How to pass arguments to function of callable-iterator in python?

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

This is quite hard to understand. What sequence of numbers do you want to generate via call_iter? — Adam Mihalcin
– Adam Mihalcin, Commented Mar 22, 2012 at 2:54
Does this have to use the callable form of iter? Is it the starting number you want to pass? — agf
– agf, Commented Mar 22, 2012 at 2:57
Perhaps a custom iterator? stackoverflow.com/questions/19151/build-a-basic-python-iterator — Theron Luhn
– Theron Luhn, Commented Mar 22, 2012 at 3:01
@AdamMihalcin I want to implement a iter, which takes its previous iter.next() return value as its argument of next iter.next(). — xiaochen
– xiaochen, Commented Mar 22, 2012 at 3:42

John La Rooy · Accepted Answer · 2012-03-22 02:59:09Z

3

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

answered Mar 22, 2012 at 2:59

John La Rooy

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8 Comments

agf Over a year ago

But since the callable is returning a function, what purpose does the sentinel value of 100 serve?

John La Rooy Over a year ago

@agf, i am still trying to guess that :)

agf Over a year ago

My thought was he meant something like call_iter = iter(lambda x=itertools.count().next: x() + 1, 100) but who knows.

John La Rooy Over a year ago

@agf, I thought something like that is really just an xrange

John La Rooy Over a year ago

@agf, I wait with bated breath for the OP to clarify the question :)

|

Community · Accepted Answer · 2017-05-23 12:33:32Z

The form of iter you're trying to use only takes a 0-argument function. The below is for illustration only; don't actually do this.

>>> x = 0
>>> def counter():
...     global x
...     x += 1
...     return x
... 
>>> list(iter(counter, 10))
[1, 2, 3, 4, 5, 6, 7, 8, 9]

In general, this form of iter isn't very useful. It requires some kind of callable that maintains state between calls. So for example, you could pass the readline method of a file object, as suggested in the docs. But generally, there are better ways to do this. So for example, say you created a class like this:

>>> class Incrementer(object):
...     def __init__(self):
...         self.state = 0
...     def __call__(self):
...         rval = self.state
...         self.state += 1
...         return rval
... 
>>> list(iter(Incrementer(), 10))
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

It's lovely and all, but if you have to create a class that's supposed to be iterable, you might as well make it a real iterator by giving it a next method and an __iter__ method. Conversely, if you aren't creating a class, just use yield

aschmid00 · Accepted Answer · 2012-03-22 02:57:57Z

-1

i think what you want is:

call_iter = iter(map(lambda x: x + 1, range(100)))
>>> call_iter.next()
1
>>> call_iter.next()
2
>>> call_iter.next()
3
>>>

to pass an argument to the lambda function you need to map the lambda to an iterable like range(100) or [2,4,5]

answered Mar 22, 2012 at 2:57

aschmid00

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1 Comment

John La Rooy Over a year ago

I think you should use imap otherwise there is little sense using an iterator

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