Pythonic way to construct a list filter

Question

I have a function for filtering a list of dict based on the value of certain keys, like a SELECT * WHERE xxx query in SQL

list_of_dict = [
{'key1':val, 'key2':val},
{'key1':val, 'key2':val},
...
]

def filter_list(list_of_dict, key1, key2=None):
    if key2:
         filtered_list = [i for i in list_of_dict if i['key1']==key1 and i['key2']==key2]
    else:
         filtered_list = [i for i in list_of_dict if i['key1']==key1]

but when I have more keys as arguments to the function the if ...else... could go really long.

Is there a more pythonic way to do this?

kaya3 · Accepted Answer · 2020-09-20 14:34:55Z

4

If you have a large or variable number of keys, you can use all to loop over them. Here's an example where the key values are provided as keyword arguments:

def filter_dicts(dicts, **keys):
    return [
        d for d in dicts
        if all(d[k] == v for k, v in keys.items())
    ]

As @juanpa.arrivillaga points out, dict_items objects behave like sets in many regards, so you can alternatively filter for dictionaries which have keys as a subset:

def filter_dicts(dicts, **keys):
    return [d for d in dicts if keys.items() <= d.items()]

Example:

>>> dicts = [{'x': 1, 'y': 2}, {'x': 1, 'y': 3}]
>>> filter_dicts(dicts, x=1)
[{'x': 1, 'y': 2}, {'x': 1, 'y': 3}]
>>> filter_dicts(dicts, x=1, y=2)
[{'x': 1, 'y': 2}]
>>> filter_dicts(dicts, y=3)
[{'x': 1, 'y': 3}]

edited Sep 20, 2020 at 14:34

answered Sep 20, 2020 at 14:14

kaya3

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3 Comments

juanpa.arrivillaga Over a year ago

consider: if keys.items() <= d.items()

kaya3 Over a year ago

@juanpa.arrivillaga I did not know you could use a dict_items object in that way; if you post it as an answer I will upvote.

juanpa.arrivillaga Over a year ago

Eh, the crux of the answer is in using **kwargs. You can just add it as an alternative. I'm trying to up the awareness of dict view objects as set-like objects.

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