All combinations of letters/numbers under specific conditions

Question

I created these vectors:

Letters <- c("A","C","E","G","H","J","K")  
Numbers <- c(0,1,2,3,4,6,7,9) 
AlphaNumeric <- c(Letters, Numbers)

I would like to receive a dataframe of all 3-element combinations (e.g. AA1, G26 etc.) using all elements mentioned above following three conditions:

1.) The first element is a letter

2.) The second element is a number or the SAME letter as the first element

3.) The third element is a number

Approach: I have tried to use expand.grid() and successfully managed to get ALL combinations with 3 elements. Then I tried expand.grid(x = Letters, y = AlphaNumeric, z = Numbers) and managed to achieve 1.) and 3.) but failed to manage 2.) so far.

Unsatisfying Solution: I have figured out a way of doing this with a for-loop, but I guess there must be a way easier way of doing it other than:

   LNN <- expand.grid(x = Letters, y = Numbers, z = Numbers)

   for ( Element in Letters) {
       currentLLN <- expand.grid(x = Element, y = Element, z = Numbers)
       LNN <- merge(LNN, currentLLN, all = TRUE)}

Any help would be greatly appreciated, thank you, Christian

Answer 1

You could create two dataframes, one where the second element is a number, and one where the second element is the same as the first element, and then rbind those. An example is given below, note that I have limited your example data for illustration purposes.

Letters <- LETTERS[1:3]  
Numbers <- c(1,2)

df1 = expand.grid(v1=Letters,v3=Numbers,stringsAsFactors = F)
df1$v2 = df1$v1
df1 = df1[,c('v1','v2','v3')]
df2 = expand.grid(v1=Letters,v2=as.character(Numbers),v3=Numbers, stringsAsFactors = F)
df = rbind(df1,df2)

Output:

> df
   v1 v2 v3
1   A  A  1
2   B  B  1
3   C  C  1
4   A  A  2
5   B  B  2
6   C  C  2
7   A  1  1
8   B  1  1
9   C  1  1
10  A  2  1
11  B  2  1
12  C  2  1
13  A  1  2
14  B  1  2
15  C  1  2
16  A  2  2
17  B  2  2
18  C  2  2

Hope this helps!

---

Although both answers run very fast and Parfait's solution is a nice solution to your problem and I certainly do not want to discredit his answer, I think it is good to point out that creating extra combinations and subsetting will become a larger issue when you data is larger. A benchmark is shown below.

Letters <- c(LETTERS[1:26],letters[1:4])
Numbers <- seq(30)
AlphaNumeric <- c(Letters, Numbers)


f_flo <- function()
{
  df1 = expand.grid(v1=Letters,v3=Numbers,stringsAsFactors = F)
  df1$v2 = df1$v1
  df1 = df1[,c('v1','v2','v3')]
  df2 = expand.grid(v1=Letters,v2=as.character(Numbers),v3=Numbers, stringsAsFactors = F)
  df = rbind(df1,df2)
}

f_parfait <- function()
{
  df <- expand.grid(x = Letters, y = AlphaNumeric, z = Numbers, stringsAsFactors = FALSE)
  sub <- subset(df,  (x == y | grepl("[0-9]", y)) &  grepl("[0-9]", z) )
  sub <- with(sub, sub[order(x, y, z),])   # SORT DATAFRAME
  rownames(sub) <- NULL                    # RESET ROWNAMES
}

library(dplyr)
one_letter <- function(l) {
  expand.grid(l, c(l, Numbers), Numbers, stringsAsFactors = FALSE)
}

f_stibu <- function(){
  df <- bind_rows(lapply(Letters, one_letter))
}


library(microbenchmark)
library(ggplot2)

run_times = microbenchmark(f_flo(),f_parfait(),f_stibu())
autoplot(run_times)

Results:

Unit: milliseconds
        expr        min         lq       mean     median         uq       max neval cld
     f_flo()   1.900719   2.047591   3.666935   2.314258   3.922053  78.74793   100  a 
 f_parfait() 138.028364 142.529904 152.876116 144.159444 146.835958 246.92318   100   b
   f_stibu()   4.130464   4.333130   5.169664   4.585028   6.209233  10.23139   100  a 

Answer 2

Simply subset your expand.grid() dataframe with grepl calls:

df <- expand.grid(x = Letters, y = AlphaNumeric, z = Numbers, stringsAsFactors = FALSE)

sub <- subset(df,  (x == y | grepl("[0-9]", y)) )

sub <- with(sub, sub[order(x, y, z),])   # SORT DATAFRAME
rownames(sub) <- NULL                    # RESET ROWNAMES

head(sub, 10)    
#    x y z
# 1  A 0 0
# 2  A 0 1
# 3  A 0 2
# 4  A 0 3
# 5  A 0 4
# 6  A 0 6
# 7  A 0 7
# 8  A 0 9
# 9  A 1 0