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I am reading the Python cookbook at the moment and am currently looking at generators. I'm finding it hard to get my head round.
As I come from a Java background, is there a Java equivalent? The book was speaking about 'Producer / Consumer', however when I hear that I think of threading.
What is a generator and why would you use it? Without quoting any books, obviously (unless you can find a decent, simplistic answer direct from a book). Perhaps with examples, if you're feeling generous!
Note: this post assumes Python 3.x syntax.†
A generator is simply a function which returns an object on which you can call next, such that for every call it returns some value, until it raises a StopIteration exception, signaling that all values have been generated. Such an object is called an iterator.
Normal functions return a single value using return, just like in Java. In Python, however, there is an alternative, called yield. Using yield anywhere in a function makes it a generator. Observe this code:
>>> def myGen(n):
... yield n
... yield n + 1
...
>>> g = myGen(6)
>>> next(g)
6
>>> next(g)
7
>>> next(g)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
StopIteration
As you can see, myGen(n) is a function which yields n and n + 1. Every call to next yields a single value, until all values have been yielded. for loops call next in the background, thus:
>>> for n in myGen(6):
... print(n)
...
6
7
Likewise there are generator expressions, which provide a means to succinctly describe certain common types of generators:
>>> g = (n for n in range(3, 5))
>>> next(g)
3
>>> next(g)
4
>>> next(g)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
StopIteration
Note that generator expressions are much like list comprehensions:
>>> lc = [n for n in range(3, 5)]
>>> lc
[3, 4]
Observe that a generator object is generated once, but its code is not run all at once. Only calls to next actually execute (part of) the code. Execution of the code in a generator stops once a yield statement has been reached, upon which it returns a value. The next call to next then causes execution to continue in the state in which the generator was left after the last yield. This is a fundamental difference with regular functions: those always start execution at the "top" and discard their state upon returning a value.
There are more things to be said about this subject. It is e.g. possible to send data back into a generator (reference). But that is something I suggest you do not look into until you understand the basic concept of a generator.
Now you may ask: why use generators? There are a couple of good reasons:
Certain concepts can be described much more succinctly using generators.
Instead of creating a function which returns a list of values, one can write a generator which generates the values on the fly. This means that no list needs to be constructed, meaning that the resulting code is more memory efficient. In this way one can even describe data streams which would simply be too large to fit in memory.
Generators allow for a natural way to describe infinite streams. Consider for example the Fibonacci numbers:
>>> def fib():
... a, b = 0, 1
... while True:
... yield a
... a, b = b, a + b
...
>>> import itertools
>>> list(itertools.islice(fib(), 10))
[0, 1, 1, 2, 3, 5, 8, 13, 21, 34]
This code uses itertools.islice to take a finite number of elements from an infinite stream. You are advised to have a good look at the functions in the itertools module, as they are essential tools for writing advanced generators with great ease.
† About Python <=2.6: in the above examples next is a function which calls the method __next__ on the given object. In Python <=2.6 one uses a slightly different technique, namely o.next() instead of next(o). Python 2.7 has next() call .next so you need not use the following in 2.7:
>>> g = (n for n in range(3, 5))
>>> g.next()
3
A generator is effectively a function that returns (data) before it is finished, but it pauses at that point, and you can resume the function at that point.
>>> def myGenerator():
... yield 'These'
... yield 'words'
... yield 'come'
... yield 'one'
... yield 'at'
... yield 'a'
... yield 'time'
>>> myGeneratorInstance = myGenerator()
>>> next(myGeneratorInstance)
These
>>> next(myGeneratorInstance)
words
and so on. The (or one) benefit of generators is that because they deal with data one piece at a time, you can deal with large amounts of data; with lists, excessive memory requirements could become a problem. Generators, just like lists, are iterable, so they can be used in the same ways:
>>> for word in myGeneratorInstance:
... print word
These
words
come
one
at
a
time
Note that generators provide another way to deal with infinity, for example
>>> from time import gmtime, strftime
>>> def myGen():
... while True:
... yield strftime("%a, %d %b %Y %H:%M:%S +0000", gmtime())
>>> myGeneratorInstance = myGen()
>>> next(myGeneratorInstance)
Thu, 28 Jun 2001 14:17:15 +0000
>>> next(myGeneratorInstance)
Thu, 28 Jun 2001 14:18:02 +0000
The generator encapsulates an infinite loop, but this isn't a problem because you only get each answer every time you ask for it.
