检查两个列表的交集

Question

我把关键词分类如下：

Keywords_33=[('File_2', ['with', 'as']),
             ('Module_2', ['from', 'import']),
             ('Constant_3', {'bool': ['False', 'True'],
                             'none': ['None']}),
             ('Operator_4', {'boolean_operation': {'or', 'and', 'not'},
                             'comparison': {'is'}}),
             ('Container_operation_2', ['in', 'del']),
             ('Klass_1', ['class']),
             ('Function_7',['lambda', 'def', 'pass',
                            'global', 'nonlocal',
                            'return', 'yield']),
             ('Repetition_4', ['while', 'for', 'continue', 'break']),
             ('Condition_3', ['if', 'elif', 'else']),
             ('Debug_2', ['assert', 'raise']),
             ('Exception_3', ['try', 'except', 'finally'])]

我打算确认每一个关键字在没有任何遗留类别的地方。 我认为最方便的方法是首先将Keywords_33转换为字符串。你知道吗

from keyword import kwlist
In [55]: print(kwlist)
['False', 'None', 'True', 'and', 'as', 'assert', 'break', 'class', 'continue', 'def', 'del', 'elif', 'else', 'except', 'finally', 'for', 'from', 'global', 'if', 'import', 'in', 'is', 'lambda', 'nonlocal', 'not', 'or', 'pass', 'raise', 'return', 'try', 'while', 'with', 'yield']
In [54]: from keyword import kwlist
...: s = str(Keywords_33)
...: [keyword for keyword in kwlist if keyword not in s]
...:
Out[54]: []
# It indicate no keyword left

如何优雅地完成这样的任务？你知道吗

Answer 1

依赖dict/list嵌套列表的字符串表示有点危险，因为您可以匹配不需要的单词/子字符串（例如elif包含if），因此如果列表包含elif，它也会匹配if。一些聪明的re.findall可以工作，在引号之间提取文本，但这仍然是一种技巧。你知道吗

相反，您可以创建dict值的列表或列表（取决于类型），从而生成：

['with', 'as', 'from', 'import', ['None'], ['False', 'True'], {'and', 'or', 'not'}, {'is'}, 'in', 'del', 'class', 'lambda', 'def', 'pass', 'global', 'nonlocal', 'return', 'yield', 'while', 'for', 'continue', 'break', 'if', 'elif', 'else', 'assert', 'raise', 'try', 'except', 'finally']

然后在不规则的项目列表（Flatten (an irregular) list of lists）上使用展开方法，转换为set并双向减法/相交：

Keywords_33=[('File_2', ['with', 'as']),
             ('Module_2', ['from', 'import']),
             ('Constant_3', {'bool': ['False', 'True'],
                             'none': ['None']}),
             ('Operator_4', {'boolean_operation': {'or', 'and', 'not'},
                             'comparison': {'is'}}),
             ('Container_operation_2', ['in', 'del']),
             ('Klass_1', ['class']),
             ('Function_7',['lambda', 'def', 'pass',
                            'global', 'nonlocal',
                            'return', 'yield']),
             ('Repetition_4', ['while', 'for', 'continue', 'break']),
             ('Condition_3', ['if', 'elif', 'else']),
             ('Debug_2', ['assert', 'raise']),
             ('Exception_3', ['try', 'except', 'finally'])]

import collections
from keyword import kwlist

def flatten(l):
    for el in l:
        if isinstance(el, collections.Iterable) and not isinstance(el, (str, bytes)):
            yield from flatten(el)
        else:
            yield el

my_keywords = set(flatten(x for _,l in Keywords_33 for x in (l if isinstance(l,list) else l.values())))

在这种情况下set(kwlist) == my_keywords

Answer 2

Keywords_33=[('File_2', ['with', 'as']),
         ('Module_2', ['from', 'import']),
         ('Constant_3', {'bool': ['False', 'True'],
                         'none': ['None']}),
         ('Operator_4', {'boolean_operation': {'or', 'and', 'not'},
                         'comparison': {'is'}}),
         ('Container_operation_2', ['in', 'del']),
         ('Klass_1', ['class']),
         ('Function_7',['lambda', 'def', 'pass',
                        'global', 'nonlocal',
                        'return', 'yield']),
         ('Repetition_4', ['while', 'for', 'continue', 'break']),
         ('Condition_3', ['if', 'elif', 'else']),
         ('Debug_2', ['assert', 'raise']),
         ('Exception_3', ['try', 'except', 'finally'])]

kwlist = ['False', 'None', 'True', 'and', 'as', 'assert', 'break', 'class', 'continue', 'def',
      'del', 'elif', 'else', 'except', 'finally', 'for', 'from', 'global', 'if', 'import',
      'in','is', 'lambda', 'nonlocal', 'not', 'or', 'pass', 'raise', 'return', 'try', 'while',
      'with', 'yield']

if __name__ == '__main__':
    result = []
    for kw in kwlist:
        for key in Keywords_33:
            if isinstance(key[1], list):
                for i in key[1]:
                    if kw == i:
                        result.append(i)
            elif isinstance(key[1], dict):
                for value in key[1].values():
                    for j in value:
                        if kw == j:
                            result.append(j)

    print(result)

我区分，如果第二个元素是列表还是字典。（注意，在本例中它是有效的，因为每个元组正好包含两个元素）。在list情况下，我可以轻松地遍历并比较元素是否在列表中。对于dictionary，如果值在kwlist中，我可以遍历dictionary的值并进行比较