invoking yield for a generator in another function - python

suppose I have some manager object. This object's API has a main_hook function, that gets another function f as it's argument, and runs the given f in a loop, doing some stuff in between each iteration:
def main_hook(self,f):
while (self.shouldContinue()):
#do some preparations
f(self)
#do some tear down
Now, I also have (more accurately, would like to have) a function stop_and_do_stuff, that once called, stops main_hook dead in it's tracks, returns the control to whichever func called main_hook, and after that func finished what's it doing, get control back to main_hook and continue. Basically the result will be the same as doing
def main_hook(self,f):
while (self.shouldContinue()):
#do some preparations
yield
#do some tear down
Except that instead yield I want to have a call to f(), while giving f the option to call self.stop_and_do_stuff()
I can't work around this by making f also a generator for 2 reasons:
1.f isn't part of my API - it's given to me by a user who uses my lib
2.Even if could ask him to use yield, the place in the code in which he will need to call stop_and_do_stuff won't be directly inside f, rather in some place in the function stack which will be inside f(), but not directly in it, e.g
def h(manager):
#do stuff
if should stop:
manager.stop_and_do_stuff()
#do more stuff
def g(manager):
#some stuff
if should stop:
manager.stop_and_do_stuff()
#more stuff
if should stop again:
manager.stop_and_do_stuff()
if should call h:
h()
def f(manager):
g(manager)
so if I choose to make f a generator, I also need to make g a generator and also h, otherwise this trick won't work.
Is there any solution to all of this? maybe I'm trying to solve it the wrong way?
(I know this question is long and ugly - it's the best I could do. If something isn't clear please tell me and I'll clarify it)
EDIT
Maybe pep 342 is the solution?

My previous answer describes how to do this in Python2, which is very ugly. But now I ran across PEP 380: Syntax for Delegating to a Subgenerator. That does exactly what you ask. The only problem is that it requires Python3. But that shouldn't really be a problem.
Here's how it works:
def worker():
yield 1
yield 2
return 3
def main():
yield 0
value = yield from worker()
print('returned %d' % value)
yield 4
for m in main():
print('generator yields %d' % m)
The result of this is:
generator yields 0
generator yields 1
generator yields 2
returned 3
generator yields 4
Exceptions are passed through the way you would expect.

I believe I should also add an answer from the other point of view, ie not trying to explain how you could achieve what we can understand of what you are trying to do, but why yield definitely couldn't possibly work.
When a function contains yield keyword it is deeply modified. It is still a callable but not a normal function any more : it becomes a factory that return an iterator.
From the caller's point of view there is no difference between the three implementations below (except that the yield one is so much simpler).
##########################################
print "Function iterator using yield",
def gen():
for x in range(0, 10):
yield x
f = gen()
try:
while True:
print f.next(),
except StopIteration:
pass
for x in gen():
print x,
print
#########################################
print "Class iterator defining iter and next",
class gen2(object):
def __init__(self):
self.index = 0;
self.limit = 10;
def __iter__(self):
return self
def next(self):
if self.index >= self.limit:
raise StopIteration
self.index += 1;
return self.index - 1;
f = gen2()
try:
while True:
print f.next(),
except StopIteration:
pass
for x in gen2():
print x,
print
#########################################
print "Function iterator using iter() and sentinel",
def gen3():
def g3():
if g3.index is None:
g3.index = 0
g3.index += 1;
return g3.index - 1
g3.index = None
return iter(g3, 10)
f = gen3()
try:
while True:
print f.next(),
except StopIteration:
pass
for x in gen3():
print x,
print
Then you should understand that yield is not much about control flow, but about keeping call context inside variables. Once it is understood you have to decide if the API of main_loop really want to provide an iterator to it's caller. Then if so, if f may loop it must should also be an iterator (and there should be a loop around calls to f() like below).
def main_hook(self,f):
while (self.shouldContinue()):
#do some preparations
for v in f(self):
yield v
#do some tear down
But you should not care if f() has to call inner functions g(), etc. That is completely irrelevant. You provide a lib and it is your user problem to call with an appropriate iterable. If you believe your lib user won't be able to, you will have to change the overall design.
Hope it helps.

