Related
I have a sample data frame like below:
data <- data.frame(matrix(sample(1:40), 4, 10, dimnames = list(1:4, LETTERS[1:10])))
I want to know how can I select multiple columns and convert them together to factors. I usually do it in the way like data$A = as.factor(data$A). But when the data frame is very large and contains lots of columns, this way will be very time consuming. Does anyone know of a better way to do it?
Choose some columns to coerce to factors:
cols <- c("A", "C", "D", "H")
Use lapply() to coerce and replace the chosen columns:
data[cols] <- lapply(data[cols], factor) ## as.factor() could also be used
Check the result:
sapply(data, class)
# A B C D E F G
# "factor" "integer" "factor" "factor" "integer" "integer" "integer"
# H I J
# "factor" "integer" "integer"
Here is an option using dplyr. The %<>% operator from magrittr update the lhs object with the resulting value.
library(magrittr)
library(dplyr)
cols <- c("A", "C", "D", "H")
data %<>%
mutate_each_(funs(factor(.)),cols)
str(data)
#'data.frame': 4 obs. of 10 variables:
# $ A: Factor w/ 4 levels "23","24","26",..: 1 2 3 4
# $ B: int 15 13 39 16
# $ C: Factor w/ 4 levels "3","5","18","37": 2 1 3 4
# $ D: Factor w/ 4 levels "2","6","28","38": 3 1 4 2
# $ E: int 14 4 22 20
# $ F: int 7 19 36 27
# $ G: int 35 40 21 10
# $ H: Factor w/ 4 levels "11","29","32",..: 1 4 3 2
# $ I: int 17 1 9 25
# $ J: int 12 30 8 33
Or if we are using data.table, either use a for loop with set
setDT(data)
for(j in cols){
set(data, i=NULL, j=j, value=factor(data[[j]]))
}
Or we can specify the 'cols' in .SDcols and assign (:=) the rhs to 'cols'
setDT(data)[, (cols):= lapply(.SD, factor), .SDcols=cols]
The more recent tidyverse way is to use the mutate_at function:
library(tidyverse)
library(magrittr)
set.seed(88)
data <- data.frame(matrix(sample(1:40), 4, 10, dimnames = list(1:4, LETTERS[1:10])))
cols <- c("A", "C", "D", "H")
data %<>% mutate_at(cols, factor)
str(data)
$ A: Factor w/ 4 levels "5","17","18",..: 2 1 4 3
$ B: int 36 35 2 26
$ C: Factor w/ 4 levels "22","31","32",..: 1 2 4 3
$ D: Factor w/ 4 levels "1","9","16","39": 3 4 1 2
$ E: int 3 14 30 38
$ F: int 27 15 28 37
$ G: int 19 11 6 21
$ H: Factor w/ 4 levels "7","12","20",..: 1 3 4 2
$ I: int 23 24 13 8
$ J: int 10 25 4 33
As of 2021 (still current in early 2023), the current tidyverse/dplyr approach would be to use across, and a <tidy-select> statement.
library(dplyr)
data %>% mutate(across(*<tidy-select>*, *function*))
across(<tidy-select>) allows very consistent and easy selection of columns to transform.
