CODEWITHSUNDEEP
rlapply
phil_t
phil_t

Reputation: 861

Create list of dataframes with each sorted on different column

I have a dataframe that is similar to - 

 Name  State  A  B  C
  Tom     NY  4  2  5
 Dick     IL  8  1  3
Harry     WI  2  6  4
 John     MD  5  4  9

I want to create a list of dataframes where each dataframe is sorted on either A or B or C, and the resulting dataframe contains the columns Name, State and the column sorted on.

lst
 $`A`
 Name  State  A
 Dick     IL  8
  Tom     NY  4
 John     MD  5
Harry     WI  2

 $`B`
 Name  State  B
Harry     WI  6
 John     MD  4
  Tom     NY  2
 Dick     IL  1


 $`C`
 Name  State  C
 John     MD  9
  Tom     NY  5
Harry     WI  4
 Dick     IL  3

I am trying to use lapply -

lst = lapply(df[sapply(df, is.numeric)], function(x){df[order(-x), ]})

but I am getting stuck on how to get the columns I want.

Upvotes: 3

Views: 45

Answers (1)

akrun
akrun

Reputation: 887221

We could use map to loop through the column names 'A', 'B', 'C', subset the columns of dataset with select and arrange (arrange_at is used here) based on the looped column in descending order

library(tidyverse)
out <- map(LETTERS[1:3], ~ df1 %>% 
                      select(Name, State, .x) %>%
                      arrange_at(.x, desc))
out
#[[1]]
#   Name State A
#1  Dick    IL 8
#2  John    MD 5
#3   Tom    NY 4
#4 Harry    WI 2

#[[2]]
#   Name State B
#1 Harry    WI 6
#2  John    MD 4
#3   Tom    NY 2
#4  Dick    IL 1

#[[3]]
#   Name State C
#1  John    MD 9
#2   Tom    NY 5
#3 Harry    WI 4
#4  Dick    IL 3

names(out) <- names(df1)[3:5]

In case, we have to automatically check numeric columns, then use select_if and extract the names

df1 %>% 
  select_if(is.numeric) %>%
  names %>%
  map(~ df1 %>% 
           select(Name, State, .x) %>% 
           arrange(desc(!! rlang::sym(.x)))) # another way to evaluate from symbol

Or using lapply from base R

lapply(names(df1)[3:5], function(x) df1[order(-df1[[x]]), c("Name", "State", x)])

Upvotes: 3

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