按嵌套字典列表中的值对字典排序

Question

This has been driving me nuts for days, I've seen several questions on sorting Dicts by value, but they are simple structures and mine is complex.

My top level Dict Key is a hash, the sub Dict Key is a sequence #, and the sub Dict value is a list. The last value in that is a number, which is what I want to sort the top level Dict by. The size of the Dict can be quite large, but here is a sample:

 {'16741b673a418af3812f6d43ea3f7daf': 
    {1: [0, '16741b673a418af3812f6d43ea3f7daf', 'data-01', 1132],
     2: [1, '16741b673a418af3812f6d43ea3f7daf', 'data-02', 1132],
     3: [2, '16741b673a418af3812f6d43ea3f7daf', 'data-03', 1132]},

 'cbef6de99cc2b9739c824db6d0246093':
    {4: [0, 'cbef6de99cc2b9739c824db6d0246093', 'data-04', 55296],
     5: [1, 'cbef6de99cc2b9739c824db6d0246093', 'data-05', 55296],
     6: [1, 'cbef6de99cc2b9739c824db6d0246093', 'data-06', 55296],
     7: [2, 'cbef6de99cc2b9739c824db6d0246093', 'data-07', 55296]},
 'a1e0f7ccdd8d38cb5ae00cdac71b6724':
    {8: [0, 'a1e0f7ccdd8d38cb5ae00cdac71b6724', 'data-08', 20125],
     9: [1, 'a1e0f7ccdd8d38cb5ae00cdac71b6724', 'data-09', 20125],
    10: [1, 'a1e0f7ccdd8d38cb5ae00cdac71b6724', 'data-10', 20125]}}

This code will give me the value I'm looking for, but only for the first iteration, then I get KeyError: 1

for item1 in mydict.items():
    print(item1[1][1][3])

item1[1] returns subkey 1's list
item1[2] returns subkey 2's list
item1[3] returns subkey 3's list
item1[1][1][3] returns subkey 1's "value"

I want to be able to sort the dict forward and reverse by that value. I've seen:

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

I can't figure out how to apply that to my problem, generally my attempts end up with KeyError: 1 or IndexError: string out of range.

What am I missing? How can I reference that value for lamba? Is that what I need to do?

I'd prefer not to use a solution that includes Pandas. I'm trying to make this fast/efficient since the data can be quite large (currently 10,000 subkeys)

Edit:

Output would look the same but sorted by last value in list:

 {'16741b673a418af3812f6d43ea3f7daf': 
    {1: [0, '16741b673a418af3812f6d43ea3f7daf', 'data-01', 1132],
     2: [1, '16741b673a418af3812f6d43ea3f7daf', 'data-02', 1132],
     3: [2, '16741b673a418af3812f6d43ea3f7daf', 'data-03', 1132]},

 'a1e0f7ccdd8d38cb5ae00cdac71b6724':
    {8: [0, 'a1e0f7ccdd8d38cb5ae00cdac71b6724', 'data-08', 20125],
     9: [1, 'a1e0f7ccdd8d38cb5ae00cdac71b6724', 'data-09', 20125],
    10: [1, 'a1e0f7ccdd8d38cb5ae00cdac71b6724', 'data-10', 20125]},

 'cbef6de99cc2b9739c824db6d0246093':
    {4: [0, 'cbef6de99cc2b9739c824db6d0246093', 'data-04', 55296],
     5: [1, 'cbef6de99cc2b9739c824db6d0246093', 'data-05', 55296],
     6: [1, 'cbef6de99cc2b9739c824db6d0246093', 'data-06', 55296],
     7: [2, 'cbef6de99cc2b9739c824db6d0246093', 'data-07', 55296]}}

Answer 1

Your question is a bit unclear, what I understand is that you have {k1: {k2: [v1, v2, v3, v4]}} , you want to sort every top-level entry by v4 which should be the same in every list (so it doesn't matter which we pick). However the the sub-entries ( k2 ) are not constant between the top-level entries.

Getting v4 from a sub-entry is easy ( [3] or [-1] ) the issue is getting an arbitrary value of the second-level dict. next(iter(d.values())) ought do: iterate the sub-values (the lists), and get the first value out of the iterator. Not that this will raise an error if a sub-entry is empty (a top-level key maps to an empty dict).

