mik.ferrucci mik.ferrucci - 1 month ago 6
Python Question

Sorting dict items by key, beyond alphanumeric sorting

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 :)

Here I have written to find original solutions!

Thanks a lot!

Answer

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)])