How to sort list of dicts by dict value in pre-defined format?

Question

I've got a list of dicts which describe certain classes in a school of which some are still taught, some are not anymore, and some are going to be taught in the future:

[
    {'status': 'DEACTIVE', 'name': 'Music'},
    {'status': 'DEACTIVE', 'name': 'Juggling'},
    {'status': 'ACTIVE', 'name': 'Chess'},
    {'status': 'ACTIVE', 'name': 'Dutch Language'},
    {'status': 'COMING', 'name': 'Python'},
    {'status': 'COMING', 'name': 'Drinking coffee like a pro'},
]

I now want to order this list so that the ACTIVE are first, then the DEACTIVE , and finally the COMING . I found solutions such as this:

sorted(a, key=dict.values, reverse=True)

but that only orders it by alphabet where I want to sort it in a custom way. Does anybody know how I can do this?

Answer 1

You can do it using dictionary with weight of all statuses:

STATUS_RANK = {"ACTIVE": 1, "DEACTIVE": 2, "COMING": 3}

courses.sort(key=lambda x: STATUS_RANK[x['status']])

Answer 2

i just made a method to handle that called it order_my_list() as you see below: all you need is just to pass your list as a parameter!

 def order_my_list(lst):
     ordered = []
     loop = 0
     finished_flag = 0
     while finished_flag == 0:
         for x in range(0,len(lst)):
             if loop == 0:
                 if lst[x]['status'] == 'ACTIVE':
                     ordered.append(lst[x])
             if loop == 1:
                 if lst[x]['status'] == 'DEACTIVE':
                     ordered.append(lst[x])
             if loop == 2:
                 if lst[x]['status'] == 'COMING':
                     ordered.append(lst[x])
             if loop == 3:
                 finished_flag = 1

         loop += 1
     return ordered

Im sure there is better/faster ways to do it but thats how i would solve it as my python knowledge isnt that great :p ...