Find third occurrence of a special character and d

Question

I have this sample vector containing URLs. My goal is to obtain the path of the URL.

sample1 <- c("http://tercihblog.com/indirisu/docugard/", "http://funerariagomez.com/js/ggogle/a201209e3f79b740337b7bdb521630fe/", 
      "http://www.t-online.de/contacts/2015/08/atlas.html/", "http://mgracetimber.ie/wp-content/themes/Banner/db/box/", 
      "http://zamartrade.com/cs/DHL/DHL%20_%20Tracking.htm/", "http://dunhamengineering.com/menu/Auto-loadgoogleDrive/Document.Index/", 
      "http://www.indiegogo.com/guide/forum/2014/09/forgot-password/", 
      "http://raetc.com/wp-admin/Service/clients/votre-compte/en-ligne/imp-rem.fr/", 
      "http://www.lidanhang.com/img/?https://secure.runescape.com/m=weblogin/loginform.ws?mod=www&amp;hwjklxlamp;ssl=0&amp;dest/", 
      "http://www.sudaener.com/wp-includes/js/crop/dropbox/", "https://zeustracker.abuse.ch/blocklist.php/", 
      "https://zeustracker.abuse.ch/blocklist.php?download=hostsdeny/", 
      "https://zeustracker.abuse.ch/blocklist.php?download=iptablesblocklist/", 
      "https://zeustracker.abuse.ch/blocklist.php?download=snort/", 
      "https://zeustracker.abuse.ch/blocklist.php?download=squiddomain/"
    )

My initial try was this:

gsub('http://[^/]+/','/',sample1)

However this won't work with URLs that have https://. A suitable solution would be to drop everything before the third occurrence of"/". I was wondering how to use regexto do this and also if there is a way to do it using substring.

Thanks

Answer 1

It is really advisable to go with gsub here since the code is cleaner and more straightforward.

If you want to remove all before the 3rd /, use

> gsub('^(?:[^/]*/){3}','/',sample1)
 [1] "/indirisu/docugard/"                                                                              
 [2] "/js/ggogle/a201209e3f79b740337b7bdb521630fe/"                                                     
 [3] "/contacts/2015/08/atlas.html/"                                                                    
 [4] "/wp-content/themes/Banner/db/box/"                                                                
 [5] "/cs/DHL/DHL%20_%20Tracking.htm/"                                                                  
 [6] "/menu/Auto-loadgoogleDrive/Document.Index/"                                                       
 [7] "/guide/forum/2014/09/forgot-password/"                                                            
 [8] "/wp-admin/Service/clients/votre-compte/en-ligne/imp-rem.fr/"                                      
 [9] "/img/?https://secure.runescape.com/m=weblogin/loginform.ws?mod=www&hwjklxlamp;ssl=0&dest/"
[10] "/wp-includes/js/crop/dropbox/"                                                                    
[11] "/blocklist.php/"                                                                                  
[12] "/blocklist.php?download=hostsdeny/"                                                               
[13] "/blocklist.php?download=iptablesblocklist/"                                                       
[14] "/blocklist.php?download=snort/"                                                                   
[15] "/blocklist.php?download=squiddomain/"   

The ^(?:[^/]*/){3} matches:

• ^ - start of string
• (?:[^/]*/){3} - exactly 3 occurrences of:
  • [^/]* - zero or more characters other than /
  • / - a literal / character.

Cath suggests a more precise your regex fix, but perhaps, you'd like to add ^ at the start to only match at the beginning of the string:

gsub('^https?://[^/]+/','/',sample1)
      ^     ^

The ? (greedy) quantifier means one or zero occurrences, thus making the s after http optional. It is identical to (but more efficient than) gsub('^(https|http)://[^/]+/','/',sample1).

You may also want to make your regex case-insensitive, add ignore.case = TRUE.