How to sort strings with integers by numeric ordering?

Question

I have a column in a data.table full of strings in the format string+integer. e.g.

string1, string2, string3, string4, string5,

When I use sort(), I put these strings in the wrong order.

string1, string10, string11, string12, string13, ..., string2, string20, 
string21, string22, string23, ....

How would I sort these to be in the order

string01, string02, string03, string04, strin0g5, ... , string10,, string11, 
string12, etc.

One method could be to add a 0 to each integer <10, 1-9? I suspect you would extract the string with str_extract(dt$string_column, "[a-z]+") and then add a 0 to each single-digit integer...somehow with sprintf()

akrun · Accepted Answer · 2017-04-11 15:49:44Z

6

We can remove the characters that are not numbers to do the sorting

dt[order(as.integer(gsub("\\D+", "", col1)))]

answered Apr 11, 2017 at 15:49

akrun

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3 Comments

ShanZhengYang Over a year ago

What would be the difference of using sort() and order() here? It appears sorts leads to the same errors

ShanZhengYang Over a year ago

The other problem I'm having with this: there are two strings which are labeled stringX and stringZ. These do not have integers, and should be put last. The above as.integer() will result in NAs if I don't somehow exclude these

akrun Over a year ago

@ShanZhengYang Sorry, in your post you didn't mention all these things and the reproducible example was incomplete

989 · Accepted Answer · 2017-04-11 16:10:15Z

1

You could go for mixedsort in gtools:

vec <- c("string1", "string10", "string11", "string12", "string13","string2", 
         "string20", "string21", "string22", "string23")

library(gtools)
mixedsort(vec)

#[1] "string1"  "string2"  "string10" "string11" "string12" "string13"
# "string20" "string21" "string22" "string23"

answered Apr 11, 2017 at 16:10

989

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Comments

d.b · Accepted Answer · 2017-04-11 18:19:31Z

1

You could use the str_extract of stringr package to obtain the digits and order according to that

x = c("string1","string3","stringZ","string2","stringX","string10")
library(stringr)
c(x[grepl("\\d+",x)][order(as.integer(str_extract(x[grepl("\\d+",x)],"\\d+")))], 
   sort(x[!grepl("\\d+",x)]))
#[1] "string1"  "string2"  "string3"  "string10" "stringX"  "stringZ"

edited Apr 11, 2017 at 18:19

answered Apr 11, 2017 at 15:50

d.b

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1 Comment

ShanZhengYang Over a year ago

(Apologies for not mentioning this in the original question) This approach would work, except for one problem: here are two strings which are labeled stringX and stringZ. These do not have integers, and should be put last. The above as.integer() will result in NAs if I don't somehow exclude these

PKumar · Accepted Answer · 2017-04-11 18:42:58Z

1

Assuming the string is something like below:

library(data.table)
library(stringr)

  xstring <- data.table(x = c("string1","string11","string2",'string10',"stringx"))
  extracts <- str_extract(xstring$x,"(?<=string)(\\d*)")
  y_string <- ifelse(nchar(extracts)==2 | extracts=="",extracts,paste0("0",extracts))
  fin_string <- str_replace(xstring$x,"(?<=string)(\\d*)",y_string)
  sort(fin_string)

Output:

> sort(fin_string)
[1] "string01" "string02" "string10" "string11"
[5] "stringx"

answered Apr 11, 2017 at 18:42

PKumar

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