在 Python 中从嵌套的 OrderedDict 中分离文本答案

【问题标题】：Separate text from nested OrderedDict in Python在 Python 中从嵌套的 OrderedDict 中分离文本
【发布时间】：2014-04-10 14:18:05
【问题描述】：

这是我的OrderedDict 对象，

a=OrderedDict([(u'p', [u'"The Exam Room" is a new series in which everyday medical questions are answered by physicians and professors from the Yale School of Medicine.', u'In our second episode: Dr. Stephen Strittmatter, Vincent Coates Professor of Neurology and director of the Adler Memory Clinic in Neurology, explains when memory loss can become a problem and what you can do to boost your brain power.', OrderedDict([(u'em', u'Produced &amp; Hosted by Noah Golden')])])])

我想要做的是从这个对象中获取文本，

>>> a.get('p')

得到输出，

[u'"The Exam Room" is a new series in which everyday medical questions are answered by physicians and professors from the Yale School of Medicine.', u'In our second episode: Dr. Stephen Strittmatter, Vincent Coates Professor of Neurology and director of the Adler Memory Clinic in Neurology, explains when memory loss can become a problem and what you can do to boost your brain power.', OrderedDict([(u'em', u'Produced &amp; Hosted by Noah Golden')])]

但结果文本也包含一个OrderedDict。

如何合并来自 OrderedDict 的文本，

预期输出：

The Exam Room" is a new series in which everyday medical questions are answered by physicians and professors from the Yale School of Medicine.', u'In our second episode: Dr. Stephen Strittmatter, Vincent Coates Professor of Neurology and director of the Adler Memory Clinic in Neurology, explains when memory loss can become a problem and what you can do to boost your brain power. Produced &amp; Hosted by Noah Golden

【问题讨论】：

标签： python ordereddictionary

【解决方案1】：

如果您事先不知道类型的嵌套，这里的关键是递归。这是一个示例（为便于阅读，对文本进行了格式化）：

#!/usr/bin/env python

import collections

a = collections.OrderedDict([(u'p', [u""" 
    "The Exam Room" is a new series in
    which everyday medical questions are answered by physicians and 
    professors from the Yale School of Medicine.""", 
    u"""In our second episode: Dr. Stephen Strittmatter,
    Vincent Coates Professor of Neurology and director of
    the Adler Memory Clinic in Neurology, explains when 
    memory loss can become a problem and what you can do to 
    boost your brain power.""", 
    collections.OrderedDict([(u'em',
        u'Produced &amp; Hosted by Noah Golden')])])])

现在展平对象，可能是映射或列表。实现了三个选项：如果找到的值是一个字符串，我们只需将其附加到我们的collector。如果是list 或Mapping，我们再次调用flatten。请注意，您可以使用 allowed kwarg 指定一些允许的标签：

def flatten(obj, allowed=(u'p', u'em')):
    collector = []

    def process(v, collector=collector):
        if isinstance(v, (list, collections.Mapping)):
            collector += flatten(v, allowed=allowed)
        elif isinstance(v, basestring):
            collector.append(v)
        else:
            raise ValueError('Cannot handle type: {t}'.format(t=v.__class__))

    if isinstance(obj, list):
        for v in obj:
            process(v)

    if isinstance(obj, collections.Mapping):
        for k, v in obj.iteritems():
            if k in allowed:
                process(v)

    return collector

if __name__ == '__main__':
    print(flatten(a))

您的示例的结果将是一个三元素列表，如下所示：

[u'"The Exam Room" is a new series ...',
 u'In our second episode: ...',
 u'Produced &amp; Hosted by Noah Golden']

现在，如果您想要一个字符串，只需 join 现在扁平化的列表：

print(''.join(flatten(a)))

【讨论】：

【解决方案2】：

这是一个奇怪的字典，但你可以像这样实现你想要的：

[a['p'][0],a['p'][1] + u' ' + a['p'][2]['em']]

结果：

[u'“The Exam Room”是一个新系列，其中包含日常医学问题由耶鲁大学的医生和教授回答医学。'，你'在我们的第二集中：斯蒂芬·斯特里特马特博士，文森特 Coates 神经病学教授和阿德勒记忆诊所主任在神经病学中，解释了记忆丧失何时会成为问题以及什么你可以这样做来提高你的脑力。由诺亚制作和主持金']

这将返回一个列表，正如您在问题中所要求的那样。如果您想使用单个字符串：

import string
string.join([a['p'][0],a['p'][1],a['p'][2]['em']])

这将导致：

“The Exam Room”是一个新系列，其中包含日常医学问题由耶鲁大学的医生和教授回答药物。在我们的第二集中：斯蒂芬·斯特里特马特博士，文森特 Coates 神经病学教授和阿德勒记忆诊所主任在神经病学中，解释了记忆丧失何时会成为问题以及什么你可以这样做来提高你的脑力。由诺亚制作和主持金色

【讨论】：