Some examples:
data %>% mutate(across(c(A, B, C, E), as.factor)) # select columns A to C, and E (by name)
data %>% mutate(across(where(is.character), as.factor)) # select character columns
data %>% mutate(across(1:5, as.factor)) # select first 5 columns (by index)
You can use mutate_if (dplyr):
For example, coerce integer in factor:
mydata=structure(list(a = 1:10, b = 1:10, c = c("a", "a", "b", "b",
"c", "c", "c", "c", "c", "c")), row.names = c(NA, -10L), class = c("tbl_df",
"tbl", "data.frame"))
# A tibble: 10 x 3
a b c
<int> <int> <chr>
1 1 1 a
2 2 2 a
3 3 3 b
4 4 4 b
5 5 5 c
6 6 6 c
7 7 7 c
8 8 8 c
9 9 9 c
10 10 10 c
Use the function:
library(dplyr)
mydata%>%
mutate_if(is.integer,as.factor)
# A tibble: 10 x 3
a b c
<fct> <fct> <chr>
1 1 1 a
2 2 2 a
3 3 3 b
4 4 4 b
5 5 5 c
6 6 6 c
7 7 7 c
8 8 8 c
9 9 9 c
10 10 10 c
and, for completeness and with regards to this question asking about changing string columns only, there's mutate_if:
data <- cbind(stringVar = sample(c("foo","bar"),10,replace=TRUE),
data.frame(matrix(sample(1:40), 10, 10, dimnames = list(1:10, LETTERS[1:10]))),stringsAsFactors=FALSE)
factoredData = data %>% mutate_if(is.character,funs(factor(.)))
Here is a data.table example. I used grep in this example because that's how I often select many columns by using partial matches to their names.
library(data.table)
data <- data.table(matrix(sample(1:40), 4, 10, dimnames = list(1:4, LETTERS[1:10])))
factorCols <- grep(pattern = "A|C|D|H", x = names(data), value = TRUE)
data[, (factorCols) := lapply(.SD, as.factor), .SDcols = factorCols]
A simple and updated solution
data <- data %>%
mutate_at(cols, list(~factor(.)))
If you have another objective of getting in values from the table then using them to be converted, you can try the following way
### pre processing
ind <- bigm.train[,lapply(.SD,is.character)]
ind <- names(ind[,.SD[T]])
### Convert multiple columns to factor
bigm.train[,(ind):=lapply(.SD,factor),.SDcols=ind]
This selects columns which are specifically character based and then converts them to factor.
Here is another tidyverse approach using the modify_at() function from the purrr package.
library(purrr)
# Data frame with only integer columns
data <- data.frame(matrix(sample(1:40), 4, 10, dimnames = list(1:4, LETTERS[1:10])))
# Modify specified columns to a factor class
data_with_factors <- data %>%
purrr::modify_at(c("A", "C", "E"), factor)
# Check the results:
str(data_with_factors)
# 'data.frame': 4 obs. of 10 variables:
# $ A: Factor w/ 4 levels "8","12","33",..: 1 3 4 2
# $ B: int 25 32 2 19
# $ C: Factor w/ 4 levels "5","15","35",..: 1 3 4 2
# $ D: int 11 7 27 6
# $ E: Factor w/ 4 levels "1","4","16","20": 2 3 1 4
# $ F: int 21 23 39 18
# $ G: int 31 14 38 26
# $ H: int 17 24 34 10
# $ I: int 13 28 30 29
# $ J: int 3 22 37 9
It appears that the use of SAPPLY on a data.frame to convert variables to factors at once does not work as it produces a matrix/ array. My approach is to use LAPPLY instead, as follows.
## let us create a data.frame here
class <- c("7", "6", "5", "3")
cash <- c(100, 200, 300, 150)
height <- c(170, 180, 150, 165)
people <- data.frame(class, cash, height)
class(people) ## This is a dataframe
## We now apply lapply to the data.frame as follows.
bb <- lapply(people, as.factor) %>% data.frame()
## The lapply part returns a list which we coerce back to a data.frame
class(bb) ## A data.frame
##Now let us check the classes of the variables
class(bb$class)
class(bb$height)
class(bb$cash) ## as expected, are all factors.
