Python : passing a dictionnary as a parameter (sl4a)

Question

I'm learning python, working with the sl4a to make a phone app. I'm having trouble with a dictionnary when passing it as a parameter. I've searched to see if someone had that problem already and found some alike questions but couldn't figure my mistake out.

Here's the the code (I removed all unrelevant parts):

def choiceForm(screenTitle, choices, posMessage, neutMessage,
               negMessage, questType, screen):

    qType = questType[screen]
    print(qType) #Not giving what it should
    # More code

def reservForm():
     questionType = {0 : 'single', 1 : 'date', 2 : 'input', 3 : 'input', 
                     4 : 'multi', 5 : 'single'}
     reservChoice, screenID, success = choiceForm(titlePack[screenID], 
         choicePack[screenID], 'Suivant', 'Précédent', 'Quitter',
          questionType, screenID)
     # More code

On execution, i get "s" for qtype which isn't what should be there. I want to access my dictionary in the called function to be able to get the value, given an index (screenID). I read a tutorial telling about unpacking the thing with a ** operator but i didn't understand a thing. Everything worked fine before i changed my parameter in choiceForm to pass the dictionary (i was just passing one of it's value 'til i realized i'd need the whole thing for a feature i've to add)

Well i'm using the Scripting layer for android, (sl4a). This runs on my phone and it's android. The rest of the code shows it, should i post it ? — LionelM, May 10 '15 at 12:11
Post any necessary code to help people resolve your problem. If the rest of the code is not, don't. — Bojan Kseneman, May 10 '15 at 12:13
What do get and what do want to get? You talk about a list in the text but the code shows a dictionary access. — Mike Müller, May 10 '15 at 12:17
I'll edit my question then, i mean dictionary. I'm at the very start sorry. I get "s" when i should get "single" (with another index i could get "date", etc. — LionelM, May 10 '15 at 12:18
Make sure you are executing the code you are showing. Maybe you have some unsaved changes or use a different file. — Mike Müller, May 10 '15 at 12:36

score 0 · Answer 1 · answered May 10 '15 at 12:39

If you are talking about passing a Dictionary as a parameter, it should be as simple as passing any other parameters type and using this dictionary inside your calling function is as usual as you would work with regular dictionary:

dic.keys(), dic.values(), dic.items(), dic[key]=value...etc

Now, if you are talking about this notation: "**", this means passing parameters as dictionary, example from Python doc should make it clear:

def cheeseshop(kind, *arguments, **keywords):
    print "-- Do you have any", kind, "?"
    print "-- I'm sorry, we're all out of", kind
    for arg in arguments:
        print arg
    print "-" * 40
    keys = sorted(keywords.keys())
    for kw in keys:
        print kw, ":", keywords[kw]

Calling this function:

cheeseshop("Limburger", "It's very runny, sir.",
           "It's really very, VERY runny, sir.",
           shopkeeper='Michael Palin',
           client="John Cleese",
           sketch="Cheese Shop Sketch")

I suggest to you to read about it from: https://docs.python.org/2.7/tutorial/controlflow.html#keyword-arguments

Thanks for your response. What i wanted to know was how to pass it as a parameter and i now know it's as simple as any other type. I looked into the built-in function of the dictionary (i wasn't aware of their existence) and i found out my "questType" is a string and not a dictionary. I'm trying to find out why know but at least i'm not stuck anymore. Thanks everybody — LionelM, May 10 '15 at 13:05

Python : passing a dictionnary as a parameter (sl4a)

1 Answers1