Someone should have asked this already, but I couldn't find an answer. Say I have:

x = data.frame(q=1,w=2,e=3, ...and many many columns...) 

what is the most elegant way to rename an arbitrary subset of columns, whose position I don't necessarily know, into some other arbitrary names?

e.g. Say I want to rename "q" and "e" into "A" and "B", what is the most elegant code to do this?

Obviously, I can do a loop:

oldnames = c("q","e")
newnames = c("A","B")
for(i in 1:2) names(x)[names(x) == oldnames[i]] = newnames[i]

But I wonder if there is a better way? Maybe using some of the packages? (plyr::rename etc.)

20 Answers

setnames from the data.tablepackage will work on data.frames or data.tables

library(data.table)
d <- data.frame(a=1:2,b=2:3,d=4:5)
setnames(d, old = c('a','d'), new = c('anew','dnew'))
d # anew b dnew # 1 1 2 4 # 2 2 3 5

Note that changes are made by reference, so no copying (even for data.frames!)

5

With dplyr you would do:

library(dplyr)
df = data.frame(q = 1, w = 2, e = 3)
df %>% rename(A = q, B = e)
# A w B
#1 1 2 3

Or if you want to use vectors, as suggested by @Jelena-bioinf:

library(dplyr)
df = data.frame(q = 1, w = 2, e = 3)
oldnames = c("q","e")
newnames = c("A","B")
df %>% rename_at(vars(oldnames), ~ newnames)
# A w B
#1 1 2 3

L. D. Nicolas May suggested a change given rename_at is being superseded by rename_with:

df %>% rename_with(~ newnames[which(oldnames == .x)], .cols = oldnames)
# A w B
#1 1 2 3

4

Another solution for dataframes which are not too large is (building on @thelatemail answer):

x <- data.frame(q=1,w=2,e=3)
> x q w e
1 1 2 3
colnames(x) <- c("A","w","B")
> x A w B
1 1 2 3

Alternatively, you can also use:

names(x) <- c("C","w","D")
> x C w D
1 1 2 3

Furthermore, you can also rename a subset of the columnnames:

names(x)[2:3] <- c("E","F")
> x C E F
1 1 2 3

Here is the most efficient way I have found to rename multiple columns using a combination of purrr::set_names() and a few stringr operations.

library(tidyverse)
# Make a tibble with bad names
data <- tibble( `Bad NameS 1` = letters[1:10], `bAd NameS 2` = rnorm(10)
)
data
# A tibble: 10 x 2 `Bad NameS 1` `bAd NameS 2` <chr> <dbl> 1 a -0.840 2 b -1.56 3 c -0.625 4 d 0.506 5 e -1.52 6 f -0.212 7 g -1.50 8 h -1.53 9 i 0.420 10 j 0.957
# Use purrr::set_names() with annonymous function of stringr operations
data %>% set_names(~ str_to_lower(.) %>% str_replace_all(" ", "_") %>% str_replace_all("bad", "good"))
# A tibble: 10 x 2 good_names_1 good_names_2 <chr> <dbl> 1 a -0.840 2 b -1.56 3 c -0.625 4 d 0.506 5 e -1.52 6 f -0.212 7 g -1.50 8 h -1.53 9 i 0.420
10 j 0.957

5

So I recently ran into this myself, if you're not sure if the columns exist and only want to rename those that do:

existing <- match(oldNames,names(x))
names(x)[na.omit(existing)] <- newNames[which(!is.na(existing))]

Update dplyr 1.0.0

The newest dplyr version became more flexible by adding rename_with() where _with refers to a function as input. The trick is to reformulate the character vector newnames into a formula (by ~), so it would be equivalent to function(x) return (newnames).

In my subjective opinion, that is the most elegant dplyr expression. Update: thanks to @desval, the oldnames vector must be wrapped by all_of to include all its elements:

# shortest & most elegant expression
df %>% rename_with(~ newnames, all_of(oldnames))
A w B
1 1 2 3

Side note:

If you reverse the order, either argument .fn must be specified as .fn is expected before .cols argument:

df %>% rename_with(oldnames, .fn = ~ newnames)
A w B
1 1 2 3

or specify argument .col:

 df %>% rename_with(.col = oldnames, ~ newnames)
A w B
1 1 2 3

1

Building on @user3114046's answer:

x <- data.frame(q=1,w=2,e=3)
x
# q w e
#1 1 2 3
names(x)[match(oldnames,names(x))] <- newnames
x
# A w B
#1 1 2 3

This won't be reliant on a specific ordering of columns in the x dataset.

