I know if I have a data frame with more than 1 column, then I can use
colnames(x) <- c("col1","col2")
to rename the columns. How to do this if it's just one column?
Meaning a vector or data frame with only one column.
Example:
trSamp <- data.frame(sample(trainer$index, 10000))
head(trSamp )
# sample.trainer.index..10000.
# 1 5907862
# 2 2181266
# 3 7368504
# 4 1949790
# 5 3475174
# 6 6062879
ncol(trSamp)
# [1] 1
class(trSamp)
# [1] "data.frame"
class(trSamp[1])
# [1] "data.frame"
class(trSamp[,1])
# [1] "numeric"
colnames(trSamp)[2] <- "newname2"
# Error in names(x) <- value :
# 'names' attribute [2] must be the same length as the vector [1]
This is a generalized way in which you do not have to remember the exact location of the variable:
# df = dataframe
# old.var.name = The name you don't like anymore
# new.var.name = The name you want to get
names(df)[names(df) == 'old.var.name'] <- 'new.var.name'
This code pretty much does the following:
names(df) looks into all the names in the df
[names(df) == old.var.name] extracts the variable name you want to check
<- 'new.var.name' assigns the new variable name.
colnames(trSamp)[2] <- "newname2"
attempts to set the second column's name. Your object only has one column, so the command throws an error. This should be sufficient:
colnames(trSamp) <- "newname2"
colnames(df)[colnames(df) == 'oldName'] <- 'newName'
This is an old question, but it is worth noting that you can now use setnames from the data.table package.
library(data.table)
setnames(DF, "oldName", "newName")
# or since the data.frame in question is just one column:
setnames(DF, "newName")
# And for reference's sake, in general (more than once column)
nms <- c("col1.name", "col2.name", etc...)
setnames(DF, nms)
This can also be done using Hadley's plyr package, and the rename function.
library(plyr)
df <- data.frame(foo=rnorm(1000))
df <- rename(df,c('foo'='samples'))
You can rename by the name (without knowing the position) and perform multiple renames at once. After doing a merge, for example, you might end up with:
letterid id.x id.y
1 70 2 1
2 116 6 5
3 116 6 4
4 116 6 3
5 766 14 9
6 766 14 13
Which you can then rename in one step using:
letters <- rename(letters,c("id.x" = "source", "id.y" = "target"))
letterid source target
1 70 2 1
2 116 6 5
3 116 6 4
4 116 6 3
5 766 14 9
6 766 14 13
I think the best way of renaming columns is by using the dplyr package like this:
require(dplyr)
df = rename(df, new_col01 = old_col01, new_col02 = old_col02, ...)
It works the same for renaming one or many columns in any dataset.
I find that the most convenient way to rename a single column is using dplyr::rename_at :
library(dplyr)
cars %>% rename_at("speed",~"new") %>% head
cars %>% rename_at(vars(speed),~"new") %>% head
cars %>% rename_at(1,~"new") %>% head
# new dist
# 1 4 2
# 2 4 10
# 3 7 4
# 4 7 22
# 5 8 16
# 6 9 10
works well in pipe chaines
convenient when names are stored in variables
works with a name or an column index
clear and compact
I like the next style for rename dataframe column names one by one.
colnames(df)[which(colnames(df) == 'old_colname')] <- 'new_colname'
where
which(colnames(df) == 'old_colname')
returns by the index of the specific column.
Let df be the dataframe you have with col names myDays and temp.
If you want to rename "myDays" to "Date",
library(plyr)
rename(df,c("myDays" = "Date"))
or with pipe, you can
dfNew <- df %>%
plyr::rename(c("myDays" = "Date"))
Try:
colnames(x)[2] <- 'newname2'
This is likely already out there, but I was playing with renaming fields while searching out a solution and tried this on a whim. Worked for my purposes.
Table1$FieldNewName <- Table1$FieldOldName
Table1$FieldOldName <- NULL
Edit begins here....
