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Groups > comp.lang.python > #3881
| Date | 2011-04-22 16:18 +0100 |
|---|---|
| From | MRAB <python@mrabarnett.plus.com> |
| Subject | Re: suggestions, comments on an "is_subdict" test |
| References | <mailman.747.1303480525.9059.python-list@python.org> <4db19756$0$81473$e4fe514c@news.xs4all.nl> |
| Newsgroups | comp.lang.python |
| Message-ID | <mailman.754.1303485531.9059.python-list@python.org> (permalink) |
On 22/04/2011 15:57, Irmen de Jong wrote:
> On 22-4-2011 15:55, Vlastimil Brom wrote:
>> Hi all,
>> I'd like to ask for comments or advice on a simple code for testing a
>> "subdict", i.e. check whether all items of a given dictionary are
>> present in a reference dictionary.
>> Sofar I have:
>>
>> def is_subdict(test_dct, base_dct):
>> """Test whether all the items of test_dct are present in base_dct."""
>> unique_obj = object()
>> for key, value in test_dct.items():
>> if not base_dct.get(key, unique_obj) == value:
>> return False
>> return True
>>
>> I'd like to ask for possibly more idiomatic solutions, or more obvious
>> ways to do this. Did I maybe missed some builtin possibility?
>
>
> I would use:
>
> test_dct.items()<= base_dct.items()
>
In Python 2:
>>> test_dct = {"foo": 0, "bar": 1}
>>> base_dct = {"foo": 0, "bar": 1, "baz": 2}
>>>
>>> test_dct.items() <= base_dct.items()
False
In Python 3:
>>> test_dct = {"foo": 0, "bar": 1}
>>> base_dct = {"foo": 0, "bar": 1, "baz": 2}
>>> test_dct.items() <= base_dct.items()
True
YMMV
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suggestions, comments on an "is_subdict" test Vlastimil Brom <vlastimil.brom@gmail.com> - 2011-04-22 15:55 +0200
Re: suggestions, comments on an "is_subdict" test Irmen de Jong <irmen.NOSPAM@xs4all.nl> - 2011-04-22 16:57 +0200
Re: suggestions, comments on an "is_subdict" test MRAB <python@mrabarnett.plus.com> - 2011-04-22 16:18 +0100
Re: suggestions, comments on an "is_subdict" test Raymond Hettinger <python@rcn.com> - 2011-04-23 00:23 -0700
Re: suggestions, comments on an "is_subdict" test Paul Rubin <no.email@nospam.invalid> - 2011-04-23 00:26 -0700
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