Python: How to assign function name dynamically in Class

Question

I have an question about how to assign function name dynamically in Class.

For example:

If a class wants to be used for "for...in" loop, similar to a list
or a tuple, you must implement an __iter__ () method

python2.x will use __iter__() and next(),
python3.x need to use __iter__() and __next__()

Code:

The sample is get fibonacci numbers within 10

class Fib(object):
    def __init__(self):
        self.a, self.b = 0, 1

    def __iter__(self):
        return self

    if sys.version_info[0] == 2:
        iter_n = 'next'  #if python version is 2.x
    else:
        iter_n = '__next__'  #if python version is 3.x

    print('iter_n value is ', iter_n)

    #for py2 I want to replace "iter_n" to "next" dynamically
    #for py3 I want to replace "iter_n" to "__next__" dynamically

    def iter_n(self):
        self.a, self.b = self.b, self.a + self.b
        if self.a > 10:
            raise StopIteration();
        return self.a

Test:

for i in Fib():
    print(i)

Expected Result should be:

('iter_n value is ', 'next')
1
1
2
3
5
8

Actual Result:

('iter_n value is ', 'next')
Traceback (most recent call last):
......
TypeError: iter() returned non-iterator of type 'Fib'

Code will be able to get the right results

for python 2, if I replace def iter_n(self) to def next(self)
for python 3, if I replace def iter_n(self) to def __next__(self)

Question:

How should I put next or __next__ to iter_n dynamically?

user707650 · Accepted Answer · 2016-09-29 05:31:27Z

5

I don't think there's a need to create this method dynamically. Just implement both; clearer and easier. And your code will be Python 2/3 compatible without needing an if-statement.

class Fib(object):
    def __init__(self):
        self.a, self.b = 0, 1

    def __iter__(self):
        return self

    def iter_n(self):
        self.a, self.b = self.b, self.a + self.b
        if self.a > 10:
            raise StopIteration();
        return self.a

    def next(self):
        return self.iter_n()

    def __next__(self):
        return self.iter_n()


if __name__ == '__main__':
    for i in Fib():
        print(i)

answered Sep 29, 2016 at 5:31

user707650

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Comments

Chris · Accepted Answer · 2016-09-29 05:45:45Z

1

Agreed that just implementing both is probably best, but something like this seems to do what you intend:

    def iter_n(self):
        self.a, self.b = self.b, self.a + self.b
        if self.a > 10:
            raise StopIteration();
        return self.a

    if sys.version_info[0] == 2:
        next = iter_n
    else:
        __next__ = iter_n
    del iter_n

The above removes the original iter_n once it is assigned to the appropriate attribute. Presumably if you're going through this effort you would want to do this cleanup, too.

answered Sep 29, 2016 at 5:45

Chris

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Comments

sachin · Accepted Answer · 2016-09-29 05:47:03Z

-1

In python there is the way to call a function ,method ,get any element by using getattr(from,'name of attribute to call').

below is the sample example of this:

class A():
    def s(self):
            print "s fun"
    def d(self):
            print "a fun"
    def run(self):
            print "in run"

def main():
    print "in main"
    classA = A()
    func_list = ['s','d','run']
    for i in func_list:
            func = getattr(classA,i)
            func()

main()

Try to use this to call function dynamically.

edited Sep 29, 2016 at 5:47

answered Sep 29, 2016 at 5:34

sachin

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