First of all, the term generator originally was somewhat ill-defined in Python, leading to lots of confusion. You probably mean iterators and iterables (see here). Then in Python there are also generator functions (which return a generator object), generator objects (which are iterators) and generator expressions (which are evaluated to a generator object).
According to the glossary entry for generator it seems that the official terminology is now that generator is short for "generator function". In the past the documentation defined the terms inconsistently, but fortunately this has been fixed.
It might still be a good idea to be precise and avoid the term "generator" without further specification.
Generators could be thought of as shorthand for creating an iterator. They behave like a Java Iterator. Example:
>>> g = (x for x in range(10))
>>> g
<generator object <genexpr> at 0x7fac1c1e6aa0>
>>> g.next()
0
>>> g.next()
1
>>> g.next()
2
>>> list(g) # force iterating the rest
[3, 4, 5, 6, 7, 8, 9]
>>> g.next() # iterator is at the end; calling next again will throw
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
StopIteration
Hope this helps/is what you are looking for.
Update:
As many other answers are showing, there are different ways to create a generator. You can use the parentheses syntax as in my example above, or you can use yield. Another interesting feature is that generators can be "infinite" -- iterators that don't stop:
>>> def infinite_gen():
... n = 0
... while True:
... yield n
... n = n + 1
...
>>> g = infinite_gen()
>>> g.next()
0
>>> g.next()
1
>>> g.next()
2
>>> g.next()
3
...
There is no Java equivalent.
Here is a bit of a contrived example:
#! /usr/bin/python
def mygen(n):
x = 0
while x < n:
x = x + 1
if x % 3 == 0:
yield x
for a in mygen(100):
print a
There is a loop in the generator that runs from 0 to n, and if the loop variable is a multiple of 3, it yields the variable.
During each iteration of the for loop the generator is executed. If it is the first time the generator executes, it starts at the beginning, otherwise it continues from the previous time it yielded.
I like to describe generators, to those with a decent background in programming languages and computing, in terms of stack frames.
In many languages, there is a stack on top of which is the current stack "frame". The stack frame includes space allocated for variables local to the function including the arguments passed in to that function.
When you call a function, the current point of execution (the "program counter" or equivalent) is pushed onto the stack, and a new stack frame is created. Execution then transfers to the beginning of the function being called.
With regular functions, at some point the function returns a value, and the stack is "popped". The function's stack frame is discarded and execution resumes at the previous location.
When a function is a generator, it can return a value without the stack frame being discarded, using the yield statement. The values of local variables and the program counter within the function are preserved. This allows the generator to be resumed at a later time, with execution continuing from the yield statement, and it can execute more code and return another value.
Before Python 2.5 this was all generators did. Python 2.5 added the ability to pass values back in to the generator as well. In doing so, the passed-in value is available as an expression resulting from the yield statement which had temporarily returned control (and a value) from the generator.
The key advantage to generators is that the "state" of the function is preserved, unlike with regular functions where each time the stack frame is discarded, you lose all that "state". A secondary advantage is that some of the function call overhead (creating and deleting stack frames) is avoided, though this is a usually a minor advantage.
It helps to make a clear distinction between the function foo, and the generator foo(n):
def foo(n):
yield n
yield n+1
foo is a function.
foo(6) is a generator object.
The typical way to use a generator object is in a loop:
for n in foo(6):
print(n)
The loop prints
# 6
# 7
Think of a generator as a resumable function.
yield behaves like return in the sense that values that are yielded get "returned" by the generator. Unlike return, however, the next time the generator gets asked for a value, the generator's function, foo, resumes where it left off -- after the last yield statement -- and continues to run until it hits another yield statement.
Behind the scenes, when you call bar=foo(6) the generator object bar is defined for you to have a next attribute.
You can call it yourself to retrieve values yielded from foo:
next(bar) # Works in Python 2.6 or Python 3.x
bar.next() # Works in Python 2.5+, but is deprecated. Use next() if possible.
When foo ends (and there are no more yielded values), calling next(bar) throws a StopInteration error.
