How to automate renaming of columns in wide data using R - r

Question

Consider the following data in the wide format
df<-data.frame("id"=c(1,2,3,4),
"ex"=c(1,0,0,1),
"aQL"=c(5,4,NA,6),
"bQL"=c(5,7,NA,9),
"cQL"=c(5,7,NA,9),
"bST"=c(3,7,8,9),
"cST"=c(8,7,5,3),
"aXY"=c(1,9,4,4),
"cXY"=c(5,3,1,4))
I want to keep the column (or variable) names "id" and "ex" and rename the remaining columns, e.g. "aQL", "bQL" and "cQL" as "QL.1", "QL.2" and "QL.3", respectively. The other columns with names ending with "ST" and "XY" are expected to be renamed in the same manner, also having the order .1, .2 and .3. Of note is "aST" and "bXY" are missing from the data set, but I want them to be included and renamed as ST.1 and XY.2, with each having NAs as their entries. The expected output would look like
df
id ex QL.1 QL.2 QL.3 ST.1 ST.2 ST.3 XY.1 XY.2 XY.3
1 1 1 5 5 5 NA 3 8 1 NA 5
2 2 0 4 7 7 NA 7 7 9 NA 3
3 3 0 NA NA NA NA 8 5 4 NA 1
4 4 1 6 9 9 NA 9 3 4 NA 4
The main data set has many variables, so I would like the renaming to be done in an automated manner. I tried the following code
renameCol <- function(x) {
setNames(x, paste0("QL.", seq_len(ncol(x))))
}
renameCol(df)
but it does not work as expected. Thus, it renames "id" and "ex" that I want to maintain and it is not flexible on the renaming of multiple variable (i.e. QL, ST, XY). Any help is greatly appreciated.

Answer 1

I would suggest a tidyverse approach where there is no need of a function. In this solution you can extract the first letter of each variable name as id and then assign a number with cur_group_id so that the order is kept. Finally, with this new number you transform the variable containing the names and then you format to wide in order to obtain the expected output:
library(tidyverse)
#Data
df<-data.frame("id"=c(1,2,3,4),
"ex"=c(1,0,0,1),
"aQL"=c(5,4,NA,6),
"bQL"=c(5,7,NA,9),
"cQL"=c(5,7,NA,9),
"bST"=c(3,7,8,9),
"cST"=c(8,7,5,3),
"aXY"=c(1,9,4,4),
"cXY"=c(5,3,1,4))
#Reshape
df %>% pivot_longer(cols = -c(1,2)) %>%
#Extract first letter as id
mutate(id2=substring(name,1,1)) %>%
#Create the number id
group_by(id2) %>%
mutate(id3=cur_group_id()) %>%
#Clean name
mutate(name=substring(name,2,nchar(name))) %>%
#Create final var
mutate(name2=paste0(name,'.',id3)) %>% ungroup() %>%
dplyr::select(-c(name,id2,id3)) %>%
#Format to wide
pivot_wider(names_from = name2,values_from=value)
Output:
# A tibble: 4 x 9
id ex QL.1 QL.2 QL.3 ST.2 ST.3 XY.1 XY.3
<dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 1 1 5 5 5 3 8 1 5
2 2 0 4 7 7 7 7 9 3
3 3 0 NA NA NA 8 5 4 1
4 4 1 6 9 9 9 3 4 4