I am using the following code, which works fine (improvement suggestions very much welcome):
WeeklySlopes <- function(Year, Week){
DynamicQuery <- paste('select DayOfYear, Week, Year, Close from SourceData where year =', Year, 'and week =', Week, 'order by DayOfYear')
SubData = sqldf(DynamicQuery)
SubData$X <- as.numeric(rownames(SubData))
lmfit <- lm(Close ~ X, data = SubData)
lmfit <- tidy(lmfit)
Slope <- as.numeric(sqldf("select estimate from lmfit where term = 'X'"))
e <- globalenv()
e$WeeklySlopesDf[nrow(e$WeeklySlopesDf) + 1,] = c(Year,Week, Slope)
}
WeeklySlopesDf <- data.frame(Year = integer(), Week = integer(), Slope = double())
WeeklySlopes(2017, 15)
WeeklySlopes(2017, 14)
head(WeeklySlopesDf)
Is there really no other way to append a row to my existing dataframe. I seem to need to access the globalenv. On the other hand, why can sqldf 'see' the 'global' dataframe SourceData?
dfrm <- data.frame(a=1:10, b=letters[1:10]) # reproducible example
myfunc <- function(new_a=20){ g <- globalenv(); g$dfrm[3,1] <- new_a; cat(dfrm[3,1])}
myfunc()
20
dfrm
a b
1 1 a
2 2 b
3 20 c # so your strategy might work, although it's unconventional.
Now try to extend dataframe outside a function:
dfrm[11, ] <- c(a=20,b="c")
An occult disaster (conversion of numeric column to character):
str(dfrm)
'data.frame': 11 obs. of 2 variables:
$ a: chr "1" "2" "20" "4" ...
$ b: Factor w/ 10 levels "a","b","c","d",..: 1 2 3 4 5 6 7 8 9 10 ...
So use a list to avoid occult coercion:
dfrm <- data.frame(a=1:10, b=letters[1:10]) # start over
dfrm[11, ] <- list(a=20,b="c")
str(dfrm)
'data.frame': 11 obs. of 2 variables:
$ a: num 1 2 3 4 5 6 7 8 9 10 ...
$ b: Factor w/ 10 levels "a","b","c","d",..: 1 2 3 4 5 6 7 8 9 10 ...
Now try within a function:
myfunc <- function(new_a=20, new_b="ZZ"){ g <- globalenv(); g$dfrm[nrow(dfrm)+1, ] <- list(a=new_a,b=new_b)}
myfunc()
Warning message:
In `[<-.factor`(`*tmp*`, iseq, value = "ZZ") :
invalid factor level, NA generated
str(dfrm)
'data.frame': 12 obs. of 2 variables:
$ a: num 1 2 3 4 5 6 7 8 9 10 ...
$ b: Factor w/ 10 levels "a","b","c","d",..: 1 2 3 4 5 6 7 8 9 10 ...
So it succeeds, but if there are any factor columns, non-existent levels will get turned into NA values (with a warning). You method of using named access to objects in the global environment is rather unconventional but there is a set of tested methods that you might want to examine. Look at ?R6. Other options are <<- and assign which allows one to specify the environment in which the assignment is to occur.
I have a sample data frame like below:
data <- data.frame(matrix(sample(1:40), 4, 10, dimnames = list(1:4, LETTERS[1:10])))
I want to know how can I select multiple columns and convert them together to factors. I usually do it in the way like data$A = as.factor(data$A). But when the data frame is very large and contains lots of columns, this way will be very time consuming. Does anyone know of a better way to do it?
Choose some columns to coerce to factors:
cols <- c("A", "C", "D", "H")
Use lapply() to coerce and replace the chosen columns:
data[cols] <- lapply(data[cols], factor) ## as.factor() could also be used
Check the result:
sapply(data, class)
# A B C D E F G
# "factor" "integer" "factor" "factor" "integer" "integer" "integer"
# H I J
# "factor" "integer" "integer"
Here is an option using dplyr. The %<>% operator from magrittr update the lhs object with the resulting value.