The only thing I can add to Stephan202's answer is a recommendation that you take a look at David Beazley's PyCon '08 presentation "Generator Tricks for Systems Programmers," which is the best single explanation of the how and why of generators that I've seen anywhere. This is the thing that took me from "Python looks kind of fun" to "This is what I've been looking for." It's at http://www.dabeaz.com/generators/.
This post will use Fibonacci numbers as a tool to build up to explaining the usefulness of Python generators.
This post will feature both C++ and Python code.
Fibonacci numbers are defined as the sequence: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, ....
Or in general:
F0 = 0
F1 = 1
Fn = Fn-1 + Fn-2
This can be transferred into a C++ function extremely easily:
size_t Fib(size_t n)
{
//Fib(0) = 0
if(n == 0)
return 0;
//Fib(1) = 1
if(n == 1)
return 1;
//Fib(N) = Fib(N-2) + Fib(N-1)
return Fib(n-2) + Fib(n-1);
}
But if you want to print the first six Fibonacci numbers, you will be recalculating a lot of the values with the above function.
For example: Fib(3) = Fib(2) + Fib(1), but Fib(2) also recalculates Fib(1). The higher the value you want to calculate, the worse off you will be.
So one may be tempted to rewrite the above by keeping track of the state in main.
// Not supported for the first two elements of Fib
size_t GetNextFib(size_t &pp, size_t &p)
{
int result = pp + p;
pp = p;
p = result;
return result;
}
int main(int argc, char *argv[])
{
size_t pp = 0;
size_t p = 1;
std::cout << "0 " << "1 ";
for(size_t i = 0; i <= 4; ++i)
{
size_t fibI = GetNextFib(pp, p);
std::cout << fibI << " ";
}
return 0;
}
But this is very ugly, and it complicates our logic in main. It would be better to not have to worry about state in our main function.
We could return a vector of values and use an iterator to iterate over that set of values, but this requires a lot of memory all at once for a large number of return values.
So back to our old approach, what happens if we wanted to do something else besides print the numbers? We'd have to copy and paste the whole block of code in main and change the output statements to whatever else we wanted to do.
And if you copy and paste code, then you should be shot. You don't want to get shot, do you?
To solve these problems, and to avoid getting shot, we may rewrite this block of code using a callback function. Every time a new Fibonacci number is encountered, we would call the callback function.
void GetFibNumbers(size_t max, void(*FoundNewFibCallback)(size_t))
{
if(max-- == 0) return;
FoundNewFibCallback(0);
if(max-- == 0) return;
FoundNewFibCallback(1);
size_t pp = 0;
size_t p = 1;
for(;;)
{
if(max-- == 0) return;
int result = pp + p;
pp = p;
p = result;
FoundNewFibCallback(result);
}
}
void foundNewFib(size_t fibI)
{
std::cout << fibI << " ";
}
int main(int argc, char *argv[])
{
GetFibNumbers(6, foundNewFib);
return 0;
}
This is clearly an improvement, your logic in main is not as cluttered, and you can do anything you want with the Fibonacci numbers, simply define new callbacks.
But this is still not perfect. What if you wanted to only get the first two Fibonacci numbers, and then do something, then get some more, then do something else?
Well, we could go on like we have been, and we could start adding state again into main, allowing GetFibNumbers to start from an arbitrary point.
But this will further bloat our code, and it already looks too big for a simple task like printing Fibonacci numbers.
We could implement a producer and consumer model via a couple of threads. But this complicates the code even more.
Instead let's talk about generators.
Python has a very nice language feature that solves problems like these called generators.
A generator allows you to execute a function, stop at an arbitrary point, and then continue again where you left off.
Each time returning a value.
Consider the following code that uses a generator:
def fib():
pp, p = 0, 1
while 1:
yield pp
pp, p = p, pp+p
g = fib()
for i in range(6):
g.next()
Which gives us the results:
0
1
1
2
3
5
The yield statement is used in conjuction with Python generators. It saves the state of the function and returns the yeilded value. The next time you call the next() function on the generator, it will continue where the yield left off.
This is by far more clean than the callback function code. We have cleaner code, smaller code, and not to mention much more functional code (Python allows arbitrarily large integers).
