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Groups > comp.lang.python > #3871

Re: suggestions, comments on an "is_subdict" test

Date 2011-04-22 15:29 +0100
From MRAB <python@mrabarnett.plus.com>
Subject Re: suggestions, comments on an "is_subdict" test
References <BANLkTinazu41uVqxmkfTns0ZOHScZ1a3nQ@mail.gmail.com>
Newsgroups comp.lang.python
Message-ID <mailman.748.1303482591.9059.python-list@python.org> (permalink)

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On 22/04/2011 14:55, 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 am unsure whether the check  against an unique object() or the
> negated comparison are usual.?
> (The builtin exceptions are ok, in case anything not dict-like is
> passed. A cornercase like>>>  is_subdict({}, 4)
>>>> True
> doesen't seem to be worth a special check just now.)
>
You could shorten it slightly to:

def is_subdict(test_dct, base_dct):
     """Test whether all the items of test_dct are present in base_dct."""
     unique_obj = object()
     return all(base_dct.get(key, unique_obj) == value for key, value in 
test_dct.items())

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Re: suggestions, comments on an "is_subdict" test MRAB <python@mrabarnett.plus.com> - 2011-04-22 15:29 +0100

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