I don't understand the whole either (what does the main_hook caller look like ?), but i would say, Throw a StopNow exception, when you should stop, just like you should throw StopIteration when your generator is finished.
here is how i understood the thing as well as what i would do.
class StopNow(Exception):
pass
def main_hook(self,f):
got_stop_now_exc = False
while (!got_stop_now_exc and self.shouldContinue()):
#do some preparations
try:
f(self)
except StopNow:
got_stop_now_exc = True
#do some compulsary tear down, exception or not
def stop_and_do_stuff()
raise StopNow()
def my_f():
if needed:
stop_and_do_stuff()
def the_main_hook_caller():
while i_should:
managerthingie.main_hook(my_f)
do_stuff()

The behavior you describe looks exactly like a simple function call. Like below.
def f(manager):
print("Entering f")
manager.stop_and_do_stuff()
print("Exiting f")
class Manager(Object):
def shouldContinue(self):
return True
def stop_and_do_stuff(self):
print("Manager stop and do stuff")
def main_hook(self,f):
while self.shouldContinue()
print("Manager Setup")
f(self)
print("Manager Tear Down")
No problem if f() is provided by another user of if stop_and_do_stuff is called from some inner function. If you also want the manager to be able to unwind stack from stop_and_do_stuff and really exit in some cases, no problem. Just raise some exception from it and you would catch it from main_hook or upper code.
You should be able to do from inside stop_and_and_do_stuff() whatever you want to do from the caller of main hook. If not you should explain why.
What is unclear in the question is what's happening on the caller side of main_hook() and why you would want to be able to exit the main_hook loop, but not really. Either the main_loop caller expect a generator either it does not. You need to explain that part if you want to get a sensible answer (some context informations would also be nice, if you really explain WTF you are trying to do, and your real restrictions - you said f is provided by some other user and main_hook is in a lib, what of main_hook caller ? - there is probably well known usual solutions).

I am not quite sure what exactly you are trying to achieve, so maybe if you can explain the problem more instead of giving solution that would be better.
From my partial understanding why don't you do something like this
def main_hook(self,f):
while (self.shouldContinue()):
#do some preparations
stop_and_do_stuff = f(self)
if stop_and_do_stuff :
yield
#do some tear down
So basically f returns a flag to stop or not, and if it says stop we yield to function which called main_hook and that function can continue after doing some stuff
e.g.
class A(object):
def main_hook(self,f):
while (self.shouldContinue()):
#do some preparations
stop = f(self)
if stop:
yield
#do some tear down
def shouldContinue(self):
return True
def f(a):
return True
a = A()
for x in a.main_hook(f):
print x

Related

Unexpected decorator behavior, "if" statement executes after if 1==2:

А few hours ago I submitted a related question and got an answer why I need to add yield to my decorator in order to function properly.
I have lately recalled that I had omitted it for a reason - a strange behavior I can not explain.
I apologize in advance if I am blind, but I spent hours staring at and playing with this code and this is what I get:
def decor(func):
def wrapper(*args, **kwargs):
if 1==2:
print ("Generator")
for item in func(*args, **kwargs):
print(item)
#yield(item)
else:
print ("Not generator")
res = func(*args, **kwargs)
print(res)
return res
return wrapper
#decor
def f():
return "a"
f()
"""
Output:
Not generator
a
"""
And if I remove the comment before yield there is no output at all.
Why is that? And how is it possible that anything which I change within a if 1==2: statement takes any effect on the script?
If a function contains yield anywhere in the body, it is a generator function. It does not matter if the yield is executed or not. The fact that 1 == 2 is false has nothing to do with it.
Consider the following function:
def addone(numbers):
for number in numbers:
yield number + 1
What happens when you call addone([])? The yield is never executed, and yet addone still returns a generator. Why should this be any different:
def addone(numbers):
if numbers:
for number in numbers:
yield number + 1
So it becomes clear that whether or not the yield is actually executed is not relevant. The only relevant fact is whether the yield exists in the body of the function.
How to fix the function
The fix is relatively simple, all you have to do is pull the part with the yield into a separate function:
import types
def wrapper(*args, **kwargs):
result = func(*args, **kwargs)
if isinstance(result, types.GeneratorType):
print("Is a generator")
return wrap_generator(result)
print("Not a generator")
return result
def wrap_generator(gen):
for item in gen:
print(item)
yield item
How to avoid in the future
In general, the problem here is that a function is either a generator (and uses yield) or is a normal function (and uses return). It is a bit confusing when you use both yield and return in the same function!
For Python, it turns out that if you use both yield and return in the same function, the function is a generator function. This may be somewhat confusing, so as a matter of style, I would generally avoid using both return and yield in the same function.

