user971102 - 2 months ago 4

R Question

I have a list of several (variable) number of letters in combination, for example this:

`vec = c("a", "b", "c")`

comb = unlist(lapply(1:length(vec), combn, x = vec, simplify = FALSE), recursive = FALSE)

# this creates all the combinations of the vector I am interested in, i.e. for three letters:

# a b c ab ac bc abc

For each combination, I am trying to fill elements depending on the position, into vectors with the same length as the number of vectors. So I am trying to get:

`a = 200`

b = 020

c = 002

ab = 220

ac = 202

bc = 022

abc = 222

Right now I am trying with loops replacing each element of an array i,j but since all values are "2" there must be a more efficient way to do this?

Thanks so much!!

Answer

Starting just from `vec`

, you can do...

```
comb_cases = do.call(expand.grid, lapply(vec, function(x) c("", x)))
Var1 Var2 Var3
1
2 a
3 b
4 a b
5 c
6 a c
7 b c
8 a b c
```

There's a blank row for the empty set, as there probably should be.

From here...

```
comb = do.call(paste0, comb_cases)
# [1] "" "a" "b" "ab" "c" "ac" "bc" "abc"
do.call(paste0, split( ifelse(nchar(as.matrix(comb_cases)), 2, 0), col(comb_cases)) )
# [1] "000" "200" "020" "220" "002" "202" "022" "222"
```

`ifelse`

is slow, but that can be fixed up later if it matters.

Source (Stackoverflow)

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