列表理解是合并这些 JSON 文件 Python 的正确方法吗？

Question

如何使用 python 列表推导将一个 JSON 文件中的值替换为单独 JSON 文件中的链接值？

一个看起来像这样并且有一个“a”值我需要用来替换另一个列表中的值，使用“b”作为连接符（a、b 和 c 值都是唯一的 id）：

{
   "records":[
      {
         "a": "7hk2k989u23lesdfsfd",
         "b":"b8",
      },
      {
         "a": "9ty562349u23lesdfsfd",
         "b":"b6",
      },
      {
         "a": "Ur233Fglesdfsfd",
         "b":"b2",
      }
   ]
}

另一个看起来像这样，其中“d”需要替换为相应的“a”值，其中“b”是键：

{
   "records":[
      {
         "c":00023414,
         "d":["b8","b6"]
      },
      {
         "c":0005814,
         "d":["b8","b2","b6"]
      }
   ]
}

所以我最终得到：

{
   "records":[
      {
         "c":00023414,
         "d":["7hk2k989u23lesdfsfd","9ty562349u23lesdfsfd"]
      },
      {
         "c":0005814,
         "d":["7hk2k989u23lesdfsfd","Ur233Fglesdfsfd","9ty562349u23lesdfsfd"]
      }
   ]
}

使用 python 解决这个问题的正确方法是什么，尤其是当我需要代码具有高性能时？

Answer 1

您的文件不是有效的 JSON。您应该使用 JSON 验证器进行检查，例如 JSON Lint

In [494]: import json

In [495]: with open('/Users/ado/Desktop/ab.json') as f:
     ...:     ab = json.load(f)
     ...:

In [496]: with open('/Users/ado/Desktop/cd.json') as f:
     ...:     cd = json.load(f)
     ...:

请注意，您可以将ab 简单地视为相关as 和bs 的集合。这是使用字典将bs 映射到as 的好时机

In [497]: d_ab = {r['b']: r['a'] for r in ab['records']}

In [498]: d_ab
Out[498]:
{'b2': 'Ur233Fglesdfsfd',
 'b6': '9ty562349u23lesdfsfd',
 'b8': '7hk2k989u23lesdfsfd'}

现在您可以遍历 cd 中的 records 并使用 list 推导式来创建新值

In [499]: for r in cd['records']:
     ...:     r['d'] = [d_ab.get(d) for d in r['d']]
     ...:

In [500]: cd
Out[500]:
{'records': [{'c': 23414,
   'd': ['7hk2k989u23lesdfsfd', '9ty562349u23lesdfsfd']},
  {'c': 5814,
   'd': ['7hk2k989u23lesdfsfd', 'Ur233Fglesdfsfd', '9ty562349u23lesdfsfd']}]}

最后，将新内容写入文件

In [502]: with open('/Users/ado/Desktop/cd-mapped.json', 'w') as f:
     ...:     f.write(json.dumps(cd))
     ...:

此解决方案假定在ab 中，每条记录中始终存在as 和bs。

对于有趣的 PS，您可以使用 map 和 dict.get 代替理解

In [505]: for r in cd['records']:
     ...:     r['d'] = list(map(d_ab.get, r['d']))
     ...:

In [506]: cd
Out[506]:
{'records': [{'c': 23414,
   'd': ['7hk2k989u23lesdfsfd', '9ty562349u23lesdfsfd']},
  {'c': 5814,
   'd': ['7hk2k989u23lesdfsfd', 'Ur233Fglesdfsfd', '9ty562349u23lesdfsfd']}]}

就性能而言，理解通常优于 maps

In [509]: %timeit for r in cd['records']: r['d'] = [d_ab.get(d) for d in r['d']]
     ...:
The slowest run took 7.19 times longer than the fastest. This could mean that an intermediate result is being cached.
1000000 loops, best of 3: 1.34 µs per loop

In [511]: %timeit for r in cd['records']: r['d'] = list(map(d_ab.get, r['d']))
The slowest run took 7.19 times longer than the fastest. This could mean that an intermediate result is being cached.
1000000 loops, best of 3: 1.74 µs per loop