I am having list of urls like ,
l=['bit.ly/1bdDlXc','bit.ly/1bdDlXc',.......,'bit.ly/1bdDlXc']
I just want to see the full url from the short one for every element in that list.
Here is my approach,
import urllib2
for i in l:
print urllib2.urlopen(i).url
But when list contains thousands of url , the program takes long time.
My question : Is there is any way to reduce execution time or any other approach I have to follow ?
First method
As suggested, one way to accomplish the task would be to use the official api to bitly, which has, however, limitations (e.g., no more than 15 shortUrl's per request).
Second method
As an alternative, one could just avoid getting the contents, e.g. by using the HEAD HTTP method instead of GET. Here is just a sample code, which makes use of the excellent requests package:
import requests
l=['bit.ly/1bdDlXc','bit.ly/1bdDlXc',.......,'bit.ly/1bdDlXc']
for i in l:
print requests.head("http://"+i).headers['location']
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