Pythonic Way of Doing this problem in lexical order

[Yozuru]

Hello good people, I am doing practice problems for python 3.3. I managed to finish it and got it to work, however, I was wondering if there is a pythonic way of doing this. I understand the basic premise of it but I want to know the inner workings of the program itself.

Here is a pastebin of what I have for the not so pythonic way: My Code (http://pastebin.com/uy3LwFKw)

So I was wondering how I would write a function that compared two lists, and skipping the longest common prefix, look at the element in both list where there is a difference. From there, it would either return 0, 1, or -1.
For example:
List1 = [0, 1, 2]
List2 = [0, 1, 3]
This would return as -1.

Many thanks in advance for any input you can provide as I really do appreciate working with you all on this.

[lukeiamyourfather]

There are a lot of ways to do this, this is my take on it. Also what do you mean by skip the longest common prefix?



>>> foo = [0,1,2]
>>> bar = [0,1,3]
>>> def compare(first,second):
... if first > second:
... return 1
... elif first < second:
... return -1
... else:
... return 0
...
>>> map(compare,foo,bar)
[0, 0, -1]

[ofnuts]

Use the build-in cmp(), which also works on lists:


cmp(foo,bar)


A possible solution to do it "by hand" (ie not using cmp() to compare lists), not optimal since it scans the shortest of the lists:



cmp(*filter(lambda x:cmp(*x),zip(foo,bar))[0])


In slow motion:


zip(foo,bar) makes a list of tuples (one tuple element from each list)
cmp(*x) compare the two elements of the tuples
taking a 0 returned from cmp(...) as False, filter(...) keep only those tuples where elements of foo and bar are unequal
[0] takes the first
cmp(*...) re-compares the tuple contents to give the final result

[lukeiamyourfather]

Use the build-in cmp(), which also works on lists:


cmp(foo,bar)



Nice! That's much less complicated than mine.

[ofnuts]

Nice! That's much less complicated than mine.

You solution wasn't complete since it didn't retrieve the first unequal items.


(filter(None,map(cmp,foo,bar))+[0])[0]


where:

filter(None,...) using identity, keeps only the non-null elements
+[0] makes sure that the filnal list is never empty (which would happen if the inputs are equal)
[0] takes the first.

[Yozuru]

Many thanks to all of you for your inputs. I have a better understanding of the material! Its interesting that there is so many different ways of doing this.

When I mean skipping the longest common prefix, it skips the numbers that are the same.