Question

有下表：

read.table(text = "route origin dest seq
    1   a b 1
    1   b c 2
    1   c d 3
    1   d e 4
    2   f g 1
    2   g h 2
    2   h i 3", header = TRUE)

我试图找到一种经过每一行的方式，按路线分组，并迭代原始目的地对的每个潜在组合，同时考虑所提到的seq变量和路线。

输出应该如下所示：

  origin   dest
    a       b 
    a       c 
    a       d 
    a       e 
    b       c
    b       d
  (...)   (...)

这背后的想法是，例如路线1的列车从a到e。但是，我想列出列车对的每一种可能性。我试过igraph但没有成功。有dplyr的任何想法吗？

Answer 1

希望这有帮助！

library(dplyr)
library(tidyr)

df %>%
  mutate_if(is.factor, as.character) %>%    #convert factor variable to character
  group_by(route) %>%
  expand(origin = paste(origin, seq, sep = "_"), dest = paste(dest, seq, sep = "_")) %>%    #all possible combination of origin & destination grouped by route
  rowwise() %>%
  filter(strsplit(origin, split = "_")[[1]][1] != strsplit(dest, split = "_")[[1]][1] & 
           strsplit(origin, split = "_")[[1]][2] <= strsplit(dest, split = "_")[[1]][2]) %>%
  mutate(origin = gsub("_.*$", "", origin),
         dest   = gsub("_.*$", "", dest))

输出为：

   route origin dest 
 1     1 a      b    
 2     1 a      c    
 3     1 a      d    
 4     1 a      e    
 5     1 b      c    
...

示例数据：

df <- structure(list(route = c(1L, 1L, 1L, 1L, 2L, 2L, 2L), origin = structure(1:7, .Label = c("a", 
"b", "c", "d", "f", "g", "h"), class = "factor"), dest = structure(1:7, .Label = c("b", 
"c", "d", "e", "g", "h", "i"), class = "factor"), seq = c(1L, 
2L, 3L, 4L, 1L, 2L, 3L)), class = "data.frame", row.names = c(NA, 
-7L))

#  route origin dest seq
#1     1      a    b   1
#2     1      b    c   2
#3     1      c    d   3
#4     1      d    e   4
#5     2      f    g   1
#6     2      g    h   2
#7     2      h    i   3

迭代分组行以获得不同的对组合

1 个答案: