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  • \$\begingroup\$ a.index(b)==0 is a bit shorter. Alternatively, you could do 0**sum(a.index(b)for a in l for b in l). \$\endgroup\$ Commented May 2, 2016 at 7:46
  • \$\begingroup\$ @Mego That doesn't work because index throws an exception when b isn't found. And because it should be ==, not >=. However, find works. (And it's shorter too!) \$\endgroup\$ Commented May 2, 2016 at 7:54
  • \$\begingroup\$ Whoops, I meant to type find. Sleepy brain is sleepy. The second version should also work with find. \$\endgroup\$ Commented May 2, 2016 at 8:13
  • \$\begingroup\$ @Mego I'm not sure if I get the second version. Wouldn't that always return 0? \$\endgroup\$ Commented May 2, 2016 at 8:14
  • \$\begingroup\$ @Mego That only works if every string is the same. The reason we compare it to len(l) is since we're iterating through all bs on each a, there will always be at least one match per a. So we check if the number of matches is the same as the number of elements. \$\endgroup\$ Commented May 2, 2016 at 8:32