Is Generator.Next() Visible in Python 3

Is generator.next() visible in Python 3?

g.next() has been renamed to g.__next__(). The reason for this is consistency: special methods like __init__() and __del__() all have double underscores (or "dunder" in the current vernacular), and .next() was one of the few exceptions to that rule. This was fixed in Python 3.0. [*]

But instead of calling g.__next__(), use next(g).

[*] There are other special attributes that have gotten this fix; func_name, is now __name__, etc.

Does next() eliminate values from a generator?

Actually this is can be implicitly deduced from next's docs and by understanding the iterator protocol/contract:

next(iterator[, default])

Retrieve the next item from the iterator by
calling its next() method. If default is given, it is returned if
the iterator is exhausted, otherwise StopIteration is raised.

Yes. Using a generator's __next__ method retrieves and removes the next value from the generator.

Getting next item from generator fails

You need to use:

next(inf_data_gen)

Rather than:

inf_data_gen.next()

Python 3 did away with .next(), renaming it as .__next__(), but its best that you use next(generator) instead.

there's no next() function in a yield generator in python 3

In Python 3, use next(uptofive) instead of uptofive.next().

The built-in next() function also works in Python 2.6 or greater.

next() method in Python file object

next() is a built-in function, not a bound method. Example:

next(open("scala.txt"))

x = (e for e in range(1, 10))
next(x)

Behind-the-scenes, calling next() on an object tends to call the bound (but hidden) method __next__(), which does exactly what you think it does.

Python generator next method

next() is a function, so it is listed in the functions documentation:

next(iterator[, default])
Retrieve the next item from the iterator by calling its __next__() method. If default is given, it is returned if the iterator is exhausted, otherwise StopIteration is raised.

The second argument then is the default, returned if iterator.__next__() raised StopIteration. If no default is set, the StopIteration exception is not caught but propagated:

>>> def gen():
...     yield 1
...
>>> g = gen()
>>> next(g, 'default')
1
>>> next(g, 'default')
'default'
>>> g = gen()
>>> next(g, 'default')
1
>>> next(g)  # no default
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
StopIteration

The default is rather redundant for the specific example you gave, as the (x**2 for x in range(0,100)) generator expression is guaranteed to have at least one result.

PyCharm can show you documentation for Python standard-library functions; just use the Quick Documentation feature (CTRL-Q).

How to get one value at a time from a generator function in Python?

Yes, or next(gen) in 2.6+.

Python-3.2 coroutine: AttributeError: 'generator' object has no attribute 'next'

You're getting thrown off by the error message; type-wise, Python doesn't make a distinction - you can .send to anything that uses yield, even if it doesn't do anything with the sent value internally.

In 3.x, there is no longer a .next method attached to these; instead, use the built-in free function next:

next(matcher)

python 3 iterator not executing next

In Python 3, use the global function next():

one(next(cnt))

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