How to calculate the number of occurrence of a given character in each row of a column of strings?

Question

I have a data.frame in which certain variables contain a text string. I wish to count the number of occurrences of a given character in each individual string.

Example:

q.data<-data.frame(number=1:3, string=c("greatgreat", "magic", "not"))

I wish to create a new column for q.data with the number of occurence of "a" in string (ie. c(2,1,0)).

The only convoluted approach I have managed is:

string.counter<-function(strings, pattern){  
  counts<-NULL
  for(i in 1:length(strings)){
    counts[i]<-length(attr(gregexpr(pattern,strings[i])[[1]], "match.length")[attr(gregexpr(pattern,strings[i])[[1]], "match.length")>0])
  }
return(counts)
}

string.counter(strings=q.data$string, pattern="a")

 number     string number.of.a
1      1 greatgreat           2
2      2      magic           1
3      3        not           0

Answer 1

Another base R answer, not so good as those by @IRTFM and @Finn (or as those using stringi/stringr), but better than the others:

sapply(strsplit(q.data$string, split=""), function(x) sum(x %in% "a"))

q.data<-data.frame(number=1:3, string=c("greatgreat", "magic", "not"))
q.data<-q.data[rep(1:NROW(q.data), 3000),]
library(rbenchmark)
library(stringr)
library(stringi)

benchmark( Dason = {str_count(q.data$string, "a") },
           Tim = {sapply(q.data$string, function(x, letter = "a"){sum(unlist(strsplit(x, split = "")) == letter) }) },
           DWin = {nchar(q.data$string) -nchar( gsub("a", "", q.data$string, fixed=TRUE))}, 
           Markus = {stringi::stri_count(q.data$string, fixed = "a")},
           Finn={nchar(gsub("[^a]", "", q.data$string))},
           tmmfmnk={lengths(lapply(q.data$string, grepRaw, pattern = "a", all = TRUE, fixed = TRUE))},
           Josh1 = {sapply(regmatches(q.data$string, gregexpr("g",q.data$string )), length)}, 
           Josh2 = {lengths(regmatches(q.data$string, gregexpr("g",q.data$string )))}, 
           Iago = {sapply(strsplit(q.data$string, split=""), function(x) sum(x %in% "a"))}, 
           replications =100, order = "elapsed")

     test replications elapsed relative user.self sys.self user.child sys.child
4  Markus          100   0.076    1.000     0.076    0.000          0         0
3    DWin          100   0.277    3.645     0.277    0.000          0         0
1   Dason          100   0.290    3.816     0.291    0.000          0         0
5    Finn          100   1.057   13.908     1.057    0.000          0         0
9    Iago          100   3.214   42.289     3.215    0.000          0         0
2     Tim          100   6.000   78.947     6.002    0.000          0         0
6 tmmfmnk          100   6.345   83.487     5.760    0.003          0         0
8   Josh2          100  12.542  165.026    12.545    0.000          0         0
7   Josh1          100  13.288  174.842    13.268    0.028          0         0

Answer 2

Another good option, using charToRaw:

sum(charToRaw("abc.d.aa") == charToRaw('.'))

Answer 3

The next expression does the job and also works for symbols, not only letters.

The expression works as follows:

1: it uses lapply on the columns of the dataframe q.data to iterate over the rows of the column 2 ("lapply(q.data[,2],"),

2: it apply to each row of the column 2 a function "function(x){sum('a' == strsplit(as.character(x), '')[[1]])}". The function takes each row value of column 2 (x), convert to character (in case it is a factor for example), and it does the split of the string on every character ("strsplit(as.character(x), '')"). As a result we have a a vector with each character of the string value for each row of the column 2.

3: Each vector value of the vector is compared with the desired character to be counted, in this case "a" (" 'a' == "). This operation will return a vector of True and False values "c(True,False,True,....)", being True when the value in the vector matches the desired character to be counted.

4: The total times the character 'a' appears in the row is calculated as the sum of all the 'True' values in the vector "sum(....)".

5: Then it is applied the "unlist" function to unpack the result of the "lapply" function and assign it to a new column in the dataframe ("q.data$number.of.a<-unlist(....")

q.data$number.of.a<-unlist(lapply(q.data[,2],function(x){sum('a' == strsplit(as.character(x), '')[[1]])}))

>q.data

#  number     string     number.of.a
#1   greatgreat         2
#2      magic           1
#3      not             0

Answer 4

Yet another base R option could be:

lengths(lapply(q.data$string, grepRaw, pattern = "a", all = TRUE, fixed = TRUE))

