Sorting dict items by key, beyond alphanumeric sorting

Question

I have written this code:

n=5
dizN={}
for q in range(0,n+1):
    h=n-q
    dizN['a'+str(q)+'p'+str(h)]=0

that creates such a dictionary:

dizN

Out[120]: {'a0p5': 0, 'a1p4': 0, 'a2p3': 0, 'a3p2': 0, 'a4p1': 0, 'a5p0': 0}

Note that "n" is the basic parameter for my code. As you can see, the sum of integers present in dict keys string is always =n (=5 in this case, where n=5).

It is important for me (for more difficult purposes in my program) that, for every n anyone can choose, the dict is ordered in this way:

{'a0p(n)': 0, 'a1p(n-1)': 0, ....., 'a(n-1)p1': 0, 'a(n)p0': 0}

My code is ok, but only for n<10. If n is >=10, this is what happens: (n=12) dizN:

Out[121]: 
  {'a0p12': 0,
   'a10p2': 0,
   'a11p1': 0,
   'a12p0': 0,
   'a1p11': 0,
   'a2p10': 0,
   'a3p9': 0,
   'a4p8': 0,
   'a5p7': 0,
   'a6p6': 0,
   'a7p5': 0,
   'a8p4': 0,
   'a9p3': 0}

As you can see the interpreter follows alphanumeric sorting;

Anybody know if there is a way to obtain the same dict sorted this way:

{'a0p12': 0,
 'a1p11': 0,
 'a2p10': 0,
 'a3p9': 0,
 'a4p8': 0,
 'a5p7': 0,
 'a6p6': 0,
 'a7p5': 0,
 'a8p4': 0,
 'a9p3': 0,
 'a10p2': 0,
 'a11p1': 0,
 'a12p0': 0}

?

I know that dictionaries are basically non sortable, but i hope somebody knows some trick to obtain my purpose anyway :)

Thanks a lot!

You write multiple times about ordering a dictionary, but Python's dictionaries cannot be ordered. Perhaps you should look at `OrderedDict`. — Rory Daulton, Oct 17 '16 at 14:05
The whole point of a dictionary is that it isn't ordered. It's a hash table. — Daniel, Oct 17 '16 at 14:11
The sorting criteria i would like to follow is "simply" by the first integer in dict keys strings, considered as an integer (not string) ascending. Or, to explain better, by the integer that follows letter 'a' in the strings that are the keys of my dict: 'a(integer)p(integer)'. — mik.ferrucci, Oct 17 '16 at 14:18

Padraic Cunningham · Accepted Answer · 2016-10-17T15:59:49.830

1

dicts are unordered, so to get the order you are going to have to sort the items and use an OrderedDict to maintain the sorted order. To get the order you want you can create tuples from the groups of integers so you sort as integers in lexicographical order:

from itertools import groupby
from collections import OrderedDict
d = {'a0p12': 0, 'a10p2': 0, 'a11p1': 0, 'a12p0': 0, 'a1p11': 0, 'a2p10': 0,
     'a3p9': 0, 'a4p8': 0, 'a5p7': 0, 'a6p6': 0, 'a7p5': 0, 'a8p4': 0, 'a9p3': 0}

def key_func(x):
    """'a0p12' -> (0, 12)"""
    return tuple(int("".join(v)) for k,v in groupby(x[0], key=str.isdigit) if k)
od = OrderedDict(sorted(d.items(), key=key_func))

print(od)

Which would give you:

OrderedDict([('a0p12', 0), ('a1p11', 0), ('a2p10', 0), ('a3p9', 0), 
('a4p8', 0), ('a5p7', 0), ('a6p6', 0), ('a7p5', 0), ('a8p4', 0), 
('a9p3', 0), ('a10p2', 0), ('a11p1', 0), ('a12p0', 0)])

You could also use a regex to find the groups of digits:

from collections import OrderedDict
import re

d = {'a0p12': 0, 'a10p2': 0, 'a11p1': 0, 'a12p0': 0, 'a1p11': 0, 'a2p10': 0,
     'a3p9': 0, 'a4p8': 0, 'a5p7': 0, 'a6p6': 0, 'a7p5': 0, 'a8p4': 0, 'a9p3': 0}



def key_func(x,patt=re.compile("\d+")):
    """'a0p12' -> (0, 12)"""
    return tuple(map(int, patt.findall(x[0])))

od = OrderedDict(sorted(d.items(), key=key_func))

print(od)

edited Oct 17 '16 at 15:59

answered Oct 17 '16 at 14:16

Padraic Cunningham

176,452
29
245
321

Thank you a lot Padraic! I will try this pretty solution in my program, then I will ask another question for the biggest problem that involves this strange sorting need! :) – mik.ferrucci Oct 17 '16 at 14:31
No worries, are the letters always the same and in the same order or do they need to be taken into account? – Padraic Cunningham Oct 17 '16 at 14:32
1

No such problem with letters. The dict keys form is always: – mik.ferrucci Oct 17 '16 at 14:33
1

'a'(integer)'p'(integer) Bye! – mik.ferrucci Oct 17 '16 at 14:34
1

Thanks again, the 2nd solution works well too, of course. – mik.ferrucci Oct 17 '16 at 16:11
I find it useful to add the conversion from Ordered dict to a list, so that access to tuples is possible ;). – mik.ferrucci Oct 17 '16 at 16:13
@mik.ferrucci, if you just want a list of tuples you just need `sorted(d.items(), key=key_func)` – Padraic Cunningham Oct 17 '16 at 16:14
lis_od = list(od.items()) : this way i obtained an accessible list. Going on programming, bye! – mik.ferrucci Oct 17 '16 at 16:14
Oh that's right! Yes my effective purpose was to obtain a list of tuples in order to be able to make iterations over it. Thanks a lot anyway! – mik.ferrucci Oct 17 '16 at 16:19

Sorting dict items by key, beyond alphanumeric sorting

1 Answers1

Linked