Source
I believe the first appearance of iterators and generators were in the Icon programming language, about 20 years ago.
You may enjoy the Icon overview, which lets you wrap your head around them without concentrating on the syntax (since Icon is a language you probably don't know, and Griswold was explaining the benefits of his language to people coming from other languages).
After reading just a few paragraphs there, the utility of generators and iterators might become more apparent.
I put up this piece of code which explains 3 key concepts about generators:
def numbers():
for i in range(10):
yield i
gen = numbers() #this line only returns a generator object, it does not run the code defined inside numbers
for i in gen: #we iterate over the generator and the values are printed
print(i)
#the generator is now empty
for i in gen: #so this for block does not print anything
print(i)
Performance difference:
macOS Big Sur 11.1
MacBook Pro (13-inch, M1, 2020)
Chip Apple M1
Memory 8gb
CASE 1
import random
import psutil # pip install psutil
import os
from datetime import datetime
def memory_usage_psutil():
# return the memory usage in MB
process = psutil.Process(os.getpid())
mem = process.memory_info().rss / float(2 ** 20)
return '{:.2f} MB'.format(mem)
names = ['John', 'Milovan', 'Adam', 'Steve', 'Rick', 'Thomas']
majors = ['Math', 'Engineering', 'CompSci', 'Arts', 'Business']
print('Memory (Before): {}'.format(memory_usage_psutil()))
def people_list(num_people):
result = []
for i in range(num_people):
person = {
'id': i,
'name': random.choice(names),
'major': random.choice(majors)
}
result.append(person)
return result
t1 = datetime.now()
people = people_list(1000000)
t2 = datetime.now()
print('Memory (After) : {}'.format(memory_usage_psutil()))
print('Took {} Seconds'.format(t2 - t1))
output:
Memory (Before): 50.38 MB
Memory (After) : 1140.41 MB
Took 0:00:01.056423 Seconds
Function which returns a list of 1 million results.
At the bottom I'm printing out the memory usage and the total time.
Base memory usage was around 50.38 megabytes and this memory after is after I created that list of 1 million records so you can see here that it jumped up by nearly 1140.41 megabytes and it took 1,1 seconds.
CASE 2
import random
import psutil # pip install psutil
import os
from datetime import datetime
def memory_usage_psutil():
# return the memory usage in MB
process = psutil.Process(os.getpid())
mem = process.memory_info().rss / float(2 ** 20)
return '{:.2f} MB'.format(mem)
names = ['John', 'Milovan', 'Adam', 'Steve', 'Rick', 'Thomas']
majors = ['Math', 'Engineering', 'CompSci', 'Arts', 'Business']
print('Memory (Before): {}'.format(memory_usage_psutil()))
def people_generator(num_people):
for i in range(num_people):
person = {
'id': i,
'name': random.choice(names),
'major': random.choice(majors)
}
yield person
t1 = datetime.now()
people = people_generator(1000000)
t2 = datetime.now()
print('Memory (After) : {}'.format(memory_usage_psutil()))
print('Took {} Seconds'.format(t2 - t1))
output:
Memory (Before): 50.52 MB
Memory (After) : 50.73 MB
Took 0:00:00.000008 Seconds
After I ran this that the memory is almost exactly the same and that's because the generator hasn't actually done anything yet it's not holding those million values in memory it's waiting for me to grab the next one.
Basically it didn't take any time because as soon as it gets to the first yield statement it stops.
I think that it is generator a little bit more readable and it also gives you big performance boosts not only with execution time but with memory.
As well and you can still use all of the comprehensions and this generator expression here so you don't lose anything in that area. So those are a few reasons why you would use generators and also some of the advantages that come along with that.
Experience with list comprehensions has shown their widespread utility throughout Python. However, many of the use cases do not need to have a full list created in memory. Instead, they only need to iterate over the elements one at a time.
For instance, the following summation code will build a full list of squares in memory, iterate over those values, and, when the reference is no longer needed, delete the list:
sum([x*x for x in range(10)])
Memory is conserved by using a generator expression instead:
sum(x*x for x in range(10))
Similar benefits are conferred on constructors for container objects:
s = Set(word for line in page for word in line.split())
d = dict( (k, func(k)) for k in keylist)
Generator expressions are especially useful with functions like sum(), min(), and max() that reduce an iterable input to a single value:
max(len(line) for line in file if line.strip())
more
When I use a generator as an iterable argument with multiprocessing.Pool.map function:
pool.map(func, iterable=(x for x in range(10)))
It seems that the generator is fully exhausted before func is ever called.
