R中的快速部分字符串匹配

Posted 2023-02-22

【中文标题】R中的快速部分字符串匹配

【英文标题】：Fast partial string matching in R

【发布时间】：2014-08-07 02:44:03

【问题描述】：

给定一个字符串向量texts 和一个模式向量patterns，我想为每个文本找到任何匹配的模式。

对于小型数据集，这可以在 R 中使用 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
  # ...
)

此解决方案是正确的，但无法按比例放大。即使有较大的数据集（约 500 个文本和模式），这段代码也非常慢，在现代机器上每秒只能解决大约 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（使用 @ 设置），这很荒谬987654325@)。即使使lapply 并行也不能解决问题。 有没有办法高效地重写这段代码？

谢谢， 木龙

【问题讨论】：

你的模式总是单字吗？您是否只是对patterns 的每个元素是否出现在texts 的一个或多个元素中感兴趣（或者您是否需要知道它们出现在texts 的哪些元素中）？

【参考方案1】：

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

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]]))

【参考方案2】：

您是否准确地描述了您的问题和您所看到的性能？这是Complete Works of William Shakespeare 和针对它们的查询

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

这似乎比你暗示的要好得多？

> 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 

我们期望通过模式和文本的长度（元素数量）进行线性缩放。我好像记错了我的莎士比亚

> 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"