为了简化我的问题,我创建了一个小型DataFrame,如下所示：

Type  From  To
A    "H1"  "U1"
A    "H9"  "I8"
A    "H1"  "IL"
B    "P2"  "P8"
B    "P2"  "P7"
C    "P9"  "O8"
C    "P9"  "I0"
C    "P7"  "O8"

在对字符串进行分组和汇编之后,我们应该得到以下期望的结果：

Type  From  To 
A    "H1"  "U1, IL"
A    "H9"  "I8"
B    "P2"  "P8, P7"
C    "P9"  "O8, I0"
C    "P7"  "O8"

我使用拆分和聚合函数做了它.对于任何想法或建议如何使用Python,我将非常感谢！

解决方法:

在R中,我们可以通过粘贴做一组. (注意,首先发布的问题中有一个R标签.否则,我们甚至不会尝试这个R解决方案)

library(tidyverse)
df1 %>%
    group_by(Type, From) %>%
    summarise(To = toString(To))
# A tibble: 5 x 3
# Groups:   Type [?]
#  Type  From  To    
#  <chr> <chr> <chr> 
#1 A     H1    U1, IL
#2 A     H9    I8    
#3 B     P2    P8, P7
#4 C     P7    O8    
#5 C     P9    O8, I0

数据

df1 <- structure(list(Type = c("A", "A", "A", "B", "B", "C", "C", "C"
), From = c("H1", "H9", "H1", "P2", "P2", "P9", "P9", "P7"), 
To = c("U1", "I8", "IL", "P8", "P7", "O8", "I0", "O8")),
 class = "data.frame", row.names = c(NA, 
 -8L))

在python中,我们可以做到

out = df2.groupby(['Type', 'From'])['To'].apply(lambda x: ','.join(x)).reset_index()
print(out)
# Type From     To
#0    A   H1  U1,IL
#1    A   H9     I8
#2    B   P2  P8,P7
#3    C   P7     O8
#4    C   P9  O8,I0

数据

import pandas as pd
df2 = pd.DataFrame({'Type': ['A', 'A', 'A', 'B', 'B', 'C', 'C', 'C'], \
                'From': ['H1', 'H9', 'H1', 'P2', 'P2', 'P9', 'P9', 'P7'], \
                'To': ['U1', 'I8', 'IL', 'P8', 'P7', 'O8', 'I0', 'O8']})