Daniel - 2 months ago 5x

R Question

Answer

```
dat <- data.frame(letter = c("A", "A", "A", "A",
"B", "B", "B", "C",
"C", "C", "C", "D"),
number = c(2, 3, 4,5, 4, 5, 6, 1, 3, 5, 7, 1))
```

I would like to provide an R base solution (maybe just for fun...), based on matrix indexing.

```
lev <- unique(dat[[1L]]); k <- length(lev) ## unique levels
x <- dat[[2L]]; p <- max(x) ## column position
z <- matrix(0L, nrow = k, ncol = p, dimnames = list(lev, seq_len(p))) ## initialization
z[cbind(match(dat[[1L]], lev), dat[[2L]])] <- 1L ## replacement
z ## display
# 1 2 3 4 5 6 7
#A 0 1 1 1 1 0 0
#B 0 0 0 1 1 1 0
#C 1 0 1 0 1 0 1
#D 1 0 0 0 0 0 0
```

Source (Stackoverflow)

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