How do I check if every consecutive list element is equal to or 1 more than the last?

Question

Input: a list a of real numbers, of length 0 or greater.
Output: the Boolean value True if for every i in the list a , a[i] <= a[i+1] ,
otherwise False .

This is what I have so far but it doesn't work:

def consecutive_elements_equal_or_one_more(a):
    for i in range(a)
        for i+1 in range(a)
            if a[i] <= a[i+1]:
                return true
            else:
                return false

[1, 2, 3] should return true. [1, 2, 2] should return true. [1, 3, 2] should return false.

Answer 1

If you're looking for an efficient way of doing this and the lists are numerical, you would probably want to use numpy and apply the diff (difference) function:

>>> numpy.diff([1,2,3,4,5,5,6])
array([1, 1, 1, 1, 0, 1])

Then to get a single result regarding whether there are any consecutive elements:

>>> numpy.any(~numpy.diff([1,2,3,4,5,5,6]).astype(bool))

This first performs the diff, inverts the answer, and then checks if any of the resulting elements are non-zero.

Similarly,

>>> 0 in numpy.diff([1, 2, 3, 4, 5, 5, 6])

also works well and is similar in speed to the np.any approach

Answer 2

Solution without third party libraries:

ls = [
    [1, 2, 3],
    [1, 2, 2],
    [1, 3, 2]
]

for l in ls:
    print(all(a - b <= 1 for a, b in zip(l, [l[0] - 1, *l])))

Answer 3

Based on these conditions you need to check if the list is already sorted.

Simple solution:

def is_sorted(a):
    return a == sorted(a)

Faster, without sorting the list first:

def is_sorted(a):
    for i in range(len(a) - 1):
        if a[i] > a[i + 1]:
            return False
    return True

Answer 4

Your question title and problem statement are inconsistent.

Checking for non decreasing order can be done with all and zip :

if all(a<=b for a,b in zip(a,a[1:])):

Checking for "one or more than the last":

if all(a+1<=b for a,b in zip(a,a[1:])):