Create samples from the output of a combn function in R, with no repeated elements

Question

I am trying to solve this problem with R: out of all possible combinations of letters, I want to randomly select a sample of 13 pairs with the condition that no LETTER is repeated.

I am trying the following:

x<- LETTERS
combi <- combn(x, 2, FUN = NULL, simplify = FALSE) #combines by 2
fulltable <- data.frame(Reduce(rbind, combi)) #Convert list to dataframe of 2 columns

It gives me 323 possible combinations:

...
X.316  V  Y
X.317  V  Z
X.318  W  X
X.319  W  Y
X.320  W  Z
X.321  X  Y
X.322  X  Z
X.323  Y  Z

I want to select a sample -lets call it SET1- of 13 pair of letters (there are 26 letters in total) where no letter is repeated. Then, once these pairs are created, I want to extract another sample with the same condition, but in this case, excluding SET1.

Desired outcome:

X.1  A  E
X.2  C  H
X.3  B  X
X.4  W  Y
X.5  F  K
…..

Until X.13 and no letter repeated either in row or column.

In the second sample extraction,again, a combination of pairs with unique letters, but in this case, previous combinations not allowed (ie AE / CH). It should also exclude permutations, such as EA / H C.

Thanks AJS

EDIT--------------------- This solution works for me:

test <- LETTERS
ctest <- combn(test, 2, FUN = NULL, simplify = FALSE)
ctabl <- data.frame(Reduce(rbind,ctest))
ctabl$row <- 1:nrow(ctabl)

for (i in 1:nrow(ctabl)){
  sname <- ctabl%>% sample_n(13)
  ctabl <- ctabl %>% subset(!row %in% sname$row)
  print(sname)
}

Answer 1

A simpler approach to obtain comparable results, hope it helps...

> x  <- sample(LETTERS, replace = F)
>  ft <- data.frame(x[1:13],x[14:26])
>  ft
   x.1.13. x.14.26.
1        X        D
2        T        Y
3        P        N
4        Z        I
5        M        E
6        K        V
7        B        J
8        R        O
9        H        C
10       S        L
11       A        W
12       G        Q
13       U        F

> # UPDATE based on comment:  
> # That probably moves the post from primarily being R, to being a math problem 
>  # It depends if you want a comprehensive ste of solutions or a couple unique solutions
>  # Couple unique solutions is easier: 
> x1 <- x[1:13]

x1 <- x[1:13]

x2 <- x[14:26];  df2 <- data.frame(x1,x2); df2
x3 <- x[c(15:26,14)]; df3 <- data.frame(x1,x3); df3
x4 <- x[c(16:26,14:15)]; df4 <- data.frame(x1,x4); df4
x5 <- x[c(17:26,14:16)]; df5 <- data.frame(x1,x5); df5
# .... and so on till x14
# Implemented code 
> x1 <- x[1:13]
> 
> x2 <- x[14:26];  df2 <- data.frame(x1,x2); df2
   x1 x2
1   X  D
2   T  Y
3   P  N
4   Z  I
5   M  E
6   K  V
7   B  J
8   R  O
9   H  C
10  S  L
11  A  W
12  G  Q
13  U  F
> x3 <- x[c(15:26,14)]; df3 <- data.frame(x1,x3); df3
   x1 x3
1   X  Y
2   T  N
3   P  I
4   Z  E
5   M  V
6   K  J
7   B  O
8   R  C
9   H  L
10  S  W
11  A  Q
12  G  F
13  U  D
> x4 <- x[c(16:26,14:15)]; df4 <- data.frame(x1,x4); df4
   x1 x4
1   X  N
2   T  I
3   P  E
4   Z  V
5   M  J
6   K  O
7   B  C
8   R  L
9   H  W
10  S  Q
11  A  F
12  G  D
13  U  Y
> x5 <- x[c(17:26,14:16)]; df5 <- data.frame(x1,x5); df5
   x1 x5
1   X  I
2   T  E
3   P  V
4   Z  J
5   M  O
6   K  C
7   B  L
8   R  W
9   H  Q
10  S  F
11  A  D
12  G  Y
13  U  N
> # .... and so on till x14
> 
> # You may need to write a loop /nested loop to get a comprehensive set
> # logic is - find N, combinations of 2/26 letters, then find the combinations of 13/length(N) those 
> # with the condition that no character is repeated in a single vector of any combination in any df.

Answer 2

对于刚刚添加的附加条件 - 您可以在 sample(LETTERS, replace = F) 之前使用 set.seed 来控制序列并确保不同的英尺。