Create data.frames out of an initial dataframe using `lapply` in R

Question

I was wondering how if I could make 7 other data.frames out of dat1 using lapply such that at each round from left to right each pair of 1s in C1 and C2 become 0 except the last two elements of C1 and C2? (see desired output below)

dat1 <- data.frame(C1 = c(1, 1, 1, 1, 1, 1, 1, 1, 0, 0),
                   C2 = c(1, 1, 1, 1, 1, 1, 1, 1, 1, 1))

dat2 <- data.frame(C1 = c(0, 1, 1, 1, 1, 1, 1, 1, 0, 0),
                   C2 = c(0, 1, 1, 1, 1, 1, 1, 1, 1, 1))

dat3 <- data.frame(C1 = c(0, 0, 1, 1, 1, 1, 1, 1, 0, 0),
                   C2 = c(0, 0, 1, 1, 1, 1, 1, 1, 1, 1))

dat4 <- data.frame(C1 = c(0, 0, 0, 1, 1, 1, 1, 1, 0, 0),
                   C2 = c(0, 0, 0, 1, 1, 1, 1, 1, 1, 1))
.
.
.
dat8 <- data.frame(C1 = c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0),  # the last desired data.frame
                   C2 = c(0, 0, 0, 0, 0, 0, 0, 0, 1, 1))

akrun · Accepted Answer · 2020-01-03 19:25:28Z

3

We can use lapply by looping over sequence of numbers (1:7), create a two column matrix with the corresponding replicate of rows and rbind with the subset of original data with the same number of rows removed the top

lst1 <- lapply(1:7, function(i) rbind(matrix(rep(c(0, 0), i), ncol = 2,
      dimnames = list(NULL, c("C1", "C2"))), dat1[-seq_len(i), ]))
names(lst1) <- paste0("dat", seq_along(lst1) + 1)

It can be made shorter by using cbind

lapply(1:7, function(i) {n <- rep(0, i)
       rbind(cbind(C1 = n, C2 = n),  dat1[-seq_len(i), ])})

Or it can be also (inspired from @Markus comments)

lapply(0:8, function(i) dat1 * rep(c(0, 1), c(i, nrow(dat1)-i)))

If it needs to be separate objects, use list2env (not recommended though)

list2env(lst1, .GlobalEnv)

edited Jan 3, 2020 at 19:25

answered Jan 3, 2020 at 18:39

akrun

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1 Comment

markus Over a year ago

Good catch @rnorouzian. Try lapply(0:8, function(x) {dat1[0:x, ] <- 0; dat1 })

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