Determine if a list is in descending order

Question

I am trying to write a function that will test whether or not a list is in decending order. This is what I have so far, but it doesn't seem to be working for all lists.

I used the list [9,8,5,1,4,3,2] and it returned 'true'.

I can't seem to figure out where my mistake is.

def ordertest(A):
    n = len(A)
    for i in range(n):
        if A[i] >= A[i+1]:
            return 'true'
        else:
            return 'false'

Answer 1

You can do this easily with a generator expression and the all() builtin:

all(earlier >= later for earlier, later in zip(seq, seq[1:]))

For example:

>>> seq = [9, 8, 5, 1, 4, 3, 2] 
>>> all(earlier >= later for earlier, later in zip(seq, seq[1:]))
False
>>> seq = [9, 8, 5, 4, 3, 2] 
>>> all(earlier >= later for earlier, later in zip(seq, seq[1:]))
True

This should be nice and fast as it avoids python-side loops, short circuits nicely (if you use itertools.izip() in 2.x), and is nice and clear and readable (avoiding looping over indices, for example).

Note that a generic solution for all iterators (not just sequences) is possible too:

first, second = itertools.tee(iterable)
next(second)
all(earlier >= later for earlier, later in zip(first, second))