Rearrange and Sort - r

Question

I have the following data
ID v1 v2 v3 v4 v5
1 1 3 6 4
2 4 2
3 3 1 8 5
4 2 5 3 1
Can I rearrange the data so that it will automatically create new columns and assign binary value (1 or 0) according to the value in each variable (v1 to v5)?
E.g. In first row, I have values of 1,3,4 and 6. Can R automatically create 6 dummy variables to have assign the value to the respective column as below:
ID dummy1 dummy2 dummy3 dummy4 dummy5 dummy6
1 1 0 1 1 0 1
To have something like this:
ID c1 c2 c3 c4 c5 c6 c7 c8
1 1 0 1 1 0 1 0 0
2 0 1 0 1 0 0 0 0
3 1 0 1 0 1 0 0 1
4 1 1 1 0 1 0 0 0
Thanks.

Answer 1

We can use base R to do this. Loop through the rows of the dataset except the first column, get the sequence of max value in the row, check how many of these are in the row and convert it to integer with as.integer, append NAs at the end to make the lengths same in the list output and cbind with the first column
lst <- apply(df[-1], 1, function(x) as.integer(seq_len(max(x, na.rm = TRUE)) %in% x))
res <- cbind(df[1], do.call(rbind, lapply(lst, `length<-`, max(lengths(lst)))))
res[is.na(res)] <- 0
colnames(res)[-1] <- paste0('c', 1:8)
res
# ID c1 c2 c3 c4 c5 c6 c7 c8
#1 1 1 0 1 1 0 1 0 0
#2 2 0 1 0 1 0 0 0 0
#3 3 1 0 1 0 1 0 0 1
#4 4 1 1 1 0 1 0 0 0

Answer 2

In base R, you can use:
table(transform(cbind(mydf[1], stack(mydf[-1]))[1:2], values = factor(values, 1:8)))
## values
## ID 1 2 3 4 5 6 7 8
## 1 1 0 1 1 0 1 0 0
## 2 0 1 0 1 0 0 0 0
## 3 1 0 1 0 1 0 0 1
## 4 1 1 1 0 1 0 0 0
Note that you need to convert the stacked values to factor if you want the "7" to be included in the output. This applies to the "data.table" and "tidyverse" approaches as well.
Alternatively, you can try the following with "data.table":
library(data.table)
melt(as.data.table(mydf), "ID", na.rm = TRUE)[
, dcast(.SD, ID ~ factor(value, 1:8), fun = length, drop = FALSE)]
Or the following with the "tidyverse":
library(tidyverse)
mydf %>%
gather(var, val, -ID, na.rm = TRUE) %>%
select(-var) %>%
mutate(var = 1, val = factor(val, 1:8)) %>%
spread(val, var, fill = 0, drop = FALSE)
Sample data:
mydf <- structure(list(ID = 1:4, v1 = c(1L, 4L, 3L, 2L), v2 = c(3L, 2L,
1L, 5L), v3 = c(6L, NA, 8L, 3L), v4 = c(4L, NA, 5L, 1L), v5 = c(NA,
NA, NA, NA)), .Names = c("ID", "v1", "v2", "v3", "v4", "v5"), row.names = c(NA,
4L), class = "data.frame")
If automation is important, you can also use syntax like factor(value, sequence(max(value)) in the "data.table" approach or val = factor(val, sequence(max(val)))) in the "tidyverse" approach.

Answer 3

Another base R answer with some similarities to akrun's is
# create matrix of values
myMat <- as.matrix(dat[-1])
# create result matrix of desired shape, filled with 0s
res <- matrix(0L, nrow(dat), ncol=max(myMat, na.rm=TRUE))
# use matrix indexing to fill in 1s
res[cbind(dat$ID, as.vector(myMat))] <- 1L
# convert to data.frame, add ID column, and provide variable names
setNames(data.frame(cbind(dat$ID, res)), c("ID", paste0("c", 1:8)))
which returns
ID c1 c2 c3 c4 c5 c6 c7 c8
1 1 1 0 1 1 0 1 0 0
2 2 0 1 0 1 0 0 0 0
3 3 1 0 1 0 1 0 0 1
4 4 1 1 1 0 1 0 0 0