How to compare two lists in python?

Question:

How to compare two lists in python?

date = "Thu Sep 16 13:14:15 CDT 2010" 
sdate = "Thu Sep 16 14:14:15 CDT 2010" 
dateArr = [] dateArr = date.split() 
sdateArr = [] sdateArr = sdate.split() 

Now I want to compare these two lists. I guess split returns a list. We can do simple comparision in Java like dateArr[i] == sdateArr[i], but how can we do it in Python?

Asked By: Umesh K

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Answers:

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
Answered By: Dominic Rodger

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
Answered By: skajfes

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.

Answered By: Björn Pollex

From your post I gather that you want to compare dates, not arrays. If this is the case, then use the appropriate object: a datetime object.

Please check the documentation for the datetime module. Dates are a tough cookie. Use reliable algorithms.

Answered By: Escualo
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.

Answered By: PyRsquared
for i in arr1:
    if i in arr2:
        return 1
    return  0
arr1=[1,2,5]
arr2=[2,4,15]
q=checkarrayequalornot(arr1,arr2)
print(q)
>>0
Answered By: ravi tanwar
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