R中的快速部分字符串匹配

Question

给定一个字符串texts向量和一个模式patterns向量，我想找到每个文本的任何匹配模式。

对于小数据集，可以使用grepl在R中轻松完成：

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
  # ...
})

此解决方案是正确的，但它不会扩展。 即使使用适度更大的数据集（约500个文本和模式），这段代码也非常慢，在现代机器上每秒仅解决大约100个案例 - 考虑到这是一个粗略的字符串部分匹配，没有正则表达式（设置为fixed=TRUE ，这是荒谬的） fixed=TRUE ）。 即使使lapply平行也无法解决问题。 有没有办法有效地重写这段代码？

谢谢，Mulone

Answer 1

使用stringi包 - 它甚至比grepl更快。 检查基准！ 我使用了来自@ Martin-Morgan帖子的文字

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]]))
}

Answer 2

您是否准确地描述了您的问题以及您所看到的性能？ 以下是威廉·莎士比亚的全集和对它们的质疑

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

这似乎比你所暗示的更高效？

> length(text)
[1] 124787
> system.time(xx <- lapply(pattern, grepl, text, fixed=TRUE))
   user  system elapsed 
  0.444   0.001   0.444 
## avoid retaining memory; 500 x 500 case; no blank lines
> text = text[nzchar(text)]
> system.time({ for (p in rep(pattern, 50)) grepl(p, text[1:500], fixed=TRUE) })
   user  system elapsed 
  0.096   0.000   0.095 

我们期望使用模式和文本的长度（元素数量）进行线性缩放。 我好像记得我的莎士比亚

> 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"