What's the fastest way to save images by URL?

Question

Server optimization

There is an array with links to pictures. It is necessary to shove them into the archive as quickly as possible.
At the moment it is done like this:

for item in datas:
        iitem = item.replace('\\','')
        resource = urllib2.urlopen(item)
        out = open('img+'+str(iterat)+'.jpg', 'wb+')
        out.write(resource.read())
        out.close()
        zf.write('img+'+str(iterat)+'.jpg')
        os.remove('img+'+str(iterat)+'.jpg')
        iterat+=1
        zf.close

Reworked via socket, the gain is quite small.
What is the best way to speed up the process in this case?

Answer 1

here
https://habrahabr.ru/post/275595/
start with
from multiprocessing import Pool

Answer 2

You can speed up by downloading in multiple streams.