R 在特殊字符之间提取最后一个单词/&引用;
我想提取“/”符号后的最后一个字符串。比如说,R 在特殊字符之间提取最后一个单词/&引用;,r,regex,stringr,stringi,R,Regex,Stringr,Stringi,我想提取“/”符号后的最后一个字符串。比如说, url<- c('https://example.com/names/ani/digitalcod-org','https://example.com/names/bmc/ambulancecod.org' ) df<- data.frame (url) 试试这个: as.data.frame(sapply(str_extract_all(df$url,"\\w{2,}(?=\\/)"),"["))[3,] # V1 V2 #3
url<- c('https://example.com/names/ani/digitalcod-org','https://example.com/names/bmc/ambulancecod.org' )
df<- data.frame (url)
试试这个:
as.data.frame(sapply(str_extract_all(df$url,"\\w{2,}(?=\\/)"),"["))[3,]
# V1 V2
#3 ani bmc
as.data.frame(sapply(str_extract_all(df$url,"\\w{2,}(?=\\/)"),"["))[2:3,]
# V1 V2
#2 names names
#3 ani bmc
使用
basename
basename(mapply(sub, pattern = basename(url), replacement = "", x = url, fixed = TRUE))
#[1] "ani" "bmc"
basename(url)
“删除最后一个路径分隔符(如果有)之前的所有路径”并返回
[1] "digitalcod-org" "ambulancecod.org"
使用
mapply
将url
中每个元素的结果替换为”
,然后再次调用basename
。您可以使用word
,但需要指定分隔符
library(stringr)
word(url, -2, sep = '/')
#[1] "ani" "bmc"
将
gsub
与
.*?([^/]+)/[^/]+$
在
R
中:
urls <- c('https://example.com/names/ani/digitalcod-org','https://example.com/names/bmc/ambulancecod.org' )
gsub(".*?([^/]+)/[^/]+$", "\\1", urls)
请参阅。这里有一个使用strsplit的解决方案
words <- strsplit(url, '/')
L <- lengths(words)
vapply(seq_along(words), function (k) words[[k]][L[k]-1], character(1))
# [1] "ani" "bmc"
words应该有更有效的方法。我只是在继续你的思路
urls <- c('https://example.com/names/ani/digitalcod-org','https://example.com/names/bmc/ambulancecod.org' )
gsub(".*?([^/]+)/[^/]+$", "\\1", urls)
[1] "ani" "bmc"
words <- strsplit(url, '/')
L <- lengths(words)
vapply(seq_along(words), function (k) words[[k]][L[k]-1], character(1))
# [1] "ani" "bmc"