查找具有相同列值的所有序列

Question

我有以下数据框:

??????????????????
? Code ? Airline ?
??????????????????
?    1 ? AF      ?
?    1 ? KL      ?
?    8 ? AR      ?
?    8 ? AZ      ?
?    8 ? DL      ?
??????????????????

dat <- structure(list(Code = c(1L, 1L, 8L, 8L, 8L), Airline = structure(c(1L, 
5L, 2L, 3L, 4L), .Label = c("AF  ", "AR  ", "AZ  ", "DL", "KL  "
), class = "factor")), .Names = c("Code", "Airline"), class = "data.frame", row.names = c(NA, 
-5L))

我的目标是让每家航空公司查找所有共享代码,即一个或多个其他航空公司使用的代码.所以输出就是

+--------------------+
| Airline SharedWith |
+--------------------+
| AF      "KL"       |
| KL      "AF"       |
| AR      "AZ","DL"  |
+--------------------+

伪代码是任何命令式语言

for each code
  lookup all rows in the table where the value = code

由于R不是那么多面向列表的,那么实现预期输出的最佳方法是什么？

Answer 1

使用该data.table包的几个选项:

1)使用strsplit,paste与由行操作:

library(data.table)
setDT(dat)[, Airline := trimws(Airline)  # this step is needed to remove the leading and trailing whitespaces
           ][, sharedwith := paste(Airline, collapse = ','), Code
            ][, sharedwith := paste(unlist(strsplit(sharedwith,','))[!unlist(strsplit(sharedwith,',')) %in% Airline], 
                                    collapse = ','), 1:nrow(dat)]

这使:

> dat
   Code Airline sharedwith
1:    1      AF         KL
2:    1      KL         AF
3:    8      AR      AZ,DL
4:    8      AZ      AR,DL
5:    8      DL      AR,AZ

2)使用strsplit与paste以mapply代替by = 1:nrow(dat):

setDT(dat)[, Airline := trimws(Airline)
           ][, sharedwith := paste(Airline, collapse = ','), Code
             ][, sharedwith := mapply(function(s,a) paste(unlist(strsplit(s,','))[!unlist(strsplit(s,',')) %in% a], 
                                                          collapse = ','),
                                      sharedwith, Airline)][]

这将给你相同的结果.

3)或者使用CJ函数paste(受expand.grid@ zx8754解决方案的启发):

library(data.table)
setDT(dat)[, Airline := trimws(Airline)
           ][, CJ(air=Airline, Airline,  unique=TRUE)[air!=V2][, .(shared=paste(V2,collapse=',')), air],
             Code]

这使:

   Code air shared
1:    1  AF     KL
2:    1  KL     AF
3:    8  AR  AZ,DL
4:    8  AZ  AR,DL
5:    8  DL  AR,AZ

使用dplyr&tidyr获得所需解决方案的解决方案(受@jaimedash启发):

library(dplyr)
library(tidyr)

dat <- dat %>% mutate(Airline = trimws(as.character(Airline)))

dat %>%
  mutate(SharedWith = Airline) %>% 
  group_by(Code) %>%
  nest(-Code, -Airline, .key = SharedWith) %>%
  left_join(dat, ., by = 'Code') %>%
  unnest() %>%
  filter(Airline != SharedWith) %>%
  group_by(Code, Airline) %>%
  summarise(SharedWith = toString(SharedWith))

这使:

   Code Airline SharedWith
  (int)   (chr)      (chr)
1     1      AF         KL
2     1      KL         AF
3     8      AR     AZ, DL
4     8      AZ     AR, DL
5     8      DL     AR, AZ

Answer 2

一种igraph方法

library(igraph)

g <- graph_from_data_frame(dat)

# Find neighbours for select nodes
ne <- setNames(ego(g,2, nodes=as.character(dat$Airline), mindist=2), dat$Airline)
ne
#$`AF  `
#+ 1/7 vertex, named:
#[1] KL  

#$`KL  `
#+ 1/7 vertex, named:
#[1] AF  
---
---

# Get final format
data.frame(Airline=names(ne), 
           Shared=sapply(ne, function(x)
                                      paste(V(g)$name[x], collapse=",")))
#   Airline Shared
# 1      AF     KL
# 2      KL     AF
# 3      AR  AZ,DL
# 4      AZ  AR,DL
# 5      DL  AR,AZ

Answer 3

我想你需要的只是一个 table

dat <- structure(list(Code = c(1L, 1L, 8L, 8L, 8L),Airline = structure(c(1L, 5L, 2L, 3L, 4L),.Label = c("AF", "AR", "AZ", "DL", "KL"),class = "factor")),.Names = c("Code", "Airline"),class = "data.frame", row.names = c(NA, -5L))

tbl <- crossprod(table(dat))
diag(tbl) <- 0

#        Airline
# Airline AF AR AZ DL KL
#      AF  0  0  0  0  1
#      AR  0  0  1  1  0
#      AZ  0  1  0  1  0
#      DL  0  1  1  0  0
#      KL  1  0  0  0  0

dd <- data.frame(Airline = colnames(tbl),
                 shared = apply(tbl, 1, function(x)
                   paste(names(x)[x > 0], collapse = ', ')))

merge(dat, dd)
#   Airline Code shared
# 1      AF    1     KL
# 2      AR    8 AZ, DL
# 3      AZ    8 AR, DL
# 4      DL    8 AR, AZ
# 5      KL    1     AF

Answer 4

可能会有更有效的路线,但这应该会飞:

# example data
d <- data.frame(code = c(1,1,8,8,8),
     airline = c("AF","KL","AR","AZ","DL"),
     stringsAsFactors = FALSE)

# merge d to itself on the code column.  This isn't necessarily efficient
d2 <- merge(d, d, by = "code")

# prune d2 to remove occasions where
# airline.x and airline.y (from the merge) are equal
d2 <- d2[d2[["airline.x"]] != d2[["airline.y"]], ]
# construct the combinations for each airline using a split, apply, combine
# then, use stack to get a nice structure for merging
d2 <- stack(
      lapply(split(d2, d2[["airline.x"]]),
        function(ii) paste0(ii$airline.y, collapse = ",")))

# merge d and d2.  "ind" is a column produced by stack
merge(d, d2, by.x = "airline", by.y = "ind")
#  airline code values
#1      AF    1     KL
#2      AR    8  AZ,DL
#3      AZ    8  AR,DL
#4      DL    8  AR,AZ
#5      KL    1     AF

Answer 5

使用expand.grid和aggregate:

do.call(rbind,
        lapply(split(dat, dat$Code), function(i){
          x <- expand.grid(i$Airline, i$Airline)
          x <- x[ x$Var1 != x$Var2, ]
          x <- aggregate(x$Var2, list(x$Var1), paste, collapse = ",")
          colnames(x) <- c("Airline", "SharedWith")
          cbind(Code = i$Code, x)
        }))

# output
#     Code Airline SharedWith
# 1.1    1      AF         KL
# 1.2    1      KL         AF
# 8.1    8      AR      AZ,DL
# 8.2    8      AZ      AR,DL
# 8.3    8      DL      AR,AZ

Answer 6

split有帮助。这是一个完全可复制的编辑，无需任何附加包即可工作。与 OP data.frame 一起使用 - 在 OP 添加可重现的数据集后更改了它。

# strip white space in Airline names:
dat$Airline <- gsub(" ","",dat$Airline)
li <- split(dat,factor(dat$Code))
do.call("rbind",lapply(li,function(x) 
data.frame(Airline = x[1,2],
         SharedWith = paste(x$Airline[-1]
                            ,collapse=",")
))
)