This works as well.
df <- rename(df, c("oldColName" = "newColName"))
You can use the rename.vars in the gdata package.
library(gdata)
df <- rename.vars(df, from = "oldname", to = "newname")
This is particularly useful where you have more than one variable name to change or you want to append or pre-pend some text to the variable names, then you can do something like:
df <- rename.vars(df, from = c("old1", "old2", "old3",
to = c("new1", "new2", "new3"))
For an example of appending text to a subset of variables names see:
https://stackoverflow.com/a/28870000/180892
You could also try 'upData' from 'Hmisc' package.
library(Hmisc)
trSamp = upData(trSamp, rename=c(sample.trainer.index..10000. = 'newname2'))
If you know that your dataframe has only one column, you can use:
names(trSamp) <- "newname2"
The OP's question has been well and truly answered. However, here's a trick that may be useful in some situations: partial matching of the column name, irrespective of its position in a dataframe:
Partial matching on the name:
d <- data.frame(name1 = NA, Reported.Cases..WHO..2011. = NA, name3 = NA)
## name1 Reported.Cases..WHO..2011. name3
## 1 NA NA NA
names(d)[grepl("Reported", names(d))] <- "name2"
## name1 name2 name3
## 1 NA NA NA
Another example: partial matching on the presence of "punctuation":
d <- data.frame(name1 = NA, Reported.Cases..WHO..2011. = NA, name3 = NA)
## name1 Reported.Cases..WHO..2011. name3
## 1 NA NA NA
names(d)[grepl("[[:punct:]]", names(d))] <- "name2"
## name1 name2 name3
## 1 NA NA NA
These were examples I had to deal with today, I thought might be worth sharing.
I would simply change a column name to the dataset with the new name I want with the following code:
names(dataset)[index_value] <- "new_col_name"
I found colnames() argument easier
https://www.rdocumentation.org/packages/base/versions/3.6.2/topics/row%2Bcolnames
select some column from the data frame
df <- data.frame(df[, c( "hhid","b1005", "b1012_imp", "b3004a")])
and rename the selected column in order,
colnames(df) <- c("hhid", "income", "cost", "credit")
check the names and the values to be sure
names(df);head(df)
I would simply add a new column to the data frame with the name I want and get the data for it from the existing column. like this:
dataf$value=dataf$Article1Order
then I remove the old column! like this:
dataf$Article1Order<-NULL
This code might seem silly! But it works perfectly...
We can use rename_with to rename columns with a function (stringr functions, for example).
Consider the following data df_1:
df_1 <- data.frame(
x = replicate(n = 3, expr = rnorm(n = 3, mean = 10, sd = 1)),
y = sample(x = 1:2, size = 10, replace = TRUE)
)
names(df_1)
#[1] "x.1" "x.2" "x.3" "y"
Rename all variables with dplyr::everything():
library(tidyverse)
df_1 %>%
rename_with(.data = ., .cols = everything(.),
.fn = str_replace, pattern = '.*',
replacement = str_c('var', seq_along(.), sep = '_')) %>%
names()
#[1] "var_1" "var_2" "var_3" "var_4"
Rename by name particle with some dplyr verbs (starts_with, ends_with, contains, matches, ...).
Example with . (x variables):
df_1 %>%
rename_with(.data = ., .cols = contains('.'),
.fn = str_replace, pattern = '.*',
replacement = str_c('var', seq_along(.), sep = '_')) %>%
names()
#[1] "var_1" "var_2" "var_3" "y"
Rename by class with many functions of class test, like is.integer, is.numeric, is.factor...
Example with is.integer (y):
df_1 %>%
rename_with(.data = ., .cols = is.integer,
.fn = str_replace, pattern = '.*',
replacement = str_c('var', seq_along(.), sep = '_')) %>%
names()
#[1] "x.1" "x.2" "x.3" "var_1"
The warning:
Warning messages:
1: In stri_replace_first_regex(string, pattern, fix_replacement(replacement), :
longer object length is not a multiple of shorter object length
2: In names[cols] <- .fn(names[cols], ...) :
number of items to replace is not a multiple of replacement length
It is not relevant, as it is just an inconsistency of seq_along(.) with the replace function.
library(dplyr)
rename(data, de=de.y)
Related
I have a list of dataframes for which I want to obtain (in a separate dataframe) the row mean of a specified column which may or may not exist in all dataframes of the list. My problem comes when the specified column does not exist in at least one of the dataframes of the list.
Assume the following example list of dataframes:
df1 <- read.table(text = 'X A B C
name1 1 2 3
name2 5 10 4',
header = TRUE)
df2 <- read.table(text = 'X B C A
name1 8 1 31
name2 9 9 8',
header = TRUE)
df3 <- read.table(text = 'X B A E
name1 9 9 29
name2 5 15 55',
header = TRUE)
mylist_old <-list(df1, df2)
mylist_new <-list(df1, df2, df3)
Assume I want to rowMeans column C the following piece of code works perfectly when the list of dataframe (mylist_old) is composed of elements df1 and df2, :
Mean_C <- rowMeans(do.call(cbind, lapply(mylist_old, "[", "C")))
Mean_C <- as.data.frame(Mean_C)
The trouble comes when the list is composed of at least one dataframe for which column C does not exist, which in my example is the case of df3, that is for list mylist_new:
Mean_C <- rowMeans(do.call(cbind, lapply(mylist_new, "[", "C")))
Leads to: "Error in [.data.frame(X[[i]], ...) : undefined columns selected
One way to circumvent this issue is to exclude df3 from mylist_new. However, my real program has a list of 64 dataframes for which I do not know whether column C exists or not. I would have like to lapply my piece of code only if column C is detected as existing, that is applying the command to the list of dataframes but only for dataframes for which existence of column C is true.
I tried this
if("C" %in% colnames(mylist_new))
{
Mean_C <- rowMeans(do.call(cbind, lapply(mylist_new, "[", "C")))
Mean_C <- as.data.frame(Mean_C)
}
But nothing happens, probably because colnames refers to the list and not to each dataframe of the list. With 64 dataframes, I cannot refer to each "manually" and need an automated procedure.
Here is one option to Filter the list elements and then apply the lapply on the filtered list
rowMeans(do.call(cbind, lapply(Filter(function(x) "C" %in% names(x),
mylist_new), `[[`, "C")))
#[1] 2.0 6.5
or using tidyverse without Filtering, but making use of select to ignore the cases where the column is not present
library(tidyverse)
map(mylist_new, ~ .x %>%
select(one_of("C"))) %>% # gives a warning
bind_cols %>%
rowMeans
#[1] 2.0 6.5
It may be better to have some warning that the column is not present
Or without a warning
map(mylist_new, ~ .x %>%
select(matches("^C$"))) %>%
bind_cols %>%
rowMeans
#[1] 2.0 6.5
We can use if to check names before we do the subset
rowMeans(do.call(cbind,
lapply(mylist_new, function(x) if('C' %in% names(x)) x['C'] else NA)),na.rm = TRUE)
Or using map_if in purrr 0.3.2
library(purrr)
rowMeans(do.call(cbind,map_if(mylist_new,
function(x) 'C' %in% names(x),
'C', .else=~return(NA))),na.rm = TRUE)
[1] 2.0 6.5
One way is to use purrr::safely, it will return for each iteration a list with a result and error element, then we can transpose, extract result and remove the NULL result with compact :
library(tidyverse)
rowMeans(do.call(cbind, transpose(
lapply(mylist_new, safely(`[`), "C"))$result %>% compact()))
# [1] 2.0 6.5
We could also use the otherwise parameter to have a NA result rather than NULL, and we can set na.rm to TRUE in rowMeans.
rowMeans(na.rm = TRUE, do.call(cbind, transpose(
lapply(mylist_new, safely(`[`, otherwise= NA), "C"))$result))
# [1] 2.0 6.5
This was to address your case with minimal modifications. If I have to solve this precise issue I would do it the following way :
map(mylist_new, "C") %>% compact() %>% pmap_dbl(~mean(c(...)))
# [1] 2.0 6.5
We extract the C element, remove it when it's NULL, and then compute mean by element.
This might be more efficient (not sure):
map(set_names(mylist_new), "C") %>% compact() %>% as_tibble() %>% rowMeans()
# [1] 2.0 6.5
One more, using reshaping this time :
map_dfr(mylist_new, ~gather(.,,,-1)) %>%
group_by(X) %>%
filter(key == "C") %>%
summarize_at("value", mean)
# # A tibble: 2 x 2
# X value
# <fct> <dbl>
# 1 name1 2
# 2 name2 6.5
And a base version, quite readable, with a somewhat awkward step where several columns have the same name, but it's on a temp object so that's not that bad:
wide <- do.call(cbind, mylist_new)
rowMeans(wide[names(wide) == "C"])
# [1] 2.0 6.5
I know if I have a data frame with more than 1 column, then I can use
colnames(x) <- c("col1","col2")
to rename the columns. How to do this if it's just one column?
Meaning a vector or data frame with only one column.
