A列的组间（组之间）组合，按B列分组 [英] inter-group (between groups) combination of column A grouped by column B

问题描述

我认为这是一个图形理论问题：我们可以在两组点之间画几条线...我不熟悉...

I think this is a graphical theory question: how many lines can we draw between two sets of points... which i'm not familiar with...

例如

df = data.frame(city = c('Boston', 'Cambridge', 'Long Island', 'NYC'),
                state = c('MA', 'MA', 'NY', 'NY'))

         city state
1      Boston    MA
2   Cambridge    MA
3 Long Island    NY
4         NYC    NY

随州吐痰/分组。

Boston - Long Island
Boston - NYC
Cambridge - Long Island
Cambridge - NYC

换句话说，我想生成两个城市不同的每个城市对

In other words, I want to generate every city pair where the two cities are in different states.

一个更通用的示例：

set.seed(123)
df = data.frame(value = 1:100,
                group = letters[sample(1:26, 100, replace=T)])

> df
    value group
1       1     e
2       2     m
3       3     g
4       4     o
5       5     p
6       6     a
7       7     i
8       8     o
9       9     i
10     10     h
11     11     p
12     12     h
...    ...    ...

我想要所有组合（值1，值2）或等效地（索引1，索引2），其中value1和value2具有不同的组标签。

I want all combination (value1, value2) or equivalently (index1, index2) where value1 and value2 has different group labels.

推荐答案

For循环，尽管在R中不鼓励使用，但可以用来获得所需的结果：

For loop, although discouraged in R, can be used to get desired result:

ddf = data.frame(value = 1:20,  group = letters[sample(1:3, 20, replace=T)])
head(ddf)
  value group
1     1     b
2     2     b
3     3     b
4     4     c
5     5     a
6     6     a

for(i in 1:20){
    tempdf = ddf[ddf$group!=ddf[i,2],]
    cat(ddf[i,1],': ',tempdf[,1], '\n')
}

1 :  4 5 6 8 9 10 13 15 17 19 20 
2 :  4 5 6 8 9 10 13 15 17 19 20 
3 :  4 5 6 8 9 10 13 15 17 19 20 
4 :  1 2 3 5 6 7 8 11 12 13 14 16 18 19 
5 :  1 2 3 4 7 9 10 11 12 14 15 16 17 18 20 
6 :  1 2 3 4 7 9 10 11 12 14 15 16 17 18 20 
7 :  4 5 6 8 9 10 13 15 17 19 20 
8 :  1 2 3 4 7 9 10 11 12 14 15 16 17 18 20 
9 :  1 2 3 5 6 7 8 11 12 13 14 16 18 19 
10 :  1 2 3 5 6 7 8 11 12 13 14 16 18 19 
11 :  4 5 6 8 9 10 13 15 17 19 20 
12 :  4 5 6 8 9 10 13 15 17 19 20 
13 :  1 2 3 4 7 9 10 11 12 14 15 16 17 18 20 
14 :  4 5 6 8 9 10 13 15 17 19 20 
15 :  1 2 3 5 6 7 8 11 12 13 14 16 18 19 
16 :  4 5 6 8 9 10 13 15 17 19 20 
17 :  1 2 3 5 6 7 8 11 12 13 14 16 18 19 
18 :  4 5 6 8 9 10 13 15 17 19 20 
19 :  1 2 3 4 7 9 10 11 12 14 15 16 17 18 20 
20 :  1 2 3 5 6 7 8 11 12 13 14 16 18 19 

每对都可以列出：

for(i in 1:20){
    tempdf = ddf[ddf$group!=ddf[i,2],]
    for(j in 1:nrow(tempdf)){
        cat(ddf[i,1], tempdf[j,1], '\n') 
    }
}

}
1 4 
1 5 
1 6 
1 8 
1 9 
1 10 
1 13 
1 15 
1 17 
1 19 
1 20 
2 4 
2 5 
2 6 
2 8 
2 9 
2 10 
2 13 
2 15 
2 17 
....

可以在另一个data.frame中轻松获得对。

The pairs can easily be obtained in another data.frame.

要创建另一个data.frame：

To create another data.frame:

outdf = data.frame(first=numeric(), second=numeric())

for(i in 1:20){
    tempdf = ddf[ddf$group!=ddf[i,2],]
    for(j in 1:nrow(tempdf)){
        outdf[nrow(outdf)+1,] = c(ddf[i,1], tempdf[j,1])
    }
}
head(outdf)
  first second
1     1      3
2     1      4
3     1      5
4     1      7
5     1      8
6     1      9

要删除重复项，请先对每对进行排序：

To remove duplicates, first sort each pair:

for(i in 1:nrow(outdf)){
    if(outdf[i,2] < outdf[i,1])
        outdf[i,] = c(outdf[i,2], outdf[i,1])
}
outdf

要对每一行进行排序，最好使用以下R代码：

For sorting each row, following R code may be preferred:

outdf = data.frame(t(apply(outdf, 1, sort)))

然后删除重复项：

outdf = outdf[!duplicated(outdf),]

唯一对的数量为：

nrow(outdf)

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