Sort a list with a custom order in Python

Question

I have a list

mylist = [[\'123\', \'BOOL\', \'234\'], [\'345\', \'INT\', \'456\'], [\'567\', \'DINT\', \'678\']]

I want to sort it with the orde

Answer 1

SORT_ORDER = {"DINT": 0, "INT": 1, "BOOL": 2}

mylist.sort(key=lambda val: SORT_ORDER[val[1]])

All we are doing here is providing a new element to sort on by returning an integer for each element in the list rather than the whole list. We could use inline ternary expressions, but that would get a bit unwieldy.

Answer 2

Since that is not in aphabetical order I don't think there is one single function that can sort it but what you could do is create a new list and then append. This is kind of a cheap method of doing it; but it gets the job done.

newlist=[];
for sub_list in mylist:
     if(sub_list[1] == 'DINT']):
          newlist.append(sub_list);

for sub_list in mylist:
     if(sub_list[1] == 'INT']):
         newlist.append(sub_list);

for sub_list in mylist:
     if(sub_list[1] == 'BOOL']):
            newlist.append(sub_list);

Answer 3

Another way could be; set your order in a list:

index = [2,1,0]

and create a new list with your order wished:

mylist = [mylist[_ind] for _ind in indx]

Out[2]: [['567', 'DINT', '678'], ['345', 'INT', '456'], ['123', 'BOOL', '234']]

Answer 4

     python 3.2

    1. sorted(mylist,key=lambda x:x[1][1])

    2. sorted(mylist,reverse=True)