How to compare two lists in python?

If you mean lists, try ==:

l1 = [1,2,3]
l2 = [1,2,3,4]

l1 == l2 # False

If you mean array:

l1 = array('l', [1, 2, 3])
l2 = array('d', [1.0, 2.0, 3.0])
l1 == l2 # True
l2 = array('d', [1.0, 2.0, 3.0, 4.0])
l1 == l2 # False

If you want to compare strings (per your comment):

date_string  = u'Thu Sep 16 13:14:15 CDT 2010'
date_string2 = u'Thu Sep 16 14:14:15 CDT 2010'
date_string == date_string2 # False

You could always do just:

a=[1,2,3]
b=['a','b']
c=[1,2,3,4]
d=[1,2,3]

a==b    #returns False
a==c    #returns False
a==d    #returns True

Given the code you provided in comments, I assume you want to do this:

>>> dateList = "Thu Sep 16 13:14:15 CDT 2010".split()
>>> sdateList = "Thu Sep 16 14:14:15 CDT 2010".split()
>>> dateList == sdataList
false

The split-method of the string returns a list. A list in Python is very different from an array. == in this case does an element-wise comparison of the two lists and returns if all their elements are equal and the number and order of the elements is the same. Read the documentation.

a = ['a1','b2','c3']
b = ['a1','b2','c3']
c = ['b2','a1','c3']

# if you care about order
a == b # True
a == c # False

# if you don't care about order AND duplicates
set(a) == set(b) # True
set(a) == set(c) # True

By casting a, b and c as a set, you remove duplicates and order doesn't count. Comparing sets is also much faster and more efficient than comparing lists.