I have a dict like this: dict_1 = {'a': 1, 'b': 2, 'c' : 3}
. I have a list of dict keys like list_1 = ['a', 'b', 'c', '*']
, where the *
could be any value. I would like to create a dict that can handle *
and assign it a value of 0
Any ideas if this is possible?
Why don't you just use the standard dict.get
method?
If you use d.get(key, 0)
instead of d[key]
it will give you the behaviour you want.
Example:
dict_1 = {'a': 1, 'b': 2, 'c' : 3}
dict_1.get('a', 0) # returns 1
dict_1.get('Z', 0) # returns 0
You seem to be describing python's inbuilt defaultdict .
In your example, you could do the following;
from collections import defaultdict
dict_1 = defaultdict(lambda: 0)
dict_1["a"] = 1
print(dict_1["a"]) # will print 1 as that's what is set.
print(dict_1["any key"]) # will print 0, as it hasn't been set before.
Because you want a default of 0
specifically, you could also use defaultdict(int)
as int()
returns 0
.
# if key not in dict_1 will print 0
for key in list_1:
print(dict_1.get(key, 0))
You can create the new dictionary for your requirement in following way:
dict_1 = {'a': 1, 'b': 2, 'c' : 3}
list_1 = ['a', 'b', 'c', '*']
# new dictonay with default value 0 for non-existing key in `dict_1`
new_dict = {l: dict_1.get(l, 0) for l in list_1}
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