I want to yield each item and pass it to each process, thanks
multiprocessing.map converts iterables without a __len__ method to a list before processing. This is done to aid the calculation of chunksize, which the pool uses to group worker arguments and reduce the round trip cost of scheduling jobs. This is not optimal, especially when chunksize is 1, but since map must exhaust the iterator one way or the other, its usually not a significant issue.
The relevant code is in pool.py. Notice its use of len:
def _map_async(self, func, iterable, mapper, chunksize=None, callback=None,
error_callback=None):
'''
Helper function to implement map, starmap and their async counterparts.
'''
if self._state != RUN:
raise ValueError("Pool not running")
if not hasattr(iterable, '__len__'):
iterable = list(iterable)
if chunksize is None:
chunksize, extra = divmod(len(iterable), len(self._pool) * 4)
if extra:
chunksize += 1
if len(iterable) == 0:
chunksize = 0
Alas, this isn't well-defined. Here's a test case I'm running under Python 3.6.1:
import multiprocessing as mp
def e(i):
if i % 1000000 == 0:
print(i)
if __name__ == '__main__':
p = mp.Pool()
def g():
for i in range(100000000):
yield i
print("generator done")
r = p.map(e, g())
p.close()
p.join()
The first thing you see is the "generator done" message, and peak memory use is unreasonably high (precisely because, as you suspect, the generator is run to exhaustion before any work is passed out).
However, replace the map() call like so:
r = list(p.imap(e, g()))
Now memory use remains small, and "generator done" appears at the output end.
However, you won't wait long enough to see that, because it's horridly slow :-( imap() not only treats that iterable as an iterable, but effectively passes only 1 item at a time across process boundaries. To get speed back too, this works:
r = list(p.imap(e, g(), chunksize=10000))
In real life, I'm much more likely to iterate over an imap() (or imap_unordered()) result than to force it into a list, and then memory use remains small for looping over the results too.
To build on the answer by Tim Peters, here is a jupyter notebook demonstrating the interplay between imap and chunksize:
https://gist.github.com/shadiakiki1986/273b3529d3ff7afe2f2cac7b5ac96fe2
It has 2 examples:
Example 1 uses chunksize=1 and has the following execution:
On CPU 1, execute item 1 from generator
On CPU 2, execute item 2 from generator
When CPU 1 done with item 1, execute item 3 from generator
When CPU 2 done with item 2, execute item 4 from generator
etc
Example 2 has chunksize=3 with the following execution
On CPU 1, execute items 1-3 from generator
On CPU 2, execute items 4-6 from generator
When CPU 1 done with items 1-3, execute on 7-9
When CPU 2 done with items 4-6, execute on 10
Notice in example 2 that item 10 is executed on CPU 2 before items 8 and 9 on CPU 1.
I have two functions that return a list of functions. The functions take in a number x and add i to it. i is an integer increasing from 0-9.
def test_without_closure():
return [lambda x: x+i for i in range(10)]
def test_with_yield():
for i in range(10):
yield lambda x: x+i
I would expect test_without_closure to return a list of 10 functions that each add 9 to x since i's value is 9.
print sum(t(1) for t in test_without_closure()) # prints 100
I expected that test_with_yield would also have the same behavior, but it correctly creates the 10 functions.
print sum(t(1) for t in test_with_yield()) # print 55
My question is, does yielding form a closure in Python?
Yielding does not create a closure in Python, lambdas create a closure. The reason that you get all 9s in "test_without_closure" isn't that there's no closure. If there weren't, you wouldn't be able to access i at all. The problem is that all closures contain a reference¹ to the same i variable, which will be 9 at the end of the function.
