如何对词典列表进行排序？

Question

I am new to python and doing an assignment to implement Lai Yang Global Snapshot Algorithm. I want to order a list of messages. I made a dictionary based on user input, but now I want to iterate for each process and sort msgs based on the time they were sent.

My dictionary structure is:

defaultdict(<class 'list'>, 
{
  1: [
      {10: {'color': 'w', 'data': 4, 'receiver': 2, 'msgId': 'N3FAVSNZGQP0'}}, 
      {1: {'color': 'w', 'data': 10, 'receiver': 2, 'msgId': '38BRJ7PPE42V'}}
  ]
  2: [
      {30: {'color': 'w', 'data': 50, 'receiver': 1, 'msgId': 'IY09Z5TH4D2G'}}
  ]
})

I want it to be sorted to:

defaultdict(<class 'list'>, 
{
  1: [
      {1: {'color': 'w', 'data': 10, 'receiver': 2, 'msgId': '38BRJ7PPE42V'}},
      {10: {'color': 'w', 'data': 4, 'receiver': 2, 'msgId': 'N3FAVSNZGQP0'}} 
  ]
  2: [
      {30: {'color': 'w', 'data': 50, 'receiver': 1, 'msgId': 'IY09Z5TH4D2G'}}
  ]
})

I googled alot and browsed and tried all possible solutions. I saw sortedcontainers module but even that gave error. Below is my last attempt of lambda, which is failing too (TypeError: '<' not supported between instances of 'dict' and 'dict')

for obj in sentMsgs:
    msg = sorted(sentMsgs[obj], key=lambda x:list(x.values()))
    sentMsgs[obj] = msgs

Other failed attempts:

for obj in sentMsgs:
    L = sentMsgs[obj]
    msgs = [collections.OrderedDict((k, d[k](v)) for (k, v) in l.items()) for l in L]

    msgs = collections.OrderedDict(sorted(sentMsgs[obj].items()))
    sentMsgs[obj] = msgs

    msgs = SortedDict(sentMsgs[obj])
    sentMsgs[obj] = msgs

Answer 1

messages = {
  1: [
      {10: {'color': 'w', 'data': 4, 'receiver': 2, 'msgId': 'N3FAVSNZGQP0'}}, 
      {1: {'color': 'w', 'data': 10, 'receiver': 2, 'msgId': '38BRJ7PPE42V'}}
  ],
  2: [
      {30: {'color': 'w', 'data': 50, 'receiver': 1, 'msgId': 'IY09Z5TH4D2G'}}
  ]
}

for message in messages:
    messages[message].sort(key=lambda x: list(x.keys())[0])

Answer 2

Sorting each list's dicts by their key(s):

for a in sentMsgs.values():
    a.sort(key=list)

Try it online!