I asked previously how to avoid code like this:
# Implement d[3][4] = 5
if not 3 in d:
d[3] = {}
d[3][4] = 5
I found the answer, and it lies in collections.defaultdict. defaultdict is just like dict, except that you pre-specify an initial default values for items that aren’t present. For example:
d = defaultdict(lambda: 1) d[5] += 3 print d
Will print out “{5: 4}” and I didn’t have to explicitly assign d[5] to 1 before doing the increment.
This leads to the question: Can you create a defaultdict whose default is another defaultdict? Yes!
d = defaultdict(defaultdict) d[3][4] = 5
Is totally valid. But, that inner defaultdict doesn’t have a good default value, so if you want more dimensions, you have to start stacking them up, like this:
d = defaultdict(lambda: defaultdict(defaultdict)) d[3][4][5] = 6
And so on, as you want more dimensions. Not a huge price to pay, I guess.
If you want arbitrarily and unevenly deep dictionaries, you can define a recursive defaultdict like this:
from collections import defaultdict
def deepDefaultDict():
return defaultdict(deepDefaultDict)
d = deepDefaultDict()
d[1] = “foo”
d[2][3][4][5] = “bar”
print d
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You can’t recursively call the dictionary but the link below contains a nice explanation.
http://stackoverflow.com/questions/2388302/best-way-to-define-multidimensional-dictionaries-in-python