dictionary – How to set default value to all keys of a dict object in python?-ThrowExceptions

Exception or error:

I know you can use setdefault(key, value) to set default value for a given key, but is there a way to set default values of all keys to some value after creating a dict ?

Put it another way, I want the dict to return the specified default value for every key I didn’t yet set.

How to solve:

You can replace your old dictionary with a defaultdict:

>>> from collections import defaultdict
>>> d = {'foo': 123, 'bar': 456}
>>> d['baz']
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
KeyError: 'baz'
>>> d = defaultdict(lambda: -1, d)
>>> d['baz']
-1

The “trick” here is that a defaultdict can be initialized with another dict. This means
that you preserve the existing values in your normal dict:

>>> d['foo']
123

Answer:

Use defaultdict

from collections import defaultdict
a = {} 
a = defaultdict(lambda:0,a)
a["anything"] # => 0

This is very useful for case like this,where default values for every key is set as 0:

results ={ 'pre-access' : {'count': 4, 'pass_count': 2},'no-access' : {'count': 55, 'pass_count': 19}
for k,v in results.iteritems():
  a['count'] += v['count']
  a['pass_count'] += v['pass_count']

Answer:

In case you actually mean what you seem to ask, I’ll provide this alternative answer.

You say you want the dict to return a specified value, you do not say you want to set that value at the same time, like defaultdict does. This will do so:

class DictWithDefault(dict):
    def __init__(self, default, **kwargs):
        self.default = default
        super(DictWithDefault, self).__init__(**kwargs)

    def __getitem__(self, key):
        if key in self:
            return super(DictWithDefault, self).__getitem__(key)
        return self.default

Use like this:

d = DictWIthDefault(99, x=5, y=3)
print d["x"]   # 5
print d[42]    # 99
42 in d        # False
d[42] = 3
42 in d        # True

Alternatively, you can use a standard dict like this:

d = {3: 9, 4: 2}
default = 99
print d.get(3, default)  # 9
print d.get(42, default) # 99

Answer:

Is this what you want:

>>> d={'a':1,'b':2,'c':3}
>>> default_val=99
>>> for k in d:
...     d[k]=default_val
...     
>>> d
{'a': 99, 'b': 99, 'c': 99}
>>> 

>>> d={'a':1,'b':2,'c':3}
>>> from collections import defaultdict
>>> d=defaultdict(lambda:99,d)
>>> d
defaultdict(<function <lambda> at 0x03D21630>, {'a': 1, 'c': 3, 'b': 2})
>>> d[3]
99

Answer:

Not after creating it, no. But you could use a defaultdict in the first place, which sets default values when you initialize it.

Answer:

defaultdict can do something like that for you.

Example:

>>> from collections import defaultdict
>>> d = defaultdict(list)
>>> d
defaultdict(<class 'list'>, {})
>>> d['new'].append(10)
>>> d
defaultdict(<class 'list'>, {'new': [10]})

Answer:

You can use the following class. Just change zero to any default value you like. The solution was tested in Python 2.7.

class cDefaultDict(dict):
    # dictionary that returns zero for missing keys
    # keys with zero values are not stored

    def __missing__(self,key):
        return 0

    def __setitem__(self, key, value):
        if value==0:
            if key in self:  # returns zero anyway, so no need to store it
                del self[key]
        else:
            dict.__setitem__(self, key, value)

Leave a Reply

Your email address will not be published. Required fields are marked *