This situation isn't much different in test_with_yield. Why, then, do you get different results? Because yield suspends the run of the function, so it's possible to use the yielded lambdas before the end of the function is reached, i.e. before i is 9. To see what this means, consider the following two examples of using test_with_yield:
[f(0) for f in test_with_yield()]
# Result: [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
[f(0) for f in list(test_with_yield())]
# Result: [9, 9, 9, 9, 9, 9, 9, 9, 9, 9]
What's happening here is that the first example yields a lambda (while i is 0), calls it (i is still 0), then advances the function until another lambda is yielded (i is now 1), calls the lambda, and so on. The important thing is that each lambda is called before the control flow returns to test_with_yield (i.e. before the value of i changes).
In the second example, we first create a list. So the first lambda is yielded (i is 0) and put into the list, the second lambda is created (i is now 1) and put into the list ... until the last lambda is yielded (i is now 9) and put into the list. And then we start calling the lambdas. So since i is now 9, all lambdas return 9.
¹ The important bit here is that closures hold references to variables, not copies of the value they held when the closure was created. This way, if you assign to the variable inside a lambda (or inner function, which create closures the same way that lambdas do), this will also change the variable outside of the lambda and if you change the value outside, that change will be visible inside the lambda.
No, yielding has nothing to do with closures.
Here is how to recognize closures in Python: a closure is
a function
in which an unqualified name lookup is performed
no binding of the name exists in the function itself
but a binding of the name exists in the local scope of a function whose definition surrounds the definition of the function in which the name is looked up.
The reason for the difference in behaviour you observe is laziness, rather than anything to do with closures. Compare and contrast the following
def lazy():
return ( lambda x: x+i for i in range(10) )
def immediate():
return [ lambda x: x+i for i in range(10) ]
def also_lazy():
for i in range(10):
yield lambda x:x+i
not_lazy_any_more = list(also_lazy())
print( [ f(10) for f in lazy() ] ) # 10 -> 19
print( [ f(10) for f in immediate() ] ) # all 19
print( [ f(10) for f in also_lazy() ] ) # 10 -> 19
print( [ f(10) for f in not_lazy_any_more ] ) # all 19
Notice that the first and third examples give identical results, as do the second and the fourth. The first and third are lazy, the second and fourth are not.
Note that all four examples provide a bunch of closures over the most recent binding of i, it's just that in the first an third case you evaluate the closures before rebinding i (even before you've created the next closure in the sequence), while in the second and fourth case, you first wait until i has been rebound to 9 (after you've created and collected all the closures you are going to make), and only then evaluate the closures.
Adding to #sepp2k's answer you're seeing these two different behaviours because the lambda functions being created don't know from where they have to get i's value. At the time this function is created all it knows is that it has to either fetch i's value from either local scope, enclosed scope, global scope or builtins.
In this particular case it is a closure variable(enclosed scope). And its value is changing with each iteration.
Check out LEGB in Python.
Now to why second one works as expected but not the first one?
It's because each time you're yielding a lambda function the execution of the generator function stops at that moment and when you're invoking it and it will use the value of i at that moment. But in the first case we have already advanced i's value to 9 before we invoked any of the functions.
To prove it you can fetch current value of i from the __closure__'s cell contents:
>>> for func in test_with_yield():
print "Current value of i is {}".format(func.__closure__[0].cell_contents)
print func(9)
...
Current value of i is 0
Current value of i is 1
Current value of i is 2
Current value of i is 3
Current value of i is 4
Current value of i is 5
Current value of i is 6
...
But instead if you store the functions somewhere and call them later then you will see the same behaviour as the first time:
from itertools import islice
funcs = []
for func in islice(test_with_yield(), 4):
print "Current value of i is {}".format(func.__closure__[0].cell_contents)
funcs.append(func)
print '-' * 20
for func in funcs:
print "Now value of i is {}".format(func.__closure__[0].cell_contents)
Output:
Current value of i is 0
Current value of i is 1
Current value of i is 2
Current value of i is 3
--------------------
Now value of i is 3
Now value of i is 3
Now value of i is 3
Now value of i is 3
Example used by Patrick Haugh in comments also shows the same thing: sum(t(1) for t in list(test_with_yield()))
Correct way:
Assign i as a default value to lambda, default values are calculated when function is created and they won't change(unless it's a mutable object). i is now a local variable to the lambda functions.
>>> def test_without_closure():
return [lambda x, i=i: x+i for i in range(10)]
...
>>> sum(t(1) for t in test_without_closure())
55
I'm playing around with generators and generator expressions and I'm not completely sure that I understand how they work (some reference material):
>>> a = (x for x in range(10))
>>> next(a)
0
>>> next(a)
1
>>> a.send(-1)
2
>>> next(a)
3
So it looks like generator.send was ignored. That makes sense (I guess) because there is no explicit yield expression to catch the sent information ...
However,
>>> a = ((yield x) for x in range(10))
>>> next(a)
0
>>> print next(a)
None
>>> print next(a)
1
>>> print next(a)
None
>>> a.send(-1) #this send is ignored, Why? ... there's a yield to catch it...
2
>>> print next(a)
None
>>> print next(a)
3
>>> a.send(-1) #this send isn't ignored
-1
I understand this is pretty far out there, and I (currently) can't think of a use-case for this (so don't ask;)
I'm mostly just exploring to try to figure out how these various generator methods work (and how generator expressions work in general). Why does my second example alternate between yielding a sensible value and None? Also, Can anyone explain why one of my generator.send's was ignored while the other wasn't?
The confusion here is that the generator expression is doing a hidden yield. Here it is in function form:
def foo():
for x in range(10):
yield (yield x)
When you do a .send(), what happens is the inner yield x gets executed, which yields x. Then the expression evaluates to the value of the .send, and the next yield yields that. Here it is in clearer form:
def foo():
for x in range(10):
sent_value = (yield x)
yield sent_value
Thus the output is very predictable:
>>> a = foo()
#start it off
>>> a.next()
0
#execution has now paused at "sent_value = ?"
#now we fill in the "?". whatever we send here will be immediately yielded.
>>> a.send("yieldnow")
'yieldnow'
#execution is now paused at the 'yield sent_value' expression
#as this is not assigned to anything, whatever is sent now will be lost
>>> a.send("this is lost")
1
#now we're back where we were at the 'yieldnow' point of the code
>>> a.send("yieldnow")
'yieldnow'
#etc, the loop continues
>>> a.send("this is lost")
2
>>> a.send("yieldnow")
'yieldnow'
>>> a.send("this is lost")
3
>>> a.send("yieldnow")
'yieldnow'
EDIT: Example usage. By far the coolest one I've seen so far is twisted's inlineCallbacks function. See here for an article explaining it. The nub of it is it lets you yield functions to be run in threads, and once the functions are done, twisted sends the result of the function back into your code. Thus you can write code that heavily relies on threads in a very linear and intuitive manner, instead of having to write tons of little functions all over the place.
See the PEP 342 for more info on the rationale of having .send work with potential use cases (the twisted example I provided is an example of the boon to asynchronous I/O this change offered).
You're confusing yourself a bit because you actually are generating from two sources: the generator expression (... for x in range(10)) is one generator, but you create another source with the yield. You can see that if do list(a) you'll get [0, None, 1, None, 2, None, 3, None, 4, None, 5, None, 6, None, 7, None, 8, None, 9, None].
Your code is equivalent to this:
>>> def gen():
... for x in range(10):
... yield (yield x)
Only the inner yield ("yield x") is "used" in the generator --- it is used as the value of the outer yield. So this generator iterates back and forth between yielding values of the range, and yielding whatever is "sent" to those yields. If you send something to the inner yield, you get it back, but if you happen to send on an even-numbered iteration, the send is sent to the outer yield and is ignored.
This generator translates into:
for i in xrange(10):
x = (yield i)
yield x
Result of second call to send()/next() are ignored, because you do nothing with result of one of yields.
The generator you wrote is equivalent to the more verbose:
def testing():
for x in range(10):
x = (yield x)
yield x
As you can see here, the second yield, which is implicit in the generator expression, does not save the value you pass it, therefore depending on where the generator execution is blocked the send may or may not work.
Indeed - the send method is meant to work with a generator object that is the result of a co-routine you have explicitly written. It is difficult to get some meaning to it in a generator expression - though it works.
-- EDIT --
I had previously written this, but it is incorrecct, as yield inside generator expressions are predictable across implementations - though not mentioned in any PEP.
generator expressions are not meant to have the yield keyword - I am
not shure the behavior is even defined in this case. We could think a
little and get to what is happening on your expression, to meet from
where those "None"s are coming from. However, assume that as a side
effect of how the yield is implemented in Python (and probably it is
even implementation dependent), not as something that should be so.