library(magrittr)
library(dplyr)
cols <- c("A", "C", "D", "H")
data %<>%
mutate_each_(funs(factor(.)),cols)
str(data)
#'data.frame': 4 obs. of 10 variables:
# $ A: Factor w/ 4 levels "23","24","26",..: 1 2 3 4
# $ B: int 15 13 39 16
# $ C: Factor w/ 4 levels "3","5","18","37": 2 1 3 4
# $ D: Factor w/ 4 levels "2","6","28","38": 3 1 4 2
# $ E: int 14 4 22 20
# $ F: int 7 19 36 27
# $ G: int 35 40 21 10
# $ H: Factor w/ 4 levels "11","29","32",..: 1 4 3 2
# $ I: int 17 1 9 25
# $ J: int 12 30 8 33
Or if we are using data.table, either use a for loop with set
setDT(data)
for(j in cols){
set(data, i=NULL, j=j, value=factor(data[[j]]))
}
Or we can specify the 'cols' in .SDcols and assign (:=) the rhs to 'cols'
setDT(data)[, (cols):= lapply(.SD, factor), .SDcols=cols]
The more recent tidyverse way is to use the mutate_at function:
library(tidyverse)
library(magrittr)
set.seed(88)
data <- data.frame(matrix(sample(1:40), 4, 10, dimnames = list(1:4, LETTERS[1:10])))
cols <- c("A", "C", "D", "H")
data %<>% mutate_at(cols, factor)
str(data)
$ A: Factor w/ 4 levels "5","17","18",..: 2 1 4 3
$ B: int 36 35 2 26
$ C: Factor w/ 4 levels "22","31","32",..: 1 2 4 3
$ D: Factor w/ 4 levels "1","9","16","39": 3 4 1 2
$ E: int 3 14 30 38
$ F: int 27 15 28 37
$ G: int 19 11 6 21
$ H: Factor w/ 4 levels "7","12","20",..: 1 3 4 2
$ I: int 23 24 13 8
$ J: int 10 25 4 33
As of 2021 (still current in early 2023), the current tidyverse/dplyr approach would be to use across, and a <tidy-select> statement.
library(dplyr)
data %>% mutate(across(*<tidy-select>*, *function*))
across(<tidy-select>) allows very consistent and easy selection of columns to transform.
Some examples:
data %>% mutate(across(c(A, B, C, E), as.factor)) # select columns A to C, and E (by name)
data %>% mutate(across(where(is.character), as.factor)) # select character columns
data %>% mutate(across(1:5, as.factor)) # select first 5 columns (by index)
You can use mutate_if (dplyr):
For example, coerce integer in factor:
mydata=structure(list(a = 1:10, b = 1:10, c = c("a", "a", "b", "b",
"c", "c", "c", "c", "c", "c")), row.names = c(NA, -10L), class = c("tbl_df",
"tbl", "data.frame"))
# A tibble: 10 x 3
a b c
<int> <int> <chr>
1 1 1 a
2 2 2 a
3 3 3 b
4 4 4 b
5 5 5 c
6 6 6 c
7 7 7 c
8 8 8 c
9 9 9 c
10 10 10 c
Use the function:
library(dplyr)
mydata%>%
mutate_if(is.integer,as.factor)
# A tibble: 10 x 3
a b c
<fct> <fct> <chr>
1 1 1 a
2 2 2 a
3 3 3 b
4 4 4 b
5 5 5 c
6 6 6 c
7 7 7 c
8 8 8 c
9 9 9 c
10 10 10 c
and, for completeness and with regards to this question asking about changing string columns only, there's mutate_if:
data <- cbind(stringVar = sample(c("foo","bar"),10,replace=TRUE),
data.frame(matrix(sample(1:40), 10, 10, dimnames = list(1:10, LETTERS[1:10]))),stringsAsFactors=FALSE)
factoredData = data %>% mutate_if(is.character,funs(factor(.)))
Here is a data.table example. I used grep in this example because that's how I often select many columns by using partial matches to their names.
library(data.table)
data <- data.table(matrix(sample(1:40), 4, 10, dimnames = list(1:4, LETTERS[1:10])))
factorCols <- grep(pattern = "A|C|D|H", x = names(data), value = TRUE)
data[, (factorCols) := lapply(.SD, as.factor), .SDcols = factorCols]
A simple and updated solution
data <- data %>%
mutate_at(cols, list(~factor(.)))