Using the return value of a function if it satisfies a condition without using an extra variable

Assuming a function some_function(x) that returns an integer, I'm currently doing the following:
def another_function():
# stuff...
result = some_function(x)
if result != 0:
return result
# else: more stuff (unrelated to result)...
What I want is to find the best way to write
if some_function(x) != 0:
return some_function(x)
without of course calling some_function(x) twice.
Is there any more "clever" idiomatic way to write that in Python 3, without using result?
Another option is to define an exception to raise when the function gives 0 as result. If that is the case, catch the error and do something else.
class ZeroResult(Exception):
def __init__(self):
raise AttributeError
def some_function(x):
#...
if result==0:
raise ZeroResult
else:
return result
def another_function():
# stuff
try:
return some_function(x)
except ZeroResult:
# more stuff
I'll echo my comment in this answer and suggest that what you are currently doing is the best approach.
Dimgolds answer, if you went ahead and implemented a C.M, pretty much boils down to doing the same thing. FLabs approach uses an exception which, if the case is not exceptional, introduces unwanted overhead.
What you are doing is simple, explicit and completely understandable; things that are idiomatic in Python. Keep doing it this way.
Since you're addressing the result in few places, you'll have to either run it twice or save into a variable.
Another option (which is similar to using result ) is by using ''with'' statement and continue the else part:
with some_function(x) as result:
if result: return result
This syntax won't keep result variable later on.
def mem(f, *args, **kwargs):
r = f(*args, **kwargs)
return r
def foo():
print("hi, i'm here once")
return 1
def bar(a,b):
print("also once")
return a+b
somefunc_with_foo = mem(foo)
print(somefunc_with_foo)
print(somefunc_with_foo)
somefunc_with_bar_2_3 = mem(bar,2,3)
print(somefunc_with_bar_2_3)
print(somefunc_with_bar_2_3)
I don't think the extra cuteness is worth it, but if you wanted to you could do with 2 functions.
return some_function(x) or my_other_stuff_that_gets_run_if_zero(x, y, z)
The boolean shortcut will return result right away on some_function(x) truthy and call (and return results for) another function if not.

Using generator send() within a for loop

I implemented graph traversal as a generator function which yields the node being visited.
Sometimes the user needs to tell the traversal function that the edges outgoing from a particular node shouldn't be followed; in order to support that, the traversal checks the value sent back to it (using generator send() method), and if it's True, regards the node as a leaf for traversal purposes.
The problem is that the simplest user loop is kinda long:
# simplified thanks to #tobias_k
# bfs is the traversal generator function
traversal = bfs(g, start_node)
try:
n = next(traversal)
while True:
# process(n) returns True if don't want to follow edges out of n
n = traversal.send(process(n))
except StopIteration:
pass
Is there any way to improve this?
I thought something like this should work:
for n in bfs(g, start_node):
???.send(process(n))
but I feel I'm missing the knowledge of some python syntax.
I don't see a way to do this in a regular for loop. However, you could create another generator, that iterates another generator, using some "follow-function" to determine whether to follow the current element, thus encapsulating the tricky parts of your code into a separate function.
def checking_generator(generator, follow_function):
try:
x = next(generator)
while True:
yield x
x = generator.send(follow_function(x))
except StopIteration:
pass
for n in checking_generator(bfs(g, start_node), process):
print(n)
I discovered that my question would have had a one-line answer, using the extended "continue" statement proposed in the earlier version of PEP 342:
for n in bfs(g, start_node):
continue process(n)
However, while PEP 342 was accepted, that particular feature was withdrawn after this June 2005 discussion between Raymond and Guido:
Raymond Hettinger said:
Let me go on record as a strong -1 for "continue EXPR". The
for-loop is our most basic construct and is easily understood in its
present form. The same can be said for "continue" and "break" which
have the added advantage of a near zero learning curve for people
migrating from other languages.
Any urge to complicate these basic statements should be seriously
scrutinized and held to high standards of clarity, explainability,
obviousness, usefulness, and necessity. IMO, it fails most of those
tests.
I would not look forward to explaining "continue EXPR" in the tutorial
and think it would stand out as an anti-feature.
[...] The correct argument against "continue EXPR" is that there
are no use cases yet; if there were a good use case, the explanation
would follow easily.
Guido
If python core developers have since changed their mind about the usefulness of extended "continue", perhaps this could be reintroduced into a future PEP. But, given a nearly identical use case as in this question was already discussed in the quoted thread, and wasn't found persuasive, it seems unlikely.
To simplify the client code, you could use an ordinary bsf() generator and check node.isleaf attribute in it:
for node in bfs(g, start_node):
node.isleaf = process(node) # don't follow if `process()` returns True
The disadvantage is that node is mutable. Or you have to pass a shared data structure that tracks leaf nodes: leaf[node] = process(node) where leaf dictionary is passed into bfs() earlier.
If you want to use .send() method explicitly; you have to handle StopIteration. See PEP 479 -- Change StopIteration handling inside generators. You could hide it in a helper function:
def traverse(tree_generator, visitor):
try:
node = next(tree_generator)
while True:
node = tree_generator.send(visitor(node))
except StopIteration:
pass
Example:
traverse(bfs(g, start_node), process)
I don't see this as a frequent use case, consider this as the original generator:
def original_gen():
for x in range(10):
should_break = yield x
if should_break:
break
If the value of should_break is always calculated based on some function call with x then why not just write the generator like this:
def processing_gen(check_f):
for x in range(10):
yield x
should_break = check_f(x)
if should_break:
break
However I usually think of the code that processes the generated values as being written inside the loop (otherwise what is the point of having a loop at all?)
What it really seems you want to do is create a generator where calling the __next__ method really implies send(process(LAST_VALUE)) which can be implemented with a class:
class Followup_generator(): #feel free to use a better name
def __init__(self,generator,following_function):
self.gen = generator
self.process_f = following_function
def __iter__(self):
return self
def __next__(self):
if hasattr(self,"last_value"):
return self.send(self.process_f(self.last_value))
else:
self.last_value = next(self.gen)
return self.last_value
def send(self,arg):
self.last_value = self.gen.send(arg)
return self.last_value
def __getattr__(self,attr):
"forward other lookups to the generator (.throw etc.)"
return getattr(self.gen, attr)
# call signature is the exact same as #tobias_k's checking_generator
traversal = Followup_generator(bfs(g, start_node), process)
for n in traversal:
print(n)
n = traversal.send(DATA) #you'd be able to send extra values to it
However this still doesn't see this as frequently used, I'd be perfectly fine with a while loop, although I'd put the .send call at the top:
traversal = bfs(g, start_node)
send_value = None
while True:
n = traversal.send(send_value)
#code for loop, ending in calculating the next send_value
send_value = process(n)
And you might wrap that in a try: ... except StopIteration:pass although I find that simply waiting for an error to raise is better expressed with a context manager:
class Catch:
def __init__(self,exc_type):
if issubclass(exc_type,BaseException):
self.catch_type = exc_type
else:
raise TypeError("can only catch Exceptions")
def __enter__(self):
return self
def __exit__(self,exc_type,err, tb):
if issubclass(exc_type, self.catch_type):
self.err = err
return True
with Catch(StopIteration):
traversal = bfs(g, start_node)
send_value = None
while True:
n = traversal.send(send_value)
#code for loop, ending in calculating the next send_value
send_value = process(n)
Probably this is the answer to the question from the thread's topic.
Take a look at the additional empty yields statements inside the traversal function and custom send function, that does the magical job.
# tested with Python 3.7
def traversal(n):
for i in range(n):
yield i, '%s[%s] %s' % (' ' * (4 - n), n, i)
stop = yield
if stop:
yield # here's the first part of the magic
else:
yield # the same as above
yield from traversal(int(n / 2))
def send(generator, value):
next(generator) # here's the second part of the magic
generator.send(value)
g = traversal(4)
for i, (num, msg) in enumerate(g):
print('>', i, msg)
stop = num % 2 == 0
send(g, stop)
I've written a small class SettableGenerator which uses a method to receive the value to be send and then forwards it to the actual generator when __next__ is invoked.
With this you can write:
gen = SettableGenerator(bfs(g, start_node))
for n in gen:
gen.set(process(n))
Let's consider the following generator. It generates numbers from 0 to 9. For every generated number, it gets an input and stores it into ret:
def count_to_nine():
# Output: numbers from 0 to 9
# Input: converted numbers
ret = []
for i in range(10):
# Yield a number, get something back
val = (yield i)
# Remember that "something"
ret.append(val)
return ret
You can, indeed, iterate it using next() + send(),
but the best way is to iterate using send() alone:
g = count_to_nine()
value = None # to make sure that the first send() gives a None
while True:
value = g.send(value) # send the previously generated value, get a new one
value = f'#{value}'
Here's the result:
StopIteration: ['#0', '#1', '#2', '#3', '#4', '#5', '#6', '#7', '#8', '#9']
If you want that output, catch the StopIteration and get the result from it.
Cheers!

Optional yield or return in python3. How to?

I would like to have a function that can, optionally, return or yield the result.
Here is an example.
def f(option=True):
...
for...:
if option:
yield result
else:
results.append(result)
if not option:
return results
Of course, this doesn't work, I have tried with python3 and I always get a generator no matter what option value I set.
As far I have understood, python checks the body of the function and if a yield is present, then the result will be a generator.
Is there any way to get around this and make a function that can return or yield at will?
You can't. Any use of yield makes the function a generator.
You could wrap your function with one that uses list() to store all values the generator produces in a list object and returns that:
def f_wrapper(option=True):
gen = f()
if option:
return gen # return the generator unchanged
return list(gen) # return all values of the generator as a list
However, generally speaking, this is bad design. Don't have your functions alter behaviour like this; stick to one return type (a generator or an object) and don't have it switch between the two.
Consider splitting this into two functions instead:
def f():
yield result
def f_as_list():
return list(f())
and use either f() if you need the generator, and f_as_list() if you want to have a list instead.
Since list(), (and next() to access just one value of a generator) are built-in functions, you rarely need to use a wrapper. Just call those functions directly:
# access elements one by one
gen = f()
one_value = next(gen)
# convert the generator to a list
all_values = list(f())
What about this?
def make_f_or_generator(option):
def f():
return "I am a function."
def g():
yield "I am a generator."
if option:
return f
else:
return g
This gives you at least the choice to create a function or a generator.
class based approach
class FunctionAndGenerator:
def __init__(self):
self.counter = 0
def __iter__(self):
return self
# You need a variable to indicate if dunder next should return the string or raise StopIteration.
# Raising StopIteration will stop the loop from iterating more.
# You'll have to teach next to raise StopIteration at some point
def __next__(self):
self.counter += 1
if self.counter > 1 :
raise StopIteration
return f"I'm a generator and I've generated {self.counter} times"
def __call__(self):
return "I'm a function"
x = FunctionAndGenerator()
print(x())
for i in x:
print(i)
I'm a function
I'm a generator and I've generated 1 times
[Program finished]