Example:
trSamp <- data.frame(sample(trainer$index, 10000))
head(trSamp )
# sample.trainer.index..10000.
# 1 5907862
# 2 2181266
# 3 7368504
# 4 1949790
# 5 3475174
# 6 6062879
ncol(trSamp)
# [1] 1
class(trSamp)
# [1] "data.frame"
class(trSamp[1])
# [1] "data.frame"
class(trSamp[,1])
# [1] "numeric"
colnames(trSamp)[2] <- "newname2"
# Error in names(x) <- value :
# 'names' attribute [2] must be the same length as the vector [1]
This is a generalized way in which you do not have to remember the exact location of the variable:
# df = dataframe
# old.var.name = The name you don't like anymore
# new.var.name = The name you want to get
names(df)[names(df) == 'old.var.name'] <- 'new.var.name'
This code pretty much does the following:
names(df) looks into all the names in the df
[names(df) == old.var.name] extracts the variable name you want to check
<- 'new.var.name' assigns the new variable name.
colnames(trSamp)[2] <- "newname2"
attempts to set the second column's name. Your object only has one column, so the command throws an error. This should be sufficient:
colnames(trSamp) <- "newname2"
colnames(df)[colnames(df) == 'oldName'] <- 'newName'
This is an old question, but it is worth noting that you can now use setnames from the data.table package.
library(data.table)
setnames(DF, "oldName", "newName")
# or since the data.frame in question is just one column:
setnames(DF, "newName")
# And for reference's sake, in general (more than once column)
nms <- c("col1.name", "col2.name", etc...)
setnames(DF, nms)
This can also be done using Hadley's plyr package, and the rename function.
library(plyr)
df <- data.frame(foo=rnorm(1000))
df <- rename(df,c('foo'='samples'))
You can rename by the name (without knowing the position) and perform multiple renames at once. After doing a merge, for example, you might end up with:
letterid id.x id.y
1 70 2 1
2 116 6 5
3 116 6 4
4 116 6 3
5 766 14 9
6 766 14 13
Which you can then rename in one step using:
letters <- rename(letters,c("id.x" = "source", "id.y" = "target"))
letterid source target
1 70 2 1
2 116 6 5
3 116 6 4
4 116 6 3
5 766 14 9
6 766 14 13
I think the best way of renaming columns is by using the dplyr package like this:
require(dplyr)
df = rename(df, new_col01 = old_col01, new_col02 = old_col02, ...)
It works the same for renaming one or many columns in any dataset.
I find that the most convenient way to rename a single column is using dplyr::rename_at :
library(dplyr)
cars %>% rename_at("speed",~"new") %>% head
cars %>% rename_at(vars(speed),~"new") %>% head
cars %>% rename_at(1,~"new") %>% head
# new dist
# 1 4 2
# 2 4 10
# 3 7 4
# 4 7 22
# 5 8 16
# 6 9 10
works well in pipe chaines
convenient when names are stored in variables
works with a name or an column index
clear and compact
I like the next style for rename dataframe column names one by one.
colnames(df)[which(colnames(df) == 'old_colname')] <- 'new_colname'
where
which(colnames(df) == 'old_colname')
returns by the index of the specific column.
Let df be the dataframe you have with col names myDays and temp.
If you want to rename "myDays" to "Date",
library(plyr)
rename(df,c("myDays" = "Date"))
or with pipe, you can
dfNew <- df %>%
plyr::rename(c("myDays" = "Date"))
Try:
colnames(x)[2] <- 'newname2'
This is likely already out there, but I was playing with renaming fields while searching out a solution and tried this on a whim. Worked for my purposes.
Table1$FieldNewName <- Table1$FieldOldName
Table1$FieldOldName <- NULL
Edit begins here....
This works as well.
df <- rename(df, c("oldColName" = "newColName"))
You can use the rename.vars in the gdata package.
library(gdata)
df <- rename.vars(df, from = "oldname", to = "newname")
This is particularly useful where you have more than one variable name to change or you want to append or pre-pend some text to the variable names, then you can do something like:
df <- rename.vars(df, from = c("old1", "old2", "old3",
to = c("new1", "new2", "new3"))
For an example of appending text to a subset of variables names see:
https://stackoverflow.com/a/28870000/180892
You could also try 'upData' from 'Hmisc' package.
