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paulhmorin
Group Title
In the following code, HOW can the "average" function work with no visible input passed to it? How does it know where "stuff" points to? The code works I just don't know why.
def average(stuff):
return sum(stuff)/len(stuff)
def getAverage(kid):
bar = average
return bar(kid["homework"])*.1 + bar(kid["quizzes"])*.3 + bar(kid["tests"])*.6
studavg = getAverage(x)
 2 years ago
 2 years ago
paulhmorin Group Title
In the following code, HOW can the "average" function work with no visible input passed to it? How does it know where "stuff" points to? The code works I just don't know why. def average(stuff): return sum(stuff)/len(stuff) def getAverage(kid): bar = average return bar(kid["homework"])*.1 + bar(kid["quizzes"])*.3 + bar(kid["tests"])*.6 studavg = getAverage(x)
 2 years ago
 2 years ago

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Screech Group TitleBest ResponseYou've already chosen the best response.0
Its not a good programming technique (at least IMHO), but the line bar = average creates an alias for the average function within getAverage(). My guess is that the formula used specified variables with a bar over the top, which in statistics traditionally means average or mean. Its pretty confusing, but within getAverage(), bar and average mean exactly the same thing.
 2 years ago

paulhmorin Group TitleBest ResponseYou've already chosen the best response.0
here is what I get when I print bar after bar = average <function average at 0x7fbc6280bc80>
 2 years ago

Screech Group TitleBest ResponseYou've already chosen the best response.0
Sounds likely. You'd probably get the exact same output from print average So, when getAverage invokes bar(...), its just as if it had invoked average() Why make it so confusing? Saves typing 4 characters. pfffft!!
 2 years ago

hanleybrand Group TitleBest ResponseYou've already chosen the best response.1
How it works  first "stuff" needs to be an iterable object (the obvious one with this code is dict), so when the line studavg = getAverage(x) is interpreted, x needs to be a dict (so getAverage can use the key (e.g. "homework") to pick the correct answer  and this brings up an additional point, it only makes sense if x is is a dict of lists. I'm going to change x to scoreDict for accuracy: studentScores = { "homework" : [25, 50, 75], "quizzes" : [25, 75], "tests" : [25,50,75,100] } def average(scoreList): return sum(scoreList) / len(scoreList) # fix getAverage for clarity def getAverage(scoreDict): h = kid["homework"] q = kid["quizzes"] t = kid["tests"] return average(h) * .1 + average(q) * .3 + average(t) * .6 print getAverage(studentScores) # 57.2 So average() does have visible input being passed to it  getAverage() invokes it 3 times, once for each key in studentScores. Is that any clearer?
 2 years ago

hanleybrand Group TitleBest ResponseYou've already chosen the best response.1
# Sorry, I pasted in the unfinished fix  this will run without error: studentScores = {"homework": [25, 50, 75], "quizzes": [25, 75], "tests": [25, 50, 75, 100]} def average(scoreList): return sum(scoreList) / len(scoreList) # fix getAverage for clarity def getAverage(scoreDict): h = scoreDict["homework"] q = scoreDict["quizzes"] t = scoreDict["tests"] return average(h) * .1 + average(q) * .3 + average(t) * .6 print getAverage(studentScores) # 57.2
 2 years ago

paulhmorin Group TitleBest ResponseYou've already chosen the best response.0
Thanks for your help. Hate it when programmers depend on "hidden" functions. Your code is self explanatory. bar = average is not.
 2 years ago
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