R count number of commas and string

Question

I have a string:

    str1 <- "This is a string, that I've written 
        to ask about a question, or at least tried to."

How would I :

1) count the number of commas

2) count the occurences of '-ion'

Any suggestions?

Solution 1

The stringr package has a function str_count that does this for you nicely.

library(stringr)

str_count(str1, ',')
[1] 2
str_count(str1, 'ion')
[1] 1

EDIT:

Cause I was curious:

vec <- paste(sample(letters, 1e6, replace=T), collapse=' ')

system.time(str_count(vec, 'a'))
   user  system elapsed 
  0.052   0.000   0.054 

system.time(length(gregexpr('a', vec, fixed=T)[[1]]))
   user  system elapsed 
  2.124   0.016   2.146 

system.time(length(gregexpr('a', vec, fixed=F)[[1]]))
   user  system elapsed 
  0.052   0.000   0.052 

Solution 2

The general problem of mathcing text requires regular expressions. In this case you just want to match specific characters, but the functions to call are the same. You want gregexpr.

matched_commas <- gregexpr(",", str1, fixed = TRUE)
n_commas <- length(matched_commas[[1]])

matched_ion <- gregexpr("ion", str1, fixed = TRUE)
n_ion <- length(matched_ion[[1]])

If you want to only match "ion" at the end of words, then you do need regular expressions. \b represents a word boundary, and you need to escape the backslash.

gregexpr(
  "ion\\b", 
  "ionisation should only be matched at the end of the word", 
  perl = TRUE
)

Solution 3

This really is an adaptation of Richie Cotton's answer. I hate having to repeat the same function over and over. This approach allows you to feed a vector of terms to match within the string:

str1 <- "This is a string, that I've written to ask about a question, 
    or at least tried to."
matches <- c(",", "ion") 
sapply(matches,  function(x) length(gregexpr(x, str1, fixed = TRUE)[[1]]))
#  , ion 
#  2   1 

Solution 4

Another option is stringi

library(stringi)
stri_count(str1,fixed=',')
#[1] 2
stri_count(str1,fixed='ion')
#[1] 1

Benchmarks

vec <- paste(sample(letters, 1e6, replace=T), collapse=' ')
f1 <- function() str_count(vec, 'a')
f2 <- function() stri_count(vec, fixed='a')
f3 <- function() length(gregexpr('a', vec)[[1]])

library(microbenchmark)
microbenchmark(f1(), f2(), f3(), unit='relative', times=20L)
#Unit: relative
#expr      min       lq     mean   median       uq      max neval cld
# f1() 18.41423 18.43579 18.37623 18.36428 18.46115 17.79397    20   b
# f2()  1.00000  1.00000  1.00000  1.00000  1.00000  1.00000    20  a 
# f3() 18.35381 18.42019 18.30015 18.35580 18.20973 18.21109    20   b

View more solutions

Comments

[Justin]

Again cause I was curious, you can also feed a vector of matches to str_count. str_count(str1, matches) will return the same 2 and 1.

[Josh O'Brien]

It's important to note that the time hit for gregexpr() is coming entirely from setting fixed=T (which is not needed here at all). You might want to add the timings for system.time(length(gregexpr('a', vec)[[1]])), which should be nearly identical to those for str_count(). This makes sense since str_count() is essentially a wrapper for gregexpr().

[Justin]

@JoshO'Brien Good point. I was a little surprised how slow the gregexpr was.

[Josh O'Brien]

Me too. That's why I checked. I really hadn't appreciated how much more slowly regular expressions are matched when fixed=TRUE. Good to know about that, so thanks for having adding those timings to your post!

[Sergei]

Thank you, this works without having to install stringr library. However note that length(gregexpr(",", "no commas", fixed = TRUE)[[1]]) and length(gregexpr(",", "one , comma", fixed = TRUE)[[1]]) are both 1. So we need to check that matched_commas[[1]][1] is greater than 0.

[webb]

In a data.table j-expression, you can do dt[,n=str_count(str,'a')] to get the number of 'a' in str for each row, but dt[,n= length(gregexpr('a',str)] doesn't work, and the workaround (Filter & unlist) takes a very long time. Switching to str_count in my data.table j-expression decreased my execution time from several hours to 30 minutes with a large dataset.