成对的独特组合忽略方向 [英] Pairwise unique combinations ignoring the direction

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本文介绍了成对的独特组合忽略方向的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

我想根据R中的一个共同的字符在一个数据框架中聚合某个值。
问题是我对成对组合的不同方向不感兴趣。
所以例如

  d = data.frame(x = LETTERS [1:5],y = LETTERS [ 5:1])$ ​​b 
$ bxy
1 AE
2 BD
3 CC
4 DB
5 EA

然后按以下计算结果:

  d $ z<  -  paste0(d $ x,d $ y,sep =_)

问题是我对成对差异不感兴趣。所以在这个简单例子中, A_E 应该与 E_A 相同。



是否有一个简单的解决方案来粘贴它们?我正在考虑在将它们组合成一个向量之前对它们进行排序。

解决方案

一个选项是使用 pmin pmax

  (d,z = paste(pmin(x,y),pmax(x,y),sep =_))
#xyz
#1 AE A_E
#2 BD B_D
#3 CC C_C
#4 DB B_D
#5 EA A_E

请注意,如果它们是因子,则可能需要将x和y转换为字符






  d<  -  data.frame(x = LETTERS [1:5],y = LETTERS [5:1],stringsAsFactors = FALSE)


I want to aggregate a certain value in a data.frame based on a common character in R. The Problem is that I am not interested in different directions of the pairwise combination. So for instance

d = data.frame( x = LETTERS[1:5], y = LETTERS[5:1] )

  x y
1 A E
2 B D
3 C C
4 D B
5 E A

The combination would be then calculated like this:

d$z <- paste0(d$x,d$y,sep="_")

The problem is that i am not interested in pairwise differences. So A_E should be the same as E_A in this simple example.

Is there a clever short solution to paste them? I am currently thinking about sorting each one before combining them into a vector.

解决方案

One option is to use pmin and pmax:

transform(d, z = paste(pmin(x,y), pmax(x,y), sep="_"))
#  x y   z
#1 A E A_E
#2 B D B_D
#3 C C C_C
#4 D B B_D
#5 E A A_E

Note that you might need to convert x and y to character if they are factors.


d <- data.frame( x = LETTERS[1:5], y = LETTERS[5:1], stringsAsFactors = FALSE)

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