So sorted(data.items(), key=lambda e: next(iter(e[1].values()))[-1]) should work:

[('16741b673a418af3812f6d43ea3f7daf',
  {1: [0, '16741b673a418af3812f6d43ea3f7daf', 'data-01', 1132],
   2: [1, '16741b673a418af3812f6d43ea3f7daf', 'data-02', 1132],
   3: [2, '16741b673a418af3812f6d43ea3f7daf', 'data-03', 1132]}),
 ('a1e0f7ccdd8d38cb5ae00cdac71b6724',
  {8: [0, 'a1e0f7ccdd8d38cb5ae00cdac71b6724', 'data-08', 20125],
   9: [1, 'a1e0f7ccdd8d38cb5ae00cdac71b6724', 'data-09', 20125],
   10: [1, 'a1e0f7ccdd8d38cb5ae00cdac71b6724', 'data-10', 20125]}),
 ('cbef6de99cc2b9739c824db6d0246093',
  {4: [0, 'cbef6de99cc2b9739c824db6d0246093', 'data-04', 55296],
   5: [1, 'cbef6de99cc2b9739c824db6d0246093', 'data-05', 55296],
   6: [1, 'cbef6de99cc2b9739c824db6d0246093', 'data-06', 55296],
   7: [2, 'cbef6de99cc2b9739c824db6d0246093', 'data-07', 55296]})]

Be aware that this will return a list of (key, value) tuples, not a dictionary. You'll have to feed it back to dict (ideally OrderedDict, possibly the regular dict in Python 3.6 or more recent) to keep the order:

{'16741b673a418af3812f6d43ea3f7daf': 
   {1: [0, '16741b673a418af3812f6d43ea3f7daf', 'data-01', 1132],
    2: [1, '16741b673a418af3812f6d43ea3f7daf', 'data-02', 1132],
    3: [2, '16741b673a418af3812f6d43ea3f7daf', 'data-03', 1132]},
 'a1e0f7ccdd8d38cb5ae00cdac71b6724': 
   {8: [0, 'a1e0f7ccdd8d38cb5ae00cdac71b6724', 'data-08', 20125],
    9: [1, 'a1e0f7ccdd8d38cb5ae00cdac71b6724', 'data-09', 20125],
    10: [1, 'a1e0f7ccdd8d38cb5ae00cdac71b6724', 'data-10', 20125]},
 'cbef6de99cc2b9739c824db6d0246093': {
    4: [0, 'cbef6de99cc2b9739c824db6d0246093', 'data-04', 55296],
    5: [1, 'cbef6de99cc2b9739c824db6d0246093', 'data-05', 55296],
    6: [1, 'cbef6de99cc2b9739c824db6d0246093', 'data-06', 55296],
    7: [2, 'cbef6de99cc2b9739c824db6d0246093', 'data-07', 55296]}}

Answer 2

Here's an ugly (and pretty inefficient) variant. It combines a dict comprehension, sorting, and getting the dict value corresponding to the 1 ^st of the keys (by (the ugly) d[list(d.keys())[0]] ):

 >>> data.keys() dict_keys(['16741b673a418af3812f6d43ea3f7daf', 'cbef6de99cc2b9739c824db6d0246093', 'a1e0f7ccdd8d38cb5ae00cdac71b6724']) >>> data_sorted = {k: v for k, v in sorted(data.items(), key=lambda x: x[1][list(x[1].keys())[0]][3])} >>> data_sorted.keys() dict_keys(['16741b673a418af3812f6d43ea3f7daf', 'a1e0f7ccdd8d38cb5ae00cdac71b6724', 'cbef6de99cc2b9739c824db6d0246093']) 

You've got KeyError in item1[1][1][3] because 1 ( the 2 ^nd one) only exists in the sub-dictionary of '16741b673a418af3812f6d43ea3f7daf' .

Your dict contains an awful amount of duplication. It could be trimmed down to something like (also making the sorting expression much simpler):

 >>> data = {("16741b673a418af3812f6d43ea3f7daf", 1132): ["data-01", "data-02", "data-03"], ... ("cbef6de99cc2b9739c824db6d0246093", 55296): ["data-04", "data-05", "data-06", "data-07"], ... ("a1e0f7ccdd8d38cb5ae00cdac71b6724", 20125): ["data-08", "data-09", "data-10"]} >>> >>> {k: v for k, v in sorted(data.items(), key=lambda x: x[0][1])}