这是我的示例行。我只想提取网站名称,例如; 3dubs或adludio 怎么做?干杯,
URL
https://www.3dhubs.com/
https://adludio.com/
https://aircall.io/
https://www.andjaro.com/en/home/
结果
3dhubs
adludio
aircall
andjaro
输入此代码后,
suffix_extract(domain(df$URL))
我得到的结果如下:当我尝试分配它时,它看起来有所不同。如何获取域并分配给列?
host subdomain domain suffix
www.3dhubs.com www 3dhubs com
adludio.com <NA> adludio com
答案 0 :(得分:1)
使用适当的URL解析器(例如TextBox
包中的URL解析器)可能是最安全的。例如
Station
经过测试
PropertyChanged
答案 1 :(得分:0)
如何获取域并分配给列?
使用{urltools},以下对我有用:
imageView.setAlpha(100);
答案 2 :(得分:0)
库urltools应该可以工作。如果您的数据位于对象url中,则返回的数据。
library(urltools)
df1 <- suffix_extract(domain(urls))
df1
host subdomain domain suffix
1 www.3dhubs.com www 3dhubs com
2 adludio.com <NA> adludio com
3 aircall.io <NA> aircall io
4 www.andjaro.com www andjaro com
df1$domain
[1] "3dhubs" "adludio" "aircall" "andjaro"
dplyr / tidyr选项如下,但使用urltools中的url_parse
来确保它是有效的url。
library(dplyr)
library(tidyr)
df <- data_frame(urls)
df %>%
mutate(url_parsed = urltools::url_parse(urls)$domain) %>%
separate(url_parsed, into = c("subdomain", "domain", "suffix"), fill = "left")
# A tibble: 4 x 4
urls subdomain domain suffix
<chr> <chr> <chr> <chr>
1 https://www.3dhubs.com/ www 3dhubs com
2 https://adludio.com/ NA adludio com
3 https://aircall.io/ NA aircall io
4 https://www.andjaro.com/en/home/ www andjaro com
数据:
urls <- c("https://www.3dhubs.com/", "https://adludio.com/", "https://aircall.io/",
"https://www.andjaro.com/en/home/")