在嵌套字典中将多个相同的键合并为一个答案

【问题标题】：Merge multiple same keys into one in the nested dictionary在嵌套字典中将多个相同的键合并为一个
【发布时间】：2020-01-14 17:59:33
【问题描述】：

我在列表中有一个嵌套字典，如下所示：

      my_list =

      [{'id': '166073',
        'ref': [{'MeSH': 'C548074'},
       {'UMLS': 'C1969084'},
       {'OMIM': '611523'},
       {'ICD-10': 'Q04.3'}]},

       {'id': '213',
       'ref': [{'MeSH': 'D003554'},
       {'UMLS': 'C0010690'},
       {'MedDRA': '10011777'},
       {'ICD-10': 'E72.0'},
       {'OMIM': '219750'},
       {'OMIM': '219800'},
       {'OMIM': '219900'}]},

       {'id': '333',
        'ref': [{'UMLS': 'C2936785'},
       {'ICD-10': 'E75.2'},
       {'MeSH': 'C537075'},
       {'MeSH': 'D055577'},
       {'UMLS': 'C0268255'},
       {'OMIM': '228000'}]}
         .
         .
         .               
                         ]

我想将嵌套字典中具有相同键的字典合并为键中的列表，如下所示：

    my_list =

    [{'id': '166073',
      'ref': [{'MeSH': 'C548074'},
       {'UMLS': 'C1969084'},
       {'OMIM': '611523'},
       {'ICD-10': 'Q04.3'}]},

     {'id': '213',
      'ref': [{'MeSH': 'D003554'},
       {'UMLS': 'C0010690'},
       {'MedDRA': '10011777'},
       {'ICD-10': 'E72.0'},
       {'OMIM': ['219750', '219800', '219900']}]},

     {'id': '333',
      'ref': [{'UMLS': 'C2936785'},
       {'ICD-10': 'E75.2'},
       {'MeSH': ['C537075', 'D055577']},
       {'UMLS': 'C0268255'},
       {'OMIM': '228000'}]}
         .
         .
         .               
                         ]

我尝试通过使用双for循环读取字典并将信息存储到另一个新字典来进行合并，但我发现该方法不是最佳的，有没有其他推荐的方法来完成这种合并？谢谢！

【问题讨论】：

标签： python dictionary

【解决方案1】：

为什么它不是最理想的？我认为这种合并应该没问题。我假设你的合并是这样的：

my_list = [{'id': '166073',
        'ref': [{'MeSH': 'C548074'},
       {'UMLS': 'C1969084'},
       {'OMIM': '611523'},
       {'ICD-10': 'Q04.3'}]},

       {'id': '213',
       'ref': [{'MeSH': 'D003554'},
       {'UMLS': 'C0010690'},
       {'MedDRA': '10011777'},
       {'ICD-10': 'E72.0'},
       {'OMIM': '219750'},
       {'OMIM': '219800'},
       {'OMIM': '219900'}]},

       {'id': '333',
        'ref': [{'UMLS': 'C2936785'},
       {'ICD-10': 'E75.2'},
       {'MeSH': 'C537075'},
       {'MeSH': 'D055577'},
       {'UMLS': 'C0268255'},
       {'OMIM': '228000'}]}]

def merge(item):
  from collections import defaultdict
  merged = defaultdict(list)
  for ref in item.get('ref', []):
    for key, val in ref.items():
      merged[key].append(val)
  return {**item, 'ref': dict(merged)}

print(list(map(merge, my_list)))

【讨论】：

【解决方案2】：

#!/usr/bin/env python                                                                                                                                                                                                                                                       

o = {'id': '213',
     'ref': [{'MeSH': 'D003554'},
             {'UMLS': 'C0010690'},
             {'MedDRA': '10011777'},
             {'ICD-10': 'E72.0'},
             {'OMIM': '219750'},
             {'OMIM': '219800'},
             {'OMIM': '219900'}]}

n = {'id': o['id'],
     'ref': {x:[] for x in set([item for sublist in o['ref'] for item in sublist])}}

for p in o['ref']:
    for k, v in p.items():
        n['ref'][k].append(v)

n['ref'] = [n['ref']]

print(n)

【讨论】：

【解决方案3】：

我发现创建一个字典来收集值比解压成所需的格式更容易：

new_list = []

for item in my_list:
    d = {'id': item['id'], 'ref': {}}
    for r in item['ref']:
        only_key = list(r.keys())[0]
        d['ref'][only_key] = d['ref'].get(only_key, []) + [r[only_key]]
    new_list.append(d)

    new_ref = []
    for k, v in d['ref'].items():
        new_ref.append({k: v if len(v) > 1 else v[0]})
    d['ref'] = new_ref



[{'id': '166073', 'ref': [{'OMIM': '611523'}, {'MeSH': 'C548074'}, {'ICD-10': 'Q04.3'}, {'UMLS': 'C1969084'}]},
 {'id': '213', 'ref': [{'MeSH': 'D003554'}, {'UMLS': 'C0010690'}, {'MedDRA': '10011777'}, {'ICD-10': 'E72.0'}, {'OMIM': ['219750', '219800', '219900']}]},
 {'id': '333', 'ref': [{'ICD-10': 'E75.2'}, {'OMIM': '228000'}, {'MeSH': ['C537075', 'D055577']}, {'UMLS': ['C2936785', 'C0268255']}]}]

【讨论】：

【解决方案4】：

使用python的列表推导：

def merge(item):
  from collections import defaultdict
  merged = defaultdict(list)

  [[merged[k].append(v) for k, v in ref.items()] for ref in item.get('ref', [])]

  return {**item, 'ref': dict(merged)}

【讨论】：