CODEWITHSUNDEEP
pythondictionary
RoadRunner
RoadRunner

Reputation: 26315

Adding keys to defaultdict(dict)

I have a defaultdict that looks like this:

my_dict = defaultdict(dict)

which will print out:

defaultdict(<class 'dict'>, {})

I also have two lists, which look like this:

list1 =  ["W", "IY", "W"]
list2 =  ["w", "ee", "w"]

I would like to create a default dict which looks like this:

defaultdict(<class 'dict'>, {'W': {'w': 2}, 'IY': {'ee': 1}}

which has list1 within a dictionary as keys, with the keys as the next list with a separate dictionary, counting the instances of list2 as values.

So far I have this:

from collections import defaultdict

d = defaultdict(dict)

list1 = ["W", "IY", "W"]
list2 = ["w", "ee", "w"]

for char in list1:
    d[char] += 1

I know that this is not correct, as the defaultdict(dict) cannot be treated this way. Is there a way a I could do this? Any help would be greatly appreciated :)

Upvotes: 16

Views: 14056

Answers (4)

RootTwo
RootTwo

Reputation: 4418

Slightly different take on a solution:

import collections

phonemes  = ["W", "IY", "O", "W", "O"]
graphemes = ["w", "ee", "o", "w", "oh"]

# count all the (phoneme, grapheme) pairs
counter = collections.Counter(zip(phonemes, graphemes))

# convert to desired data structure
d = collections.defaultdict(dict)
for (phoneme, grapheme), count in counter.items():
    d[phoneme][grapheme] = count

print(d)

prints:

defaultdict(<class 'dict'>, {'W': {'w': 2}, 'O': {'oh': 1, 'o': 1}, 'IY': {'ee': 1}})

Upvotes: 2

user2390182
user2390182

Reputation: 73450

You can also use a nested defaultdict and zip like so:

d = defaultdict(lambda: defaultdict(int))
for k, v in zip(list1, list2):
    d[k][v] += 1
# d['TH']['th']: 1
# d['O']['o']: 2

or, if you want to keep your data structure:

d = defaultdict(dict)
for k, v in zip(list1, list2):
    d[k][v] = d[k].get(v, 0) + 1  
    # use dict.get(key, default=None) and specify an appropriate default value (0)

Using dict.get(key, default=None) allows you to access key-values of a common dict much like those a defaultdict, however, updating is a little more clunky.

Upvotes: 4

aberger
aberger

Reputation: 2397

EDITED based on the comment on my original answer.

You'll need a mapping of all possible phonemes to all possible spellings (graphemes).

phonemes = {TH : [th], O : [o], OH : [oh, oo]}

for char in set(list1):
    if char not in d:
        d[char] = {char.lower() : {phone : list2.count(phone) for phone in phonemes[char]}}

Upvotes: 3

wwii
wwii

Reputation: 23743

Here is a solution using collections.Counter.

import collections
d = collections.defaultdict(collections.Counter)

list1 = ["O", "TH", "O", "O"]
list2 = ["o", "th", "o", "o1"]

for key, value in zip(list1, list2):
    d[key].update([value])

>>> d
defaultdict(<class 'collections.Counter'>, {'TH': Counter({'th': 1}), 'O': Counter({'o': 2, 'o1': 1})})
>>>

While this doesn't strictly follow your requirements, collections.Counter inherits from dict so it has all of dict's attributes

Upvotes: 6

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