按键排序字典列表。 如果 key 缺失，假设一个连续的编号

Question

I want so sort a list of dictionaries by the key "pos". However, if "pos" is missing in the dict, I want to keep the order of the item(s) and assume that "pos" is the item's 1-based index in the list.

This is working fine, as long as all list items are different:

L = [
    {   "id": "1" }, # assume pos: 1
    {   "id": "2" }, # assume pos: 2
    {   "id": "3" }, # assume pos: 3
    {   "id": "4" }, # assume pos: 4
    {   "id": "ZZZ" }, # assume pos: 5
    {   "id": "AAA" }, # assume pos: 6
    {   "id": "ABC", "pos": 3.2 },
    {   "id": "XYZ", "pos": 3.1 },
]

s = sorted(L,key=lambda i:i.get("pos",L.index(i)+1))
print(s)

Output:

[{'id': '1'}, {'id': '2'}, {'id': '3'}, {'id': 'XYZ', 'pos': 3.1}, {'id': 'ABC', 'pos': 3.2}, {'id': '4'}, {'id': 'ZZZ'}, {'id': 'AAA'}]

But it fails if I have multiple of the same items, because then list.index will return the first occurence, rather than the "assumed position".

L = [
    {   "id": "1" }, # assume pos: 1
    {   "id": "1" }, # assume pos: 2
    {   "id": "1" }, # assume pos: 3
    {   "id": "1" }, # assume pos: 4
    {   "id": "1" }, # assume pos: 5
    {   "id": "AAA" }, # assume pos: 6
    {   "id": "ABC", "pos": 3.2 },
    {   "id": "XYZ", "pos": 3.1 },
]

s = sorted(L,key=lambda i:i.get("pos",L.index(i)+1))
print(s)

Actual output:

[{'id': '1'}, {'id': '1'}, {'id': '1'}, {'id': '1'}, {'id': '1'}, {'id': 'XYZ', 'pos': 3.1}, {'id': 'ABC', 'pos': 3.2}, {'id': 'AAA'}]

Expected output:

[{'id': '1'}, {'id': '1'}, {'id': '1'}, {'id': 'XYZ', 'pos': 3.1}, {'id': 'ABC', 'pos': 3.2}, {'id': '1'}, {'id': '1'}, {'id': 'AAA'}]

How can the sorting be changed to return the expected output?

Note: the item IDs are not guaranteed to be in any order, that means 1,2,3,4,AAA,ABC,XYZ have been chosen arbitrarily.

Answer 1

Use enumerate :

L = [
    {"id": "1"},  # assume pos: 1
    {"id": "2"},  # assume pos: 2
    {"id": "3"},  # assume pos: 3
    {"id": "4"},  # assume pos: 4
    {"id": "ZZZ"},  # assume pos: 5
    {"id": "AAA"},  # assume pos: 6
    {"id": "ABC", "pos": 3.2},
    {"id": "XYZ", "pos": 3.1},
]

result = [e for _, e in sorted(enumerate(L, 1), key=lambda x: x[1].get("pos", x[0]))]

print(result)

Output

[{'id': '1'}, {'id': '2'}, {'id': '3'}, {'id': 'XYZ', 'pos': 3.1}, {'id': 'ABC', 'pos': 3.2}, {'id': '4'}, {'id': 'ZZZ'}, {'id': 'AAA'}]

For the duplicates example:

L = [
    {"id": "1"},  # assume pos: 1
    {"id": "1"},  # assume pos: 2
    {"id": "1"},  # assume pos: 3
    {"id": "1"},  # assume pos: 4
    {"id": "1"},  # assume pos: 5
    {"id": "AAA"},  # assume pos: 6
    {"id": "ABC", "pos": 3.2},
    {"id": "XYZ", "pos": 3.1},
]

result = [e for _, e in sorted(enumerate(L, 1), key=lambda x: x[1].get("pos", x[0]))]

print(result)

Output

[{'id': '1'}, {'id': '1'}, {'id': '1'}, {'id': 'XYZ', 'pos': 3.1}, {'id': 'ABC', 'pos': 3.2}, {'id': '1'}, {'id': '1'}, {'id': 'AAA'}]

A perhaps cleaner alternative is to use itertools.count :

from itertools import count

counter = count(1)

result = sorted(L, key=lambda x: x.get("pos", next(counter)))
print(result)

Answer 2

We can do something like this:

d = [
    {"id": "1"}, # assume pos: 1
    {"id": "1"}, # assume pos: 2
    {"id": "1"}, # assume pos: 3
    {"id": "1"}, # assume pos: 4
    {"id": "1"}, # assume pos: 5
    {"id": "AAA"}, # assume pos: 6
    {"id": "ABC", "pos": 3.2},
    {"id": "XYZ", "pos": 3.1},
]


def my_compare(x):
    if 'pos' in x[1]:
        return x[1]['pos']
    return x[0] + 1

sorted_d = [x[1] for x in sorted(enumerate(d), key=my_compare)]

expected_output = [
    {'id': '1'}, 
    {'id': '1'}, 
    {'id': '1'}, 
    {'id': 'XYZ','pos': 3.1}, 
    {'id': 'ABC','pos': 3.2}, 
    {'id': '1'}, 
    {'id': '1'}, 
    {'id': 'AAA'},
]
assert sorted_d == expected_output