[1] 2 1 0

Answer 5

A variation of https://stackoverflow.com/a/12430764/589165 is

> nchar(gsub("[^a]", "", q.data$string))
[1] 2 1 0

Answer 6

The question below has been moved here, but it seems this page doesn't directly answer to Farah El's question. How to find number 1s in 101 in R

So, I'll write an answer here, just in case.

library(magrittr)
n %>% # n is a number you'd like to inspect
  as.character() %>%
  str_count(pattern = "1")

https://stackoverflow.com/users/8931457/farah-el

Answer 7

The stringi package provides the functions stri_count and stri_count_fixed which are very fast.

stringi::stri_count(q.data$string, fixed = "a")
# [1] 2 1 0

benchmark

Compared to the fastest approach from @42-'s answer and to the equivalent function from the stringr package for a vector with 30.000 elements.

library(microbenchmark)

benchmark <- microbenchmark(
  stringi = stringi::stri_count(test.data$string, fixed = "a"),
  baseR = nchar(test.data$string) - nchar(gsub("a", "", test.data$string, fixed = TRUE)),
  stringr = str_count(test.data$string, "a")
)

autoplot(benchmark)

data

q.data <- data.frame(number=1:3, string=c("greatgreat", "magic", "not"), stringsAsFactors = FALSE)
test.data <- q.data[rep(1:NROW(q.data), 10000),]

Answer 8

If you don't want to leave base R, here's a fairly succinct and expressive possibility:

x <- q.data$string
lengths(regmatches(x, gregexpr("a", x)))
# [1] 2 1 0

Answer 9

You could just use string division

require(roperators)
my_strings <- c('apple', banana', 'pear', 'melon')
my_strings %s/% 'a'

Which will give you 1, 3, 1, 0. You can also use string division with regular expressions and whole words.

Answer 10

The easiest and the cleanest way IMHO is :

q.data$number.of.a <- lengths(gregexpr('a', q.data$string))

#  number     string number.of.a`
#1      1 greatgreat           2`
#2      2      magic           1`
#3      3        not           0`

Answer 11

s <- "aababacababaaathhhhhslsls jsjsjjsaa ghhaalll"
p <- "a"
s2 <- gsub(p,"",s)
numOcc <- nchar(s) - nchar(s2)

May not be the efficient one but solve my purpose.

Answer 12

nchar(as.character(q.data$string)) -nchar( gsub("a", "", q.data$string))
[1] 2 1 0

Notice that I coerce the factor variable to character, before passing to nchar. The regex functions appear to do that internally.

Here's benchmark results (with a scaled up size of the test to 3000 rows)

 q.data<-q.data[rep(1:NROW(q.data), 1000),]
 str(q.data)
'data.frame':   3000 obs. of  3 variables:
 $ number     : int  1 2 3 1 2 3 1 2 3 1 ...
 $ string     : Factor w/ 3 levels "greatgreat","magic",..: 1 2 3 1 2 3 1 2 3 1 ...
 $ number.of.a: int  2 1 0 2 1 0 2 1 0 2 ...

 benchmark( Dason = { q.data$number.of.a <- str_count(as.character(q.data$string), "a") },
 Tim = {resT <- sapply(as.character(q.data$string), function(x, letter = "a"){
                            sum(unlist(strsplit(x, split = "")) == letter) }) }, 

 DWin = {resW <- nchar(as.character(q.data$string)) -nchar( gsub("a", "", q.data$string))},
 Josh = {x <- sapply(regmatches(q.data$string, gregexpr("g",q.data$string )), length)}, replications=100)
#-----------------------
   test replications elapsed  relative user.self sys.self user.child sys.child
1 Dason          100   4.173  9.959427     2.985    1.204          0         0
3  DWin          100   0.419  1.000000     0.417    0.003          0         0
4  Josh          100  18.635 44.474940    17.883    0.827          0         0
2   Tim          100   3.705  8.842482     3.646    0.072          0         0

Answer 13

I'm sure someone can do better, but this works:

sapply(as.character(q.data$string), function(x, letter = "a"){
  sum(unlist(strsplit(x, split = "")) == letter)
})
greatgreat      magic        not 
     2          1          0 

or in a function:

countLetter <- function(charvec, letter){
  sapply(charvec, function(x, letter){
    sum(unlist(strsplit(x, split = "")) == letter)
  }, letter = letter)
}
countLetter(as.character(q.data$string),"a")

Answer 14

The stringr package provides the str_count function which seems to do what you're interested in

# Load your example data
q.data<-data.frame(number=1:3, string=c("greatgreat", "magic", "not"), stringsAsFactors = F)
library(stringr)

# Count the number of 'a's in each element of string
q.data$number.of.a <- str_count(q.data$string, "a")
q.data
#  number     string number.of.a
#1      1 greatgreat           2
#2      2      magic           1
#3      3        not           0