计算 R 中的组合和排列

组合学是数据分析和统计的一个重要方面。它用于解决许多基于能力的和现实生活中的问题。虽然排列确实考虑了顺序，但组合是独立的。因此，排列被认为是有序组合。 R 语言使我们能够调用许多包来计算组合和排列。

方法一：组合包

R 编程语言中的 Combinat 包可用于计算数字的排列和组合。它提供了执行组合学的例程和方法。

属于该包的 R 语言中的 combn()方法用于生成一次取 m 个 x 元素的所有组合。如果 x 是正整数，则返回 seq(x) 一次取 m 个元素的所有组合。

Syntax:

combn(x, m, fun=NULL, simplify=TRUE, …)

Parameters : 

• x – vector source for combinations
• m – number of elements to be taken
• fun – function to be applied to each combination (may be null)
• simplify – logical, if FALSE, returns a list, otherwise returns vector or array

示例 1：

R

# using required libraries
library(combinat)
  
# generating combinations of the 
# alphabets taking 2 at a time
print ("Combination of letters two at a time")
combn(letters[1:4], 2)

输出

[1] "Combination of letters two at a time" 
> combn(letters[1:4], 2)     
       [,1] [,2] [,3] [,4] [,5] [,6] 
[1,]   "a"  "a"  "a"  "b"  "b"  "c" 
[2,]   "b"  "c"  "d"  "c"  "d"  "d" 

示例 2：

电阻

# using required libraries
library(combinat)
  
# generating combinations of the
# alphabets taking 2 at a time
print ("Combination where x=m")
  
# selecting 8 items out of 8 
combn(8,8)

输出

[1] "Combination where x=m" 
> #selecting 8 items out of 8  
> combn(8,8) 
[1] 1 2 3 4 5 6 7 8 

下面的代码说明了 m=1 的方法，即选择的项数等于 1。执行此操作的方法数等于项的总数，因为每个项都可以选择一次。

示例 3：

电阻

# using required libraries
library(combinat)
  
# generating combinations of 
# the alphabets taking 2 at a time
print ("Combination where m=1")
  
# selecting 1 items out of 10 
combn(10,1)

输出

[1] “Combination where m=1”    

      [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] 

[1,]    1    2    3    4    5    6    7    8    9    10

combn() 方法还提供将作为输出生成的组合数量的计数。由于输出是以矩阵的形式得到的，所以最终的结果就是矩阵的列数，由ncol(res)给出，其中res是combn()方法应用的结果。

示例 4：

电阻

# using required libraries
library(combinat)
  
# generating combinations of the
# alphabets taking 2 at a time
print ("Combination of letters two at a time")
res <- combn(letters[1:4], 2)
  
# calculate number of columns in 
# the result 
print ("Number of combinations : ")
print (ncol(res))

输出

[1] "Number of combinations : " 
> print (ncol(res)) 
[1] 6

R 中的 permn()方法生成 x 元素的所有排列。如果 x 是正整数，则返回 seq(x) 元素的所有排列。数字 0 的排列是 1。

Syntax:

permn(x, fun=NULL, …)

Parameters : 

• x – vector source for combinations
• fun – function to be applied to each permutation (may be null)

示例 1：

电阻

# using required libraries
library(combinat)
  
# generating permutations 
# of the numbers 3
print ("Permutations of 3")
permn(3)

输出

[1] "Permutations of 3"
[[1]] [1] 1 2 3  
[[2]] [1] 1 3 2 
[[3]] [1] 3 1 2 
[[4]] [1] 3 2 1  
[[5]] [1] 2 3 1  
[[6]] [1] 2 1 3

我们也可以计算向量或列表的排列。

示例 2：

电阻

# using required libraries
library(combinat)
  
# declaring a list
x <- c('red', 'blue', 'green', 'violet')
print ("Permutations of vector x")
permn(x)

输出：

[1] "Permutations of vector x" 
> permn(x)
[[1]] [1] "red"    "blue"   "green"  "violet" 
[[2]] [1] "red"    "blue"   "violet" "green"   
[[3]] [1] "red"    "violet" "blue"   "green"  
[[4]] [1] "violet" "red"    "blue"   "green"   
[[5]] [1] "violet" "red"    "green"  "blue"    
[[6]] [1] "red"    "violet" "green"  "blue"    
[[7]] [1] "red"    "green"  "violet" "blue"    
[[8]] [1] "red"    "green"  "blue"   "violet"  
[[9]] [1] "green"  "red"    "blue"   "violet"  
[[10]] [1] "green"  "red"    "violet" "blue"    
[[11]] [1] "green"  "violet" "red"    "blue"    
[[12]] [1] "violet" "green"  "red"    "blue"    
[[13]] [1] "violet" "green"  "blue"   "red"     
[[14]] [1] "green"  "violet" "blue"   "red"     
[[15]] [1] "green"  "blue"   "violet" "red"     
[[16]] [1] "green"  "blue"   "red"    "violet"  
[[17]] [1] "blue"   "green"  "red"    "violet"  
[[18]] [1] "blue"   "green"  "violet" "red"     
[[19]] [1] "blue"   "violet" "green"  "red"     
[[20]] [1] "violet" "blue"   "green"  "red"     
[[21]] [1] "violet" "blue"   "red"    "green"   
[[22]] [1] "blue"   "violet" "red"    "green"   
[[23]] [1] "blue"   "red"    "violet" "green"   
[[24]] [1] "blue"   "red"    "green"  "violet" 

