为什么有序列表不等于list.sort（）？

Question

I have this code where I check if the list is ordered. So I used array == array.sort() to verify this condition. Where did I go wrong?

Code:

def isOrdered(t):
    """
    t: list
    return True if the list is ordered in a growing sense
    """
    array = t
    if array == array.sort(): #This doesn't work
        return True

    else:
        return False

print(isOrdered([1, 2, 3, 4, 5]))
print(isOrdered([1, 3, 2, 4, 2]))

Answer 1

You should use sorted instead, as it returns a copy of the list that is sorted. You can also simplify your code a little bit, as comparison returns a boolean:

def isOrdered(t):
    """
    t: list
    return True if the list is ordered in a growing sense
    """
    return t == sorted(t)

Answer 2

array.sort()
works inline, meaning it doesn't return anything.

Instead, make a copy, sort the copy then check for similarity.

def isOrdered(t):
    """
    t: list
    return True if the list is ordered in a growing sense
    """
    t_copy = t[:]
    t_copy.sort()
    return t_copy == t

Answer 3

OP : return True if the list is ordered in a growing sense

Hence :

def isOrdered(t):
    """
    t: list
    return True if the list is ordered in a growing sense
    """
    res = sorted(t)                # check if the lst is soreted    
    if res:   
        if len(t) == len(set(t)):  # ensure no duplicates in the lst
            return True
        else:
            return False           # duplicate means list is not in growing sense
    else:
        return False

print(isOrdered([1, 2, 3, 4, 5]))     # Valid list with Growing sense
print(isOrdered([1, 1, 2, 3, 4, 5]))  # Invalid list with dupes
print(isOrdered([1, 3, 2, 4, 2]))     # Invalid list not sorted
print(isOrdered([1, 2, 3, 4, 5, 1]))  # Invalid list with declining 

OUTPUT :

True
False
False
False