How to calculate the number of occurrence of a given character in each row of a column of strings?
RegexRDataframeRegex Problem Overview
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$string, pattern="a")
number string number.of.a
1 1 greatgreat 2
2 2 magic 1
3 3 not 0
Regex Solutions
Solution 1 - Regex
The stringr package provides the str_count function which seems to do what you're interested in
# Load your example data
q.data<-data.frame(number=1:3, string=c("greatgreat", "magic", "not"), stringsAsFactors = F)
library(stringr)
# Count the number of 'a's in each element of string
q.data$number.of.a <- str_count(q.data$string, "a")
q.data
# number string number.of.a
#1 1 greatgreat 2
#2 2 magic 1
#3 3 not 0
Solution 2 - Regex
If you don't want to leave base R, here's a fairly succinct and expressive possibility:
x <- q.data$string
lengths(regmatches(x, gregexpr("a", x)))
# [1] 2 1 0
Solution 3 - Regex
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$string )), length)}, replications=100)
#-----------------------
test replications elapsed relative user.self sys.self user.child sys.child
1 Dason 100 4.173 9.959427 2.985 1.204 0 0
3 DWin 100 0.419 1.000000 0.417 0.003 0 0
4 Josh 100 18.635 44.474940 17.883 0.827 0 0
2 Tim 100 3.705 8.842482 3.646 0.072 0 0
Solution 4 - Regex
Another good option, using charToRaw:
sum(charToRaw("abc.d.aa") == charToRaw('.'))
Solution 5 - Regex
The stringi package provides the functions stri_count and stri_count_fixed which are very fast.
stringi::stri_count(q.data$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),]
Solution 6 - Regex
A variation of https://stackoverflow.com/a/12430764/589165 is
> nchar(gsub("[^a]", "", q.data$string))
[1] 2 1 0
Solution 7 - Regex
I'm sure someone can do better, but this works:
sapply(as.character(q.data$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$string),"a")
Solution 8 - Regex
You could just use string division
require(roperators)
my_strings <- c('apple', banana', 'pear', 'melon')
my_strings %s/% 'a'
Which will give you 1, 3, 1, 0. You can also use string division with regular expressions and whole words.
Solution 9 - Regex
The question below has been moved here, but it seems this page doesn't directly answer to Farah El's question. https://stackoverflow.com/questions/55233227/how-to-find-number-1s-in-101-in-r
So, I'll write an answer here, just in case.
library(magrittr)
n %>% # n is a number you'd like to inspect
as.character() %>%
str_count(pattern = "1")
Solution 10 - Regex
Yet another base R option could be:
lengths(lapply(q.data$string, grepRaw, pattern = "a", all = TRUE, fixed = TRUE))
[1] 2 1 0
Solution 11 - Regex
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$string, split=""), function(x) sum(x %in% "a"))},
replications =100, order = "elapsed")
test replications elapsed relative user.self sys.self user.child sys.child
4 Markus 100 0.076 1.000 0.076 0.000 0 0
3 DWin 100 0.277 3.645 0.277 0.000 0 0
1 Dason 100 0.290 3.816 0.291 0.000 0 0
5 Finn 100 1.057 13.908 1.057 0.000 0 0
9 Iago 100 3.214 42.289 3.215 0.000 0 0
2 Tim 100 6.000 78.947 6.002 0.000 0 0
6 tmmfmnk 100 6.345 83.487 5.760 0.003 0 0
8 Josh2 100 12.542 165.026 12.545 0.000 0 0
7 Josh1 100 13.288 174.842 13.268 0.028 0 0
Solution 12 - Regex
The easiest and the cleanest way IMHO is :
q.data$number.of.a <- lengths(gregexpr('a', q.data$string))
# number string number.of.a`
#1 1 greatgreat 2`
#2 2 magic 1`
#3 3 not 0`
Solution 13 - Regex
The next expression does the job and also works for symbols, not only letters.
The expression works as follows: >1: it uses lapply on the columns of the dataframe q.data to iterate over the rows of the column 2 ("lapply(q.data[,2],"),
>2: it apply to each row of the column 2 a function "function(x){sum('a' == strsplit(as.character(x), '')[[1]])}". The function takes each row value of column 2 (x), convert to character (in case it is a factor for example), and it does the split of the string on every character ("strsplit(as.character(x), '')"). As a result we have a a vector with each character of the string value for each row of the column 2.
>3: Each vector value of the vector is compared with the desired character to be counted, in this case "a" (" 'a' == "). This operation will return a vector of True and False values "c(True,False,True,....)", being True when the value in the vector matches the desired character to be counted.
>4: The total times the character 'a' appears in the row is calculated as the sum of all the 'True' values in the vector "sum(....)".
>5: Then it is applied the "unlist" function to unpack the result of the "lapply" function and assign it to a new column in the dataframe ("q.data$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
Solution 14 - Regex
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.