String R语言中的快速部分字符串匹配

给定一个字符串向量

文本

和一个模式向量

模式

，我想为每个文本找到任何匹配模式

对于小型数据集，这可以在R中通过

grepl

轻松完成：

patterns = c("some","pattern","a","horse")
texts = c("this is a text with some pattern", "this is another text with a pattern")

# for each x in patterns
lapply( patterns, function(x){
  # match all texts against pattern x
  res = grepl( x, texts, fixed=TRUE )
  print(res)
  # do something with the matches
  # ...
})

这个解决方案是正确的，但它不能放大。即使有中等规模的数据集（约500个文本和模式），这段代码的速度也令人尴尬，在现代机器上每秒只能解决大约100个案例——考虑到这是一个粗糙的字符串部分匹配，没有正则表达式（设置为

fixed=TRUE

），这是荒谬的。即使使

lappy

并行也不能解决问题。 有没有办法有效地重新编写此代码

谢谢，

---

Mulone

您是否准确描述了您的问题和您看到的性能？以下是对他们的调查和质疑

text = readLines("~/Downloads/pg100.txt")
pattern <- 
    strsplit("all the world's a stage and all the people players", " ")[[1]]

---

使用

stringi

package——它甚至比grepl更快。检查基准！ 我使用了@Martin Morgan post的文本

require(stringi)
require(microbenchmark)

text = readLines("~/Desktop/pg100.txt")
pattern <-  strsplit("all the world's a stage and all the people players", " ")[[1]]

grepl_fun <- function(){
    lapply(pattern, grepl, text, fixed=TRUE)
}

stri_fixed_fun <- function(){
    lapply(pattern, function(x) stri_detect_fixed(text,x,NA))
}

#        microbenchmark(grepl_fun(), stri_fixed_fun())
#    Unit: milliseconds
#                 expr      min       lq   median       uq      max neval
#          grepl_fun() 432.9336 435.9666 446.2303 453.9374 517.1509   100
#     stri_fixed_fun() 213.2911 218.1606 227.6688 232.9325 285.9913   100

# if you don't believe me that the results are equal, you can check :)
xx <- grepl_fun()
stri <- stri_fixed_fun()

for(i in seq_along(xx)){
    print(all(xx[[i]] == stri[[i]]))
}

require（stringi）
要求（微基准）
text=readLines（“~/Desktop/pg100.txt”）
你的句型都是单字吗？您是否只对
模式的每个元素是否出现在
文本的一个或多个元素中感兴趣（或者您需要知道它们出现在
文本的哪些元素中）？
> idx = Reduce("+", lapply(pattern, grepl, text, fixed=TRUE))
> range(idx)
[1] 0 7
> sum(idx == 7)
[1] 8
> text[idx == 7]
[1] "    And all the men and women merely players;"                       
[2] "    cicatrices to show the people when he shall stand for his place."
[3] "    Scandal'd the suppliants for the people, call'd them"            
[4] "    all power from the people, and to pluck from them their tribunes"
[5] "    the fashion, and so berattle the common stages (so they call"    
[6] "    Which God shall guard; and put the world's whole strength"       
[7] "    Of all his people and freeze up their zeal,"                     
[8] "    the world's end after my name-call them all Pandars; let all"    

require(stringi)
require(microbenchmark)

text = readLines("~/Desktop/pg100.txt")
pattern <-  strsplit("all the world's a stage and all the people players", " ")[[1]]

grepl_fun <- function(){
    lapply(pattern, grepl, text, fixed=TRUE)
}

stri_fixed_fun <- function(){
    lapply(pattern, function(x) stri_detect_fixed(text,x,NA))
}

#        microbenchmark(grepl_fun(), stri_fixed_fun())
#    Unit: milliseconds
#                 expr      min       lq   median       uq      max neval
#          grepl_fun() 432.9336 435.9666 446.2303 453.9374 517.1509   100
#     stri_fixed_fun() 213.2911 218.1606 227.6688 232.9325 285.9913   100

# if you don't believe me that the results are equal, you can check :)
xx <- grepl_fun()
stri <- stri_fixed_fun()

for(i in seq_along(xx)){
    print(all(xx[[i]] == stri[[i]]))
}