How can I run the initialization code for a generator function immediately, rather than at the first call?

问题

I have a generator function that goes something like this:

def mygenerator():
    next_value = compute_first_value() # Costly operation
    while next_value != terminating_value:
        yield next_value
        next_value = compute_next_value()

I would like the initialization step (before the while loop) to run as soon as the function is called, rather than only when the generator is first used. What is a good way to do this?

I want to do this because the generator will be running in a separate thread (or process, or whatever multiprocessing uses) and I won't be using the return for a short while, and the initialization is somewhat costly, so I would like it to do the initialization while I'm getting ready to use the values.

回答1:

class mygenerator(object):
    def __init__(self):
        next_value = compute_first_value()
    def __iter__(self):
        return self
    def next(self):
        if next_value == terminating_value:
            raise StopIteration()
        return next_value

回答2:

I needed something similar. This is what I landed on. Push the generator function into an inner and return it's call.

def mygenerator():
    next_value = compute_first_value()

    def generator():
        while next_value != terminating_value:
            yield next_value
            next_value = compute_next(next_value)

    return generator()

回答3:

You can create a "preprimed" iterator fairly easily by using itertools.chain:

from itertools import chain

def primed(iterable):
    """Preprimes an iterator so the first value is calculated immediately
       but not returned until the first iteration
    """
    itr = iter(iterable)
    try:
        first = next(itr)  # itr.next() in Python 2
    except StopIteration:
        return itr
    return chain([first], itr)

>>> def g():
...     for i in range(5):
...         print("Next called")
...         yield i
...
>>> x = primed(g())
Next called
>>> for i in x: print(i)
...
0
Next called
1
Next called
2
Next called
3
Next called
4

回答4:

I suppose you can yield None after that first statement is completed, then in your calling code:

gen = mygenerator()
next(gen) # toss the None
do_something(gen)

回答5:

For my use case I used a modified version of @ncoghlan answer but wrapped in a factory function to decorate the generating function:

import collections, functools, itertools

def primed_generator(generating_function):
    @functools.wraps(generating_function)
    def get_first_right_away_wrapper(*args,**kw):
        "call the generator function, prime before returning"
        gen = generating_function(*args,**kw)
        assert isinstance(gen,collections.Iterator)
        first_value = next(gen)
        return itertools.chain((first_value,),gen)
    return get_first_right_away_wrapper

Then just decorate the function:

@primed_generator
def mygenerator():
    next_value = compute_first_value() # Costly operation
    while next_value != terminating_value:
        yield next_value
        next_value = compute_next_value()

and the first value will be calculated immediately, and the result is transparent.

来源：https://stackoverflow.com/questions/5724009/how-can-i-run-the-initialization-code-for-a-generator-function-immediately-rath

标签

python

generator

multiprocessing

eager