如何将IPython.parallel map（）与生成器一起用作函数的输入

我正在尝试使用IPython.parallel映射。我希望并行化的函数的输入是生成器。由于大小/内存的原因，我无法将生成器转换为列表。见下面的代码：

from itertools import product
from IPython.parallel import Client

c = Client()
v = c[:]
c.ids

def stringcount(longstring, substrings):
    scount = [longstring.count(s) for s in substrings]
    return scount

substrings = product('abc', repeat=2)
longstring = product('abc', repeat=3)

# This is what I want to do in parallel
# I should be 'for longs in longstring' I use range() because it can get long.
for num in range(10): 
    longs = longstring.next()
    subs = substrings.next()
    print(subs, longs)
    count = stringcount(longs, subs)
    print(count)

# This does not work, and I understand why.
# I don't know how to fix it while keeping longstring and substrings as
# generators  
v.map(stringcount, longstring, substrings)

for r in v:
    print(r.get())

如果不先遍历整个生成器，则无法将View.map与生成器一起使用。但是，您可以编写自己的自定义函数来提交生成器中的成批任务，并以增量方式等待它们。我没有一个更有趣的例子，但我可以用一个基本搜索的糟糕实现来说明

从我们的令牌“数据生成器”开始：

from math import sqrt

def generate_possible_factors(N):
    """generator for iterating through possible factors for N

    yields 2, every odd integer <= sqrt(N)
    """
    if N <= 3:
        return
    yield 2
    f = 3
    last = int(sqrt(N))
    while f <= last:
        yield f
        f += 2

以及使用生成器和我们的因子函数的基本检查的完整实现：

def dumb_prime(N):
    """dumb implementation of is N prime?"""
    for f in generate_possible_factors(N):
        if is_factor(f, N):
            return False
    return True

from IPython import parallel

rc = parallel.Client()
view = rc.load_balanced_view()

for N in range(900,1000):
    if parallel_dumb_prime(N, view, 10):
        print N

一次只提交有限数量任务的并行版本：

def parallel_dumb_prime(N, v, max_outstanding=10, dt=0.1):
    """dumb_prime where each factor is checked remotely

    Up to `max_outstanding` factors will be checked in parallel.

    Submission will halt as soon as we know that N is not prime.
    """
    tasks = set()
    # factors is a generator
    factors = generate_possible_factors(N)
    while True:
        try:
            # submit a batch of tasks, with a maximum of `max_outstanding`
            for i in range(max_outstanding-len(tasks)):
                f = factors.next()
                tasks.add(v.apply_async(is_factor, f, N))
        except StopIteration:
            # no more factors to test, stop submitting
            break
        # get the tasks that are done
        ready = set(task for task in tasks if task.ready())
        while not ready:
            # wait a little bit for some tasks to finish
            v.wait(tasks, timeout=dt)
            ready = set(task for task in tasks if task.ready())

        for t in ready:
            # get the result - if True, N is not prime, we are done
            if t.get():
                return False
        # update tasks to only those that are still pending,
        # and submit the next batch
        tasks.difference_update(ready)
    # check the last few outstanding tasks
    for task in tasks:
        if t.get():
            return False
    # checked all candidates, none are factors, so N is prime
    return True

它一次提交有限数量的任务，一旦我们知道N不是素数，我们就停止使用生成器

要使用此功能，请执行以下操作：

def dumb_prime(N):
    """dumb implementation of is N prime?"""
    for f in generate_possible_factors(N):
        if is_factor(f, N):
            return False
    return True

from IPython import parallel

rc = parallel.Client()
view = rc.load_balanced_view()

for N in range(900,1000):
    if parallel_dumb_prime(N, view, 10):
        print N

更完整的说明。

如果不先浏览整个生成器，则无法将View.map与生成器一起使用。但是，您可以编写自己的自定义函数来提交生成器中的成批任务，并以增量方式等待它们。我没有一个更有趣的例子，但我可以用一个基本搜索的糟糕实现来说明

从我们的令牌“数据生成器”开始：

from math import sqrt

def generate_possible_factors(N):
    """generator for iterating through possible factors for N

    yields 2, every odd integer <= sqrt(N)
    """
    if N <= 3:
        return
    yield 2
    f = 3
    last = int(sqrt(N))
    while f <= last:
        yield f
        f += 2

以及使用生成器和我们的因子函数的基本检查的完整实现：

def dumb_prime(N):
    """dumb implementation of is N prime?"""
    for f in generate_possible_factors(N):
        if is_factor(f, N):
            return False
    return True

from IPython import parallel

rc = parallel.Client()
view = rc.load_balanced_view()

for N in range(900,1000):
    if parallel_dumb_prime(N, view, 10):
        print N

一次只提交有限数量任务的并行版本：

def parallel_dumb_prime(N, v, max_outstanding=10, dt=0.1):
    """dumb_prime where each factor is checked remotely

    Up to `max_outstanding` factors will be checked in parallel.

