Question

我有以下嵌套字典变量dict2，我只需打印VarCharValue中的值（我使用的是python 2.7）

我尝试使用

print (" {0[ResultSet][Rows]}".format(dict2))

但无法获取Data

中的值

print dict2

{u'ResultSet': {u'Rows': [{u'Data': [{u'VarCharValue': u'CNT'}]}, {u'Data': [{u'VarCharValue': u'1'}]}], u'ResultSetMetadata': {u'ColumnInfo': [{u'Scale': 0, u'Nullable': u'UNKNOWN', u'TableName': u'', u'Precision': 19, u'CatalogName': u'hive'}]}}, 'ResponseMetadata': {'RetryAttempts': 0, 'HTTPStatusCode': 200,'HTTPHeaders': {'date': 'Fr GMT', 'connection': 'keep-alive'}}}

必需输出

CNT
1

Answer 1

你的字典里面有很多细分（但也有列表），因此你需要将所有字符编入索引，直到找到所需的元素：

print ("{0[ResultSet][Rows][0][Data][0][VarCharValue]}".format(dict2)) 
print ("{0[ResultSet][Rows][1][Data][0][VarCharValue]}".format(dict2))

打印：

CNT
1

Answer 2

递归函数将完成这项工作。它可能远非快速但仍适用于大多数情况。您可以向其提供任何结构的字典，传递您想要的密钥的名称，然后瞧。

    dict2 = {u'ResultSet': {u'Rows': [{u'Data': [{u'VarCharValue': u'CNT'}]}, {u'Data': [{u'VarCharValue': u'1'}]}], u'ResultSetMetadata': {u'ColumnInfo': [{u'Scale': 0, u'Nullable': u'UNKNOWN', u'TableName': u'', u'Precision': 19, u'CatalogName': u'hive'}]}}, 'ResponseMetadata': {'RetryAttempts': 0, 'HTTPStatusCode': 200,'HTTPHeaders': {'date': 'Fr GMT', 'connection': 'keep-alive'}}}

    def get_dict_values(from_dict, get_key):
        '''
        Return the values of specified keys from a nested dictionary.
        '''

        # A list to collect resulting values in.
        list_collect = list()

        # For each key in a dict.
        for dict_key in from_dict.keys():
            # If it's value is an instance of dict.
            if isinstance(from_dict[dict_key], dict):
                # Call self recursively.
                get_list = get_dict_values(from_dict[dict_key], get_key)

                # Extend the list_collect with what we've got from recursive call.
                list_collect.extend(get_list)
            # If a value is anything iterable.
            elif isinstance(from_dict[dict_key], (list, set, tuple, frozenset)):
                # Call self for each element.
                for list_elem in from_dict[dict_key]:
                    # Extend the list_collect with what we've got from recursive call.
                    get_list = get_dict_values(list_elem, get_key)
                    list_collect.extend(get_list)
            elif dict_key == get_key:
                list_collect.extend([from_dict[dict_key]])
        return list_collect

    if __name__ == '__main__':
        varchar_values = get_dict_values(dict2, 'VarCharValue')
        print(varchar_values)

它会返回一个列表，例如['CNT', '1']因此，如果您希望将值作为字符串，请执行'\n'.join(varchar_values)之类的操作。

如何从深层嵌套字典

2 个答案: