更好的方法来处理dict列表或提高我的cod

Question

我有个口述

my_dict =
    { '123@com' : [ { 'event': 'event1', 'oldval' = 1, 'newval = 2}, { 'event': 'event2', 'oldval' = 11, 'newval': 22}, { 'event': 'event3', 'oldval' = 33, 'newval': 44}] 
    {'1234@com' : { 'event1': 'oldval': 45, 'newval': 66},
    {'12345@com': [ {'event':'event2', 'oldval': 77, 'newval': 99}, { 'event': 'event3', 'oldval': '99', 'newval': 100'}]

所以基本上我的目标是建立一个列表

 [ ['123@com', (2,1,True), (22,11, True), (44,33,True)], ['1234@com', (66,45,True), ('','',False),('','',False)], ['12345@com', ('',''False), (99,77,True), (100,99)]]

因此，我的目标是在事件发生更改时获取值，并将其标记为true。你知道吗

所以我写的代码如下：

for key,value in my_dict.items():
  if str(my_dict).find('event1') >- 1 or str(my_dict).find('even2') > 1 or str(my_dict).find('event3') > -1: 
  tmp_list = []
  tmp_list.append(key)
  if type(value) is dict:
    if value['event'] == 'event1': 
       @ get all the value and append to tmp_list
    if value['event'] == 'event2':
      @ do the same thing as above
  else:
     for val in value:
      if val['name'] == 'event':
        @ do the same stuff. 
      @do the same stuff again for rest of event. 

所以基本上我可以通过遵循上面的代码模式来实现我的目标。你知道吗

所以任何人都可以帮助我更好地编码，我会非常感激。 我不喜欢我编码的方式。你知道吗

谢谢

Answer 1

这是我能想到的最好的编码方式。你知道吗

def value_finder(val):
    return lambda ev,li: next((di[val] for di in li if di['event']==ev),'')

def event_finder(ev,li):
    return any(di['event']==ev for di in li)

def tuple_return(ev,li,nv,ov):
    return (nv(ev,li),ov(ev,li),event_finder(ev,li))

def main(my_dict):
    lis = []
    nv = value_finder('newval')
    ov = value_finder('oldval')
    events = ['event1','event2','event3']
    for key,value in my_dict.items():
        temp_list = [key]
        temp_list += [tuple_return(ev,value) for ev in events]
        lis.append(temp_list)
    print lis

my_dict = {'123@com':[{'event':'event1','oldval':1,'newval':2},
                      {'event':'event2','oldval':11,'newval':22},
                      {'event':'event3','oldval':33,'newval':44}],
          '1234@com':[{'event':'event1','oldval':45,'newval':66}],
          '12345@com':[{'event':'event2','oldval': 77,'newval': 99},
                      {'event':'event3','oldval':'99','newval':100}]}
main(my_dict)

Answer 2

拥有像['123@com', (2,1,True), (22,11, True), (44,33,True)]这样的异构列表是一种代码味道。这几乎总是意味着您试图对简单的数据类型做太多的工作，而实际使用类可能会更好。假设python 2.x：

class Event(object):
    def __init__(self, event, oldval, newval):
        self.event = event
        self.oldval = oldval
        self.newval = newval

class EventThing(object):
    def __init__(self, key, event_list):
        self.key = key
        self.event_list = [Event(**e) for e in event_list]

    def produceCondensedList(self):
        return [(e.oldval, e.newval, True) for e in self.event_list]

thing_list = [EventThing(key, event_list) for key, event_list in my_dict.items()]

这与您给出的示例并不完全相同，但是如果不了解您的具体用例，就很难给出准确的答案。以上这些应该足以让你开始。你知道吗