如何在多级字典中附加到列表？

Question

I'm traversing through a file and want to build a dynamic dictionary of multiple levels. The last level needs to store a list of values.

myDict = defaultdict(dict)

for key in lvlOneKeys: # this I know ahead of time so I set up my dictionary with first level. I'm not sure if this is necessary.
    myDict[key] = {}

with open(myFile, "rb") as fh:
    for line in fh:
        # found something, which will match lvlOneKey and dynamically determine lvlTwoKey and valueFound
        # ...
        myDict[lvlOneKey][lvlTwoKey].append(valueFound)

I need to do this because lvlTwoKey will be found multiple times with different valueFound's.

Unfortunately this code results in a KeyError for lvlOneKey. What am I doing wrong?

Answer 1

This is an almost foolproof way of ensuring you won't get an error. The way we have defined myDict , you can have any keys for dictionary "level 1" and dictionary "level 2". By default, an empty list is assumed at the end of the dictionary tree.

myDict = defaultdict(lambda: defaultdict(list))

with open(myFile, "rb") as fh:
    for line in fh:
        # found something, which will match lvlOneKey and dynamically determine lvlTwoKey and valueFound
        # ...
        myDict[lvlOneKey][lvlTwoKey].append(valueFound)

Answer 2

Replacing the code under the for loop with the following should solve your problem:

# if there is no `lvlTwoKey` in `myDict`, initialize a list with `valueFound` in it  
if not myDict[lvlOneKey].get(lvlTwoKey, None):
    myDict[lvlOneKey][lvlTwoKey] = [valueFound]
# otherwise, append to the existing list 
else: 
    myDict[lvlOneKey][lvlTwoKey].append(valueFound)

This uses the get() method on the dictionary, which you can read about here . Besides that, it's just a standard dictionary, which I find often to be more readable / intuitive than complex defaultdicts.

Answer 3

This can get a bit annoying if you get much further nested, but using this pattern will work.

l1keys = [1, 2, 1]
l2keys = ['foo', 'bar', 'spangle']

base = {}
for l1key, l2key in zip(l1keys, l2keys):
    for i in range(5):
        l1 = base.get(l1key, {})
        l2 = l1.get(l2key, [])
        l2.append(i)
        l1[l2key] = l2
        base[l1key] = l1

assert base == {
    1: {'foo': [0, 1, 2, 3, 4], 'spangle': [0, 1, 2, 3, 4]},
    2: {'bar': [0, 1, 2, 3, 4]}
}