[FIXED] How do I check if an array is in ascending order in python?

Issue

suppose that I have the following NumPy array:

x = np.array([0,1,2,3,4,5])

I'd like to check if the numbers are in ascending order, i.e., if exists a number larger than the number before, it should warn me. I believe that I can do it with the any syntax, but I'm not sure how. For instance, I've been trying:

if any(x[i] < x[i-1] for i in range(len(x))):
    print('Warning!')

But it's not working properly. It should activate the warning in arrays like [0,2,1,3,4,5], and so on. Could someone please help me fix this issue? Thank you in advance!

Solution

You can use np.diff and check if it is all positive.

>>> import numpy as np
>>> arr = np.array([0,1,2,3,4,5])
>>> np.diff(arr)
array([1, 1, 1, 1, 1])
>>> (np.diff(arr) >= 0).all()
True

Or

>>> arr = np.array([0,2,1,3,4,5])
>>> np.diff(arr)
array([ 2, -1,  2,  1,  1])
>>> (np.diff(arr) >= 0).all()
False

Note, you could use any/all here as well, but you should try to use numpy operations on numpy arrays. But just for clarification, you basically wanted all(arr[i] <= arr[i+1] for i in range(len(arr) -1)).

And note, I'm expressing things in terms of "all", but you can use "any" equivalently:

(np.diff(arr) < 0).any()

And of course, any(arr[i] > arr[i+1] for i in range(len(arr) -1)).

Finally, you must decide how to treat the case of no change, i.e. when the difference between subsequent elements is 0. Since you used the terminology "ascending order", typically, [1, 1, 2, 3] i considered in ascending order so I've used that convention above (edited since originally posted))

Answered By - juanpa.arrivillaga