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我在 python 2.6.6 中收到以下错误:

python: Objects/dictobject.c:501: insertdict_clean: Assertion 'ep->me_value == ((void *)0)' failed.

以下是蒸馏的来源:

pool = multiprocessing.Pool(32)
mutation_counts = pool.map(mutate, enumerate(data))
mutation_count = reduce(lambda a, b: a + b, mutation_counts)
args = [(interval, mutation_count) for interval in sub_intervals]
pool_results = pool.map(validate, enumerate(args)) # Issue occurs here

我怀疑mutation_count是导致问题的原因,它是由以下定义的计数器:

from operator import itemgetter
from heapq import nlargest
from itertools import repeat, ifilter

class Counter(dict):
    '''Dict subclass for counting hashable objects.  Sometimes called a bag
    or multiset.  Elements are stored as dictionary keys and their counts
    are stored as dictionary values.

    >>> Counter('zyzygy')
    Counter({'y': 3, 'z': 2, 'g': 1})

    '''

    def __init__(self, iterable=None, **kwds):
        '''Create a new, empty Counter object.  And if given, count elements
        from an input iterable.  Or, initialize the count from another mapping
        of elements to their counts.

        >>> c = Counter()                           # a new, empty counter
        >>> c = Counter('gallahad')                 # a new counter from an iterable
        >>> c = Counter({'a': 4, 'b': 2})           # a new counter from a mapping
        >>> c = Counter(a=4, b=2)                   # a new counter from keyword args

        '''        
        self.update(iterable, **kwds)

    def __missing__(self, key):
        return 0

    def most_common(self, n=None):
        '''List the n most common elements and their counts from the most
        common to the least.  If n is None, then list all element counts.

        >>> Counter('abracadabra').most_common(3)
        [('a', 5), ('r', 2), ('b', 2)]

        '''        
        if n is None:
            return sorted(self.iteritems(), key=itemgetter(1), reverse=True)
        return nlargest(n, self.iteritems(), key=itemgetter(1))

    def elements(self):
        '''Iterator over elements repeating each as many times as its count.

        >>> c = Counter('ABCABC')
        >>> sorted(c.elements())
        ['A', 'A', 'B', 'B', 'C', 'C']

        If an element's count has been set to zero or is a negative number,
        elements() will ignore it.

        '''
        for elem, count in self.iteritems():
            for _ in repeat(None, count):
                yield elem

    # Override dict methods where the meaning changes for Counter objects.

    @classmethod
    def fromkeys(cls, iterable, v=None):
        raise NotImplementedError(
            'Counter.fromkeys() is undefined.  Use Counter(iterable) instead.')

    def update(self, iterable=None, **kwds):
        '''Like dict.update() but add counts instead of replacing them.

        Source can be an iterable, a dictionary, or another Counter instance.

        >>> c = Counter('which')
        >>> c.update('witch')           # add elements from another iterable
        >>> d = Counter('watch')
        >>> c.update(d)                 # add elements from another counter
        >>> c['h']                      # four 'h' in which, witch, and watch
        4

        '''        
        if iterable is not None:
            if hasattr(iterable, 'iteritems'):
                if self:
                    self_get = self.get
                    for elem, count in iterable.iteritems():
                        self[elem] = self_get(elem, 0) + count
                else:
                    dict.update(self, iterable) # fast path when counter is empty
            else:
                self_get = self.get
                for elem in iterable:
                    self[elem] = self_get(elem, 0) + 1
        if kwds:
            self.update(kwds)

    def copy(self):
        'Like dict.copy() but returns a Counter instance instead of a dict.'
        return Counter(self)

    def __delitem__(self, elem):
        'Like dict.__delitem__() but does not raise KeyError for missing values.'
        if elem in self:
            dict.__delitem__(self, elem)

    def __repr__(self):
        if not self:
            return '%s()' % self.__class__.__name__
        items = ', '.join(map('%r: %r'.__mod__, self.most_common()))
        return '%s({%s})' % (self.__class__.__name__, items)

    # Multiset-style mathematical operations discussed in:
    #       Knuth TAOCP Volume II section 4.6.3 exercise 19
    #       and at http://en.wikipedia.org/wiki/Multiset
    #
    # Outputs guaranteed to only include positive counts.
    #
    # To strip negative and zero counts, add-in an empty counter:
    #       c += Counter()

    def __add__(self, other):
        '''Add counts from two counters.

        >>> Counter('abbb') + Counter('bcc')
        Counter({'b': 4, 'c': 2, 'a': 1})


        '''
        if not isinstance(other, Counter):
            return NotImplemented
        result = Counter()
        for elem in set(self) | set(other):
            newcount = self[elem] + other[elem]
            if newcount > 0:
                result[elem] = newcount
        return result

    def __sub__(self, other):
        ''' Subtract count, but keep only results with positive counts.

        >>> Counter('abbbc') - Counter('bccd')
        Counter({'b': 2, 'a': 1})

        '''
        if not isinstance(other, Counter):
            return NotImplemented
        result = Counter()
        for elem in set(self) | set(other):
            newcount = self[elem] - other[elem]
            if newcount > 0:
                result[elem] = newcount
        return result

    def __or__(self, other):
        '''Union is the maximum of value in either of the input counters.

        >>> Counter('abbb') | Counter('bcc')
        Counter({'b': 3, 'c': 2, 'a': 1})

        '''
        if not isinstance(other, Counter):
            return NotImplemented
        _max = max
        result = Counter()
        for elem in set(self) | set(other):
            newcount = _max(self[elem], other[elem])
            if newcount > 0:
                result[elem] = newcount
        return result

    def __and__(self, other):
        ''' Intersection is the minimum of corresponding counts.

        >>> Counter('abbb') & Counter('bcc')
        Counter({'b': 1})

        '''
        if not isinstance(other, Counter):
            return NotImplemented
        _min = min
        result = Counter()
        if len(self) < len(other):
            self, other = other, self
        for elem in ifilter(self.__contains__, other):
            newcount = _min(self[elem], other[elem])
            if newcount > 0:
                result[elem] = newcount
        return result


if __name__ == '__main__':
    import doctest
    print doctest.testmod()

只有当我有大约 1M 键并且在大约 20% 的运行中出现时,我才会看到这个问题。

4

1 回答 1

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如果你去dictobject.c源码(在 Python 主干上,我没有费心去寻找 2.6),你可以读到这个函数insertdict_clean是一个

dictresize() 使用的内部例程插入一个已知在 dict 中不存在的项目。此例程还假定字典不包含已删除的条目。

异常告诉您的是,assert(ep->me_value == NULL);当前主干中第 574 行的 ,实际上是在字典中查找应该丢失的元素。可能,您之所以遇到这一行,是因为您将 Python 的内置子类化,而不是在您的 Counter 类dict中保存 a 。dict

我会怀疑__missing__魔术方法的实施。它返回一个0而不是一个无。另外,你为什么不使用这个collections.Counter类?它已经在 Python 2.6 中了。

于 2014-02-24T19:54:36.760 回答