以元组为值的字典比较

Question

I'm creating a simple dictionary comparison to search and match items. In this basic forms it works as expected.

def findOrders(lst, **kwargs):
    return [k for k in lst if kwargs.items() <= k.items()]

lst = [i for i in [ {'type':'esimene', 'id':1},
                    {'type':'teine', 'id':2},
                    {'type':'kolmas', 'id':3}]]

print(findOrders(lst, type='esimene'))

Now I would like to compare/match by multiple 'types' using a tuple as a keyword/values:

findOrders(lst, type=('esimene', 'teine'))

What would be the best way to do this?

Could I create a generator that flattens the tuple to 'type':'esimene' and 'type':'teine' and then runs the comparison? What would it look like?

Efficiency is also important this function will have heavy usage.

Answer 1

You could check if the the type parameter contains the type property of each item:

In [8]: def findOrders(lst, **kwargs):
   ...:     return [k for k in lst if k["type"] in kwargs["type"]]

In [9]: print(findOrders(lst, type='esimene'))
[{'type': 'esimene', 'id': 1}]

In [10]: print(findOrders(lst, type=('esimene',  'teine')))
[{'type': 'esimene', 'id': 1}, {'type': 'teine', 'id': 2}]

If you want to keep the current logic with the id comparison as well:

In [32]: def findOrders(lst, **kwargs):
    ...:     return [k for k in lst if k["type"] in kwargs.get("type", (k["type"], )) and k["id"] ==
    ...: kwargs.get("id", k["id"])]
    ...:
    ...:
    ...:

In [33]: print(findOrders(lst, id=1))
[{'type': 'esimene', 'id': 1}]

In [34]: print(findOrders(lst, type=('esimene',  'teine')))
[{'type': 'esimene', 'id': 1}, {'type': 'teine', 'id': 2}]

In [35]: print(findOrders(lst, type=('esimene', 'teine'), id=2))
[{'type': 'teine', 'id': 2}]