The other 2 answers have covered it, but for the sake of clarity, remember that 2D lists don't exist. That is a list of lists, and thinking about it that way should have helped you come to a solution. It's always worth being very specific in your own mind about different types (for example, the difference between a 2D array and a matrix in numpy, or the difference between a list and a numpy array) because colloquially referring to a type as something it's not will often give you faulty assumptions how it ought to work.

>>> map(sum,zip(*my_list))
[3, 6, 8, 10]

Or the itertools equivalent

>>> from itertools import imap, izip
>>> imap(sum,izip(*my_list))
<itertools.imap object at 0x00D20370>
>>> list(_)
[3, 6, 8, 10]

You could rewrite that function as,

def sum1(input):
    return sum(map(sum, input))

Basically, map(sum, input) will return a list with the sums across all your rows, then, the outer most sum will add up that list.

Example:

>>> a=[[1,2],[3,4]]
>>> sum(map(sum, a))
10

Other alternatives for summing the columns are

numpy.einsum('ij->j', a)

and

numpy.dot(a.T, numpy.ones(a.shape[0]))

If the number of rows and columns is in the same order of magnitude, all of the possibilities are roughly equally fast:

If there are only a few columns, however, both the einsum and the dot solution significantly outperform numpy's sum (note the log-scale):

Code to reproduce the plots:

import numpy
import perfplot


def numpy_sum(a):
    return numpy.sum(a, axis=1)


def einsum(a):
    return numpy.einsum('ij->i', a)


def dot_ones(a):
    return numpy.dot(a, numpy.ones(a.shape[1]))


perfplot.save(
    "out1.png",
    # setup=lambda n: numpy.random.rand(n, n),
    setup=lambda n: numpy.random.rand(n, 3),
    n_range=[2**k for k in range(15)],
    kernels=[numpy_sum, einsum, dot_ones],
    logx=True,
    logy=True,
    xlabel='len(a)',
    )

M = [[1, 2, 3],
     [4, 5, 6],
     [7, 8, 9]]

M[:2] = [[a + b for a, b in zip(M[0], M[1])]]

print(M)  # [[5, 7, 9], [7, 8, 9]]

Things to google to understand this:

• M[:2] =: python slice assignment
• [... for .. in ...]: python list comprehension
• for a, b in ...: python tuple unpacking loop

If array is a variable containing your 2d array, try

[sum([row[i] for row in array]) for i in range(len(array[0]))]