实际上你不需要“排序”，只需要比较每一对。你需要做两次，翻转条件从一对中得到另一个：

library(tidyverse)

d %>% 
  mutate(
    CityCodeBidirectionalFrom = if_else(CityFrom < CityTo, CityFrom, CityTo),
    CityCodeBidirectionalTo = if_else(CityFrom > CityTo, CityFrom, CityTo))

(as你没有提供任何测试数据，我的代码是一个未经测试的伪代码，而不是一个复制粘贴解决方案…）

另一个选项是使用pmin()和pmax()。

cbind(df, sapply(c(CityCodeBidirectionalFrom = pmin,
                   CityCodeBidirectionalTo = pmax), do.call, df[1:2]))

    CityFrom     CityTo Count CityCodeBidirectionalFrom CityCodeBidirectionalTo
1      Paris       Lyon     2                      Lyon                   Paris
2       Lyon      Paris     4                      Lyon                   Paris
3     London Manchester     5                    London              Manchester
4 Manchester     London     6                    London              Manchester

对于tidyverse解决方案，我们可以将它们放入一个列表中，对其进行排序，然后将它们解嵌套。

library(dplyr)
library(tidyr)
library(purrr)

df1 %>% 
  mutate(CityCodeBidirectional = map2(CityFrom, CityTo, ~sort(c(.x , .y)))) %>% 
  unnest_wider(data = ., col = CityCodeBidirectional, 
                         names_sep = c("From", "To"), names_repair = "universal")

#> # A tibble: 4 x 5
#>   CityFrom   CityTo     Count CityCodeBidirectionalF~ CityCodeBidirectiona~
#>   <chr>      <chr>      <int> <chr>                   <chr>                
#> 1 Paris      Lyon           2 Lyon                    Paris                
#> 2 Lyon       Paris          4 Lyon                    Paris                
#> 3 London     Manchester     5 London                  Manchester           
#> 4 Manchester London         6 London                  Manchester

数据：

df1 <- read.table(text="CityFrom    CityTo  Count
                        Paris   Lyon    2
                        Lyon    Paris   4
                        London  Manchester  5
                        Manchester London       6",
                  header = T, stringsAsFactors = F)

您可以对值进行排序，而不需要连接，如下所示：

df = data.frame(
    CityFrom = c("Paris", "Lyon", "London", "Manchester"),
    CityTo = c("Lyon", "Paris", "Manchester", "London"),
    Count = c(2, 4, 5, 6),
    stringsAsFactors = FALSE
)

mysort = apply(df[,1:2], 1, sort)
rownames(mysort) = c("CityCodeBidirectionalFrom", "CityCodeBidirectionalTo")
cbind(df, t(mysort), stringsAsFactors=FALSE)
##     CityFrom     CityTo Count CityCodeBidirectionalFrom CityCodeBidirectionalTo
## 1      Paris       Lyon     2                      Lyon                   Paris
## 2       Lyon      Paris     4                      Lyon                   Paris
## 3     London Manchester     5                    London              Manchester
## 4 Manchester     London     6                    London              Manchester