181
votes

In python, how do I check if an object is a generator object?

Trying this -

>>> type(myobject, generator)

gives the error -

Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
NameError: name 'generator' is not defined

(I know I can check if the object has a next method for it to be a generator, but I want some way using which I can determine the type of any object, not just generators.)

9
What actual problem are you trying to solve? Post more context, there may be a smarter way. Why do you need to know if it's a generator?Daenyth
from types import GeneratorType;type(myobject, GeneratorType) will give you the proper result for objects of class 'generator'. But as Daenyth implies, that isn't necessarily the right way to go.JAB
If you're checking for __next__, you're actually accepting any iterator, not just generators - which is very likely what you want.user395760
As often as not, the real point of knowing whether something is a generator is to be able to avoid them, on account of desiring to iterate over the same collection multiple times.Ian
For people wondering about the use case, this could be useful when you need to know if the iterator will be consumed (eg if your function accepts any iterator but needs to iterate more than once, you'll want to materialize it before iterating)wbadart

9 Answers

253
votes

You can use GeneratorType from types:

>>> import types
>>> types.GeneratorType
<class 'generator'>
>>> gen = (i for i in range(10))
>>> isinstance(gen, types.GeneratorType)
True
42
votes

You mean generator functions ? use inspect.isgeneratorfunction.

EDIT :

if you want a generator object you can use inspect.isgenerator as pointed out by JAB in his comment.

28
votes

I think it is important to make distinction between generator functions and generators (generator function's result):

>>> def generator_function():
...     yield 1
...     yield 2
...
>>> import inspect
>>> inspect.isgeneratorfunction(generator_function)
True

calling generator_function won't yield normal result, it even won't execute any code in the function itself, the result will be special object called generator:

>>> generator = generator_function()
>>> generator
<generator object generator_function at 0x10b3f2b90>

so it is not generator function, but generator:

>>> inspect.isgeneratorfunction(generator)
False

>>> import types
>>> isinstance(generator, types.GeneratorType)
True

and generator function is not generator:

>>> isinstance(generator_function, types.GeneratorType)
False

just for a reference, actual call of function body will happen by consuming generator, e.g.:

>>> list(generator)
[1, 2]

See also In python is there a way to check if a function is a "generator function" before calling it?

14
votes

The inspect.isgenerator function is fine if you want to check for pure generators (i.e. objects of class "generator"). However it will return False if you check, for example, a izip iterable. An alternative way for checking for a generalised generator is to use this function:

def isgenerator(iterable):
    return hasattr(iterable,'__iter__') and not hasattr(iterable,'__len__')
6
votes

You could use the Iterator or more specifically, the Generator from the typing module.

from typing import Generator, Iterator
g = (i for i in range(1_000_000))
print(type(g))
print(isinstance(g, Generator))
print(isinstance(g, Iterator))

result:

<class 'generator'>
True
True
3
votes

(I know it's an old post.) There is no need to import a module, you can declare an object for comparison at the beginning of the program:

gentyp= type(1 for i in "")                                                                                          
       ...
type(myobject) == gentyp
2
votes
>>> import inspect
>>> 
>>> def foo():
...   yield 'foo'
... 
>>> print inspect.isgeneratorfunction(foo)
True
2
votes

I know I can check if the object has a next method for it to be a generator, but I want some way using which I can determine the type of any object, not just generators.

Don't do this. It's simply a very, very bad idea.

Instead, do this:

try:
    # Attempt to see if you have an iterable object.
    for i in some_thing_which_may_be_a_generator:
        # The real work on `i`
except TypeError:
     # some_thing_which_may_be_a_generator isn't actually a generator
     # do something else

In the unlikely event that the body of the for loop also has TypeErrors, there are several choices: (1) define a function to limit the scope of the errors, or (2) use a nested try block.

Or (3) something like this to distinguish all of these TypeErrors which are floating around.

try:
    # Attempt to see if you have an iterable object.
    # In the case of a generator or iterator iter simply 
    # returns the value it was passed.
    iterator = iter(some_thing_which_may_be_a_generator)
except TypeError:
     # some_thing_which_may_be_a_generator isn't actually a generator
     # do something else
else:
    for i in iterator:
         # the real work on `i`

Or (4) fix the other parts of your application to provide generators appropriately. That's often simpler than all of this.

1
votes

If you are using tornado webserver or similar you might have found that server methods are actually generators and not methods. This makes it difficult to call other methods because yield is not working inside the method and therefore you need to start managing pools of chained generator objects. A simple method to manage pools of chained generators is to create a help function such as

def chainPool(*arg):
    for f in arg:
      if(hasattr(f,"__iter__")):
          for e in f:
             yield e
      else:
         yield f

Now writing chained generators such as

[x for x in chainPool(chainPool(1,2),3,4,chainPool(5,chainPool(6)))]

Produces output

[1, 2, 3, 4, 5, 6]

Which is probably what you want if your looking to use generators as a thread alternative or similar.