 & row['TICKET_NUMBER'] ==df.loc[i-1,'TICKET_NUMBER'] & row['PLDATE'] !=df.loc[i-1,'PLDATE']:
        df.OND_ORIGIN = df.DEP_FROM
        if  df.loc[i+1,'PLDATE'] == row['PLDATE'] & row['TICKET_NUMBER'] ==df.loc[i-1,'TICKET_NUMBER']:
            df.OND_DEST = df.loc[i+1,'ARR_TO']
        else:
            df.OND_DEST = df.ARR_TO
    elif row['COUPON_NUMBER'] == 4 & row['TICKET_NUMBER'] ==df.loc[i-1,'TICKET_NUMBER']& row['PLDATE'] ==df.loc[i-1,'PLDATE']:
        df.OND_ORIGIN = df.loc[i-1,'DEP_FROM']
        df.OND_DEST = df.ARR_TO

df.to_csv('out.csv', sep=',',index = False)

以下列的输出是

COUPON_NUMBER TICKET_NUMBER DEP_FROM    ARR_TO  OND_ORIGIN  OND_DEST  PLDATE   STOPOVER
    1          1054737998    HRE             NBO    HRE     NBO       20170419  O
    2          1054737998    NBO             KGL    NBO     KGL       20170419  X   
    3          1054737998    KGL             NBO    KGL     NBO       20170519  O   
    4          1054737998    NBO             HRE    NBO     HRE       20170419  X

所需的输出是

COUPON_NUMBER TICKET_NUMBER DEP_FROM    ARR_TO  OND_ORIGIN  OND_DEST  PLDATE   STOPOVER
    1          1054737998    HRE         NBO    HRE         KGL       20170419  O
    2          1054737998    NBO         KGL    HRE         KGL       20170419  X   
    3          1054737998    KGL         NBO    KGL         HRE       20170519  O   
    4          1054737998    NBO         HRE    KGL         HRE       20170419  X

逻辑是，对于属于特定故障单的给定coupon_number，我们会检查pldate，如果同一天有多张优惠券，ond_origin和{ {1}}应该是平等的。 ond_dest是通过检查某个城市是否有停留来确定的。如果有，ond_dest成为arr_to，ond_dest成为第一个没有停留的ond_origin。

Answer 1

您可以使用groupby，Grouper和transform执行此操作，而不是遍历每一行。要获取每个组的第一个和最后一个，您可以使用this

如果PLDATE是一个日期时间列，你可以做类似的事情

df['OND_ORIGIN'] = df.groupby(['TICKET_NUMBER', pd.Grouper(key='PLDATE', freq='1M')])['DEP_FROM'].transform(first)   
df['OND_DEST'] = df.groupby(['TICKET_NUMBER', pd.Grouper(key='PLDATE', freq='1M')])['ARR_TO'].transform(last)

仅在您希望每月分组时才需要Grouper。如果是按日期，则可以df.groupby(['TICKET_NUMBER', 'PLDATE', freq='1M'])

如何迭代pandas数据帧中的行

1 个答案: