How yield catches StopIteration exception?
Question:
Why in the example function terminates:
def func(iterable):
while True:
val = next(iterable)
yield val
but if I take off yield statement function will raise StopIteration exception?
EDIT: Sorry for misleading you guys. I know what generators are and how to use them. Of course when I said function terminates I didn’t mean eager evaluation of function. I just implied that when I use function to produce generator:
gen = func(iterable)
in case of func it works and returns the same generator, but in case of func2:
def func2(iterable):
while True:
val = next(iterable)
it raises StopIteration instead of None return or infinite loop.
Let me be more specific. There is a function tee in itertools which is equivalent to:
def tee(iterable, n=2):
it = iter(iterable)
deques = [collections.deque() for i in range(n)]
def gen(mydeque):
while True:
if not mydeque: # when the local deque is empty
newval = next(it) # fetch a new value and
for d in deques: # load it to all the deques
d.append(newval)
yield mydeque.popleft()
return tuple(gen(d) for d in deques)
There is, in fact, some magic, because nested function gen has infinite loop without break statements. gen function terminates due to StopIteration exception when there is no items in it. But it terminates correctly (without raising exceptions), i.e. just stops loop. So the question is: where is StopIteration is handled?
Answers:
Without the yield, you iterate over the entire iterable without stopping to do anything with val. The while loop does not catch the StopIteration exception. An equivalent for loop would be:
def func(iterable):
for val in iterable:
pass
which does catch the StopIteration and simply exit the loop and thus return from the function.
You can explicitly catch the exception:
def func(iterable):
while True:
try:
val = next(iterable)
except StopIteration:
break
When a function contains yield, calling it does not actually execute anything, it merely creates a generator object. Only iterating over this object will execute the code. So my guess is that you’re merely calling the function, which means the function doesn’t raise StopIteration because it is never being executed.
Given your function, and an iterable:
def func(iterable):
while True:
val = next(iterable)
yield val
iterable = iter([1, 2, 3])
This is the wrong way to call it:
func(iterable)
This is the right way:
for item in func(iterable):
# do something with item
You could also store the generator in a variable and call next() on it (or iterate over it in some other way):
gen = func(iterable)
print(next(gen)) # prints 1
print(next(gen)) # prints 2
print(next(gen)) # prints 3
print(next(gen)) # StopIteration
By the way, a better way to write your function is as follows:
def func(iterable):
for item in iterable:
yield item
Or in Python 3.3 and later:
def func(iterable):
yield from iter(iterable)
Of course, real generators are rarely so trivial. 🙂
yield doesn’t catch the StopIteration. What yield does for your function is it causes it to become a generator function rather than a regular function. Thus, the object returned from the function call is an iterable object (which calculates the next value when you ask it to with the next function (which gets called implicitly by a for loop)). If you leave the yield statement out of it, then python executes the entire while loop right away which ends up exhausting the iterable (if it is finite) and raising StopIteration right when you call it.
consider:
x = func(x for x in [])
next(x) #raises StopIteration
A for loop catches the exception — That’s how it knows when to stop calling next on the iterable you gave it.
Note: This question (and the original part of my answer to it) are only really meaningful for Python versions prior to 3.7. The behavior that was asked about no longer happens in 3.7 and later, thanks to changes described in PEP 479. So this question and the original answer are only really useful as historical artifacts. After the PEP was accepted, I added an additional section at the bottom of the answer which is more relevant to modern versions of Python.
To answer your question about where the StopIteration gets caught in the gen generator created inside of itertools.tee: it doesn’t. It is up to the consumer of the tee results to catch the exception as they iterate.
First off, it’s important to note that a generator function (which is any function with a yield statement in it, anywhere) is fundamentally different than a normal function. Instead of running the function’s code when it is called, instead, you’ll just get a generator object when you call the function. Only when you iterate over the generator will you run the code.
A generator function will never finish iterating without raising StopIteration (unless it raises some other exception instead). StopIteration is the signal from the generator that it is done, and it is not optional. If you reach a return statement or the end of the generator function’s code without raising anything, Python will raise StopIteration for you!
This is different from regular functions, which return None if they reach the end without returning anything else. It ties in with the different ways that generators work, as I described above.
Here’s an example generator function that will make it easy to see how StopIteration gets raised:
def simple_generator():
yield "foo"
yield "bar"
# StopIteration will be raised here automatically
Here’s what happens when you consume it:
>>> g = simple_generator()
>>> next(g)
'foo'
>>> next(g)
'bar'
>>> next(g)
Traceback (most recent call last):
File "<pyshell#6>", line 1, in <module>
next(g)
StopIteration
Calling simple_generator always returns a generator object immediately (without running any of the code in the function). Each call of next on the generator object runs the code until the next yield statement, and returns the yielded value. If there is no more to get, StopIteration is raised.
Now, normally you don’t see StopIteration exceptions. The reason for this is that you usually consume generators inside for loops. A for statement will automatically call next over and over until StopIteration gets raised. It will catch and suppress the StopIteration exception for you, so you don’t need to mess around with try/except blocks to deal with it.
