How to control the order of a variable mixed with string and numbers in R

Question

Suppose I have a data.frame df

library(tidyverse)
df<-data.frame(x=paste0("S", 1:15), y=letters[1:15], stringsAsFactors=F)%>%
              arrange(x)
     x y
1   S1 a
2  S10 j
3  S11 k
4  S12 l
5  S13 m
6  S14 n
7  S15 o
8   S2 b
9   S3 c
10  S4 d
11  S5 e
12  S6 f
13  S7 g
14  S8 h
15  S9 i

How can I arrange the x as an output like:

     x y
1   S1 a
2   S2 b
3   S3 c
4   S4 d
5   S5 e
6   S6 f
7   S7 g
8   S8 h
9   S9 i
10 S10 j
11 S11 k
12 S12 l
13 S13 m
14 S14 n
15 S15 o

`df[gtools::mixedorder(df$x),]` seems to work fine. Also, what happens If you have something like `S01`? The accepted `nchar` answer might fail in this case — Sotos, Jun 13 '19 at 14:09

score 1 · Answer 1 · edited Jun 13 '19 at 14:11

1

You can you the mixedorder() function in the gtools package to do this pretty easily.

library(gtools)

df<-data.frame(x=paste0("S", 1:15 ), y=letters[1:15], stringsAsFactors=F)


# mix it up
set.seed(121)
df <- df[sample(row.names(df), nrow(df)),]

# reorder it
df[mixedorder(df$x),]

That goes from this:

     x y
6   S6 f
3   S3 c
2   S2 b
15 S15 o
5   S5 e
12 S12 l
9   S9 i
10 S10 j
11 S11 k
8   S8 h
7   S7 g
1   S1 a
14 S14 n
13 S13 m
4   S4 d

to this

     x y
1   S1 a
2   S2 b
3   S3 c
4   S4 d
5   S5 e
6   S6 f
7   S7 g
8   S8 h
9   S9 i
10 S10 j
11 S11 k
12 S12 l
13 S13 m
14 S14 n
15 S15 o

In tidyverse you can do this -

df %>% 
  slice(mixedorder(x))

edited Jun 13 '19 at 14:11

Shree

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answered Jun 13 '19 at 13:49

mfidino

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If the rownames already 1:15 and the x ordered as S1, S10, S2. It doesn't work. – David Z Jun 13 '19 at 13:52
Why not? What do rownames have to do with your sorting? What `mixedorder` does it strips the numbers from the strings and orders those numbers. – Sotos Jun 13 '19 at 14:00
You may try the df I generated (ending with arrange()) – David Z Jun 13 '19 at 14:03
This seems to work `df[gtools::mixedorder(df$x),]` – Sotos Jun 13 '19 at 14:06
1

@DavidZ you need to use `slice()` with `mixedorder()` and not `arrange()`. So - `df %>% slice(gtools::mixedorder(x))` gives desired result. – Shree Jun 13 '19 at 14:10
Thanks! That's a great point – David Z Jun 13 '19 at 19:38

score 1 · Accepted Answer · answered Jun 13 '19 at 13:55

1

Here's one possible way within dplyr -

df %>% 
  arrange(nchar(x), x)

     x y
1   S1 a
2   S2 b
3   S3 c
4   S4 d
5   S5 e
6   S6 f
7   S7 g
8   S8 h
9   S9 i
10 S10 j
11 S11 k
12 S12 l
13 S13 m
14 S14 n
15 S15 o

answered Jun 13 '19 at 13:55

Shree

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@DavidZ Just make sure there are no white-spaces. – Shree Jun 13 '19 at 13:59

score 0 · Answer 3 · answered Jun 13 '19 at 13:43

Well you could also add a pure numerical sequence column to your data frame and sort by that. Then, just scope the columns you actually want to see:

df<-data.frame(ord=c(1:15), x=paste0("S", 1:15), y=letters[1:15], stringsAsFactors=FALSE)
df[order(ord), c("x", "y")]

score 0 · Answer 4 · edited Jun 13 '19 at 14:37

0

Why not this?

df[order(nchar(df$x),df$x),]

edited Jun 13 '19 at 14:37

Rahul Agarwal

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answered Jun 13 '19 at 14:14

TPArrow

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How to control the order of a variable mixed with string and numbers in R

4 Answers4