如何计算字符串列每行中给定字符的出现次数？

Question

我有一个data.frame,其中某些变量包含文本字符串.我希望计算每个字符串中给定字符的出现次数.

例:

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

我希望为q.data创建一个新列,其中包含字符串中"a"的出现次数(即c(2,1,0)).

我管理的唯一令人费解的方法是:

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

stringr包提供了str_count似乎做你感兴趣的功能

# 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

Answer 2

如果你不想离开基地R,这里有一个相当简洁和富有表现力的可能性:

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

Answer 3

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

请注意,在传递给nchar之前,我将factor变量强制转换为character.正则表达式函数似乎在内部执行此操作.

这是基准测试结果(测试的大小扩大到3000行)

 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 4

该stringi软件包提供的功能stri_count和stri_count_fixed它的速度非常快。

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

基准

与@42- 的答案中的最快方法以及包中具有 30.000 个元素的向量的等效函数相比stringr。

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)

数据

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 5

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

是个不错的选择.

Answer 6

/sf/answers/870153511/的变体是

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