Shape data frame by adding zero in R - r

Question

I have a data frame with distance in the first colomn and class in the second:
data.tab <- read.table(text = "
644 1
76 1
78 1
350 1
45 1
37 2
366 2
46 2
71 3
28 3
97 3
30 3
55 3
65 3
116 3
30 3
18 4
143 4
99 4")
I want to shape it into a new data frame by adding zero according to the longest class. The result will be:
data.tab <- read.table(text = "
1 644 76 78 350 45 0 0 0
2 37 366 46 0 0 0 0 0
3 71 28 97 30 55 65 116 30
4 18 143 99 0 0 0 0 0")

Answer 1

This essentially boils down to a simple long to wide reshape
library(tidyverse)
data.tab %>%
group_by(V2) %>%
mutate(col = paste0("V", 1:n())) %>%
spread(col, V1, fill = 0) %>%
ungroup()
## A tibble: 4 x 8
# V1 V2 V3 V4 V5 V6 V7 V8
# <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
#1 644 76 78 350 45 0 0 0
#2 37 366 46 0 0 0 0 0
#3 71 28 97 30 55 65 116 30
#4 18 143 99 0 0 0 0 0

Answer 2

Using df as name instead of data.tab:
MAX <- max(table(df$V2))
t(sapply(split(df$V1, df$V2), function(x) c(x, rep(0, MAX-length(x)))))
(The idea is to split V1 into groups defined by V2, making the vectors equal in length by adding 0's at the end when necessary, and then combining that into a single matrix. sapply does the last bit automatically but columnwise, so t is needed.)
another way using length<-
U <- unstack(df) # a hack learned from G.Grothendieck's answer
U <- with(df, split(V1,V2)) # more readable version of the above
M <- max(lengths(U))
R <- t(sapply(U, "length<-", M)) # setting all lengths equal
replace(R, is.na(R), 0) # replacing NAs by zeroes
And a (rather unreadable) one-liner doing the same thing:
"[<-"(R<-t(sapply(U<-unstack(df),"length<-",max(lengths(U)))),is.na(R),0)

Answer 3

1) xtabs Using only base R create a sequence number column within class and then use xtabs to rearrange it into a table. Finally convert that to data frame. Omit the last line of code if a table is sufficient.
data.tab2 <- transform(data.tab, seq = ave(V2, V2, FUN = seq_along))
xt <- xtabs(V1 ~ V2 + seq, data.tab2)
as.data.frame.matrix(xt)
giving:
1 2 3 4 5 6 7 8
1 644 76 78 350 45 0 0 0
2 37 366 46 0 0 0 0 0
3 71 28 97 30 55 65 116 30
4 18 143 99 0 0 0 0 0
2) ts Another base R solution is to convert the elements of each class to a ts series giving tt a multivariate time series with NAs at the ends of the shorter ones. Convert those NAs to 0 in the second line of code and then convert that to a data frame in the last line.
tt <- do.call("cbind", lapply(unstack(data.tab), ts))
tt[] <- ifelse(is.na(tt), 0, tt)
as.data.frame(t(tt))
3) Using data.tab2 from (1) use tapply to create the matrix mat and then convert that to a data.frame. Omit the last line of code if a matrix is sufficient.
mat <- with(data.tab2, tapply(V1, list(V2, seq), c, default = 0))
as.data.frame(mat)
Note
A comment claimed ifelse would be slower than a suggested alternative but benchmarking it showed no overall difference on the data in the question. Of course performance may not be very important here in the first place.
library(rbenchmark)
benchmark(
ifelse = {
tt <- do.call("cbind", lapply(unstack(data.tab), ts))
tt[] <- ifelse(is.na(tt), 0, tt)
as.data.frame(t(tt))
},
replace = {
tt <- do.call("cbind", lapply(unstack(data.tab), ts))
tt[is.na(tt)] <- 0
as.data.frame(t(tt))
}
)[1:4]
giving:
test replications elapsed relative
1 ifelse 100 0.25 1
2 replace 100 0.25 1

Answer 4

using data.table's transpose
cbind(sort(unique(data.tab$V2)),do.call(rbind,transpose(transpose(split(data.tab$V1, data.tab$V2), 0))))
# [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9]
#[1,] 1 644 76 78 350 45 0 0 0
#[2,] 2 37 366 46 0 0 0 0 0
#[3,] 3 71 28 97 30 55 65 116 30
#[4,] 4 18 143 99 0 0 0 0 0