python - 比较嵌套字典和列表中的值

Question

我想比较两个变量(字典和列表)的值。字典有一个嵌套结构，所以我必须循环所有项目。我发现了简单的解决方案，但我很确定我可以用更好的方式(使用 python)来做到这一点。简而言之，我想从 user_from_database 中查找 user_from_client 变量中不存在的项目。

我的解决方案:

#variable containing users from client side
users_from_client = {
  "0": {
    "COL1": "whatever",
    "COL2": "val1",
    "COL3": "whatever",
  },
  "1": {
    "COL1": "whatever",
    "COL2": "val2",
    "COL3": "whatever",
  },
  "3": {
    "COL1": "whatever",
    "COL2": "val3",
    "COL3": "whatever",
  }    
} 

#variable containing users from the database
users_from_database = [
  ["val1"],
  ["val2"],
  ["val5"],
  ["val7"]
]

#This function is used to find element from the nested dictionaries(d)
def _check(element, d, pattern = 'COL2'):
  exist = False
  for k, user in d.iteritems():
    for key, item in user.iteritems():
      if key == pattern and item == element:
        exist = True
  return exist

#Finding which users should be removed from the database  
to_remove = []
for user in users_from_db:
  if not _check(user[0], users_from_agent):
    if user[0] not in to_remove:
      to_remove.append(user[0])

#to_remove list contains: [val5, val7"] 

使用 python 方法给出相同结果的更好方法是什么？也许我不必补充说我是 python 新手(我假设您能够通过观看上面的代码看到这一点)。

Answer 1

只需使用 error-safe dictionary lookup :

def _check(element, d, pattern = 'COL2'):
    for user in d.itervalues():
        if user.get(pattern) == element:
            return True
    return False

或者作为单行:

def _check(element, d, pattern = 'COL2'):
    return any(user.get(pattern) == element for user in d.itervalues())

或者尝试将整个工作作为一句台词来完成:

#Finding which users should be removed from the database  
to_remove = set(
    name
    for name in users_from_database.itervalues()
    if not any(user.get('COL2') == name for (user,) in users_from_client)
)

assert to_remove == {"val5", "val7"}

set可以使其更加简洁(高效):

to_remove = set(
    user for (user,) in users_from_database
) - set(
    user.get('COL2') for user in users_from_client
)

<小时/>

你的数据结构有点奇怪。考虑使用:

users_from_client = [
  {
    "COL1": "whatever",
    "COL2": "val1",
    "COL3": "whatever",
  }, {
    "COL1": "whatever",
    "COL2": "val2",
    "COL3": "whatever",
  }, {
    "COL1": "whatever",
    "COL2": "val3",
    "COL3": "whatever",   
  }
] 

#variable containing users from the database
users_from_database = set(
  "val1",
  "val2",
  "val5",
  "val7"
)

这会将您的代码减少为:

to_remove = users_from_database - set(
    user.get('COL2') for user in users_from_client
)