'tuple' object is not callable on my code

Question

def make_sequence(seq):
    def filter1(*func):
        new_filter = filter(func, seq)
        return tuple(new_filter)

    def filter_iterator(*func):
        new_seq = tuple(filter(func, seq))
        index = 0

        def next1():
            nonlocal index
            if index >= 0 and index < len(new_seq):
                ele = new_seq[index]
                index += 1
            else:
                index = 0
                ele = new_seq[index]
                index += 1
            return ele

        def reverse():
            nonlocal index
            index -= 1
            if index >= 0 and index < len(new_seq):
                ele = new_seq[index]
            else:
                index = len(new_seq) - 1
                ele = new_seq[index]
            return ele

        return {'next': next1, 'reverse': reverse}

    def reverse():
        return tuple(seq[::-1])

    def extend(new_seq):
        nonlocal seq
        seq += new_seq

    return {'filter': filter1, 'filter_iterator': filter_iterator, 'reverse': reverse, 'extend': extend}


s1 = make_sequence((1, 2, 3, 4, 5))
print(s1['filter'](lambda x: x % 2 == 0))

When I try to return tuple from fitlter1 I get the error:

'tuple' object is not callable.

What I try to do is the filter1 get the argument *args because when I will call filiter1() with nothing function he returns the sequence without change.

Ice Bear · Accepted Answer · 2020-12-19 11:51:45Z

1

I removed the * on your def filter1(func): try that, I noticed that your parameter is a *func with * where in fact it is a tuple when you put some arguments on the function, It seems that you do not need the *. I think you need to search about *args and **kwargs

I've run it and got this output

(2, 4)
[Finished in 0.3s]

Have a quick sample here why you got that message. Because when you put * to a parameter in a function it's data type is a tuple.

def foo(*args):
    print(args)
    print(type(args))


foo(1)

Output:

(1,)
<class 'tuple'>

Why you are getting the error?

filter(func, seq)

The filter function is expecting that your func is a function or at least lambda but it's not, what it actually is , is a tuple once you put * on your def filter1(*func).

So base on what you've said here.

What I try to do is the filter1 get the argument *args because when I will call filiter1() with nothing function he returns the sequence without change.

What you can do with your function is

def filter1(func=None):
    if callable(func):
        new_filter = filter(func, seq)
        return tuple(new_filter)

    return seq

Change your filter1 function to that and you'll get this.

print(s1['filter'](lambda x: x % 2 == 0))
print(s1['filter']())
print(s1['filter'](lambda x: x>2))

Output:

(2, 4)
(1, 2, 3, 4, 5)
(3, 4, 5)

edited Dec 19, 2020 at 11:51

answered Dec 19, 2020 at 11:25

Ice Bear

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5 Comments

Roni Jack Vituli Over a year ago

yeah i know without * it will work but when i try to do print(s1['filter']()) it will say filter1() missing 1 required positional argument: 'func'

Ice Bear Over a year ago

Yes it will throw that error and the reason because in your function filter1 you have a parameter there or in other words an argument that you need to provide on. As you said on your code here print(s1['filter']()) you are not providing any, the () is blank so you need to provide there a function since the parameter there as I see is you're going to put it on the filter function.

Roni Jack Vituli Over a year ago

i understand why i get error but the mission i need to do is : when i call the key of 'filter' i need to send function to filter my seq or not send noting to filter1 and it not change my seq

Ice Bear Over a year ago

You need to do what? have *? object is not callable means tuple is not callable so it's not a function.

Roni Jack Vituli Over a year ago

for example : s1 = make_seq((1,2,3,4,5)), >>> s1['filter'](lambda x: x>2) >>>(3,4,5) >>>s1['filter']() >>>(1,2,3,4,5)

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