如何按嵌套字典中comb_indx键的值进行分组？

原始数据结构 #

original_dict = {
    'comb_indx': {0: '3925220EE', 1: '3925220EE', 2: '3925220EE', 
                  3: '66478EE', 4: '66478EE', 5: '66478EE', 
                  6: '42300EE', 7: '42300EE', 8: '42300EE'},
    'country': {0: 'EE', 1: 'EE', 2: 'EE', 3: 'EE', 4: 'EE', 5: 'EE', 6: 'EE', 7: 'EE', 8: 'EE'},
    'type': {0: 'CREDIT_ACCOUNT', 1: 'CREDIT_ACCOUNT', 2: 'CREDIT_ACCOUNT', 
             3: 'SMALL_LOAN', 4: 'SMALL_LOAN', 5: 'SMALL_LOAN', 
             6: 'SMALL_LOAN', 7: 'SMALL_LOAN', 8: 'SMALL_LOAN'}
}

期望输出结构 #

[
    {'comb_indx': '3925220EE', 
     'country': {0: 'EE', 1: 'EE', 2: 'EE'}, 
     'type': {0: 'CREDIT_ACCOUNT', 1: 'CREDIT_ACCOUNT', 2: 'CREDIT_ACCOUNT'}},
    {'comb_indx': '66478EE', 
     'country': {3: 'EE', 4: 'EE', 5: 'EE'}, 
     'type': {3: 'SMALL_LOAN', 4: 'SMALL_LOAN', 5: 'SMALL_LOAN'}},
    {'comb_indx': '42300EE', 
     'country': {6: 'EE', 7: 'EE', 8: 'EE'}, 
     'type': {6: 'SMALL_LOAN', 7: 'SMALL_LOAN', 8: 'SMALL_LOAN'}}
]

实现代码 #

使用Python标准库itertools.groupby即可完成分组逻辑，代码如下：

from itertools import groupby

original_dict = {
    'comb_indx': {0: '3925220EE', 1: '3925220EE', 2: '3925220EE', 
                  3: '66478EE', 4: '66478EE', 5: '66478EE', 
                  6: '42300EE', 7: '42300EE', 8: '42300EE'},
    'country': {0: 'EE', 1: 'EE', 2: 'EE', 3: 'EE', 4: 'EE', 5: 'EE', 6: 'EE', 7: 'EE', 8: 'EE'},
    'type': {0: 'CREDIT_ACCOUNT', 1: 'CREDIT_ACCOUNT', 2: 'CREDIT_ACCOUNT', 
             3: 'SMALL_LOAN', 4: 'SMALL_LOAN', 5: 'SMALL_LOAN', 
             6: 'SMALL_LOAN', 7: 'SMALL_LOAN', 8: 'SMALL_LOAN'}
}

# 先对comb_indx的键值对按值排序，确保groupby能正确分组
sorted_items = sorted(original_dict['comb_indx'].items(), key=lambda x: x[1])
result = []

for key, group in groupby(sorted_items, key=lambda x: x[1]):
    # 收集当前分组对应的所有索引
    indices = [idx for idx, _ in group]
    # 组装当前分组的字典
    group_dict = {
        'comb_indx': key,
        'country': {idx: original_dict['country'][idx] for idx in indices},
        'type': {idx: original_dict['type'][idx] for idx in indices}
    }
    result.append(group_dict)

# 输出结果
print(result)

代码说明 #

• groupby仅会将连续的相同值归为一组，因此需要先对comb_indx的键值对按值排序，保证同值项连续
• 遍历分组时，先收集该组对应的所有索引，再通过字典推导式从country和type中提取对应索引的键值对
• 最终将每个分组的字典加入结果列表，即可得到目标结构

内容的提问来源于stack exchange，提问作者Dmytro Kulish