如果iterable抛出

Question

考虑文件sample.py包含以下代码：

from multiprocessing import Pool

def sample_worker(x):
    print "sample_worker processes item", x
    return x

def get_sample_sequence():
    for i in xrange(2,30):
        if i % 10 == 0:
            raise Exception('That sequence is corrupted!')
        yield i

if __name__ == "__main__":
    pool = Pool(24)
    try:
        for x in pool.imap_unordered(sample_worker, get_sample_sequence()):
            print "sample_worker returned value", x
    except:
        print "Outer exception caught!"
    pool.close()
    pool.join()
    print "done"

当我执行它时，我得到以下输出：

^{pr2}$

之后，应用程序挂断。我怎样才能不挂断电话呢？在

Answer 1

正如塞普蒂提到的，你的压痕（仍然）是错误的。缩进yield语句，使i在其范围内。我不完全确定代码中实际发生了什么，但生成超出范围的变量似乎不是一个好主意：

from multiprocessing import Pool

def sample_worker(x):
    print "sample_worker processes item", x
    return x

def get_sample_sequence():
    for i in xrange(2,30):
      if i % 10 == 0:
          raise Exception('That sequence is corrupted!')
      yield i # fixed

if __name__ == "__main__":
    pool = Pool(24)
    try:
        for x in pool.imap_unordered(sample_worker, get_sample_sequence()):
            print "sample_worker returned value", x
    except:
        print "Outer exception caught!"
    pool.close()
    pool.join()
    print "done"

要处理生成器中的异常，可以使用如下包装：

^{pr2}$