If you have another objective of getting in values from the table then using them to be converted, you can try the following way
### pre processing
ind <- bigm.train[,lapply(.SD,is.character)]
ind <- names(ind[,.SD[T]])
### Convert multiple columns to factor
bigm.train[,(ind):=lapply(.SD,factor),.SDcols=ind]
This selects columns which are specifically character based and then converts them to factor.
Here is another tidyverse approach using the modify_at() function from the purrr package.
library(purrr)
# Data frame with only integer columns
data <- data.frame(matrix(sample(1:40), 4, 10, dimnames = list(1:4, LETTERS[1:10])))
# Modify specified columns to a factor class
data_with_factors <- data %>%
purrr::modify_at(c("A", "C", "E"), factor)
# Check the results:
str(data_with_factors)
# 'data.frame': 4 obs. of 10 variables:
# $ A: Factor w/ 4 levels "8","12","33",..: 1 3 4 2
# $ B: int 25 32 2 19
# $ C: Factor w/ 4 levels "5","15","35",..: 1 3 4 2
# $ D: int 11 7 27 6
# $ E: Factor w/ 4 levels "1","4","16","20": 2 3 1 4
# $ F: int 21 23 39 18
# $ G: int 31 14 38 26
# $ H: int 17 24 34 10
# $ I: int 13 28 30 29
# $ J: int 3 22 37 9
It appears that the use of SAPPLY on a data.frame to convert variables to factors at once does not work as it produces a matrix/ array. My approach is to use LAPPLY instead, as follows.
## let us create a data.frame here
class <- c("7", "6", "5", "3")
cash <- c(100, 200, 300, 150)
height <- c(170, 180, 150, 165)
people <- data.frame(class, cash, height)
class(people) ## This is a dataframe
## We now apply lapply to the data.frame as follows.
bb <- lapply(people, as.factor) %>% data.frame()
## The lapply part returns a list which we coerce back to a data.frame
class(bb) ## A data.frame
##Now let us check the classes of the variables
class(bb$class)
class(bb$height)
class(bb$cash) ## as expected, are all factors.
I have a sample data frame like below:
data <- data.frame(matrix(sample(1:40), 4, 10, dimnames = list(1:4, LETTERS[1:10])))
I want to know how can I select multiple columns and convert them together to factors. I usually do it in the way like data$A = as.factor(data$A). But when the data frame is very large and contains lots of columns, this way will be very time consuming. Does anyone know of a better way to do it?
Choose some columns to coerce to factors:
cols <- c("A", "C", "D", "H")
Use lapply() to coerce and replace the chosen columns:
data[cols] <- lapply(data[cols], factor) ## as.factor() could also be used
Check the result:
sapply(data, class)
# A B C D E F G
# "factor" "integer" "factor" "factor" "integer" "integer" "integer"
# H I J
# "factor" "integer" "integer"
Here is an option using dplyr. The %<>% operator from magrittr update the lhs object with the resulting value.