Python twisted: iterators and yields/inlineCallbacks

Folks,
Am thoroughly confused, so it's possible I am not even asking things correctly, but here goes:
I have a twisted application using inlineCallbacks. Now I need to define an iterator which will mean a generator is returned to the caller. However, the iterator cannot be inlineCallbacks decorated, can it be? If not, then how I do I code something like this.
Just to clarify: the goal is process_loop needs to be called every, say 5, seconds, it can process only ONE chunk of, say 10, and then it has to let go. However, to know that chunk of 10 (stored in cached, which is a dict of a dict), it needs to call a function that returns deferred.
#inlineCallbacks ### can\'t have inlineCallbacks here, right?
def cacheiter(cached):
for cachename,cachevalue in cached.items():
result = yield (call func here which returns deferred)
if result is True:
for k,v in cachedvalue.items():
yield cachename, k, v
#inlineCallbacks
def process_chunk(myiter, num):
try:
for i in xrange(num):
nextval = myiter.next()
yield some_processing(nextval)
returnValue(False)
except StopIteration:
returnValue(True)
#inlineCallbacks
def process_loop(cached):
myiter = cacheiter(cached)
result = yield process_chunk(myiter, 10)
if not result:
print 'More left'
reactor.callLater(5, process_loop, cached)
else:
print 'All done'
You're right that you can't express what you want to express in cacheiter. The inlineCallbacks decorator won't let you have a function that returns an iterator. If you decorate a function with it, then the result is a function that always returns a Deferred. That's what it is for.
Part of what makes this difficult is that iterators don't work well with asynchronous code. If there's a Deferred involved in producing the elements of your iterator, then the elements that come out of your iterator are going to be Deferreds first.
You might do something like this to account for that:
#inlineCallbacks
def process_work():
for element_deferred in some_jobs:
element = yield element_deferred
work_on(element)
This can work, but it looks particularly weird. Since generators can only yield to their caller (not, for example, to their caller's caller), the some_jobs iterator can't do anything about this; only code lexically within process_work can yield a Deferred to the inlineCallbacks-provided trampoline to wait on.
If you don't mind this pattern, then we could imaging your code being written something like:
from twisted.internet.task import deferLater
from twisted.internet.defer import inlineCallbacks, returnValue
from twisted.internet import reactor
class cacheiter(object):
def __init__(self, cached):
self._cached = iter(cached.items())
self._remaining = []
def __iter__(self):
return self
#inlineCallbacks
def next(self):
# First re-fill the list of synchronously-producable values if it is empty
if not self._remaining:
for name, value in self._cached:
# Wait on this Deferred to determine if this cache item should be included
if (yield check_condition(name, value)):
# If so, put all of its values into the value cache so the next one
# can be returned immediately next time this method is called.
self._remaining.extend([(name, k, v) for (k, v) in value.items()])
# Now actually give out a value, if there is one.
if self._remaining:
returnValue(self._remaining.pop())
# Otherwise the entire cache has been visited and the iterator is complete.
# Sadly we cannot signal completion with StopIteration, because the iterator
# protocol isn't going to add an errback to this Deferred and check for
# StopIteration. So signal completion with a simple None value.
returnValue(None)
#inlineCallbacks
def process_chunk(myiter, num):
for i in xrange(num):
nextval = yield myiter.next()
if nextval is None:
# The iterator signaled completion via the special None value.
# Processing is complete.
returnValue(True)
# Otherwise process the value.
yield some_processing(nextval)
# Indicate there is more processing to be done.
returnValue(False)
def sleep(sec):
# Simple helper to delay asynchronously for some number of seconds.
return deferLater(reactor, sec, lambda: None)
#inlineCallbacks
def process_loop(cached):
myiter = cacheiter(cached)
while True:
# Loop processing 10 items from myiter at a time, until process_chunk signals
# there are no values left.
result = yield process_chunk(myiter, 10)
if result:
print 'All done'
break
print 'More left'
# Insert the 5 second delay before starting on the next chunk.
yield sleep(5)
d = process_loop(cached)
Another approach you might be able to take, though, is to use twisted.internet.task.cooperate. cooperate takes an iterator and consumes it, assuming that consuming it is potentially costly, and splitting up the job over multiple reactor iterations. Taking the definition of cacheiter from above:
from twisted.internet.task import cooperate
def process_loop(cached):
finished = []
def process_one(value):
if value is None:
finished.append(True)
else:
return some_processing(value)
myiter = cacheiter(cached)
while not finished:
value_deferred = myiter.next()
value_deferred.addCallback(process_one)
yield value_deferred
task = cooperate(process_loop(cached))
d = task.whenDone()
I think you're trying to do this:
#inlineCallbacks
def cacheiter(cached):
for cachename,cachevalue in cached.items():
result = yield some_deferred() # some deferred you'd like evaluated
if result is True:
# here you want to return something, so you have to use returnValue
# the generator you want to return can be written as a generator expression
gen = ((cachename, k, v) for k,v in cachedvalue.items())
returnValue(gen)
When a genexp can't express what you're trying to return you can write a closure:
#inlineCallbacks
def cacheiter(cached):
for cachename,cachevalue in cached.items():
result = yield some_deferred()
if result is True:
# define the generator, saving the current values of the cache
def gen(cachedvalue=cachedvalue, cachename=cachename):
for k,v in cachedvalue.items():
yield cachename, k, v
returnValue(gen()) # return it
Try writing your iterator as a DeferredGenerator.

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