> ok so assuming I had a dictionary with 1key that contained a list like
> so... dictionary[key1][0]
>
> How would I increment it or assign it if it didn't exist. I assumed like
> this. dict[key1][0] = dictionary.get(key1[0],0) + X
Hi Paul,
Given a dictionary d and some arbitrary key k, let's assume two
possibilities:
1. k is a key in d. If so, no problem, and we'll assume that d[k] is
associated to some twelve-element list.
2. k isn't yet a key in d. This is the situation we want to work on.
We can go the uncomplicated route and just say something like this:
if k not in d:
d[k] = [0] * 12
in which case we change Possibility 2 to Possibility 1: the end result is
that we make sure that d[k] points to a twelve-element list of zeros.
Once we guarantee this, we can go on our merry way.
For example:
if k not in d:
d[k] = [0] * 12
d[k][0] += 1
There is a one-liner way of doing this: we can use the setdefault() method
of dictionaries.
d.setdefault(k, [0] * 12)[0] += 1
This is a bit dense. There's no shame in not using setdefault() if you
don't want to. *grin* I actually prefer the longer approach unless I'm
really in a hurry.
This being said, what data are you modeling? It almost sounds like you're
implementing some kind of 3d matrix, even a sparse one. More information
about the data you're modelling might lead to a different representation
For example, would using a dictionary whose keys are three-tuples be
approrpriate?
######
def increment(d, x, y, z):
"""Given a dictionary whose keys are 3-tuples, increments at the
position (x, y, z)."""
d[(x, y, z)] = d.get((x, y, z), 0) + 1
######
Here's this code in action:
###
>>> map = {}
>>> increment(map, 3, 1, 4)
>>> map
{(3, 1, 4): 1}
>>> increment(map, 2, 7, 1)
>>> map
{(2, 7, 1): 1, (3, 1, 4): 1}
>>> increment(map, 2, 7, 1)
>>> map
{(2, 7, 1): 2, (3, 1, 4): 1}
###
Best of wishes to you!
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