Ggtcf

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

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

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

is an good option.

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")

I count characters same way as Amarjeet. However I prefer to do it in a single line.

HowManySpaces<-nchar(DF$string)-nchar(gsub(" ","",DF$string)) # count spaces in DF$string

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`

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.

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),]

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

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.