在熊猫数据栏中的不同列之间满足条件时，显示带有指示符/标志的列(不合并)

Question

你们大家好。我有一个地区的数据，客户和一些交货，加上他们的价格。这一栏被用作购买类型，第一次和最后一次采购被标记为“第一次”和“最后一次”，有时我们有中间交货标记为“交付”。我需要将具有相同的第一次和最后一次交付价格的客户和区域标记为，作为期望输出中的一列。必须显示整个数据。

我已经使用merge解决了这个问题，但是我想知道是否有一种不使用合并的方法，因为它看起来并不那么有效。耽误您时间，实在对不起。

样本数据：

import pandas as pd  
data = [['NY', 'A','FIRST', 25], ['NY', 'A','DELIVERY', 20], ['NY', 'A','DELIVERY', 30], ['NY', 'A','LAST', 25],
       ['NY', 'B','FIRST', 15], ['NY', 'B','DELIVERY', 10], ['NY', 'B','LAST', 20],
       ['FL', 'A','FIRST', 15], ['FL', 'A','DELIVERY', 10], ['FL', 'A','DELIVERY', 12], ['FL', 'A','DELIVERY', 25], ['FL', 'A','LAST', 15],
       ['FL', 'C','FIRST', 15], ['FL', 'C','LAST', 10],
       ['FL', 'D','FIRST', 10], ['FL', 'D','DELIVERY', 20], ['FL', 'D','LAST', 30],
       ['FL', 'E','FIRST', 20], ['FL', 'E','LAST', 20]
       ] 
  
# Create the pandas DataFrame 
df = pd.DataFrame(data, columns = ['region', 'customer', 'purchaseType', 'price']) 
  
# print dataframe. 
print(df)

   region customer purchaseType  price
0      NY        A        FIRST     25
1      NY        A     DELIVERY     20
2      NY        A     DELIVERY     30
3      NY        A         LAST     25
4      NY        B        FIRST     15
5      NY        B     DELIVERY     10
6      NY        B         LAST     20
7      FL        A        FIRST     15
8      FL        A     DELIVERY     10
9      FL        A     DELIVERY     12
10     FL        A     DELIVERY     25
11     FL        A         LAST     15
12     FL        C        FIRST     15
13     FL        C         LAST     10
14     FL        D        FIRST     10
15     FL        D     DELIVERY     20
16     FL        D         LAST     30
17     FL        E        FIRST     20
18     FL        E         LAST     20

预期输出：

   region customer purchaseType  price  firstLastEqual
0      NY        A        FIRST     25            True
1      NY        A     DELIVERY     20            True
2      NY        A     DELIVERY     30            True
3      NY        A         LAST     25            True
4      NY        B        FIRST     15           False
5      NY        B     DELIVERY     10           False
6      NY        B         LAST     20           False
7      FL        A        FIRST     15            True
8      FL        A     DELIVERY     10            True
9      FL        A     DELIVERY     12            True
10     FL        A     DELIVERY     25            True
11     FL        A         LAST     15            True
12     FL        C        FIRST     15           False
13     FL        C         LAST     10           False
14     FL        D        FIRST     10           False
15     FL        D     DELIVERY     20           False
16     FL        D         LAST     30           False
17     FL        E        FIRST     20            True
18     FL        E         LAST     20            True

Answer 1

用“合并”回答：

df_first = df[df['purchaseType'] == 'FIRST']
df_last = df[df['purchaseType'] == 'LAST']
df_compare = df_first.merge(df_last, how='inner', left_on=['region','customer'], right_on=['region','customer'])
df_compare = df_compare[df_compare['price_x'] == df_compare['price_y']]
df_compare['firstLastEqual'] = True
df = df.merge(df_compare, how='left', left_on=['region','customer'], right_on=['region','customer'])
df['firstLastEqual'] = df['firstLastEqual'].fillna(False)
df = df.drop(['purchaseType_x', 'price_x', 'purchaseType_y', 'price_y'], axis=1)
print(df)

想知道如果可能的话不合并。