3

names(x)[names(x) %in% c("q","e")]<-c("A","B")

1

This would change all the occurrences of those letters in all names:

 names(x) <- gsub("q", "A", gsub("e", "B", names(x) ) )

2

There are a few answers mentioning the functions dplyr::rename_with and rlang::set_names already. By they are separate. this answer illustrates the differences between the two and the use of functions and formulas to rename columns.

rename_with from the dplyr package can use either a function or a formula to rename a selection of columns given as the .cols argument. For example passing the function name toupper:

library(dplyr)
rename_with(head(iris), toupper, starts_with("Petal"))

Is equivalent to passing the formula ~ toupper(.x):

rename_with(head(iris), ~ toupper(.x), starts_with("Petal"))

When renaming all columns, you can also use set_names from the rlang package. To make a different example, let's use paste0 as a renaming function. pasteO takes 2 arguments, as a result there are different ways to pass the second argument depending on whether we use a function or a formula.

rlang::set_names(head(iris), paste0, "_hi")
rlang::set_names(head(iris), ~ paste0(.x, "_hi"))

The same can be achieved with rename_with by passing the data frame as first argument .data, the function as second argument .fn, all columns as third argument .cols=everything() and the function parameters as the fourth argument .... Alternatively you can place the second, third and fourth arguments in a formula given as the second argument.

rename_with(head(iris), paste0, everything(), "_hi")
rename_with(head(iris), ~ paste0(.x, "_hi"))

rename_with only works with data frames. set_names is more generic and can also perform vector renaming

rlang::set_names(1:4, c("a", "b", "c", "d"))

You can use a named vector. Below two options (with base R and dplyr).

base R, via subsetting:

x = data.frame(q = 1, w = 2, e = 3)
rename_vec <- c(q = "A", e = "B")
## vector of same length as names(x) which returns NA if there is no match to names(x)
which_rename <- rename_vec[names(x)]
## simple ifelse where names(x) will be renamed for every non-NA
names(x) <- ifelse(is.na(which_rename), names(x), which_rename)
x
#> A w B
#> 1 1 2 3

Or a dplyr option with !!!:

library(dplyr)
rename_vec <- c(A = "q", B = "e") # the names are just the other way round than in the base R way!
x %>% rename(!!!rename_vec)
#> A w B
#> 1 1 2 3

The latter works because the 'big-bang' operator !!! is forcing evaluation of a list or a vector.

?`!!`

!!! forces-splice a list of objects. The elements of the list are spliced in place, meaning that they each become one single argument.

2

You can get the name set, save it as a list, and then do your bulk renaming on the string. A good example of this is when you are doing a long to wide transition on a dataset:

names(labWide) Lab1 Lab10 Lab11 Lab12 Lab13 Lab14 Lab15 Lab16
1 35.75366 22.79493 30.32075 34.25637 30.66477 32.04059 24.46663 22.53063
nameVec <- names(labWide)
nameVec <- gsub("Lab","LabLat",nameVec)
names(labWide) <- nameVec
"LabLat1" "LabLat10" "LabLat11" "LabLat12" "LabLat13" "LabLat14""LabLat15" "LabLat16" " 

0

Sidenote, if you want to concatenate one string to all of the column names, you can just use this simple code.

colnames(df) <- paste("renamed_",colnames(df),sep="")

If the table contains two columns with the same name then the code goes like this,

rename(df,newname=oldname.x,newname=oldname.y)

Lot's of sort-of-answers, so I just wrote the function so you can copy/paste.

rename <- function(x, old_names, new_names) { stopifnot(length(old_names) == length(new_names)) # pull out the names that are actually in x old_nms <- old_names[old_names %in% names(x)] new_nms <- new_names[old_names %in% names(x)] # call out the column names that don't exist not_nms <- setdiff(old_names, old_nms) if(length(not_nms) > 0) { msg <- paste(paste(not_nms, collapse = ", "), "are not columns in the dataframe, so won't be renamed.") warning(msg) } # rename names(x)[names(x) %in% old_nms] <- new_nms x
} x = data.frame(q = 1, w = 2, e = 3) rename(x, c("q", "e"), c("Q", "E")) Q w E 1 1 2 3

1

If one row of the data contains the names you want to change all columns to you can do

names(data) <- data[row,]

Given data is your dataframe and row is the row number containing the new values.