library(Hmisc)
trSamp = upData(trSamp, rename=c(sample.trainer.index..10000. = 'newname2'))
If you know that your dataframe has only one column, you can use:
names(trSamp) <- "newname2"
The OP's question has been well and truly answered. However, here's a trick that may be useful in some situations: partial matching of the column name, irrespective of its position in a dataframe:
Partial matching on the name:
d <- data.frame(name1 = NA, Reported.Cases..WHO..2011. = NA, name3 = NA)
## name1 Reported.Cases..WHO..2011. name3
## 1 NA NA NA
names(d)[grepl("Reported", names(d))] <- "name2"
## name1 name2 name3
## 1 NA NA NA
Another example: partial matching on the presence of "punctuation":
d <- data.frame(name1 = NA, Reported.Cases..WHO..2011. = NA, name3 = NA)
## name1 Reported.Cases..WHO..2011. name3
## 1 NA NA NA
names(d)[grepl("[[:punct:]]", names(d))] <- "name2"
## name1 name2 name3
## 1 NA NA NA
These were examples I had to deal with today, I thought might be worth sharing.
I would simply change a column name to the dataset with the new name I want with the following code:
names(dataset)[index_value] <- "new_col_name"
I found colnames() argument easier
https://www.rdocumentation.org/packages/base/versions/3.6.2/topics/row%2Bcolnames
select some column from the data frame
df <- data.frame(df[, c( "hhid","b1005", "b1012_imp", "b3004a")])
and rename the selected column in order,
colnames(df) <- c("hhid", "income", "cost", "credit")
check the names and the values to be sure
names(df);head(df)
I would simply add a new column to the data frame with the name I want and get the data for it from the existing column. like this:
dataf$value=dataf$Article1Order
then I remove the old column! like this:
dataf$Article1Order<-NULL
This code might seem silly! But it works perfectly...
We can use rename_with to rename columns with a function (stringr functions, for example).
Consider the following data df_1:
df_1 <- data.frame(
x = replicate(n = 3, expr = rnorm(n = 3, mean = 10, sd = 1)),
y = sample(x = 1:2, size = 10, replace = TRUE)
)
names(df_1)
#[1] "x.1" "x.2" "x.3" "y"
Rename all variables with dplyr::everything():
library(tidyverse)
df_1 %>%
rename_with(.data = ., .cols = everything(.),
.fn = str_replace, pattern = '.*',
replacement = str_c('var', seq_along(.), sep = '_')) %>%
names()
#[1] "var_1" "var_2" "var_3" "var_4"
Rename by name particle with some dplyr verbs (starts_with, ends_with, contains, matches, ...).
Example with . (x variables):
df_1 %>%
rename_with(.data = ., .cols = contains('.'),
.fn = str_replace, pattern = '.*',
replacement = str_c('var', seq_along(.), sep = '_')) %>%
names()
#[1] "var_1" "var_2" "var_3" "y"
Rename by class with many functions of class test, like is.integer, is.numeric, is.factor...
Example with is.integer (y):
df_1 %>%
rename_with(.data = ., .cols = is.integer,
.fn = str_replace, pattern = '.*',
replacement = str_c('var', seq_along(.), sep = '_')) %>%
names()
#[1] "x.1" "x.2" "x.3" "var_1"
The warning:
Warning messages:
1: In stri_replace_first_regex(string, pattern, fix_replacement(replacement), :
longer object length is not a multiple of shorter object length
2: In names[cols] <- .fn(names[cols], ...) :
number of items to replace is not a multiple of replacement length
It is not relevant, as it is just an inconsistency of seq_along(.) with the replace function.
library(dplyr)
rename(data, de=de.y)
dplyr's rename functions require the new column name to be passed in as unquoted variable names. However I have a function where the column name is constructed by pasting a string onto an argument passed in and so is a character string.
For example say I had this function
myFunc <- function(df, col){
new <- paste0(col, '_1')
out <- dplyr::rename(df, new = old)
return(out)
}
If I run this
df <- data.frame(a = 1:3, old = 4:6)
myFunc(df, 'x')
I get
a new
1 1 4
2 2 5
3 3 6
Whereas I want the 'new' column to be the name of the string I constructed ('x_1'), i.e.
a x_1
1 1 4
2 2 5
3 3 6
Is there anyway of doing this?