A for loop like for item in iterable: do_suff(item) is almost exactly equivalent to this while loop (the only difference being that a real for doesn’t need a temporary variable to hold the iterator):
iterator = iter(iterable)
try:
while True:
item = next(iterator)
do_stuff(item)
except StopIteration:
pass
finally:
del iterator
The gen generator function you showed at the top is one exception. It uses the StopIteration exception produced by the iterator it is consuming as it’s own signal that it is done being iterated on. That is, rather than catching the StopIteration and then breaking out of the loop, it simply lets the exception go uncaught (presumably to be caught by some higher level code).
Unrelated to the main question, there is one other thing I want to point out. In your code, you’re calling next on an variable called iterable. If you take that name as documentation for what type of object you will get, this is not necessarily safe.
next is part of the iterator protocol, not the iterable (or container) protocol. It may work for some kinds of iterables (such as files and generators, as those types are their own iterators), but it will fail for others iterables, such as tuples and lists. The more correct approach is to call iter on your iterable value, then call next on the iterator you receive. (Or just use for loops, which call both iter and next for you at appropriate times!)
I just found my own answer in a Google search for a related question, and I feel I should update to point out that the answer above is not true in modern Python versions.
PEP 479 has made it an error to allow a StopIteration to bubble up uncaught from a generator function. If that happens, Python will turn it into a RuntimeError exception instead. This means that code like the examples in older versions of itertools that used a StopIteration to break out of a generator function needs to be modified. Usually you’ll need to catch the exception with a try/except and then return.
Because this was a backwards incompatible change, it was phased in gradually. In Python 3.5, all code worked as before by default, but you could get the new behavior with from __future__ import generator_stop. In Python 3.6, unmodified code would still work, but it would give a warning. In Python 3.7 and later, the new behavior applies all the time.
Tested on Python 3.8, chunk as lazy generator
def split_to_chunk(size: int, iterable: Iterable) -> Iterable[Iterable]:
source_iter = iter(iterable)
while True:
batch_iter = itertools.islice(source_iter, size)
try:
yield itertools.chain([next(batch_iter)], batch_iter)
except StopIteration:
return
Why handling StopInteration error: https://www.python.org/dev/peps/pep-0479/
def sample_gen() -> Iterable[int]:
i = 0
while True:
yield i
i += 1
for chunk in split_to_chunk(7, sample_gen()):
pprint.pprint(list(chunk))
time.sleep(2)
Output:
[0, 1, 2, 3, 4, 5, 6]
[7, 8, 9, 10, 11, 12, 13]
[14, 15, 16, 17, 18, 19, 20]
[21, 22, 23, 24, 25, 26, 27]
............................
Why in the example function terminates:
def func(iterable):
while True:
val = next(iterable)
yield val
but if I take off yield statement function will raise StopIteration exception?
EDIT: Sorry for misleading you guys. I know what generators are and how to use them. Of course when I said function terminates I didn’t mean eager evaluation of function. I just implied that when I use function to produce generator:
gen = func(iterable)
in case of func it works and returns the same generator, but in case of func2:
def func2(iterable):
while True:
val = next(iterable)
it raises StopIteration instead of None return or infinite loop.
Let me be more specific. There is a function tee in itertools which is equivalent to:
def tee(iterable, n=2):
it = iter(iterable)
deques = [collections.deque() for i in range(n)]
def gen(mydeque):
while True:
if not mydeque: # when the local deque is empty
newval = next(it) # fetch a new value and
for d in deques: # load it to all the deques
d.append(newval)
yield mydeque.popleft()
return tuple(gen(d) for d in deques)
There is, in fact, some magic, because nested function gen has infinite loop without break statements. gen function terminates due to StopIteration exception when there is no items in it. But it terminates correctly (without raising exceptions), i.e. just stops loop. So the question is: where is StopIteration is handled?
Without the yield, you iterate over the entire iterable without stopping to do anything with val. The while loop does not catch the StopIteration exception. An equivalent for loop would be:
def func(iterable):
for val in iterable:
pass
which does catch the StopIteration and simply exit the loop and thus return from the function.
You can explicitly catch the exception:
def func(iterable):
while True:
try:
val = next(iterable)
except StopIteration:
break
When a function contains yield, calling it does not actually execute anything, it merely creates a generator object. Only iterating over this object will execute the code. So my guess is that you’re merely calling the function, which means the function doesn’t raise StopIteration because it is never being executed.
Given your function, and an iterable:
def func(iterable):
while True:
val = next(iterable)
yield val
iterable = iter([1, 2, 3])
This is the wrong way to call it:
func(iterable)
This is the right way:
for item in func(iterable):
# do something with item
You could also store the generator in a variable and call next() on it (or iterate over it in some other way):
gen = func(iterable)
print(next(gen)) # prints 1
print(next(gen)) # prints 2
print(next(gen)) # prints 3
print(next(gen)) # StopIteration
By the way, a better way to write your function is as follows:
def func(iterable):
for item in iterable:
yield item
Or in Python 3.3 and later:
def func(iterable):
yield from iter(iterable)
Of course, real generators are rarely so trivial. 🙂
yield doesn’t catch the StopIteration. What yield does for your function is it causes it to become a generator function rather than a regular function. Thus, the object returned from the function call is an iterable object (which calculates the next value when you ask it to with the next function (which gets called implicitly by a for loop)). If you leave the yield statement out of it, then python executes the entire while loop right away which ends up exhausting the iterable (if it is finite) and raising StopIteration right when you call it.
consider:
x = func(x for x in [])
next(x) #raises StopIteration
A for loop catches the exception — That’s how it knows when to stop calling next on the iterable you gave it.
