CODEWITHSUNDEEP
pythonpython-3.x
Analovespugs
Analovespugs

Reputation: 21

Python - Find sum of all even numbers in 2 Dimensional List

The problem is:

Given a "two-dimensional list" of integers – that is, a list where each
value in it is a list of integers – find all the even numbers and sum them together.

The first one is my first attempt where I seem to have successfully made it work with 1 list. Then I try to make it work with a 2D list, and this code gives me an error 'int object is not subscriptible' and I think it's to do with my while statement. Any help would be appreciated, thanks!

def sum_even(xss):
    result = 0
    for item in xss:
        if item % 2 == 0:
            result = result + item
    return result
ans = sum_even([4,8])
print (ans)

def sum_even(xss):
    result = 0
    n = 0
    i = 0
    j = 0
    while xss[i][0] in xss:
        if xss[i][0] % 2 == 0:
            result = result + xss [i][0]
        while xss[0][j] in xss:
            if xss[0][j] % 2 == 0:
                result = result +xss[0][j]
    i +=1   
    return result
ans = sum_even([4,8])
print (ans)

Upvotes: 2

Views: 1599

Answers (5)

user2390182
user2390182

Reputation: 73450

There are multiple issues with your code. The simplest would be using sum with a nested conditional generator expression:

def sum_even(xss):
    return sum(x for sub_lst in xss for x in sub_lst if not x % 2)

The scopes of the nested for-expressions seem counter-intuitive at times. This is roughly equivalent to:

def sum_even(xss):
    result = 0
    for sub_lst in xss:
        for x in sub_lst:
            if not x % 2:
                result += x
    return result

For arbitrarily nested lists of integers, you would have to use recursion:

def sum_even(xss):
    if isinstance(xss, int):
        return (not xss % 2) * xss  # the bool expr is neatly coerced to 1 or 0 ;)
    return sum(sum_even(sub) for sub in xss)

Upvotes: 4

kaza
kaza

Reputation: 2327

>>> a=[[1,2],[3,4],[5,6]]
>>> reduce(lambda s,x:s+x,[i for l in a for i in l if not i%2],0)
12

without itertools and using just list comprehension.

Much simpler thanks to @juanpa.arrivillaga, much simpler version

>>> a=[[1,2],[3,4],[5,6,7,8]]
>>> sum([i for l in a for i in l if not i%2])
20

Upvotes: 1

Yberê Campelo
Yberê Campelo

Reputation: 13

Specifically this error means you are trying to use an int as if it were a list.

The cause for this is that your second sum_even function was built to handle 2D lists, but you are using it on a 1D list ([4,8]) instead of on a 2D list ([[4],[8]]).

Despite of that, your code has another error:

def sum_even(xss):
    result = 0
    n = 0
    i = 0
    j = 0
    while xss[i][0] in xss:
        if xss[i][0] % 2 == 0:
            result = result + xss [i][0]
        while xss[0][j] in xss:
            if xss[0][j] % 2 == 0:
                result = result +xss[0][j]
    i +=1   
    #there is no update for j
    return result

In python, for statements dont use indexes with the iterator. The result is something more like a phrase:

For each 1d_list within the 2D list xss, check the elements. If the element is even, add it to result.

def sum_even(xss):
    result = 0
    for 1d_list in xss:
        for element in 1d_list:
            if element%2 == 0:
                result = result + element
    return result

In this way you dont make any assumption on the length of the lists (each 1D sublist may have a different size).

Upvotes: 1

whackamadoodle3000
whackamadoodle3000

Reputation: 6748

You could try this. It works with any dimension list (not just 2):

list2d=[[1,2],[2,3,4]]
result=sum([int(elem) for elem in str(list2d).replace('[','').replace(']','').split(',') if not int(elem)%2])

Upvotes: 1

Ajax1234
Ajax1234

Reputation: 71451

You can try this:

import itertools
even_numbers = sum(i for i in itertools.chain.from_iterable(first_list) if i%2 == 0)

Upvotes: 2

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