Column name in a dataframe inside a list from lapply-CodePudding

I am transforming this dataframe into a list with separate dataframes

df <- data.frame(1:10, 11:20, letters[seq(from=1, to=10)])
colnames(df) <- c(letters[seq(from=1,to=3)])

list.df <- lapply(df[,1:2], function(j) {
  data.frame(j, df$c)
})

but I want to set the name the first column name to be the same as the name of the list (a and b respectively). While in this exemple, inside each list, the first column of the dataframe is always "j".

How could I do that???

Also, how could I change the second column from "df.c" to anything else??

Thanks

CodePudding user response：

Iterate over the column number and subset the dataframe to get the same name as original column names. Use cbind to bind c column and assign name of your choice.

lapply(1:2, function(i) cbind(df[i], new_name = df$c))

#[[1]]
#    a new_name
#1   1        a
#2   2        b
#3   3        c
#4   4        d
#5   5        e
#6   6        f
#7   7        g
#8   8        h
#9   9        i
#10 10        j

#[[2]]
#    b new_name
#1  11        a
#2  12        b
#3  13        c
#4  14        d
#5  15        e
#6  16        f
#7  17        g
#8  18        h
#9  19        i
#10 20        j

CodePudding user response：

we can use parent.frame() to get the current column index and set its name;

df <- data.frame(1:10, 11:20, letters[seq(from=1, to=10)])
colnames(df) <- c(letters[seq(from=1,to=3)])

list.df <- lapply(df[,1:2], function(j) {
  name_index <- parent.frame()$i
  name_to_set <- colnames(df[,1:2])[name_index] 

  subdf <- data.frame(j,the_name_you_wish = df$c)

  colnames(subdf) <- gsub('j',name_to_set,colnames(subdf))
    
  subdf
  
})

also set a new name to df$c

output;

$a
    a the_name_you_wish
1   1                 a
2   2                 b
3   3                 c
4   4                 d
5   5                 e
6   6                 f
7   7                 g
8   8                 h
9   9                 i
10 10                 j

$b
    b the_name_you_wish
1  11                 a
2  12                 b
3  13                 c
4  14                 d
5  15                 e
6  16                 f
7  17                 g
8  18                 h
9  19                 i
10 20                 j

CodePudding user response：

df <- data.frame(1:10, 11:20, letters[seq(from=1, to=10)])
colnames(df) <- c(letters[seq(from=1,to=3)])

library(dplyr, warn.conflicts = FALSE)

list.df <- lapply(1:2, function(j) {
  df %>% select(!!!j, c)
})

list.df
#> [[1]]
#>     a c
#> 1   1 a
#> 2   2 b
#> 3   3 c
#> 4   4 d
#> 5   5 e
#> 6   6 f
#> 7   7 g
#> 8   8 h
#> 9   9 i
#> 10 10 j
#> 
#> [[2]]
#>     b c
#> 1  11 a
#> 2  12 b
#> 3  13 c
#> 4  14 d
#> 5  15 e
#> 6  16 f
#> 7  17 g
#> 8  18 h
#> 9  19 i
#> 10 20 j

^{Created on 2022-02-25 by the reprex package (v2.0.1)}

Or if you want to rename

df <- data.frame(1:10, 11:20, letters[seq(from=1, to=10)])
colnames(df) <- c(letters[seq(from=1,to=3)])

library(dplyr, warn.conflicts = FALSE)

list.df <- lapply(1:2, function(j) {
  df %>% select(!!!j, newname = c)
})

list.df
#> [[1]]
#>     a newname
#> 1   1       a
#> 2   2       b
#> 3   3       c
#> 4   4       d
#> 5   5       e
#> 6   6       f
#> 7   7       g
#> 8   8       h
#> 9   9       i
#> 10 10       j
#> 
#> [[2]]
#>     b newname
#> 1  11       a
#> 2  12       b
#> 3  13       c
#> 4  14       d
#> 5  15       e
#> 6  16       f
#> 7  17       g
#> 8  18       h
#> 9  19       i
#> 10 20       j

^{Created on 2022-02-25 by the reprex package (v2.0.1)}