Note: This question (and the original part of my answer to it) are only really meaningful for Python versions prior to 3.7. The behavior that was asked about no longer happens in 3.7 and later, thanks to changes described in PEP 479. So this question and the original answer are only really useful as historical artifacts. After the PEP was accepted, I added an additional section at the bottom of the answer which is more relevant to modern versions of Python.
To answer your question about where the StopIteration gets caught in the gen generator created inside of itertools.tee: it doesn’t. It is up to the consumer of the tee results to catch the exception as they iterate.
First off, it’s important to note that a generator function (which is any function with a yield statement in it, anywhere) is fundamentally different than a normal function. Instead of running the function’s code when it is called, instead, you’ll just get a generator object when you call the function. Only when you iterate over the generator will you run the code.
A generator function will never finish iterating without raising StopIteration (unless it raises some other exception instead). StopIteration is the signal from the generator that it is done, and it is not optional. If you reach a return statement or the end of the generator function’s code without raising anything, Python will raise StopIteration for you!
This is different from regular functions, which return None if they reach the end without returning anything else. It ties in with the different ways that generators work, as I described above.
Here’s an example generator function that will make it easy to see how StopIteration gets raised:
def simple_generator():
yield "foo"
yield "bar"
# StopIteration will be raised here automatically
Here’s what happens when you consume it:
>>> g = simple_generator()
>>> next(g)
'foo'
>>> next(g)
'bar'
>>> next(g)
Traceback (most recent call last):
File "<pyshell#6>", line 1, in <module>
next(g)
StopIteration
Calling simple_generator always returns a generator object immediately (without running any of the code in the function). Each call of next on the generator object runs the code until the next yield statement, and returns the yielded value. If there is no more to get, StopIteration is raised.
Now, normally you don’t see StopIteration exceptions. The reason for this is that you usually consume generators inside for loops. A for statement will automatically call next over and over until StopIteration gets raised. It will catch and suppress the StopIteration exception for you, so you don’t need to mess around with try/except blocks to deal with it.
A for loop like for item in iterable: do_suff(item) is almost exactly equivalent to this while loop (the only difference being that a real for doesn’t need a temporary variable to hold the iterator):
iterator = iter(iterable)
try:
while True:
item = next(iterator)
do_stuff(item)
except StopIteration:
pass
finally:
del iterator
The gen generator function you showed at the top is one exception. It uses the StopIteration exception produced by the iterator it is consuming as it’s own signal that it is done being iterated on. That is, rather than catching the StopIteration and then breaking out of the loop, it simply lets the exception go uncaught (presumably to be caught by some higher level code).
Unrelated to the main question, there is one other thing I want to point out. In your code, you’re calling next on an variable called iterable. If you take that name as documentation for what type of object you will get, this is not necessarily safe.
next is part of the iterator protocol, not the iterable (or container) protocol. It may work for some kinds of iterables (such as files and generators, as those types are their own iterators), but it will fail for others iterables, such as tuples and lists. The more correct approach is to call iter on your iterable value, then call next on the iterator you receive. (Or just use for loops, which call both iter and next for you at appropriate times!)
I just found my own answer in a Google search for a related question, and I feel I should update to point out that the answer above is not true in modern Python versions.
PEP 479 has made it an error to allow a StopIteration to bubble up uncaught from a generator function. If that happens, Python will turn it into a RuntimeError exception instead. This means that code like the examples in older versions of itertools that used a StopIteration to break out of a generator function needs to be modified. Usually you’ll need to catch the exception with a try/except and then return.
Because this was a backwards incompatible change, it was phased in gradually. In Python 3.5, all code worked as before by default, but you could get the new behavior with from __future__ import generator_stop. In Python 3.6, unmodified code would still work, but it would give a warning. In Python 3.7 and later, the new behavior applies all the time.
Tested on Python 3.8, chunk as lazy generator
def split_to_chunk(size: int, iterable: Iterable) -> Iterable[Iterable]:
source_iter = iter(iterable)
while True:
batch_iter = itertools.islice(source_iter, size)
try:
yield itertools.chain([next(batch_iter)], batch_iter)
except StopIteration:
return
Why handling StopInteration error: https://www.python.org/dev/peps/pep-0479/
def sample_gen() -> Iterable[int]:
i = 0
while True:
yield i
i += 1
for chunk in split_to_chunk(7, sample_gen()):
pprint.pprint(list(chunk))
time.sleep(2)
Output:
[0, 1, 2, 3, 4, 5, 6]
[7, 8, 9, 10, 11, 12, 13]
[14, 15, 16, 17, 18, 19, 20]
[21, 22, 23, 24, 25, 26, 27]
............................