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 order of 1. DINT 2. INT 3. BOOL

Result:

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

I've seen other similar questions in stackoverflow but nothing similar or easily applicable to me.

Answer 1

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

indx = [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 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

     python 3.2

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

    2. sorted(mylist,reverse=True)