返回所有可能的值组合

I have the following table:

我有以下表格:

Group Value
----  ----
1     A
1     B
1     C
1     D
2     A
2     B
2     C

For each of the two groups, I want to return all possible combinations of values. For group 1, e.g., the possible combinations of are (A,B), (A,C), (A,D), (B,C), (B,D), (C,D), (A,B,C), (B,D,C), (D,C,A), (C,A,B). Analogous, for group 2 it is (A,B), (A,C), (B,C) [Remark: I don't want to consider (1) the combintions with just one value, (2) the combination with all values and (3) the combination with no values. Thus I have 2^(n)-n-1-1 combinations for n different values].

对于这两组中的每一组，我希望返回所有可能的值组合。组1,例如,可能的组合(A,B)(A、C)(A,D)、(B,C)、(B,D),(C,D)(A,B,C)、(B、D、C)(D C),(C,A,B)。类似的，对于第2组，它是(A,B)， (A,C)， (B,C)[注:我不想考虑(1)只有一个值的组合，(2)与所有值的组合，(3)没有值的组合。因此我有2 ^ n(n)-n-1-1组合不同的值)。

I want to list all those combinations with the help of an additional column "Combi". This column numbers the different combinations consecutively.

我想在另外一个专栏“Combi”的帮助下列出所有这些组合。这个列连续地编号不同的组合。

Group Combi Value
----  ----  ----
1     1     A
1     1     B
1     2     A
1     2     C
1     3     A
1     3     D
1     4     B
1     4     C
1     5     B
1     5     D
1     6     C
1     6     C
1     7     A
1     7     B
1     7     C
1     8     B
1     8     C
1     8     D
1     9     C
1     9     D
1     9     A
1     10    D
1     10    A
1     10    B
2     11    A
2     11    B
2     12    A
2     12    C
2     13    B
2     13    C

How do I do this in R?

在R中怎么做呢?

3 个解决方案

#1

Here is a general tidyverse solution, should work with sets of values that have more than 3 items.

这里有一个通用的tidyverse解决方案，应该使用具有3个以上项的值集。

The idea is to use combn (with m = 2 then 3 etc) and format the output as a tibblefor different Group and m values. From there we can use tidyverse functions map_dfr and unnest. Finally as we have multiple ids rather than one, we build a table of unique ids, build the unique combi id and join it back to our result.

其思想是使用combn (m = 2，然后是3)并将输出格式化为针对不同组和m值的tibblefor。从那里我们可以使用tidyverse函数map_dfr和unnest。最后，当我们拥有多个id而不是一个id时，我们构建一个惟一id表，构建惟一的combi id并将其连接到结果。

# convenience fonction to store combinations in a long format
combi_as_tibble <- function(n,values) combn(values,n) %>%
  {tibble(id = rep(seq(ncol(.)),each=nrow(.)),Value=c(.))}    
combi_as_tibble(2,letters[1:3]) # example
# # A tibble: 6 x 2
#      id Value
#   <chr> <chr>
# 1     1     a
# 2     1     b
# 3     2     a
# 4     2     c
# 5     3     b
# 6     3     c


df1 %>% group_by(Group) %>%
  summarize(combis = list(
    map_dfr(2:(length(unique(Value))-1),combi_as_tibble,Value,.id="id2")
  ))     %>% # by Group, build a long tibble with all combinations
  unnest %>% # unnest to get a long unnested table
  left_join(.,select(.,Group,id2,id) %>% distinct %>% mutate(combi=row_number())
  )      %>% # build combi ids
  select(Group,Value,combi) %>%
  as.data.frame

#    Group Value combi
# 1      1     A     1
# 2      1     B     1
# 3      1     A     2
# 4      1     C     2
# 5      1     A     3
# 6      1     D     3
# 7      1     B     4
# 8      1     C     4
# 9      1     B     5
# 10     1     D     5
# 11     1     C     6
# 12     1     D     6
# 13     1     A     7
# 14     1     B     7
# 15     1     C     7
# 16     1     A     8
# 17     1     B     8
# 18     1     D     8
# 19     1     A     9
# 20     1     C     9
# 21     1     D     9
# 22     1     B    10
# 23     1     C    10
# 24     1     D    10
# 25     2     A    11
# 26     2     B    11
# 27     2     A    12
# 28     2     C    12
# 29     2     B    13
# 30     2     C    13

data

数据

df1 <- read.table(text="Group Value
1     A
1     B
1     C
1     D
2     A
2     B
2     C",h=T,strin=F)

#2

A possible solution with data.table:

有数据的可能解决方案。

library(data.table)
setDT(dat)[, .(Value = {n <- 2:(uniqueN(Value)-1);
                        unlist(lapply(n, function(x) combn(Value, x)))})
           , by = Group
           ][, Combi := cumsum(c(1, diff(match(Value, LETTERS)) < 0))][]

which gives:

这使:

    Group Value Combi
 1:     1     A     1
 2:     1     B     1
 3:     1     A     2
 4:     1     C     2
 5:     1     A     3
 6:     1     D     3
 7:     1     B     4
 8:     1     C     4
 9:     1     B     5
10:     1     D     5
11:     1     C     6
12:     1     D     6
13:     1     A     7
14:     1     B     7
15:     1     C     7
16:     1     A     8
17:     1     B     8
18:     1     D     8
19:     1     A     9
20:     1     C     9
21:     1     D     9
22:     1     B    10
23:     1     C    10
24:     1     D    10
25:     2     A    11
26:     2     B    11
27:     2     A    12
28:     2     C    12
29:     2     B    13
30:     2     C    13

#3

Here is a base solution. Comments inline.

这是一个基本解。内联的评论。

#for each length (not incl. 1 and number of unique values), create all possible combinations using combn
combiLs <- by(dat, dat$Group, function(x) {
    #number of elements to choose excl. 1 and all values
    idx <- seq_along(x$Value)[-c(1, nrow(x))]

    do.call(rbind, lapply(idx, function(m) {
        #for each number of elements, generate all combinations
        sets <- combn(x$Value, m, simplify=FALSE)

        #get OP's desired format
        combi <- rep(seq_along(sets), each=m)
        data.frame(
            Group=x$Group[1],
            Combi=paste(x$Group[1], combi, sep="."),
            Value=unlist(sets))
    }))
})

#final output
do.call(rbind, combiLs)

Another possible data.table implementation using both choose and combn simultaneously:

另一个可能的数据。表实现同时使用choose和combn:

res <- setDT(dat)[, {
        idx <- seq_along(Value)[-c(1, .N)]
        list(
            Set=paste0(Group[1], ".", unlist(lapply(idx, function(m) rep(seq_len(choose(.N, m)), each=m)))),
            Value=unlist(lapply(idx, function(m) as.vector(combn(Value, m))))
        )
    }, by=Group]

res[, Combi := unlist(Map(rep, x=seq_along(rle(Set)$values), each=rle(Set)$lengths))]

data:

数据:

dat <- data.frame(Group=c(rep(1,4), rep(2,3)), Value=c("A","B","C","D","A","B","C"))
dat

#1