library(magrittr)
library(dplyr)
cols <- c("A", "C", "D", "H")
data %<>%
mutate_each_(funs(factor(.)),cols)
str(data)
#'data.frame': 4 obs. of 10 variables:
# $ A: Factor w/ 4 levels "23","24","26",..: 1 2 3 4
# $ B: int 15 13 39 16
# $ C: Factor w/ 4 levels "3","5","18","37": 2 1 3 4
# $ D: Factor w/ 4 levels "2","6","28","38": 3 1 4 2
# $ E: int 14 4 22 20
# $ F: int 7 19 36 27
# $ G: int 35 40 21 10
# $ H: Factor w/ 4 levels "11","29","32",..: 1 4 3 2
# $ I: int 17 1 9 25
# $ J: int 12 30 8 33
Or if we are using data.table, either use a for loop with set
setDT(data)
for(j in cols){
set(data, i=NULL, j=j, value=factor(data[[j]]))
}
Or we can specify the 'cols' in .SDcols and assign (:=) the rhs to 'cols'
setDT(data)[, (cols):= lapply(.SD, factor), .SDcols=cols]
The more recent tidyverse way is to use the mutate_at function:
library(tidyverse)
library(magrittr)
set.seed(88)
data <- data.frame(matrix(sample(1:40), 4, 10, dimnames = list(1:4, LETTERS[1:10])))
cols <- c("A", "C", "D", "H")
data %<>% mutate_at(cols, factor)
str(data)
$ A: Factor w/ 4 levels "5","17","18",..: 2 1 4 3
$ B: int 36 35 2 26
$ C: Factor w/ 4 levels "22","31","32",..: 1 2 4 3
$ D: Factor w/ 4 levels "1","9","16","39": 3 4 1 2
$ E: int 3 14 30 38
$ F: int 27 15 28 37
$ G: int 19 11 6 21
$ H: Factor w/ 4 levels "7","12","20",..: 1 3 4 2
$ I: int 23 24 13 8
$ J: int 10 25 4 33
As of 2021 (still current in early 2023), the current tidyverse/dplyr approach would be to use across, and a <tidy-select> statement.
library(dplyr)
data %>% mutate(across(*<tidy-select>*, *function*))
across(<tidy-select>) allows very consistent and easy selection of columns to transform.
Some examples:
data %>% mutate(across(c(A, B, C, E), as.factor)) # select columns A to C, and E (by name)
data %>% mutate(across(where(is.character), as.factor)) # select character columns
data %>% mutate(across(1:5, as.factor)) # select first 5 columns (by index)
You can use mutate_if (dplyr):
For example, coerce integer in factor:
mydata=structure(list(a = 1:10, b = 1:10, c = c("a", "a", "b", "b",
"c", "c", "c", "c", "c", "c")), row.names = c(NA, -10L), class = c("tbl_df",
"tbl", "data.frame"))
# A tibble: 10 x 3
a b c
<int> <int> <chr>
1 1 1 a
2 2 2 a
3 3 3 b
4 4 4 b
5 5 5 c
6 6 6 c
7 7 7 c
8 8 8 c
9 9 9 c
10 10 10 c
Use the function:
library(dplyr)
mydata%>%
mutate_if(is.integer,as.factor)
# A tibble: 10 x 3
a b c
<fct> <fct> <chr>
1 1 1 a
2 2 2 a
3 3 3 b
4 4 4 b
5 5 5 c
6 6 6 c
7 7 7 c
8 8 8 c
9 9 9 c
10 10 10 c
and, for completeness and with regards to this question asking about changing string columns only, there's mutate_if:
data <- cbind(stringVar = sample(c("foo","bar"),10,replace=TRUE),
data.frame(matrix(sample(1:40), 10, 10, dimnames = list(1:10, LETTERS[1:10]))),stringsAsFactors=FALSE)
factoredData = data %>% mutate_if(is.character,funs(factor(.)))
Here is a data.table example. I used grep in this example because that's how I often select many columns by using partial matches to their names.
library(data.table)
data <- data.table(matrix(sample(1:40), 4, 10, dimnames = list(1:4, LETTERS[1:10])))
factorCols <- grep(pattern = "A|C|D|H", x = names(data), value = TRUE)
data[, (factorCols) := lapply(.SD, as.factor), .SDcols = factorCols]
A simple and updated solution
data <- data %>%
mutate_at(cols, list(~factor(.)))
If you have another objective of getting in values from the table then using them to be converted, you can try the following way
### pre processing
ind <- bigm.train[,lapply(.SD,is.character)]
ind <- names(ind[,.SD[T]])
### Convert multiple columns to factor
bigm.train[,(ind):=lapply(.SD,factor),.SDcols=ind]
This selects columns which are specifically character based and then converts them to factor.
