删除重复的 2 列排列 [英] Remove duplicated 2 columns permutations

问题描述

我找不到这个问题的好标题，所以请随时编辑它.

I can't find a good title for this question so feel free to edit it please.

我有这个data.frame

I have this data.frame

  section time to from
1       a    9  1    2
2       a    9  2    1
3       a   12  2    3
4       a   12  2    4
5       a   12  3    2
6       a   12  3    4
7       a   12  4    2
8       a   12  4    3

我想同时删除具有相同 to 和 from 的重复行，而不计算 2 列的排列:例如 (1,2) 和 (2,1) 重复.

I want to remove duplicated rows that have the same to and from simultaneously, without computing permutations of the 2 columns: e.g (1,2) and (2,1) are duplicated.

所以最终输出是:

  section time to from
1       a    9  1    2
3       a   12  2    3
4       a   12  2    4
6       a   12  3    4

我有一个解决方案，方法是构造一个新的列键，例如

I have a solution by constructing a new column key e.g

  key <- paste(min(to,from),max(to,from))

并使用 duplicated 删除重复的密钥，但我认为这是肮脏的解决方案.

and remove duplicated key using duplicated, but I think this is dirty solution.

这里是我的数据输入

structure(list(section = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 
1L, 1L), .Label = "a", class = "factor"), time = c(9L, 9L, 12L, 
12L, 12L, 12L, 12L, 12L), to = c(1L, 2L, 2L, 2L, 3L, 3L, 4L, 
4L), from = c(2L, 1L, 3L, 4L, 2L, 4L, 2L, 3L)), .Names = c("section", 
"time", "to", "from"), row.names = c(NA, -8L), class = "data.frame")

推荐答案

mn <- pmin(s$to, s$from)
mx <- pmax(s$to, s$from)
int <- as.numeric(interaction(mn, mx))
s[match(unique(int), int),]
  section time to from
1       a    9  1    2
3       a   12  2    3
4       a   12  2    4
6       a   12  3    4

这个想法的功劳归于这个问题:从数据帧中删除连续重复项和特别是@MatthewPlourde 的回答.

Credit for the idea goes to this question: Remove consecutive duplicates from dataframe and specifically @MatthewPlourde's answer.

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