Custom generator for file filter

Question

I'm writing a small wrapper class around open that will filter out particular lines from a text file and then split them into name/value pairs before passing them back to the user. Naturally, this process lends itself to being implemented using generators.

My "file" class

class special_file:
    def __init__(self, fname):
        self.fname = fname

    def __iter__(self):
        return self

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

    def next(self):
        with open(self.fname, 'r') as file:
            for line in file:
                line = line.strip()
                if line == '':
                    continue
                name,value = line.split()[0:2]
                if '%' in name:
                    continue
                yield name,value
            raise StopIteration()

Userland code

for g in special_file('input.txt'):
    for n,v in g:
        print(n,v)

My code, sadly, has two enormous problems: 1) special_file returns a generator when it really needs to return a tuple , and 2) the StopIteration() exception is never raised so the file is read repeatedly ad infinitum . I have a sneaking suspicion that these two issues are related, but my understanding of generators and iterable sequences is fairly limited. Have I missed something painfully obvious about implementing a generator?

Edit:

I fixed my infinite reading problem by moving the first generator outside of the loop and then just looping over it.

g = special_file('input.txt')
k = next(g)
for n,v in k:
    print(n,v)

However, I would like the user to be able to use it like a normal call to open :

for n,v in special_file('input.txt'):
    print(n,v)

Answer 1

You've implemented an iterator, in terms of using a generator. Just write the generator directly.

def special_file(filename):
    with open(filename, 'r') as file:
        for line in file:
            line = line.strip()
            if line == '':
                continue
            name, value, *_ = line.split()
            if '%' in name:
                continue
            yield name, value

See here for an overview of what it means to be iterable, what an iterator is, and python's protocols for using them.

Answer 2

Just change

def __iter__(self):
    return self

to

def __iter__(self):
    return next(self)

and it works as expected!

Thanks to @Leva7 for the suggestion.