通過變量 id 獲取兩列的所有可能組合

Question

給定“組 ID”值，我需要獲得 2 個值的所有排列

我有這個：

group id    value
   1         a
   1         b
   1         c
   2         b
   2         c
   2         d

並想要這個：

group id    value1   value2
   1         a        b
   1         a        c
   1         b        a
   1         b        c  
   1         c        a
   1         c        b
   2         b        c
   2         b        d
   2         c        b
   2         c        d
   2         d        b
   2         d        c

Answer 1

使用split from data.table和expand.grid可以實現一種可能的解決方案。

步驟是：

  library(data.table)

  setDT(df)

  #list will be generated for each group
  ll <- lapply(split(df, by="group_id"), 
     function(x)cbind(group_id = unique(x$group_id), 
     expand.grid(x$value, x$value, stringsAsFactors = F)))

  #Combine data frames from list and then filter those with 
  # having same value for both columns
  do.call("rbind", ll) %>% filter(Var1 != Var2)

#Result
   group_id Var1 Var2
1         1    b    a
2         1    c    a
3         1    a    b
4         1    c    b
5         1    a    c
6         1    b    c
7         2    c    b
8         2    d    b
9         2    b    c
10        2    d    c
11        2    b    d
12        2    c    d

數據

df <- read.table(text = "group_id    value
1         a
1         b
1         c
2         b
2         c
2         d", header = TRUE, stringsAsFactors = FALSE)

Answer 2

下面是快速和簡單的

library(gtools)
library(data.table)

indices <- c(1,1,1,2,2,2)
variables <- c("a", "b", "c", "b", "c", "d")
dt <- data.table(indices, variables)

get_permutations <- function(df){
    perm <- permutations(nrow(unique(df[,1])), 2, df$variables)
    as.data.table(perm)
}

ds <- dt[, get_permutations(.SD), by = indices]

    indices V1 V2
 1:       1  a  b
 2:       1  a  c
 3:       1  b  a
 4:       1  b  c
 5:       1  c  a
 6:       1  c  b
 7:       2  b  c
 8:       2  b  d
 9:       2  c  b
10:       2  c  d
11:       2  d  b
12:       2  d  c

Answer 3

您正在尋找來自 gtools 的permutations 。

## In general

library(gtools)

char.var <- c('a','b','c')
df = as.data.frame(permutations(n=length(char.var), r=2, v=char.var))
df

   V1 V2
1  a  b
2  a  c
3  b  a
4  b  c
5  c  a
6  c  b

## answer for question

library(data.table)
library(gtools)

df <- data.frame(groupid = c(1,1,1,2,2,2), value = c('a','b','c','b','c','d'))
df$value <- as.character(df$value)

setDT(df)

output <- data.table()

for(i in unique(df$groupid))
{
    temp_df = df[groupid == eval(i)] # this gets group
    temp_df2 <- as.data.table(permutations(length(temp_df$value), r=2, temp_df$value)) # this creates combinations
    temp_df2[, groupid := i]
    colnames(temp_df2)[1:2] <- c('value1','value2')
    output <- rbind(output, temp_df2) # this appends value in output df

}

print(output)

        value1 value2 groupid
 1:      a      b       1
 2:      a      c       1
 3:      b      a       1
 4:      b      c       1
 5:      c      a       1
 6:      c      b       1
 7:      b      c       2
 8:      b      d       2
 9:      c      b       2
10:      c      d       2
11:      d      b       2
12:      d      c       2

Answer 4

幾乎像@João Machado

df <- data.frame(group_id = c(rep(1,3),rep(2,3)), value = c(letters[1:3],letters[2:4]))
df <- split(x= df, f= df$group_id)
df <- lapply(df, function(i)
{
  library(gtools)
  a<- data.frame(gtools::permutations(n = length(as.vector(i[,"value"])), r= 2,v = as.vector(i[,"value"])))
  colnames(a) <- c("value1", "value2")
  a$group_id <- unique(as.vector(i[,"group_id"]))
  a <- a[,c("group_id","value1","value2")]
})
df <- do.call(rbind, df)