Given two lists:
a = [[1,2],[3,4]]
b = [[1,2],[3,4]]
How would I write compare such that:
compare(a,b) => true
2 Answers
Do you want this:
>>> a = [[1,2],[3,4]]
>>> b = [[1,2],[3,4]]
>>> a == b
True
Note: == not useful when List are unordered e.g (notice order in a, and in b)
>>> a = [[3,4],[1,2]]
>>> b = [[1,2],[3,4]]
>>> a == b
False
See this question for further reference: How to compare a list of lists/sets in python?
Edit: Thanks to @dr jimbob
If you want to compare after sorting you can use sorted(a)==sorted(b).
But again a point, if c = [[4,3], [2,1]] then sorted(c) == sorted(a) == False because, sorted(c) being different [[2,1],[4,3]] (not in-depth sort)
for this you have to use techniques from linked answer. Since I am learning Python too :)
1 Comment
dr jimbob
The last part isn't right.
a.sort() sorts a in place and then returns None. So a.sort() == None will be True, and your a.sort()==b.sort() evaluates to None == None. Try getting a False; e.g., let b=[]. (Note its better style to use is comparison when checking for None). Now, sorted(a) == sorted(b) will work. Granted note if you have a = [[1,2],[3,4]], b = [[3,4],[1,2]] and c = [[4,3], [2,1]], then only sorted(a) and sorted(b) will be equal (both being [[1,2],[3,4]]), with sorted(c) being different ([[2,1],[4,3]]).Simple:
def compare(a, b): return a == b
Another way is using lambda to create an anonymous function:
compare = lambda a, b: a == b
a == bserve your purpose?==wouldn't work. Depends on your expected types.==tests equality for Python objects. It's the same if testing the equality of two normal lists.