Can I sort text by its numeric value in Python?

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)]

Answer 2

Python's sorting functions can take a custom compare function, so you just need to define a function that compares keys the way you like:

def version_cmp(a, b):
  '''These keys just look like version numbers to me....'''
  ai = map(int, a.split('.'))
  bi = map(int, b.split('.'))
  return cmp(ai, bi)

for k in sorted(mydict.keys(), version_cmp):
  print k, mydict[k]

In this case you should better to use the key parameter to sorted(), though. See Ian Clelland's answer for an example for that.

Answer 3

I would do a search on "sorting a python dictionary" and take a look at the answers. I would give PEP-265 a read as well. The sorted() function is what you are looking for.

Answer 4

As an addendum to Ian Clelland's answer, the map() call can be replaced with a list comprehension... if you prefer that style. It may also be more efficient (though negligibly in this case I suspect).

sorted(mydict.keys(), key=lambda a: [int(i) for i in a.split('.')])

Answer 5

For fun & usefulness (for googling ppl, mostly):

f = lambda i: [int(j) if re.match(r"[0-9]+", j) else j for j in re.findall(r"([0-9]+|[^0-9]+)", i)]
cmpg = lambda x, y: cmp(f(x), f(y))

use as sorted(list, cmp=cmpg). Additionally, regexes might be pre-compiled (rarely necessary though, actually, with re module's caching). And, it may be (easily) modified, for example, to include negative values (add -? to num regex, probably) and/or to use float values.

It might be not very efficient, but even with that it's quite useful.

And, uhm, it can be used as key= for sorted() too.