Question

This is basically what I’m attempting to do, which isn’t working. Is there a way to single out a value of the dictionary and mathify it? For the sake of a crap example:

dictionaryNumbers = {'a':10,'b':10,'c':1,'d':1,'e':5,'f':1}
dictionaryNumbers['a'] += 5
#The goal would be dictionaryNumbers['a'] would equal 15.

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EDIT:

Guys thanks for the feedback. It seems there was a flaw in the order in which I was calling the functions to modify the collection. I was printing the output before the math took place. Thanks again.

Answer 1

You are mostly doing it right, and your code is working fine:

>>> dictionaryNumbers = {'a':10,'b':10,'c':1,'d':1,'e':5,'f':1}
>>> dictionaryNumbers['a'] += 5
>>> dictionaryNumbers['a']
15

but for any key not yet in the dict you’d have to test first (if key not in dictionaryNumbers) or use .get():

>>> dictionaryNumbers['z'] = dictionaryNumbers.get('z', 0) + 3

which gets old fast.

But I’d use a collections.Counter() class instead:

>>> from collections import Counter
>>> counter = Counter()
>>> counter.update({'a':10,'b':10,'c':1,'d':1,'e':5,'f':1})
>>> counter
Counter({'a': 10, 'b': 10, 'e': 5, 'c': 1, 'd': 1, 'f': 1})
>>> counter['a'] += 5
>>> counter['a'] 
15
>>> counter.most_common(3)
[('a', 15), ('b', 10), ('e', 5)]

Advantages:

• No need to test for new keys; accessing a nonexisting key automatically assigns a count of 0
• Creating a new counter from a list of items to count is as easy as Counter(items_to_count).
• You can sum counters; counter1 + counter2 returns a new Counter with all values summed.
• You can subtract counters (negative values are removed), intersect them (find the minimum of either count), or create a union (maximum counts).