使用元组的嵌套列表创建Python defaultdict

Question

The scenario is that I have a 2-D list. Each item of the inner list is tuple (key, value pair). The key might repeat in the list. I want to create a default-dict on the fly, in such a way that finally, the dictionary stores the key, and the cumulative sum of all the values of that key from the 2-D list.

To put the code :

listOfItems = [[('a', 1), ('b', 3)], [('a', 6)], [('c', 0), ('d', 5), ('b', 2)]]
finalDict = defaultdict(int)
for eachItem in listOfItems:
    for key, val in eachItem:
        finalDict[key] += val
print(finalDict)

This is giving me what I want : defaultdict(<class 'int'>, {'a': 7, 'b': 5, 'c': 0, 'd': 5}) but I am looking for a more 'Pythonic' way using comprehensions. So I tried the below :

finalDict = defaultdict(int)
finalDict = {key : finalDict[key]+val for eachItem in listOfItems for key, val in eachItem}
print(finalDict)

But the output is : {'a': 6, 'b': 2, 'c': 0, 'd': 5} What is it that I am doing wrong? Or is it that when using comprehension the Dictionary is not created and modified on the fly?

Answer 1

Yes a comprehension can't be updated on-the-fly. Anyway, this task might be better suited to collections.Counter() with .update() calls:

>>> from collections import Counter
>>> c = Counter()
>>> for eachItem in listOfItems:
...     c.update(dict(eachItem))
... 
>>> c
Counter({'a': 7, 'b': 5, 'd': 5, 'c': 0})

Answer 2

Simple solution without using additional modules:

inp_list = [[('a', 1), ('b', 3)], [('a', 6)], [('c', 0), ('d', 5), ('b', 2)]]

l = [item for sublist in inp_list for item in sublist] # flatten the list

sums = [(key, sum([b for (a,b) in l if a == key])) for key in dict(l)]

print(sums)

Answer 3

This is because you do not assign any value to your finalDict inside your dict in comprehension.

In your dict in comprehension you are literally changing the type of finalDict

As far as I know you cannot assign value to your dict inside a dict in comprehension.

Here is a way to get the dictionnary you want

from functools import reduce

listOfItems = [[('a', 1), ('b', 3)], [('a', 6)], [('c', 0), ('d', 5), ('b', 2)]]

list_dict = [{key: val} for eachItem in listOfItems for key, val in eachItem]

def sum_dict(x, y):
    return {k: x.get(k, 0) + y.get(k, 0) for k in set(x) | set(y)}
print(reduce(sum_dict, list_dict))

Answer 4

trying to use python's built-in methods instead of coding the functionality myself:

The long and explained solution

from itertools import chain, groupby
from operator import itemgetter

listOfItems = [[('a', 1), ('b', 3)], [('a', 6)], [('c', 0), ('d', 5), ('b', 2)]]

# just flat the list of lists into 1 list..
flatten_list = chain(*listOfItems)

# get all elements grouped by the key, e.g 'a', 'b' etc..
first = itemgetter(0)
groupedByKey = groupby(sorted(flatten_list, key=first), key=first))

#sum
summed_by_key = ((k, sum(item[1] for item in tups_to_sum)) for k, tups_to_sum in groupedByKey)

# create a dict
d = dict(summed_by_key)

print(d) # {'a': 7, 'b': 5, 'c': 0, 'd': 5}

~one line solution

from itertools import chain, groupby
from operator import itemgetter

first = itemgetter(0)
d = dict((k, sum(item[1] for item in tups_to_sum)) for k, tups_to_sum in groupby(sorted(chain(*listOfItems), key=first), key=first))

print(d) # {'a': 7, 'b': 5, 'c': 0, 'd': 5}