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 在 'red' 中有 3 个重复元素，在 'green' 中有 2 个重复元素。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" ...

所有唯一组合

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"   ...

尽管严格来说，这些组合并不都是唯一的，因为其中一些组合是彼此的排列组合。

正确独特的组合

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"

从具有重复元素的向量生成所有唯一组合

Generate all unique combinations from a vector with repeating elements

combinations

r

permutation

combn