python - 将字典列表合并到一个字典中，并将项目附加到列表中

Question

def merge_dicts(list_of_dicts: list, missval=None):
    '''Merges a list of dicts, having common keys into a single dict
    with items appended to a list

    >>> d1 = {'a' : 1, 'b': 2, 'c': 3}
    >>> d2 = {'a':4, 'b':5 }
    >>> d3 = {'d': 5}
    >>> merge_dicts([d1, d2, d3], 'NA')
    {'a': [1, 4, 'NA'], 'b': [2, 5, 'NA'],
    'c': [3, 'NA', 'NA'], 'd': ['NA', 'NA', 5]}
    '''
    all_keys = []
    for d in list_of_dicts:
        for k in d.keys():
            if k not in all_keys:
                all_keys.append(k)

    merged = {}
    for k in all_keys:
        for d in list_of_dicts:
            try:
                merged[k].append(d.get(k, missval))
            except KeyError:
                merged[k] = [d.get(k)]

    return(merged)

函数 docstring 是不言自明的。有没有更有效的方法来做到这一点而不必编写两个 for 循环？一个查找所有字典中的所有键，另一个创建一个合并的字典？

标签： python

Answer 1

set如果您不关心它们的顺序，您应该使用 a创建键列表。您可以使用理解来创建它。

对于第二部分，您可以使用字典推导，并使用列表推导创建每个列表：

def merge_dicts(list_of_dicts: list, missval=None):
    '''Merges a list of dicts, having common keys into a single dict
    with items appended to a list

    >>> d1 = {'a' : 1, 'b': 2, 'c': 3}
    >>> d2 = {'a':4, 'b':5 }
    >>> d3 = {'d': 5}
    >>> merge_dicts([d1, d2, d3], 'NA')
    {'a': [1, 4, 'NA'], 'b': [2, 5, 'NA'],
    'c': [3, 'NA', 'NA'], 'd': ['NA', 'NA', 5]}
    '''
    all_keys = {key for d in list_of_dicts for key in d.keys()}
    merged = {k: [d.get(k, missval) for d in list_of_dicts] for k in all_keys}

    return(merged)


d1 = {'a' : 1, 'b': 2, 'c': 3}
d2 = {'a':4, 'b':5 }
d3 = {'d': 5}
merge_dicts([d1, d2, d3], 'NA')


#{'a': [1, 4, 'NA'],
# 'b': [2, 5, 'NA'],
# 'c': [3, 'NA', 'NA'],
# 'd': ['NA', 'NA', 5]}