看起来很简单,
考虑以下网址
[1] "scripts.iucr.org/cgi-bin/paper?S1600536812045886"
[2] "cpa-seoadvisors.com/cvv/auth/auth/view/pdf/index.html/"
[3] "www.scirp.org/journal/PaperDownload.aspx?DOI=10.4236/csta.2012.13014"
[4] "www.google.com.cy/search?q=DNS+traffic&es_..."
[5] "seesaa.net/pede/lobortis/ligula/sit/amet.png?semper=vitae&est=..."
我希望得到第一个'/'
和将令牌与?
分开的那个之间的部分。
我写了以下函数
get_directory <- function(x){
dir <- sapply(strsplit(x, '/'), function(i)sum(grepl('\\?', i)))
ifelse(dir > 0, sapply(strsplit(x, '/'), function(i) paste(i[-c(1, length(i))], collapse = '/')), 0)
}
但它在[3]和[4] URL失败了。
预期输出应为
"cgi-bin"
"0"
"journal"
"0"
"pede/lobortis/liguls/sit"
数据
dput(df)
structure(list(V1 = c("scripts.iucr.org/cgi-bin/paper?S1600536812045886",
"cpa-seoadvisors.com/cvv/auth/auth/view/pdf/index.html/", "www.scirp.org/journal/PaperDownload.aspx?DOI=10.4236/csta.2012.13014",
"www.google.com.cy/search?q=DNS+traffic&es_...", "seesaa.net/pede/lobortis/ligula/sit/amet.png?semper=vitae&est=..."
)), .Names = "V1", row.names = c(NA, -5L), class = "data.frame")
答案 0 :(得分:4)
我们可以使用str_extract
。使用正则表达式外观,我们匹配一个或多个字符(.*
),其后成功/
后跟/
,一个或多个字符不是?
({{ 1}})后跟[^?]+
。
?