The correct form for a generator expression, in a simplified manner is:
(<expr> for <variable> in <sequence> [if <expr>])
so, <expr> is evaluated for each value in the <sequence: - not only is yield uneeded, as you should not use it.
Both yield and the send methods are meant to be used in full co-routines, something like:
def doubler():
value = 0
while value < 100:
value = 2 * (yield value)
And you can use it like:
>>> a = doubler()
>>> # Next have to be called once, so the code will run up to the first "yield"
...
>>> a.next()
0
>>> a.send(10)
20
>>> a.send(20)
40
>>> a.send(23)
46
>>> a.send(51)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
StopIteration
>>>
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Possible Duplicate:
The Python yield keyword explained
Can someone explain to me what the yield statement actually does in this bit of code here:
def fibonacci():
a, b = 0, 1
while True:
yield a
a, b = b, a+b
for number in fibonacci(): # Use the generator as an iterator; print number
What I understand so far is, we are defining a function finonacci(), with no parameters?
inside the function we are defining a and b equal to 0 and 1, next, while this is true, we are yielding a. What is this actually doing? Furthermore, while yielding a? a is now equal to b, while b is now equal to a + b.
Next question, for number in fibonacci(), does this mean for every number in the function or what? I'm equally stumped on what yield and 'for number' are actually doing. Obviously I am aware that it means for every number in fibonacci() print number. Am I actually defining number without knowing it?
Thanks, sorry if I'm not clear. BTW, it's for project Euler, if I knew how to program well this would be a breeze but I'm trying to learn this on the fly.
Using yield makes the function a generator.
The generator will continue to yield the a variable on each loop, waiting until the generator's next() method is called to continue on to the next loop iteration.
Or, until you return or StopIteration is raised.
Slightly modified to show use of StopIteration:
>>> def fib():
... a = 0
... b = 1
... while True:
... yield a
... a = b
... b += a
... if a > 100:
... raise StopIteration
...
>>>
>>> for value in fib():
... print value
...
0
1
2
4
8
16
32
64
>>>
>>> # assign the resulting object to 'generator'
>>> generator = fib()
>>> generator.next()
0
>>> generator.next()
1
>>> for value in generator:
... print value
...
2
4
8
16
32
64
>>>
Generators have a special property of being iterables which do not consume memories for their values.
They do this by calculating the new value, when it is required while being iterated.
i.e.
def f():
a = 2
yield a
a += 1
for ele in f():
print ele
would print
2
So you are using a function as an iterable that keeps returning values.
This is especially useful when you require heavy memory usage, and so you cannot afford the use of a list comprehension
i.e.
li = [ele*10 for ele in range(10)]
takes 10 memory spaces for ints as a list
but if you simple want to iterate over it, not access it individually
it would be very memory efficient to instead use
def f():
i=0
while i<10
yield i*10
i += 1
which would use 1 memory space as i keeps being reused
a short cut for this is
ge = (i*10 for i in range(10))
you can do any of the following
for ele in f():
for ele in li:
for ele in ge:
to obtain equivalent results
When the code calls fibonacci a special generator object is created. Please note, that no code gets executed - only a generator object is returned. When you are later calling its next method, the function executes until it encounters a yield statement. The object that is supplied to yield is returned. When you call next method again the function executes again until it encounters a yield. When there are no more yield statements and the end of function is reached, a StopIteration exception is raised.
Please note that the objects inside the function are preserved between the calls to next. It means, when the code continues execution on the next loop, all the objects that were in the scope from which yield was called have their values from the point where a previous next call returned.
The cool thing about generators is that they allow convenient iteration with for loops.
The for loop obtains a generator from the result of fibonacci call and then executes the loop retrieving elements using next method of generatior object until StopIteration exception is encountered.
This answer is a great explanation of the yield statement, and also of iterators and generators.
Specifically here, the first call to fibonaci() will initialize a to 0, b to 1, enter the while loop and return a.
Any next call will start after the yield statement, affect b to a, a+b to b, and then go to the next iteration of the while statement, reach again the yield statement, and return a again.