    Submission will halt as soon as we know that N is not prime.
    """
    tasks = set()
    # factors is a generator
    factors = generate_possible_factors(N)
    while True:
        try:
            # submit a batch of tasks, with a maximum of `max_outstanding`
            for i in range(max_outstanding-len(tasks)):
                f = factors.next()
                tasks.add(v.apply_async(is_factor, f, N))
        except StopIteration:
            # no more factors to test, stop submitting
            break
        # get the tasks that are done
        ready = set(task for task in tasks if task.ready())
        while not ready:
            # wait a little bit for some tasks to finish
            v.wait(tasks, timeout=dt)
            ready = set(task for task in tasks if task.ready())

        for t in ready:
            # get the result - if True, N is not prime, we are done
            if t.get():
                return False
        # update tasks to only those that are still pending,
        # and submit the next batch
        tasks.difference_update(ready)
    # check the last few outstanding tasks
    for task in tasks:
        if t.get():
            return False
    # checked all candidates, none are factors, so N is prime
    return True

它一次提交有限数量的任务，一旦我们知道N不是素数，我们就停止使用生成器

要使用此功能，请执行以下操作：

def dumb_prime(N):
    """dumb implementation of is N prime?"""
    for f in generate_possible_factors(N):
        if is_factor(f, N):
            return False
    return True

from IPython import parallel

rc = parallel.Client()
view = rc.load_balanced_view()

for N in range(900,1000):
    if parallel_dumb_prime(N, view, 10):
        print N

---

更完整的说明。

我对你的问题采取了一种稍微不同的方法，这可能对其他人有用。下面，我试图通过包装IPython.parallel.map来模拟multiprocessing.pool.pool.imap方法的行为。这需要我稍微重新编写您的函数

import IPython
from itertools import product


def stringcount((longstring, substrings)):
    scount = [longstring.count(s) for s in substrings]
    return (longstring, substrings, scount)

def gen_pairs(long_string, sub_strings):
    for l in long_string:
        s = sub_strings.next()
        yield (l, s)

def imap(function, generator, view, preprocessor=iter, chunksize=256):
    num_cores = len(view.client.ids)
    queue = []
    for i, n in enumerate(preprocessor(generator)):
        queue.append(n)
        if not i % (chunksize * num_cores):
            for result in view.map(function, queue):
                yield result
            queue = []
    for result in view.map(function, queue):
        yield result


client = IPython.parallel.Client()
lbview = client.load_balanced_view()

longstring = product('abc', repeat=3)
substrings = product('abc', repeat=2)

for result in imap(stringcount, gen_pairs(longstring, substrings), lbview):
    print result

---

我看到的结果在这个笔记本上：

我对你的问题采取了一种稍微不同的方法，这可能对其他人有用。下面，我试图通过包装IPython.parallel.map来模拟multiprocessing.pool.pool.imap方法的行为。这需要我稍微重新编写您的函数

import IPython
from itertools import product


def stringcount((longstring, substrings)):
    scount = [longstring.count(s) for s in substrings]
    return (longstring, substrings, scount)

def gen_pairs(long_string, sub_strings):
    for l in long_string:
        s = sub_strings.next()
        yield (l, s)

def imap(function, generator, view, preprocessor=iter, chunksize=256):
    num_cores = len(view.client.ids)
    queue = []
    for i, n in enumerate(preprocessor(generator)):
        queue.append(n)
        if not i % (chunksize * num_cores):
            for result in view.map(function, queue):
                yield result
            queue = []
    for result in view.map(function, queue):
        yield result


client = IPython.parallel.Client()
lbview = client.load_balanced_view()

longstring = product('abc', repeat=3)
substrings = product('abc', repeat=2)

for result in imap(stringcount, gen_pairs(longstring, substrings), lbview):
    print result

---

我看到的输出在这个笔记本上：

关于内存中可以存在多少项，您能更具体地说明您的要求吗？由于执行是异步的，如果你浏览一个生成器，你可能会在内存中有几乎所有的输入，除非你在提交新任务之前开始等待结果。由于我运行的是64位，我想我的限制是系统内存，即8GB或可以使用32GB的机器。例如，产品'abcd'，repeat=10，变得非常大，基本上，一旦我根据计数找到满足我要求的结果，我就可以停止。我假设/希望我的地图会根据需要从发电机上获取。等待结果是可以的。关于内存中可以存在多少项，您能否更具体地说明您的需求？由于执行是异步的，如果你浏览一个生成器，你可能会在内存中有几乎所有的输入，除非你在提交新任务之前开始等待结果。由于我运行的是64位，我想我的限制是系统内存，即8GB或可以使用32GB的机器。例如，产品'abcd'，repeat=10，变得非常大，基本上，一旦我根据计数找到满足我要求的结果，我就可以停止。我假设/希望我的地图会根据需要从发电机上获取。等待结果是好的。谢谢你的回答，我正试图腾出时间来看看。其他的东西变得更重要了。谢谢你的回答，我正试着抽出时间来看看。其他事情变得更加重要。