Question

Python 2.6.2

>>> call_iter = iter(lambda x: x + 1, 100)
>>> call_iter.next()
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: <lambda>() takes exactly 1 argument (0 given)

I want to pass argument to lambda x:x + 1.

Update: I think the example above is hard to understand.

I wonder whether there is a builtin func like myiter in python:

class myiter:
    def __init__(self, callable, initial, sentinel):
        self.value = initial
        self.callable = callable
        self.sentinel = sentinel

    def __iter__(self):
        return self

    def next(self):
        if self.value == self.sentinel:
            raise StopIteration
        else:
            # calculate next value from prev value
            self.value = self.callable(self.value) 
            return self.value

if __name__ == '__main__':
    call_iter = myiter(lambda x:x + 1, 0, 100)
    for i in call_iter:
        print i

Answer 1

I’m not sure what you are trying to accomplish here, but

>>> call_iter = iter(lambda:lambda x: x + 1, 100)
>>> next(call_iter)(1)
2