python - Grab following index from tuple: 2 try/except or 1 if/elif -

couldn't find information on try vs. if when you're checking more 1 thing. have tuple of strings, , want grab index , store index follows it. there 2 cases.

mylist = ('a', 'b') 

or

mylist = ('y', 'z') 

i want @ a or y, depending on exists, , grab index follows. currently, i'm doing this:

item['index'] = none try:     item['index'] = mylist[mylist.index('a') + 1] except valueerror:     pass try:     item['index'] = mylist[mylist.index('y') + 1] except valueerror:     pass 

after reading using try vs if in python, think may better/more efficient write way, since half time, raise exception, (valueerror), more expensive if/else, if i'm understanding correctly:

if 'a' in mylist:     item['index'] = mylist[mylist.index('a') + 1] elif 'y' in mylist:     item['index'] = mylist[mylist.index('y') + 1] else:     item['index'] = none 

am right in assumption if better/more efficient here? or there better way of writing entirely i'm unaware of?

well opinion close rohith subramanyam's. think loop more readable, in case have more 2 elements test !

but still think if block far more logical use (pun intended!). strictly speaking less redundant in terms of lines of code:

accepted_elements = ['a', 'y'] item['index'] = none accepted_element in accepted_elements:     if accepted_element in mylist:         item['index'] = mylist[mylist.index(accepted_element) + 1]         break 

i think solution use in end you, depends on code habits (except loop must).

edit: actually, after time measurements seems rohith subramanyam's version faster @ first sight. (640 ns per loop vs 740 ns per loop)