我想不通这个 function 有什么问题？

Question

This is what it should return

Returns True if all items are truthy.

Examples:
>>> all_true([1, True, 'oregonstate'])
True
>>> all_true([0, 1, 2])
False
"""

This is what I have written

def all_true(items):
    for i in items:
        if i == True:
            return True
        else:
          return False
    pass

Not sure why it is not working, can someone help?

Answer 1

Your function just returns the truthiness of the first item in the list, since it always returns on the first iteration.

You want to do something like this:

def all_true(items):
    for i in items:
        if not i:
            return False
    return True

Note that this all_true function already exists in Python. It is the the built-in all function.

Answer 2

Two problems. First, you are returning on the first True thing. In fact, you are returning either True or False on the first thing, regardless. Since you return on both the if and its else clauses, the loop will always return on the first iteration. A bit more subtle, but if there aren't any values in the list, it will return None .

Second, if you want "truthy", don't compare specifically to True . foo == True is False . Instead, just let the if do the boolean test

def all_true(items):
    for i in items:
        if not i:
            return False
    return True

all and any are builtin functions that do the same thing. It would generally be preferable to use them.

Answer 3

"All items are truthy" is logically equivalent to "no items are falsy."

def all_true(items):
    for i in items:
        if not i:
            return False
    return True

Answer 4

Because 0 is falsy. Also there's no need to implement your own function for that, use Python's built–in all() function instead!