从具有重复元素的向量生成所有唯一组合

Question

之前有人问过这个问题，但仅针对具有非重复元素的向量。我无法找到一个简单的解决方案来从具有重复元素的向量中获取所有组合。为了说明，我在下面列出了一个示例。

x <- c('red', 'blue', 'green', 'red', 'green', 'red')

向量 x 有 3 个重复元素代表“红色”，2 个重复元素代表“绿色”。所有独特组合的预期结果都是这样的。

# unique combinations with one element
'red'
'blue'
'green'
# unique combination with two elements
'red', 'blue' # same as 'blue','red'
'red', 'green' 
'red', 'red'
'blue', 'green'
'green', 'green'
# unique combination with three elements
'red', 'blue', 'green'
'red', 'red', 'blue'
'red', 'red', 'green'
'red', 'red', 'red' # This is valid because there are three 'red's
'green', 'green', 'red'
'green', 'green', 'blue'
# more unique combinations with four, five, and six elements

Answer 1

使用combn() 和lapply() 应该可以解决问题。

x <- c('red', 'blue', 'green', 'red', 'green', 'red')

lapply(1:3, function(y) combn(x, y))

# [[1]]
     # [,1]  [,2]   [,3]    [,4]  [,5]    [,6] 
# [1,] "red" "blue" "green" "red" "green" "red"

# [[2]]
     # [,1]   [,2]    [,3]  [,4]    [,5]  [,6]    ...
# [1,] "red"  "red"   "red" "red"   "red" "blue"  ...
# [2,] "blue" "green" "red" "green" "red" "green" ...

# [[3]]
     # [,1]    [,2]   [,3]    [,4]   [,5]    [,6]    ...
# [1,] "red"   "red"  "red"   "red"  "red"   "red"   ...
# [2,] "blue"  "blue" "blue"  "blue" "green" "green" ...
# [3,] "green" "red"  "green" "red"  "red"   "green" ...

所有独特的组合

lapply(cc, function(y)
  y[,!duplicated(apply(y, 2, paste, collapse="."))]
)

[[1]]
[1] "red"   "blue"  "green"

[[2]]
     [,1]   [,2]    [,3]  [,4]    [,5]   [,6]    [,7]   
[1,] "red"  "red"   "red" "blue"  "blue" "green" "green"
[2,] "blue" "green" "red" "green" "red"  "red"   "green"

[[3]]
     [,1]    [,2]   [,3]    [,4]    [,5]    [,6]  [,7]    ...
[1,] "red"   "red"  "red"   "red"   "red"   "red" "blue"  ...
[2,] "blue"  "blue" "green" "green" "red"   "red" "green" ...
[3,] "green" "red"  "red"   "green" "green" "red" "red"   ...

虽然严格来说这些并不是所有独特的组合，因为它们中的一些是彼此的排列。

适当的独特组合

lapply(cc, function(y)
  y[,!duplicated(apply(y, 2, function(z) paste(sort(z), collapse=".")))]
)

# [[1]]
# [1] "red"   "blue"  "green"

# [[2]]
     # [,1]   [,2]    [,3]  [,4]    [,5]   
# [1,] "red"  "red"   "red" "blue"  "green"
# [2,] "blue" "green" "red" "green" "green"

# [[3]]
     # [,1]    [,2]   [,3]    [,4]    [,5]  [,6]   
# [1,] "red"   "red"  "red"   "red"   "red" "blue" 
# [2,] "blue"  "blue" "green" "green" "red" "green"
# [3,] "green" "red"  "red"   "green" "red" "green"

Answer 2

library(DescTools)
x <- c('red', 'blue', 'green', 'red', 'green', 'red')

allSets <- lapply(1:length(x), function(i){
      unique(t(apply(CombSet(x,i,repl = F),1,sort)))
    })


#[1]]
#[,1]  [,2]   [,3]    [,4]  [,5]    [,6] 
#[1,] "red" "blue" "green" "red" "green" "red"

##[[2]]
#[,1]    [,2]   
#[1,] "blue"  "red"  
#[2,] "green" "red"  
#[3,] "red"   "red"  
#[4,] "blue"  "green"
#[5,] "green" "green"

#[[3]]
#[,1]    [,2]    [,3]   
#[1,] "blue"  "green" "red"  
#[2,] "blue"  "red"   "red"  
#[3,] "green" "red"   "red"  
#[4,] "green" "green" "red"  
#[5,] "red"   "red"   "red"  
#[6,] "blue"  "green" "green"

#[[4]]
#[,1]    [,2]    [,3]    [,4] 
#[1,] "blue"  "green" "red"   "red"
#[2,] "blue"  "green" "green" "red"
#[3,] "blue"  "red"   "red"   "red"
#[4,] "green" "green" "red"   "red"
#[5,] "green" "red"   "red"   "red"

#[[5]]
#[,1]    [,2]    [,3]    [,4]  [,5] 
#[1,] "blue"  "green" "green" "red" "red"
#[2,] "blue"  "green" "red"   "red" "red"
#[3,] "green" "green" "red"   "red" "red"

#[[6]]
#[,1]   [,2]    [,3]    [,4]  [,5]  [,6] 
#[1,] "blue" "green" "green" "red" "red" "red"

Answer 3

library(arrangements)

combinations(c("red", "blue", "green"), k = 2, freq = c(3, 1, 2))
#      [,1]    [,2]   
# [1,] "red"   "red"  
# [2,] "red"   "blue" 
# [3,] "red"   "green"
# [4,] "blue"  "green"
# [5,] "green" "green"

combinations(c("red", "blue", "green"), k = 3, freq = c(3, 1, 2))
#      [,1]   [,2]    [,3]   
# [1,] "red"  "red"   "red"  
# [2,] "red"  "red"   "blue" 
# [3,] "red"  "red"   "green"
# [4,] "red"  "blue"  "green"
# [5,] "red"  "green" "green"
# [6,] "blue" "green" "green"

如果您不想手动输入频率：

x <- c('red', 'blue', 'green', 'red', 'green', 'red')
tx <- table(x)
combinations(names(tx), k = 2, freq = tx)
#       [,1]    [,2]   
# [1,] "blue"  "green"
# [2,] "blue"  "red"  
# [3,] "green" "green"
# [4,] "green" "red"  
# [5,] "red"   "red" 

或者使用RcppAlgos:

library(RcppAlgos)
comboGeneral(names(tx), m=2, freqs = tx)
#       [,1]    [,2]   
# [1,] "blue"  "green"
# [2,] "blue"  "red"  
# [3,] "green" "green"
# [4,] "green" "red"  
# [5,] "red"   "red"