use*_*423 5 regex text r readlines
我想从一行中提取多个字符串.
假设我有以下文本行(使用'readLines'函数形成一个网站):
line <- "abc:city1-street1-long1-lat1,ldjad;skj//abc:city2-street2-long2-lat2,ldjad;skj//abc:city3-street3-long3-lat3,ldjad;skj//abc:city3-street3-long3-lat3,ldjad;skj//"
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我想提取以下内容分开:
[1] city1-street1-long1-lat1
[2] city2-street2-long2-lat2
[3] city3-street3-long3-lat3
[4] city4-street4-long4-lat4
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我希望有人可以给我一个如何执行此任务的提示.
regmatches 救援:
regmatches(line,gregexpr("city\\d+-street\\d+-long\\d+-lat\\d+",line))
#[[1]]
#[1] "city1-street1-long1-lat1"
#[2] "city2-street2-long2-lat2"
#[3] "city3-street3-long3-lat3"
#[4] "city3-street3-long3-lat3"
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使用stringi包的解决方案:
library(stringi)
stri_extract_all_regex(line, "(?<=:).+?(?=,)")[[1]]
## [1] "city1-street1-long1-lat1" "city2-street2-long2-lat2" "city3-street3-long3-lat3" "city3-street3-long3-lat3"
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并使用stringr包:
library(stringr)
str_extract_all(line, perl("(?<=:).+?(?=,)"))[[1]]
## [1] "city1-street1-long1-lat1" "city2-street2-long2-lat2" "city3-street3-long3-lat3" "city3-street3-long3-lat3"
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在这两种情况下我们都使用正则表达式。在这里,我们匹配和之间出现的所有字符(非贪婪地,即与.+?)。表示积极的后视:将匹配,但不包含在结果中。另一方面,是积极的前瞻:必须匹配但不会出现在输出中。:,(?<=:):(?=,),
一些基准:
lines <- stri_dup(line, 250) # duplicate line 250 times
library(microbenchmark)
microbenchmark(
stri_extract_all_regex(lines, "(?<=:).+?(?=,)")[[1]],
str_extract_all(lines, perl("(?<=:).+?(?=,)"))[[1]],
regmatches(lines, gregexpr("city\\d+-street\\d+-long\\d+-lat\\d+", lines)),
lapply(unlist(strsplit(lines,',')),
function(x)unlist(strsplit(x,':'))[2]),
lapply(strsplit(lines,'//'),
function(x)
sub('.*:(.*),.*','\\1',x))
)
## Unit: milliseconds
## expr min lq median uq max neval
## gagolews-stri_extract_all_regex 4.722515 4.811009 4.835948 4.883854 6.080912 100
## gagolews-str_extract_all 103.514964 103.824223 104.387175 106.246773 117.279208 100
## thelatemail-regmatches 36.049106 36.172549 36.342945 36.967325 47.399339 100
## agstudy-lapply 21.152761 21.500726 21.792979 22.809145 37.273120 100
## agstudy-lapply2 8.763783 8.854666 8.930955 9.128782 10.302468 100
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如您所见,stringi基于 - 的解决方案是最快的。
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