I think this is what you were looking for. It is the use of rename_ as #Henrik suggested, but the argument has an, lets say, interesting, name:
> myFunc <- function(df, col){
+ new <- paste0(col, '_1')
+ out <- dplyr::rename_(df, .dots=setNames(list(col), new))
+ return(out)
+ }
> myFunc(data.frame(x=c(1,2,3)), "x")
x_1
1 1
2 2
3 3
>
Note the use of setNames to use the value of new as name in the list.
Recent updates to tidyr and dplyr allow you to use the rename_with function.
Say you have a data frame:
library(tidyverse)
df <- tibble(V0 = runif(10), V1 = runif(10), V2 = runif(10), key=letters[1:10])
And you want to change all of the "V" columns. Usually, my reference for columns like this comes from a json file, which in R is a labeled list. e.g.,
colmapping <- c("newcol1", "newcol2", "newcol3")
names(colmapping) <- paste0("V",0:2)
You can then use the following to change the names of df to the strings in the colmapping list:
df <- rename_with(.data = df, .cols = starts_with("V"), .fn = function(x){colmapping[x]})
In the following example, why is there an additional column in the unite_() output vs the unite() output?
library(tidyr)
x1 <- data.frame(Sample=c("A", "B"), "1"=c("-", "y"),
"2"=c("-", "z"), "3"=c("x", "a"), check.names=F)
# Sample 1 2 3
# 1 A - - x
# 2 B y z a
Here we see the desired output:
unite(x1, mix, 2:ncol(x1), sep=",")
# Sample mix
# 1 A -,-,x
# 2 B y,z,a
Why is there an additional column here (the 1 column)? The default is to remove the columns used by unite_().
unite_(x1, "mix", 2:ncol(x1), sep=",")
# Sample 1 mix
# 1 A - -,-,x
# 2 B y y,z,a
Note: tidyr version 0.5.1
The syntax are slightly different between the two usages:
#unite(data, col, ..., sep = "_", remove = TRUE)
#unite_(data, col, from, sep = "_", remove = TRUE)
From the unite_ help page, the from option is defined as: "Names of existing columns as character vector."
Use column names as opposed to the column numbers provided the desired results:
unite_(x1, "mix", names(x1[,2:ncol(x1)]), sep=",")
# Sample mix
#1 A -,-,x
#2 B y,z,a
I tried with "Unite" but it did not work. However, It worked very well with "paste" function.
df$new_col <- paste(df$col1,df$col2,sep="-")
or if you have more columns to join,
df$new_col <- paste(df$col1,df$col2,df$col3,....,sep="-")
I have a data frame that looks like this:
> mydf
val1 val2
hsa-let-7a 2.139890 -0.03477569
hsa-let-7b 2.102590 0.04108795
hsa-let-7c 2.061705 0.02375882
hsa-let-7d 1.938950 -0.04364545
hsa-let-7e 1.889000 -0.10575235
hsa-let-7f 2.264296 0.08465690
Note that from 3 columns only 2nd and 3rd are names.
What I want to do is to name the first column (plus rename the 2nd and 3rd).
But why this command failed?
colnames(mydf) <- c("COL1","VAL1","VAL2");
What's the right way to do it?
It gave me:
Error in `colnames<-`(`*tmp*`, value = c("COL1", "VAL1", "VAL2" :
'names' attribute [3] must be the same length as the vector [2]
You could join the row names to the dataframe, like this:
mydf <- cbind(rownames(mydf), mydf)
rownames(mydf) <- NULL
colnames(mydf) <- c("COL1","VAL1","VAL2")
Or, in one step:
setNames(cbind(rownames(mydf), mydf, row.names = NULL),
c("COL1", "VAL1", "VAL2"))
# COL1 VAL1 VAL2
# 1 hsa-let-7a 2.139890 -0.03477569
# 2 hsa-let-7b 2.102590 0.04108795
# 3 hsa-let-7c 2.061705 0.02375882
# 4 hsa-let-7d 1.938950 -0.04364545
# 5 hsa-let-7e 1.889000 -0.10575235
# 6 hsa-let-7f 2.264296 0.08465690
this may also work in your case,
mydf <- cbind(rownames(mydf),mydf)
rownames(mydf) <- NULL
colnames(mydf) <- c(names(mydf)) #to not write all the column names
colnames(mydf)[1] <- "name.of.the.first.column"
names(mydf)
If one wants to use a tidyverse solution within his/her pipeline, this works
rownames_to_column(mydf, var = "var_name")
The function is contained in the tibble package