I have a list containing multiple data frames, and each list element has a unique name. The structure is similar to this dummy data
a <- data.frame(z = rnorm(20), y = rnorm(20))
b <- data.frame(z = rnorm(30), y = rnorm(30))
c <- data.frame(z = rnorm(40), y = rnorm(40))
d <- data.frame(z = rnorm(50), y = rnorm(50))
my.list <- list(a,b,c,d)
names(my.list) <- c("a","b","c","d")
I want to create a column in each of the data frames that has the name of it's respective list element. My goal is to merge all the list element into a single data frame, and know which data frame they came from originally. The end result I'm looking for is something like this:
z y group
1 0.6169132 0.09803228 a
2 1.1610584 0.50356131 a
3 0.6399438 0.84810547 a
4 1.0878453 1.00472105 b
5 -0.3137200 -1.20707112 b
6 1.1428834 0.87852556 b
7 -1.0651735 -0.18614224 c
8 1.1629891 -0.30184443 c
9 -0.7980089 -0.35578381 c
10 1.4651651 -0.30586852 d
11 1.1936547 1.98858128 d
12 1.6284174 -0.17042835 d
My first thought was to use mutate to assign the list element name to a column in each respective data frame, but it appears that when used within lapply, names() refers to the column names, not the list element names
test <- lapply(my.list, function(x) mutate(x, group = names(x)))
Error: Column `group` must be length 20 (the number of rows) or one, not 2
Any suggestions as to how I could approach this problem?
there is no need to mutate just bind using dplyr's bind_rows
library(tidyverse)
my.list %>%
bind_rows(.id = "groups")
Obviously requires that the list is named.
We can use Map from base R
Map(cbind, my.list, group = names(my.list))
Or with imap from purrr
library(dplyr)
library(purrr)
imap(my.list, ~ .x %>% mutate(group = .y))
Or if the intention is to create a single data.frame
library(data.table)
rbindlist(my.list. idcol = 'groups')
I would like to write a function which uses dplyr::filter() within the function. When writing the function I ran into an issue with using a parameter name in the function that is also a name of one of the columns of the data frame I am filtering.
Suppose I call the data frame to be filtered dat:
library(dplyr)
dat <- data.frame(
a = c(1:10),
b = c(2,2,2,2,2,3,1,1,4,4)
)
and name the function test.filter(),
test.filter <- function(b, test.data = dat){
dat.t <- filter(test.data,
b == b)
return(dat.t)
}
Here I am passing a value b to the function and asking it to filter the column b based on the value b. I believe the function
test.filter(b = 4,
test.data = dat)
should produce the same result as
filter(dat,
b == 4)
However this is not the case. I am wondering if there is something I am not considering in terms of the scope of a function. Any help is appreciated!
It is a case where the argument 'b' of the function is the same as the column name. One option is to do !! inside the function argument
test.filter <- function(b, test.data = dat){
filter(test.data,
b == !!b)
}
test.filter(b = 4,
test.data = dat)
If the argument passed to the function is similar to one of the column name in the dataframe, we can use the curly-curly ({{ }}) operator from rlang to evaluate column name
library(rlang)
test.filter <- function(b, test.data = dat) {
dplyr::filter(dat,{{b}} == b)
}
test.filter(b = 4,test.data = dat)
# a b
#1 9 4
#2 10 4
test.filter(b = 2,test.data = dat)
# a b
#1 1 2
#2 2 2
#3 3 2
#4 4 2
#5 5 2
Thanks for the helpful answers. A friend let me know that the underlying reason for the issue is dplyr uses lazy eval, so b==b evaluates to all true.
Is it possible in R to create argument names in a function call dynamically?
For example, if we start with
name <- "variable"
I would like to create a new data frame like this
a.new.data.frame <- data.frame(name = c(1, 2))
which of course does not work.
The only solution I could invent was
arg <- list(c(1, 2))
names(arg) <- name
a.new.data.frame <- do.call(data.frame, arg)
a.new.data.frame
# variable
#1 1
#2 2
I don't like this code, since it seems not to be elegant.
Is there a better way to do it?
PS Important! This is a more general problem I have when writing R-programmes (e.g. when I use ggplot, and many other cases). So, I expect general solutions to this (creation of data.frame is only an example).
A more compact code for dynamic args could look like this:
df <- do.call(data.frame, list(name = c(1, 2)))
You could use the ?dotsMethods to encapsulate the do.call in a generic function like this to save the noisy list() part of the call:
call.with.dyn.args <- function(f, ...) {
args <- list(...)
do.call(f, args)
}
df1 <- call.with.dyn.args(data.frame, a = 1:2, b = letters[1:2])
df1
# a b
# 1 1 a
# 2 2 b
But you also have other options for dynamic argument passing to functions without a do.call, eg.:
dyn.values <- c(1:2)
name = "dyn.values"
df2 <- data.frame(dyn.values, # values from a variable
name = dyn.values, # values from a variable + new name
static.arg = letters[1:2], # usual direct passing of an arg
name.from.variable = get(name)) # get the values from a variable whose name is stored in another variable
df2
# dyn.values name static.arg name.from.variable
# 1 1 1 a 1
# 2 2 2 b 2
An option using tidyverse
library(tibble)
library(dplyr)
tibble(!! name := c(1, 2))
# A tibble: 2 x 1
# variable
# <dbl>
#1 1
#2 2
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)
Is there any way to use string stored in variable as a column name in a new data frame? The expected result should be:
col.name <- 'col1'
df <- data.frame(col.name=1:4)
print(df)
# Real output
col.name
1 1
2 2
3 3
4 4
# Expected output
col1
1 1
2 2
3 3
4 4
I'm aware that I can create data frame and then use names() to rename column or use df[, col.name] for existing object, but I'd like to know if there is any other solution which could be used during creating data frame.
You cannot pass a variable into the name of an argument like that.
Instead what you can do is:
df <- data.frame(placeholder_name = 1:4)
names(df)[names(df) == "placeholder_name"] <- col.name
or use the default name of "V1":
df <- data.frame(1:4)
names(df)[names(df) == "V1"] <- col.name
or assign by position:
df <- data.frame(1:4)
names(df)[1] <- col.name
or if you only have one column just replace the entire names attribute:
df <- data.frame(1:4)
names(df) <- col.name
There's also the set_names function in the magrittr package that you can use to do this last solution in one step:
library(magrittr)
df <- set_names(data.frame(1:4), col.name)
But set_names is just an alias for:
df <- `names<-`(data.frame(1:4), col.name)
which is part of base R. Figuring out why this expression works and makes sense will be a good exercise.
In addition to ssdecontrol's answer, there is a second option.
You're looking for mget. First assign the name to a variable, then the value to the variable that you have previously assigned. After that, mget will evaluate the string and pass it to data.frame.
assign("col.name", "col1")
assign(paste(col.name), 1:4)
df <- data.frame(mget(col.name))
print(df)
col1
1 1
2 2
3 3
4 4
I don't recommend you do this, but:
col.name <- 'col1'
eval(parse(text=paste0('data.frame(', col.name, '=1:4)')))