不存在键的情况下字典的排序列表

Question

I have list of dictionary and want to sort by price in ascending order. But few dictionary objects have no price key available to sort. I want to move those object to end of the list.

I have tried below code

sorted(listObj, key=partial(sort_func, ['price']))

def sort_func(sort_by, listObj):
    fields = []
    for key in sort_by:
        if key in listObj:
            fields.append(listObj[key])
    return fields

And I want to sort below kind of data:

data = [
        {
           u'errorCode': u'500',
           u'errorMsg': u'Pfffbt. server down.',
        },
        {
           u'errorCode': u'500',
           u'errorMsg': u'Pfffbt. server down.',
        },
        { 
            u'Name': u'United Kingdon',
            u'price': 5222.0,
        },
        {
            u'price': 5237.0,
            u'Name': u'US',
        },
        {
            u'Name': u'General',
            u'price': 5283.0,
        }
    ]

while sorting these data using price ascending order, 5222.0 price should come first instead of errorcode object.

Note: I don't want to use lambda function because it could have multiple keys with some condition to sort using sort_func So I've to use sort_func anyhow.

I found one solution but it is not reliable.

def sort_func(sort_by, listObj):
    fields = []
    for key in sort_by:
        if key in listObj:
            fields.append(listObj[key])
        else:
            fields.append(99999999999999999999999999) # infinity number
    return fields

Any other suggestion with using sort_func ?

Answer 1

If you do not want to use lambda , you can try this:

new_data = sorted([i for i in data if "price" in i], reverse=True)+[i for i in data if "errorCode" in i]

Output:

[{u'price': 5222.0, u'Name': u'United Kingdon'}, {u'price': 5237.0, u'Name': u'US'}, {u'price': 5283.0, u'Name': u'General'}, {u'errorCode': u'500', u'errorMsg': u'Pfffbt. server down.'}, {u'errorCode': u'500', u'errorMsg': u'Pfffbt. server down.'}]

Answer 2

You could add a 'missing_key' counter and increment it for each missing key. return a tuple of (missing_keys, fields) and they'll get sorted at the end with the most missing keys going last.

def sort_func(sort_by, listObj):
    missing_keys = 0
    fields = []
    for key in sort_by:
        if key in listObj:
            fields.append(listObj[key])
        else:
            missing_keys += 1
    return missing_keys, fields

edit: request for detailed explanation

tuples(and lists) sort as follows: compare first elements. If first elements aren't equal, use that comparison, else compare 2nd elements etc.... an empty tuple/list is always less than one with elements. so:

lst = [
    (), (10,), (1, []), (2, [10, 11]), (2, [100]), (0, [10, 9, 8]), (0, [11]), (1, [1, 1, 2])
]
print(sorted(lst))

prints:

[
    (),
    (0, [10, 9, 8]),
    (0, [11]),
    (1, []),
    (1, [1, 1, 2]),
    (2, [10, 11]),
    (2, [100]),
    (10,)
]

This sort_func returns a tuple. If the first element (missing_keys) is equal, it's evaluated by the next element (fields). But it first sorts by the number of missing keys.