Here is another tidyverse approach using the modify_at() function from the purrr package.
library(purrr)
# Data frame with only integer columns
data <- data.frame(matrix(sample(1:40), 4, 10, dimnames = list(1:4, LETTERS[1:10])))
# Modify specified columns to a factor class
data_with_factors <- data %>%
purrr::modify_at(c("A", "C", "E"), factor)
# Check the results:
str(data_with_factors)
# 'data.frame': 4 obs. of 10 variables:
# $ A: Factor w/ 4 levels "8","12","33",..: 1 3 4 2
# $ B: int 25 32 2 19
# $ C: Factor w/ 4 levels "5","15","35",..: 1 3 4 2
# $ D: int 11 7 27 6
# $ E: Factor w/ 4 levels "1","4","16","20": 2 3 1 4
# $ F: int 21 23 39 18
# $ G: int 31 14 38 26
# $ H: int 17 24 34 10
# $ I: int 13 28 30 29
# $ J: int 3 22 37 9
It appears that the use of SAPPLY on a data.frame to convert variables to factors at once does not work as it produces a matrix/ array. My approach is to use LAPPLY instead, as follows.
## let us create a data.frame here
class <- c("7", "6", "5", "3")
cash <- c(100, 200, 300, 150)
height <- c(170, 180, 150, 165)
people <- data.frame(class, cash, height)
class(people) ## This is a dataframe
## We now apply lapply to the data.frame as follows.
bb <- lapply(people, as.factor) %>% data.frame()
## The lapply part returns a list which we coerce back to a data.frame
class(bb) ## A data.frame
##Now let us check the classes of the variables
class(bb$class)
class(bb$height)
class(bb$cash) ## as expected, are all factors.
This does not work
> dfi=data.frame(v1=c(1,1),v2=c(2,2))
> dfi
v1 v2
1 1 2
2 1 2
> df$df=dfi
Error in `$<-.data.frame`(`*tmp*`, "df", value = list(v1 = c(1, 1), v2 = c(2, :
replacement has 2 rows, data has 0
df$df=I(dfi) has the same error. Please help.
Thank you.
Moved this from comments for formatting reasons:
What exactly are you trying to achieve? If you want the contents of dfi passed to df you can use this code:
df <- data.frame(matrix(vector(), 0, 2, dimnames=list(c(), c("V1", "V2"))), stringsAsFactors=F)
df=dfi
As #joran says, it is unclear why you would ever want to do this. Nevertheless, it is possible.
One of the requirements of a data frame is that all the columns have the same number of rows. This is why you are getting the error. Something like this will work:
dfi <- data.frame(v1=c(1,1),v2=c(2,2)) # 2 rows
df <- data.frame(x=1:2) # also 2 rows
df$df <- dfi # works now
Printing would lead you to believe that df has three columns...
df
# x df.v1 df.v2
# 1 1 1 2
# 2 2 1 2
but it does not!
str(df)
# 'data.frame': 2 obs. of 2 variables:
# $ x : int 1 2
# $ df:'data.frame': 2 obs. of 2 variables:
# ..$ v1: num 1 1
# ..$ v2: num 2 2
Since df$df is a data frame
class(df$df)
# [1] "data.frame"
you can use the standard data frame accessors...
df$df$v1
# [1] 1 1
df$df[1,]
# v1 v2
# 1 1 2
Incidentally, RStudio has trouble displaying this type of data structure; view(df) gives an inaccurate display of the structure.
Finally, you are probably better off creating a list of data frames, rather than a data frame containing data frames:
df <- data.frame(grp=rep(LETTERS[1:3],each=5),x=rnorm(15),y=rpois(15,5))
df.lst <- split(df,df$grp) # creates a list of data frames
df.lst$A
# grp x y
# 1 A -1.3606420 10
# 2 A -0.4511408 5
# 3 A -1.1951950 4
# 4 A -0.8017765 5
# 5 A -0.2816298 9
df.lst$A$x
# [1] -1.3606420 -0.4511408 -1.1951950 -0.8017765 -0.2816298