Reputation: 107
I have a data frame df
whose first column is a character vector and the rest numeric.
Example data frame:
df <- data.frame(my_names=sample(LETTERS,4,replace=F),
column2=sample(1.3:100.3,4,replace=T),
column3=sample(1.3:100.3,4,replace=T),
column4=sample(1.3:100.3,4,replace=T),
column5=sample(1.3:100.3,4,replace=T))
> df
my_names column2 column3 column4 column5
1 A 8.3 1.3 19.3 91.3
2 E 18.3 42.3 8.3 76.3
3 O 6.3 46.3 26.3 91.3
4 M 73.3 6.3 59.3 93.3
Now I want to create 4 different data frames like this:
And store them into a list.
d1
would look like:
> d1
my_names column2
1 A 8.3
2 E 18.3
3 O 6.3
4 M 73.3
I have tried:
>the_list <- vector("list",ncol(df)-1)
> for(i in 1:length(the_list)){ for(j in 2:ncol(df)){
+ the_list[[i]] <- select(df, my_names,j)
+ }
+ }
Note: Using an external vector in selections is ambiguous.
ℹ Use `all_of(j)` instead of `j` to silence this message.
But I get a list where all data frames are with column5
:
> str(the_list)
List of 4
$ :'data.frame': 4 obs. of 2 variables:
..$ my_names: chr [1:4] "A" "E" "O" "M"
..$ column5 : num [1:4] 91.3 76.3 91.3 93.3
$ :'data.frame': 4 obs. of 2 variables:
..$ my_names: chr [1:4] "A" "E" "O" "M"
..$ column5 : num [1:4] 91.3 76.3 91.3 93.3
$ :'data.frame': 4 obs. of 2 variables:
..$ my_names: chr [1:4] "A" "E" "O" "M"
..$ column5 : num [1:4] 91.3 76.3 91.3 93.3
$ :'data.frame': 4 obs. of 2 variables:
..$ my_names: chr [1:4] "A" "E" "O" "M"
..$ column5 : num [1:4] 91.3 76.3 91.3 93.3
I take the recommendation from the error (using all_of(j)) and write:
> for(i in 1:length(the_list)){
for(j in 2:ncol(df)){
the_list[[i]] <- select(df, my_names,all_of(j))
}
}
But the result is the same as above.
I have read that one could use split
, but I have nothing to group by, it´s each column separately.
e.g this does not work:
new_list<-list(split(df, colnames(df))
I get a wird list of 1.
Upvotes: 3
Views: 1389
Reputation: 39858
One option using purrr
and dplyr
could be:
map(2:length(df),
~ df %>%
select(1, all_of(.x)))
[[1]]
my_names column2
1 N 21.3
2 S 91.3
3 T 50.3
4 F 34.3
[[2]]
my_names column3
1 N 84.3
2 S 20.3
3 T 1.3
4 F 61.3
[[3]]
my_names column4
1 N 4.3
2 S 9.3
3 T 93.3
4 F 58.3
[[4]]
my_names column5
1 N 33.3
2 S 61.3
3 T 12.3
4 F 91.3
If you are interested in a named list:
set_names(map(2:length(df),
~ df %>%
select(1, all_of(.x))),
paste0("df", 2:length(df) - 1))
Upvotes: 0
Reputation: 35554
A base
solution with lapply()
.
lapply(seq_along(df)[-1], function(x) df[c(1, x)])
or
lapply(names(df)[-1], function(x) df[c("my_names", x)])
Afterward, you can use setNames()
to assign names to the list.
Upvotes: 0
Reputation: 388982
Using lapply
:
data <- lapply(seq_along(df[-1]), function(x) cbind(df[1], df[x+1]))
data
would have list of dataframes. If you want them in separate dataframes, name them and use list2env
.
names(data) <- paste0('d', seq_along(data))
list2env(data, .GlobalEnv)
Upvotes: 0
Reputation: 101335
Maybe you can try list2env
list2env(
setNames(
lapply(seq_along(df)[-1], function(k) cbind(df[c(1, k)])),
paste0("d", seq_along(df[-1]))
),
envir = .GlobalEnv
)
If you want a list of dataframes only, you can remove list2env
, i.e.,
setNames(
lapply(seq_along(df)[-1], function(k) cbind(df[c(1, k)])),
paste0("d", seq_along(df[-1]))
)
which gives
$d1
my_names column2
1 C 45.3
2 M 89.3
3 G 35.3
4 T 48.3
$d2
my_names column3
1 C 41.3
2 M 56.3
3 G 34.3
4 T 95.3
$d3
my_names column4
1 C 78.3
2 M 7.3
3 G 60.3
4 T 19.3
$d4
my_names column5
1 C 76.3
2 M 51.3
3 G 96.3
4 T 96.3
Upvotes: 0
Reputation: 39595
Try this tidyverse
approach. You can format your data to long to transform columns into rows. Then, with split()
you can create a list based on the column name. Finally, you can apply a function to transform your data to wide at each dataframe in the list and reach the desired output. Here the code:
library(tidyverse)
#Data
df <- data.frame(my_names=sample(LETTERS,4,replace=F),
column2=sample(1.3:100.3,4,replace=T),
column3=sample(1.3:100.3,4,replace=T),
column4=sample(1.3:100.3,4,replace=T),
column5=sample(1.3:100.3,4,replace=T))
#Reshape to long
df2 <- df %>% pivot_longer(cols = -1)
#Split into a list
List <- split(df2,df2$name)
#Now reshape function for wide format
List2 <- lapply(List,function(x){x<-pivot_wider(x,names_from = name,values_from = value);return(x)})
names(List2) <- paste0('df',1:length(List2))
Output:
List2
$df1
# A tibble: 4 x 2
my_names column2
<fct> <dbl>
1 N 21.3
2 H 35.3
3 X 42.3
4 U 89.3
$df2
# A tibble: 4 x 2
my_names column3
<fct> <dbl>
1 N 94.3
2 H 54.3
3 X 2.3
4 U 38.3
$df3
# A tibble: 4 x 2
my_names column4
<fct> <dbl>
1 N 75.3
2 H 94.3
3 X 87.3
4 U 100.
$df4
# A tibble: 4 x 2
my_names column5
<fct> <dbl>
1 N 60.3
2 H 88.3
3 X 14.3
4 U 99.3
Upvotes: 1