Then you can remove the row containing the names with

data <- data[-row,]

This is the function that you need: Then just pass the x in a rename(X) and it will rename all values that appear and if it isn't in there it won't error

rename <-function(x){ oldNames = c("a","b","c") newNames = c("d","e","f") existing <- match(oldNames,names(x)) names(x)[na.omit(existing)] <- newNames[which(!is.na(existing))] return(x)
}

1

Many good answers above using specialized packages. This is a simple way of doing it only with base R.

df.rename.cols <- function(df, col2.list) { tlist <- transpose(col2.list) names(df)[which(names(df) %in% tlist[[1]])] <- tlist[[2]] df
} 

Here is an example:

df1 <- data.frame(A = c(1, 2), B = c(3, 4), C = c(5, 6), D = c(7, 8))
col.list <- list(c("A", "NewA"), c("C", "NewC"))
df.rename.cols(df1, col.list) NewA B NewC D
1 1 3 5 7
2 2 4 6 8

I recently built off of @agile bean's answer (using rename_with, formerly rename_at) to build a function which changes column names if they exist in the data frame, such that one can make the column names of heterogeneous data frames match each other when applicable.

The looping can surely be improved, but figured I'd share for posterity.

create example data frame:

x= structure(list(observation_date = structure(c(18526L, 18784L,
17601L), class = c("IDate", "Date")), year = c(2020L, 2021L,
2018L)), sf_column = "geometry", agr = structure(c(id = NA_integer_,
common_name = NA_integer_, scientific_name = NA_integer_, observation_count = NA_integer_,
country = NA_integer_, country_code = NA_integer_, state = NA_integer_,
state_code = NA_integer_, county = NA_integer_, county_code = NA_integer_,
observation_date = NA_integer_, time_observations_started = NA_integer_,
observer_id = NA_integer_, sampling_event_identifier = NA_integer_,
protocol_type = NA_integer_, protocol_code = NA_integer_, duration_minutes = NA_integer_,
effort_distance_km = NA_integer_, effort_area_ha = NA_integer_,
number_observers = NA_integer_, all_species_reported = NA_integer_,
group_identifier = NA_integer_, year = NA_integer_, checklist_id = NA_integer_,
yday = NA_integer_), class = "factor", .Label = c("constant",
"aggregate", "identity")), row.names = c("3", "3.1", "3.2"), class = "data.frame")

function

match_col_names <- function(x){ col_names <- list(date = c("observation_date", "date"), C = c("observation_count", "count","routetotal"), yday = c("dayofyear"), latitude = c("lat"), longitude = c("lon","long") ) for(i in seq_along(col_names)){ newname=names(col_names)[i] oldnames=col_names[[i]] toreplace = names(x)[which(names(x) %in% oldnames)] x <- x %>% rename_with(~newname, toreplace)
}
return(x)
}

apply function

x <- match_col_names(x)

For execution time purposes , I would like to suggest to use data tables structure:

> df = data.table(x = 1:10, y = 3:12, z = 4:13)
> oldnames = c("x","y","z")
> newnames = c("X","Y","Z")
> library(microbenchmark)
> library(data.table)
> library(dplyr)
> microbenchmark(dplyr_1 = df %>% rename_at(vars(oldnames), ~ newnames) ,
+ dplyr_2 = df %>% rename(X=x,Y=y,Z=z) ,
+ data_tabl1= setnames(copy(df), old = c("x","y","z") , new = c("X","Y","Z")),
+ times = 100)
Unit: microseconds expr min lq mean median uq max neval dplyr_1 5760.3 6523.00 7092.538 6864.35 7210.45 17935.9 100 dplyr_2 2536.4 2788.40 3078.609 3010.65 3282.05 4689.8 100 data_tabl1 170.0 218.45 368.261 243.85 274.40 12351.7 100