Answer 2

in base R you could do:
names(df) <- sub("(\\d)([A-Z]{2})$","\\2.\\1", chartr("abc","123",names(df)))
df
id ex QL.1 QL.2 QL.3 ST.2 ST.3 XY.1 XY.3
1 1 1 5 5 5 3 8 1 5
2 2 0 4 7 7 7 7 9 3
3 3 0 NA NA NA 8 5 4 1
4 4 1 6 9 9 9 3 4 4
If you need the NA columns:
names(df) <- sub("(\\d)([A-Z]{2})$","\\2.\\1", chartr("abc","123",names(df)))
a <- read.table(text=grep("\\.\\d",names(df),value = TRUE), sep=".")
b <- subset(aggregate(.~V1, a, function(x) setdiff(1:3,x)), V2>0)
df[do.call(paste, c(sep = ".", b))] <- NA
(df1 <- df[c(1, 2, order(names(df)[-(1:2)]) + 2)])
id ex QL.1 QL.2 QL.3 ST.1 ST.2 ST.3 XY.1 XY.2 XY.3
1 1 1 5 5 5 NA 3 8 1 NA 5
2 2 0 4 7 7 NA 7 7 9 NA 3
3 3 0 NA NA NA NA 8 5 4 NA 1
4 4 1 6 9 9 NA 9 3 4 NA 4

Answer 3

Another way you can try
colnames(df)[grepl("QL", colnames(df))] <- str_c("QL.", 1:3)
colnames(df)[grepl("ST", colnames(df))] <- str_c("ST.", 2:3)
colnames(df)[grepl("XY", colnames(df))] <- str_c("XY.", c(1,3))
# id ex QL.1 QL.2 QL.3 ST.2 ST.3 XY.1 XY.3
# 1 1 1 5 5 5 3 8 1 5
# 2 2 0 4 7 7 7 7 9 3
# 3 3 0 NA NA NA 8 5 4 1
# 4 4 1 6 9 9 9 3 4 4

Answer 4

Here is a solution that uses regular expressions via the stringr package:
library(stringr)
df<-data.frame("id"=c(1,2,3,4),
"ex"=c(1,0,0,1),
"aQL"=c(5,4,NA,6),
"bQL"=c(5,7,NA,9),
"cQL"=c(5,7,NA,9),
"bST"=c(3,7,8,9),
"cST"=c(8,7,5,3),
"aXY"=c(1,9,4,4),
"cXY"=c(5,3,1,4))
renameCol <- function(x) {
col_names <- colnames(x)
index_ql <- str_detect(col_names,
"^[a-z]{1}QL")
index_st <- str_detect(col_names,
"^[a-z]{1}ST")
index_xy <- str_detect(col_names,
"^[a-z]{1}XY")
replace_fun <- function(x) {which(letters %in% x)}
col_names[index_ql] <- paste0("QL.", str_replace(substr(col_names[index_ql], 1, 1),
"[a-z]", replace_fun))
col_names[index_st] <- paste0("ST.", str_replace(substr(col_names[index_st], 1, 1),
"[a-z]", replace_fun))
col_names[index_xy] <- paste0("XY.", str_replace(substr(col_names[index_xy], 1, 1),
"[a-z]", replace_fun))
col_names
}
colnames(df) <- renameCol(df)
df
#> id ex QL.1 QL.2 QL.3 ST.2 ST.3 XY.1 XY.3
#> 1 1 1 5 5 5 3 8 1 5
#> 2 2 0 4 7 7 7 7 9 3
#> 3 3 0 NA NA NA 8 5 4 1
#> 4 4 1 6 9 9 9 3 4 4
Created on 2020-09-07 by the reprex package (v0.3.0)
Edit
The function above was adapted so that it takes the order into account.

Answer 5

using base pattern matching:
you need to define a function that does what you want on one single column name:
f = function(x){
beg <- str_extract(x,"[a-z](?=[A-Z]{2})")
num <- which(letters == beg)
output <- paste0(str_extract(x,"(?<=[a-z])[A-Z]{2}"),".",num)
return(output)
}
here extract the lower case letter if you have two upper case letters after, find it position in alphabet, and paste the found number back to the upper case letters.
> f("cQL")
[1] "QL.3"
You can then use regmatches and regular expression directly on the name of your data frame:
m <- gregexpr("[a-z][A-Z]{2}", names(df),perl = T)
regmatches(names(df), m) <- lapply(regmatches(names(df), m), f)
names(df)
> names(df)
[1] "id" "ex" "QL.1" "QL.2" "QL.3" "ST.2" "ST.3" "XY.1" "XY.3"
It solves only the renaming part, not the the "including missing column number" part of your question