A real simple one, just a question of understanding the python syntax I suppose. I won't waste your time by throwing the whole program at you, I think my question can be boiled down to this:

a = ["cat"]
b = ["bark"]
wordinsa = input("Word: ")
wordinsb = input("Word: ")
wordq = input("Try: ")
if wordq in a:
    c = a.index(wordq)
    print wordq
    print b[c]
    print "Nope"

Basically a dictionary built out of lists; the real thing jumps to back to main menu, works real nice actually.
But I was thinking just now, when the program checks for matches in list a, it compares the entire strings of wordq and the ones in the list. If one were to rewrite it a little, perhaps like this:
maybe add in another if-statement checking if the first letter of the word is present?
maybe add a loop checking letter for letter?

...would it be possible to gain a little extra effiency this way, though the program is longer?
After all, the lists in the real thing can be as long as any Britannica Encyclopedia you'd care to name.
I know about dictionaries and tuples and so on, this is more of an attempt to get a general grasp of the flow in a computer program!

You are assuming the word at index in list 'a' makes sense with the word with the same index in list 'b'. Somewhat dangerous. It would be simpler to make a dictionary with a:b (key:value) pairs, this way things that belong together stay together. Also, dictionary searches are highly optimized and very fast.

Definitely use a dictionary or SQL database if there are a lot of entries. It is hundreds or thousands of times faster depending. The time consuming part of the code is looping through the list twice. Using a try/except to check the list once with no "in" statement is a workaround, but just a workaround and not better than using a dictionary. As for checking letter by letter, that's what the existing or any compare/find/index has to do. There is no way to find something comparing an entire word to an entire word. The only shortcut that I know of would be to group words according to length. Thus, a five letter word would only look at other 5 letter words, and not words of other lengths.

##if wordq in  a:     ## omit this line - duplicate list checking
   c = a.index(wordq)
   print wordq
   print b[c]
   print wordq, "wasn't found in 'a'"

Edit: There is also sets if dictionaries are more than you want to do at this time. Both dictionaries and sets use a hash as an index, so the execution times should be similiar for both.

list1 = ['a', 'b', 'c', 'd']
list2 = ['b']
set3= set1.intersection(set2)
print "set3 =", set3

print "set6 =", set6

Yea, I realise using lists for this purpose isn't exactly the optimal way of going about it, but it was part of a task a while back and it just got me thinking in those tracks... how does this work, how could you make this particular method work more efficently.

Thanks for the input woooee, I wasn't aware of that "intersect"-command. So much to learn...