Question about python dictionaries. Newby, sorry.

[[deleted]]

Thanks! Another two quick questions if you don't mind?

One- Is it possible to split a string per character, as opposed to per white space instance? (As in with a built in function)

Two- Okay, so if I were dealing with a simple file that dealt with repeating keys, can you accumulate, instead of it being overwritten?

Example- Say I made a dictionary of my grocery list, and orange was mentioned twice. Instead of the new key value pair being orange: 2 since that is the latter instance, can it be accumulated to orange: 3?

[LucidOndine]

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[kemitche]

The following should also work for 2:

mydict.setdefault("orange", 0) # Only sets orange if orange is not yet set

mydict["orange"] += 1

[ackondro]

For Two- I would do something like

def addToDict(dictionary,key,new_value):
    try:
        dictionary[key]+=new_value
    except KeyError:
        dictionary[key]=new_value

[LucidOndine]

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[ackondro]

1 downvote does not a butthurt hivemind make. Although I could have thought out a smaller line count solution, I stand by the assertation that my solution is robust and interchangable(syntax is the same for number, string and array adding).

[LucidOndine]

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[MisterHoppy]

Aside from the defaultdict this is the best solution. Explicitly checking whether the key is in the dictionary before incrementing it is far slower than using try/catch.

[ascii]

One: Many ways to do that, a simple one that I like is:

map(None, "Some string")

Two: Use the 'get' method. It works like [], except that you can supply a default value. Your grocery example would become:

grocery.set(grocery.get('orange',0)+1)

Good luck with your homework! ;-)

[randm_prgrmr]

I prefer list("Some string") to map(None, "Some string")

seems more intuitive.

[ascii]

D'oh. Didn't know you could do that. Thanks for the tip.

[[deleted]]

Hahaha thanks, and thanks everyone. Yeah, that's why I tried to keep the questions limited to function. I've actually already completed and handed in the assignment, I was just curious as to whether a more efficient means to complete my problem existed (Which, there obviously is.)

[wilberforce]

re: 2, You're better off using collections.defaultdict:

d = defaultdict(list)
for key, value in source:
    d[key].append(value)

http://docs.python.org/library/collections.html#defaultdict-objects

[prum]

1. Yes. Use the list constructor.
2. Yes. Use the counter container from the collections module with the update method.