Permutation with two vectors

[yos83]

Hi all,

I am trying to get the all possible permutations from two categories. The raw data is:

3 categories as labels: Penalty No-penalty Money Fine
2 categories Yes or No that could be assigned to any of the categories. For example:

Penalty No-Penalty Money fine
Y Y N
N N Y

I am using this code:

s <- c('penalty','no penalty','money fine')
n <- c('Y','N')
nrow(permutations(n=3,r=3,v=s,repeats.allowed=F ))
Number of permutations should be: 6

To get the combinations, I use this code:
f<- apply(expand.grid(s, n), 1, paste, collapse=".")
result <- t(combn(f,4))
result

The result I get is no correct: It is giving me a mix of all categories with Y and No.

The final table should look like this:
Penalty No-Penalty Money fine
Y N Y
Y Y N
Y N N
N Y Y
N Y N
N N Y

Could you please advise on the code to create this summary table?

[mishabalyasin]

Hi, can you please format your code as code? Also, it'll be helpful if you can make your question into a reprex:

FAQ: What's a reproducible example (`reprex`) and how do I create one?

Why reprex? Getting unstuck is hard. Your first step here is usually to create a reprex, or reproducible example. The goal of a reprex is to package your code, and information about your problem so that others can run it and feel your pain. Then, hopefully, folks can more easily provide a solution. What's in a Reproducible Example? Parts of a reproducible example: background information - Describe what you are trying to do. What have you already done? complete set up - include any library() calls and data to reproduce your issue. data for a reprex: Here's a discussion on setting up data for a reprex make it run - include the minimal code required to reproduce your error on the data…

Right now it is difficult for me to see what you want exactly since as I read it, you have 4 groups with money, not 3. But in general, you can take a look at purrr::cross:

library(magrittr)

purrr::cross2(letters[1:3], LETTERS[1:3]) %>%
  purrr::map_chr(purrr::lift(paste, sep = "."))
#> [1] "a.A" "b.A" "c.A" "a.B" "b.B" "c.B" "a.C" "b.C" "c.C"

^{Created on 2018-11-19 by the reprex package (v0.2.1)}

[cderv]

Are you seeking for all the combination of Y and N ?
I have this:

s <- c('penalty','no penalty','money fine')
n <- c('Y','N')

setNames(expand.grid(n, n, n), s)
#>   penalty no penalty money fine
#> 1       Y          Y          Y
#> 2       N          Y          Y
#> 3       Y          N          Y
#> 4       N          N          Y
#> 5       Y          Y          N
#> 6       N          Y          N
#> 7       Y          N          N
#> 8       N          N          N

^{Created on 2018-11-19 by the reprex package (v0.2.1)}

It is more a guess as we need a reprex as @mishabalyasin explained. Or at least a more precise output wish. Thank !

[yos83]

Hi Misha,

Thank you for your reply.

Apologies for my poorly presented code. I am fairly new using R.

The idea is to seek for all Y and N combinations for penalty, no penalty and money fine.

In the below post from Chirstophe, the following code provides the structure that I require.

s <- c('penalty','no penalty','money fine')
n <- c('Y','N')

setNames(expand.grid(n, n, n), s)

#>   penalty no penalty money fine

#> 1       Y          Y          Y
#> 2       N          Y          Y
#> 3       Y          N          Y
#> 4       N          N          Y
#> 5       Y          Y          N
#> 6       N          Y          N
#> 7       Y          N          N
#> 8       N          N          N

However, when I add a new category in vector s, I get only 16 combinations instead of 26. I wonder what should I add in this code for making work?

s <- c('penalty','no penalty','money fine', 'Community_service')
n <- c('Y','N')

setNames(expand.grid(n, n, n,n), s)

[yos83]

Hi Christophe,

Thank you so much for providing this code.

It works perfectly for 3 categories in vector s. However, I tried to replicate the same with 4 categories and I am only getting 16 combinations out of 26.

s <- c('penalty','no penalty','money fine', 'Community_service')

n <- c('Y','N')

setNames(expand.grid(n, n, n,n), s)

   penalty no penalty money fine Community_service

1        Y          Y          Y                 Y
2        N          Y          Y                 Y
3        Y          N          Y                 Y
4        N          N          Y                 Y
5        Y          Y          N                 Y
6        N          Y          N                 Y
7        Y          N          N                 Y
8        N          N          N                 Y
9        Y          Y          Y                 N
10       N          Y          Y                 N
11       Y          N          Y                 N
12       N          N          Y                 N
13       Y          Y          N                 N
14       N          Y          N                 N
15       Y          N          N                 N
16       N          N          N                 N


nrow(permutations(n=4,r=4,v=s,repeats.allowed=F))
24

Should I an extra command to your existing code for the rest of the combinations to show up?.

Regards,
Yos

[cderv]

If you are looking for a table with all the permutations possible for yes and no repeated 4 times as you want 4 columns at the end, I believe there are no missing data

s <- c('penalty','no penalty','money fine', 'Community_service')
n <- c('Y','N')

gtools::permutations(n=2,r=4,v=n,repeats.allowed=TRUE)
#>       [,1] [,2] [,3] [,4]
#>  [1,] "N"  "N"  "N"  "N" 
#>  [2,] "N"  "N"  "N"  "Y" 
#>  [3,] "N"  "N"  "Y"  "N" 
#>  [4,] "N"  "N"  "Y"  "Y" 
#>  [5,] "N"  "Y"  "N"  "N" 
#>  [6,] "N"  "Y"  "N"  "Y" 
#>  [7,] "N"  "Y"  "Y"  "N" 
#>  [8,] "N"  "Y"  "Y"  "Y" 
#>  [9,] "Y"  "N"  "N"  "N" 
#> [10,] "Y"  "N"  "N"  "Y" 
#> [11,] "Y"  "N"  "Y"  "N" 
#> [12,] "Y"  "N"  "Y"  "Y" 
#> [13,] "Y"  "Y"  "N"  "N" 
#> [14,] "Y"  "Y"  "N"  "Y" 
#> [15,] "Y"  "Y"  "Y"  "N" 
#> [16,] "Y"  "Y"  "Y"  "Y"

setNames(expand.grid(n, n, n, n), s)
#>    penalty no penalty money fine Community_service
#> 1        Y          Y          Y                 Y
#> 2        N          Y          Y                 Y
#> 3        Y          N          Y                 Y
#> 4        N          N          Y                 Y
#> 5        Y          Y          N                 Y
#> 6        N          Y          N                 Y
#> 7        Y          N          N                 Y
#> 8        N          N          N                 Y
#> 9        Y          Y          Y                 N
#> 10       N          Y          Y                 N
#> 11       Y          N          Y                 N
#> 12       N          N          Y                 N
#> 13       Y          Y          N                 N
#> 14       N          Y          N                 N
#> 15       Y          N          N                 N
#> 16       N          N          N                 N

^{Created on 2018-11-20 by the reprex package (v0.2.1)}

permutations(n=4,r=4,v=s,repeats.allowed=F) is not the same as the source vector is 4 in length and the target is four also. c("Y", "F") is 2 in length - so length combination possible.

[yos83]

You are amazing Christophe. Thank you so much.

[cderv]

Glad it works for you !

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