Test if dict contained in dict

Question

Testing for equality works fine like this for python dicts:

first  = {\"one\":\"un\", \"two\":\"deux\", \"three\":\"trois\"}
second = {\"one\":\"un\", \"two\         


        

Answer 1

You can use a dictionary view:

# Python 2
if first.viewitems() <= second.viewitems():
    # true only if `first` is a subset of `second`

# Python 3
if first.items() <= second.items():
    # true only if `first` is a subset of `second`

Dictionary views are the standard in Python 3, in Python 2 you need to prefix the standard methods with view. They act like sets, and <= tests if one of those is a subset of (or is equal to) another.

Demo in Python 3:

>>> first  = {"one":"un", "two":"deux", "three":"trois"}
>>> second = {"one":"un", "two":"deux", "three":"trois", "foo":"bar"}
>>> first.items() <= second.items()
True
>>> first['four'] =  'quatre'
>>> first.items() <= second.items()
False

This works for non-hashable values too, as the keys make the key-value pairs unique already. The documentation is a little confusing on this point, but even with mutable values (say, lists) this works:

>>> first_mutable = {'one': ['un', 'een', 'einz'], 'two': ['deux', 'twee', 'zwei']}
>>> second_mutable = {'one': ['un', 'een', 'einz'], 'two': ['deux', 'twee', 'zwei'], 'three': ['trois', 'drie', 'drei']}
>>> first_mutable.items() <= second_mutable.items()
True
>>> first_mutable['one'].append('ichi')
>>> first_mutable.items() <= second_mutable.items()
False

You could also use the all() function with a generator expression; use object() as a sentinel to detect missing values concisely:

sentinel = object()
if all(first[key] == second.get(key, sentinel) for key in first):
    # true only if `first` is a subset of `second`

but this isn't as readable and expressive as using dictionary views.