CODEWITHSUNDEEP
pythonnested-lists
noor
noor

Reputation: 1

comparing data in lists Python

how to compare more then one list, below works ok for just 2 lists but what if more needed

while (list1[0:1] == list2[0:1]):
    if list1[1:2] > list2[1:2]:
        print('list 1 wins')
    elif list1[2:3] > list2[2:3]:
        print('list 1 wins!')
    elif list1[3:4] > list2[3:4]:
        print('list 1 wins')
    elif list1[4:5] > list2[4:5]:
        print('list 1 wins')
    elif list1[5:5] == list2[5:5]:
        print('its a tie')
    else:
        print('list 2 wins')
    break

Upvotes: 0

Views: 106

Answers (4)

AChampion
AChampion

Reputation: 30258

As pointed out lists can be directly compared without iterating through them.
But when you want to work with multiple lists at the same time it is idiomatic to use zip(), e.g.:

for i1, i2 in zip(list1, list2):
    if i1 > i2:
        print("List 1 wins")
        break
    if i2 > i1:
        print("List 2 wins")
        break
else:
    print("It's a tie")

Upvotes: 1

PM 2Ring
PM 2Ring

Reputation: 55469

As bereal says, you can just use list comparison for this. If ties weren't an issue, you could just use max() to determine a winner, but to handle ties you need to sort the whole collection of lists.

Eg, this code will handle any number of lists.

list1 = [50,9,5,4,5]
list2 = [50,9,8,1,6]
list3 = [50,9,8,1,5]

lists = [(seq, i) for i, seq in enumerate((list1, list2, list3), 1)]
lists.sort(reverse=True)

if lists[0][0] == lists[1][0]:
    print("it's a tie")
else:
    print("list {0} wins: {1}".format(lists[0][1], lists[0][0]))

output

list 2 wins: [50, 9, 8, 1, 6]

If you change list2 back to [50, 9, 8, 1, 6], then it's a tie will get printed.


lists is a list of tuples, with each tuple containing one of the original lists and its index number, so we can identify lists after the tuples have been sorted. The sorting will compare the lists, and if two lists are identical, then the indices will also be compared, but that doesn't affect the desired outcome here.

BTW, list1[1:2] > list2[1:2] is wasteful. It creates a new pair of single element lists & compares them. It's easier to just compare the elements themselves, eg list1[1] > list2[1].

Upvotes: 0

bereal
bereal

Reputation: 34272

Lists and tuples can be compared just like that assuming that the matching elements are comparable:

>>> list1=[50,9,5,4,5]
>>> list2=[50,9,8,1,5]
>>> list1 > list2
False
>>> list1 < list2
True

That also means that you can order any number of lists:

>>> lists = [list1, list2, list3]
>>> lists.sort()

Upvotes: 1

donbunkito
donbunkito

Reputation: 588

Did you consider using a for-loop?

The syntax is:

for value in list:
    # do something
    print value

and then there is the (zip function)

list1 = [1,2]
list2 = [a,b]
zip(list1, list2) # = [(1,a),(2,b)]

With these two tools one is able to provide a really pythonic solution to your problem.

Upvotes: 0

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