Question

I have dict in Python with keys of the following form:

mydict = {'0'     : 10,
          '1'     : 23,
          '2.0'   : 321,
          '2.1'   : 3231,
          '3'     : 3,
          '4.0.0' : 1,
          '4.0.1' : 10,
          '5'     : 11,
          # ... etc
          '10'    : 32,
          '11.0'  : 3,
          '11.1'  : 243,
          '12.0'  : 3,
          '12.1.0': 1,
          '12.1.1': 2,
          }

Some of the indices have no sub-values, some have one level of sub-values and some have two. If I only had one sub-level I could treat them all as numbers and sort numerically. The second sub-level forces me to handle them all as strings. However, if I sort them like strings I’ll have 10 following 1 and 20 following 2.

How can I sort the indices correctly?

Note: What I really want to do is print out the dict sorted by index. If there’s a better way to do it than sorting it somehow that’s fine with me.

Answer 1

You can sort the keys the way that you want, by splitting them on ‘.’ and then converting each of the components into an integer, like this:

sorted(mydict.keys(), key=lambda a:map(int,a.split('.')))

which returns this:

['0',
 '1',
 '2.0',
 '2.1',
 '3',
 '4.0.0',
 '4.0.1',
 '5',
 '10',
 '11.0',
 '11.1',
 '12.0',
 '12.1.0',
 '12.1.1']

You can iterate over that list of keys, and pull the values out of your dictionary as needed.

You could also sort the result of mydict.items(), very similarly:

sorted(mydict.items(), key=lambda a:map(int,a[0].split('.')))

This gives you a sorted list of (key, value) pairs, like this:

[('0', 10),
 ('1', 23),
 ('2.0', 321),
 ('2.1', 3231),
 ('3', 3),
 # ...
 ('12.1.1', 2)]