CODEWITHSUNDEEP
rdplyr
SCDCE
SCDCE

Reputation: 1643

Arrange numbers by different descending orders

Fairly simple problem I can't seem to come up with an elegant solution.

I'd like to arrange a column of data by differing descending levels:

library(dplyr)
test <- data.frame(ID=c(19000,19001,19002,1,2))

test %>% 
  arrange(desc(ID)) %>%
  mutate(ID = formatC(ID,width=5,format="d",flag="0"))

     ID
1 19002
2 19001
3 19000
4 00002
5 00001

I want:

     ID
1 00002
2 00001
3 19002
4 19001
5 19000

This is for a pipeline so more IDs will be added, e.g. 00003, 00004....

Here's something I came up with:

test %>% 
  mutate(ID = formatC(ID,width=5,format="d",flag="0")) %>% 
  group_by(group=substr(ID,1,1)) %>% 
  arrange(desc(ID)) %>% 
  arrange(group) %>% 
  ungroup() %>% 
  select(ID)

Anything better than this?

EDIT--

library(microbenchmark)

test <- data.frame(ID=c(1:29999))

microbenchmark(group = test %>% 
                 mutate(ID = formatC(ID,width=5,format="d",flag="0"),
                        group = substr(ID,1,1)) %>% 
                 arrange(group, desc(ID)) %>% 
                 select(ID),

               mod = test %>% 
                 arrange(ID %/% 1000, desc(ID %% 1000)) %>%
                 mutate(ID = formatC(ID,width=5,format="d",flag="0")))

Unit: milliseconds
expr      min        lq     mean    median       uq      max neval cld
group 138.0480 152.21025 168.7705 160.41305 176.6362 352.4736   100   b
mod  27.7697  29.94265  34.1312  31.92085  35.5323  88.8065   100  a

Thanks all! Looks like I have my answer.

Upvotes: 4

Views: 95

Answers (2)

Allan Cameron
Allan Cameron

Reputation: 173793

You could just sort by number of thousands then descending-sort by modulo 1000. That way you don't need to add a groups column.

library(dplyr)
test <- data.frame(ID=c(19000,19001,19002,1,2))

test %>% 
  arrange(ID %/% 1000, desc(ID %% 1000)) %>%
  mutate(ID = formatC(ID,width=5,format="d",flag="0"))

#>     ID
#> 1 00002
#> 2 00001
#> 3 19002
#> 4 19001
#> 5 19000

Upvotes: 6

camnesia
camnesia

Reputation: 2323

Here is a small edit to your solution:

library(dplyr)

test <- data.frame(ID=c(19000,19001,19002,1,2))

test %>% 
mutate(ID = formatC(ID,width=5,format="d",flag="0")) %>%
arrange(substr(ID,1,1), desc(ID))

Upvotes: 0

Related Questions