Add Counters keeping zero values

Question

I try to sum the value of a key present in other dictionaries with this code:

import functools
import operator
import collections

my_dict = [{'a':0, 'b':1, 'c':5}, {'b':3, 'c':2}, {'b':1, 'c':1}]
sum_key_value = functools.reduce(operator.add, map(collections.Counter, my_dict))

print(sum_key_value)

# Output
# Counter({'c': 8, 'b': 5})

My question is if I want the output to keep all dictionary keys, even if the key does not appear in all the dictionaries like a in my case, what is the best way to do that without using a loop ?

https://stackoverflow.com/questions/3490738/how-to-sum-dict-elements --- DRY — Thomas John, Jul 24 '20 at 07:43
It is stated in the Python documentation that operating an addition on `Counter` obects (which is what is happening in your code) will remove the values with 0 : *Each operation can accept inputs with signed counts, but the output will exclude results with counts of zero or less* (https://docs.python.org/3/library/collections.html#collections.Counter) — mgc, Jul 24 '20 at 07:47

Ayush · Accepted Answer · 2020-07-24T08:04:32.503

2

Well there's a lot of nice ways to do it with a for loop, but since you specifially want to avoid a for loop, here's one way:

sum_key_value = dict(functools.reduce(lambda a, b: a.update(b) or a,
                     my_dict, collections.Counter()))

So what happens here is you create a single Counter, and use it to accumulate the values.

edited Jul 24 '20 at 08:04

answered Jul 24 '20 at 07:46

Ayush

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Can you explain what the `or a` is for? – Tomer Cagan Dec 25 '22 at 10:24
1

Sure, the lambda takes two params and returns a value. However, `.update()` doesn't return anything, or rather returns `None`. So, `a.update(b)` will fill `a` with `b`'s content, return `None`, so the "or a" will then finally return the updated `a` – Ayush Dec 26 '22 at 17:11

yatu · Answer 2 · 2020-07-24T08:46:33.850

As mentioned in the comments, adding Counter objects will remove non positive keys.

So the issue is not really about not ending up with the union of all keys (as well as adding common values), since that is indeed the behaviour, see if we set a:2:

my_dict = [{'a':2, 'b':1, 'c':5}, {'b':3, 'c':2}, {'b':1, 'c':1}]
functools.reduce(operator.add, map(Counter, my_dict))
# Counter({'a': 2, 'b': 5, 'c': 8})

However, as shown in the question, as per the current implementation when adding Counter objects, non positive values (a:0) get removed.

If you really wanted to use Counter for this, you could tweak a little the current implementation overriding __add__ to get the expected behaviour:

class Counter_tweaked(Counter):
    def __add__(self, other):
            if not isinstance(other, Counter):
                return NotImplemented
            result = Counter_tweaked()
            for elem, count in self.items():
                newcount = count + other[elem]
                result[elem] = newcount
            for elem, count in other.items():
                if elem not in self:
                    result[elem] = count
            return result

functools.reduce(operator.add, map(Counter_tweaked, my_dict))
# Counter_tweaked({'a': 0, 'b': 5, 'c': 8})

score 0 · Answer 3 · answered Jul 24 '20 at 07:56

The most straightforward approach, here, would be a loop. You might have keys appearing in dictionaries anywhere in the list (e.g.: the third one could have the key "e"), so you would need at least one loop to get the total of keys. And then you can just loop through all dictionaries again to sum up the values. Make an own function of it and you can call it without ever caring about loops again.

def sum_it_up(dictlist):
    outdic = {}
    for d in dictlist:
        for k in d.keys():
            outdic[k] = 0
    for d in dictlist:
        for k in d.keys():
            outdic[k]+=d[k]
    return outdic

my_dict = [{'a':0, 'b':1, 'c':5}, {'b':3, 'c':2}, {'b':1, 'c':1}]
sum_key_value = sum_it_up(my_dict)

CN_Cabbage · Answer 4 · 2022-04-09T02:52:29.587

Subtract（or Update) a zero value Counter to keep zero value items

my_dict = [{'a':0, 'b':1, 'c':5}, {'b':3, 'c':2}, {'b':1, 'c':1}]

sum_my_dict = Counter()
zero_dict = Counter()

my_dict_keys = set()

# get the keys of all dict
for dic in my_dict:
    sum_my_dict += Counter(dic)
    my_dict_keys.update(dic.keys())

# create a dict with all zero values
for key in my_dict_keys:
    zero_dict[key] = 0

# in-place subtract zero dict (alter sum_my_dict)
sum_my_dict.subtract(zero_dict)

# in-plact update is the same with subtract
# sum_my_dict.update(zero_dict)


print(sum_my_dict)

Counter({'c': 8, 'b': 5, 'a': 0})

Add Counters keeping zero values

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