与 combn() 方法类似，也可以通过对 permn() 方法生成的输出向量使用 length() 方法来确定排列数。

示例 3：

电阻

# using required libraries
library(combinat)
  
# declaring a list
x <- c('red', 'blue', 'green', 'violet')
print ("Permutations of vector x")
  
res <- permn(x)
print ("Number of possible permutations : ")
print (length(res))

输出

[1] “Number of possible permutations : ” 

> print (length(res)) 

[1] 24

方法二：gtools包

使用 R 编程语言中的 gtools 包也可以轻松计算无重复和有重复的排列和组合。使用 R 中的 gtools 包可以轻松实现组合数学。

R 中的 permutations() 方法可用于轻松计算有替换和无替换的排列。它返回一个数据帧或一个矩阵，该矩阵在对受约束的指定向量的元素进行混洗时形成的可能排列。可以使用 nrow() 方法捕获此类可能排列的数量，该方法返回获得的输出数据帧中的行数。

Syntax:

permutations(n , r, vec, repeats.allowed=F)

Parameters : 

• n – no. of objects to choose from
• r – no. of objects to choose
• vec – the atomic vector or matrix to shuffle
• repeats.allowed – By default : false. If true, the permutations are generated with repetition allowed

Return : 

A data frame or matrix with plausible permutations. The number of rows in the data frame is equivalent to the value of r.

示例 1：

电阻

library(gtools)
# describing the input vector
  
vec<- LETTERS[4:7]
  
# getting permutations on choose 2 
# letters out of 4 letters without 
# replacement
res <- permutations(n=4,r=2,v=vec)
print ("Permutations without replacement")
print (res)
  
print ("Number of permutations without replacement")
print (nrow(res))
  
# calculating permutations with replacement
res1 <- permutations(n=4,r=2,v=vec,repeats.allowed=T)
print ("Permutations with replacement")
print (res1)
  
print ("Number of permutations with replacement")
print (nrow(res1))

输出

[1] "Permutations without replacement"
     [,1] [,2]
[1,] "D"  "E"
[2,] "D"  "F"
[3,] "D"  "G"
[4,] "E"  "D"
[5,] "E"  "F"
[6,] "E"  "G"
[7,] "F"  "D"
[8,] "F"  "E"
[9,] "F"  "G"
[10,] "G"  "D"
[11,] "G"  "E"
[12,] "G"  "F"
[1] "Number of permutations without replacement"
[1] 12
[1] "Permutations with replacement"
     [,1] [,2]
[1,] "D"  "D"
[2,] "D"  "E"
[3,] "D"  "F"
[4,] "D"  "G"
[5,] "E"  "D"
[6,] "E"  "E"
[7,] "E"  "F"
[8,] "E"  "G"
[9,] "F"  "D"
[10,] "F"  "E"
[11,] "F"  "F"
[12,] "F"  "G"
[13,] "G"  "D"
[14,] "G"  "E"
[15,] "G"  "F"
[16,] "G"  "G"
[1] "Number of permutations with replacement"
[1] 16

类似地，组合方法可用于从指定的源向量生成可能的组合。所有参数都类似于 permutations() 方法。

句法：

combinations(n , r, vec, repeats.allowed=F)

指定输入向量不是强制性的。

示例 2：

电阻

library(gtools)
  
# generating combinations of the 
# alphabets taking 2 at a time
print ("Combination of five objects taken two at a time")
combinations(5, 2)

输出

[1] "Combination of five objects taken two at a time"
     [,1] [,2]
[1,]    1    2
[2,]    1    3
[3,]    1    4
[4,]    1    5
[5,]    2    3
[6,]    2    4
[7,]    2    5
[8,]    3    4
[9,]    3    5
[10,]    4    5

示例 3：

电阻

library(gtools)
vec <- c(1:4)
  
# generating combinations of the 
# digits taking 2 at a time
print ("Combination of four objects taken two at\
a time without repetition")
res<- combinations(n= 4, r = 2, v = vec)
print (res)
  
print ("Number of combinations without repetition")
print (nrow(res))
  
print ("Combination of four objects taken two at a \
time with repetition")
res1 <- combinations(n= 4, r = 2, v = vec, repeats.allowed=T)
print (res1)
  
print ("Number of combinations with repetition")
print (nrow(res1))

输出

[1] "Combination of four objects taken two at a time without repetition"
    [,1] [,2]
[1,]    1    2
[2,]    1    3
[3,]    1    4
[4,]    2    3
[5,]    2    4
[6,]    3    4
[1] "Number of combinations without repetition"
[1] 6
[1] "Combination of four objects taken two at a time with repetition"
     [,1] [,2]
[1,]    1    1
[2,]    1    2
[3,]    1    3
[4,]    1    4
[5,]    2    2
[6,]    2    3
[7,]    2    4
[8,]    3    3
[9,]    3    4
[10,]    4    4
[1] "Number of combinations with repetition"
[1] 10