从具有重复元素的向量生成所有唯一组合

Question

这个问题之前已经被问过，但是只涉及到没有重复元素的向量。我没有找到一个简单的解决方案来获取包含有重复元素的向量的所有组合。为了举例说明，我在下面列出了一个例子。

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

向量x有3个'red'重复元素和2个'green'重复元素。所有独特组合的预期结果如下所示。

# 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

- Jian

Answer 1

使用 lapply()与combn()函数可以解决这个问题。

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"   ...

尽管严格来说，它们并不是所有独特的组合，因为其中一些是彼此的排列组合。

正确的唯一组合。

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"

- AkselA

---

或者 lapply(1:3, function(k) matrix(x[combn(i, k)], nrow=k))。OP 还想要 1 和 2 的组合。然后应用 unique。 - Rui Barradas

@RuiBarradas：啊，我明白了。看起来我不需要使用索引技巧。 - AkselA

这个解决方案将生成所有可能的组合，但不是所有唯一的组合。 - Jian

@Jian：我想我现在明白了。 - AkselA

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"

- LocoGris

---

OP在问题末尾说：“使用四、五和六个元素进行更多独特的组合。”我认为他想要所有六个元素。我是否误解了问题？ - LocoGris

其实我认为你是对的。我的错。没有看到最后一句话。 :) - Sotos

谢谢，这个解决方案除了一个组件[[1]]的情况之外都有效。 - Jian

我可以将其保存为数据框或将每个连接成一列吗？ - Bruce Wayne

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" 

- Stéphane Laurent

---

谢谢，如果在comboGeneral中迭代m 1:6，这个方法是有效的。感谢您的回答。 - Jian