How to add columns to a dataframe based on indexes in R? (See example) - r

Question

I'm working with a self made infix function which simply calculates the
percentage growth between observations in columns.
options(digits=3)
`%grow%` <- function(x,y) {
(y-x) / x * 100
}
test <- data.frame(a=c(101,202,301), b=c(123,214,199), h=c(134, 217, 205))
Then I use lapply to my toy database in order to add two new columns.
test[,4:5] <- lapply(1:(ncol(test)-1), function(i) test[,i] %grow% test[,(i+1)])
test
#Output
a b h V4 V5
1 101 123 134 21.78 8.94
2 202 214 217 5.94 1.40
3 301 199 205 -33.89 3.02
This is easy considering I just have three columns and I just can write test[,4:5]. Now talking in general terms: How to do this if we have n columns using column indexes?
What I mean is I want to create n-1 columns to a given database starting from the last one. Something like:
test[,(last_current_column+1):(last_column_created_using_function)]
Considering what I've read in some other posts, using my example, test[,(last_current_column+1): could be written as:
test[,(ncol(test)+1):]
but second part is still missing and I have no idea how to write it.
I hope I made myself clear. I fully appreciate any comment or advise.
Happy 2019 :)

Answer 1

Another way would be:
#options(digits=3)
`%grow%` <- function(x,y) {
(y-x) / x * 100
}
test <- data.frame(a=c(101,202,301),
b=c(123,214,199),
h=c(134, 217, 205),
d=c(156,234,235))
# a b h d
# 1 101 123 134 156
# 2 202 214 217 234
# 3 301 199 205 235
seqcols <- seq_along(test) # saved just to improve readability
test[,seqcols[-length(seqcols)] + max(seqcols)] <- lapply(seqcols[-length(seqcols)],
function(i) test[,i] %grow% test[,(i+1)])
test
# a b h d V5 V6 V7
# 1 101 123 134 156 21.78 8.94 16.42
# 2 202 214 217 234 5.94 1.40 7.83
# 3 301 199 205 235 -33.89 3.02 14.63
Similar to the second solution from #Ronak Shah, just with the use of map2_df from purrr:
cbind(test,
new=purrr::map2_df(test[seqcols[-length(seqcols)]], test[seqcols[-1]], `%grow%`),
deparse.level=1)
# a b h d new.a new.b new.h
# 1 101 123 134 156 21.78 8.94 16.42
# 2 202 214 217 234 5.94 1.40 7.83
# 3 301 199 205 235 -33.89 3.02 14.63

Answer 2

You would always ncol(test) - 1 new columns. Now using this logic there are multiple ways to do this.
One way would be to construct a character vector with some prefix value.
test[paste0("new_col", seq_len(ncol(test) - 1))] <- lapply(1:(ncol(test)-1),
function(i) test[,i] %grow% test[,(i+1)])
test
# a b h new_col1 new_col2
#1 101 123 134 21.782178 8.943089
#2 202 214 217 5.940594 1.401869
#3 301 199 205 -33.887043 3.015075
Another option using mapply and transform by creating subsets of dataframe
transform(test,
new_col = mapply(`%grow%`, test[1:(ncol(test)- 1)], test[2:ncol(test)]))
# a b h new_col.a new_col.b
#1 101 123 134 21.782178 8.943089
#2 202 214 217 5.940594 1.401869
#3 301 199 205 -33.887043 3.015075