How to get each row in Python?

Question

I need to do this:

"""
Program a function

    def increasing (m)

that will be able, for any matrix m of positive integers, to check whether the sums of the rows in this array are increasing.

examples
1 3 2 5        The sums of the rows        2 8 4 1        The sums of the rows
7 9 4 1        are 11, 21 and 23.          6 2 8 5        are 15, 21 and 15.
3 5 6 9        --> increasing              8 4 2 1        --> not increasing
"""

So, I want to use sum(), which is perfectly doable, I guess.

I started like this:

def increasing (m):
    for row in m:
        row[1]

But I know that row[1] will just output the numbers in index 1 of each row. What I have in my mind is this:

def increasing (m):
    for row in m:
        if sum(row)[first_row] > sum(row)[second_row]:
           return False

But that is just slicing it, so I don't know how to count the rows so that I can compare them.

I don't want to use any module or whatsoever, just plain simple Python. Can someone point me in the right direction? I just need it to be as simple as possible.

Input format sample:

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

not_increasing_l = [
    [2, 8, 4, 1],
    [6, 2, 8, 5],
    [8, 4, 2, 1]
]

test1 = increasing(increasing_l)
test2 = increasing(not_increasing_l)

print "should be True: %s" % test1
print "should be False: %s" % test2

Answer 1

You can do the following:

def increasing(m):
    return all(sum(r1) < sum(r2) for r1, r2 in zip(m, m[1:]))

This uses zip to pair adjacent rows and all to efficiently do the pairwise sum comparison.

Without zip:

return all(sum(m[i-1]) < sum(m[i]) for i in range(1, len(m)))