define a function for max score from a dictionary in python

Question

I am working on the following, I created a dictionary object as follows and I need to define 3 different functions. A function for maxScore, a function for minScore, and function for average as well as print all 3.

My dictionary object code is as follows

scores = {}
scores['Andy'] = 78
scores['Bill'] = 82
scores['Cindy'] = 94
scores['Dave'] = 77
scores['Emily'] = 82
scores['Frank'] = 94
scores['Gene'] = 87

I can't seem to define the function(s) correctly for it to print the results.

Possible duplicate of [Getting key with maximum value in dictionary?](https://stackoverflow.com/questions/268272/getting-key-with-maximum-value-in-dictionary) — Dadep, Jul 12 '17 at 17:50
@AGNGazer Without it (or something similar), `scores['Andy'] = 78` will raise a `NameError`. — chepner, Jul 12 '17 at 17:55

Dadep · Answer 1 · 2017-07-12T17:59:03.540

5

use min() and max :

>>> scores = {}
>>> scores['Andy'] = 78
>>> scores['Bill'] = 82
>>> scores['Cindy'] = 94
>>> scores['Dave'] = 77
>>> scores['Emily'] = 82
>>> scores['Frank'] = 94
>>> scores['Gene'] = 87

>>> max(scores, key=scores.get)
'Frank'
>>> min(scores, key=scores.get)
'Dave'

and for the average :

>>> sum(scores[i] for i in scores)/len(scores)
84

edited Jul 12 '17 at 17:59

answered Jul 12 '17 at 17:52

Dadep

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rma · Answer 2 · 2017-07-12T17:56:34.403

1

For max:

def max_val(scores):
    return max(scores.values())

min

def min_val(scores):
    return min(scores.values())

and avg

    def avg_val(scores):
        return sum(scores.values())/len(scores)

edited Jul 12 '17 at 17:56

answered Jul 12 '17 at 17:51

rma

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If `min` and `max` are both just returning a value, you don't need to know the key. Just use `max(scores.values())`. (Also, you've inadvertently created an infinite recursion situation by re-using `min` and `max` as your function names.) – chepner Jul 12 '17 at 17:53
You're completely right. That's embarrassing. Edited answer to reflect changes. – rma Jul 12 '17 at 17:55
Ditto for `avg_val`; you don't need the key. You don't need the loop, either: `sum(scores.values())` gets the sum. Although, it is debatable which is better: compute a running calculation of the length while iterating, or iterate twice implicitly (in `sum` and `len`). – chepner Jul 12 '17 at 17:56
I noticed that as well and fixed it. Thanks for the feedback! – rma Jul 12 '17 at 17:57
Thank you, that worked, I am just starting out and this one tripped me up. – E-Arsenal Jul 12 '17 at 18:21
No problem! Make sure to accept someone's answer so people know the question is resolved :) – rma Jul 12 '17 at 18:24

anugrah · Answer 3 · 2017-07-12T18:02:59.717

Max Score:

def maxScore(scDict):
  max = scDict[0]
  for i in scDict:
    if scDict[i]>max:
      max=scDict[i]
  return max

Min Score:

def minScore(scDict):
  min = scDict[0]
  for i in scDict:
    if scDict[i]<max:
      min=scDict[i]
  return min

Avg Score:

def avgScore(scDict):
  avg = 0. # python 2
  for i in scDict:
    avg+=scDict[i]
  return avg/len(scDict)

You can also use something like sys.maxsize in case of minScore (Py 3.0) if you desire.

efirvida · Answer 4 · 2017-07-12T18:03:09.853

0

try this:

best_scores = sorted(scores, key=scores.get, reverse=True)
maxScore = scores[best_scores[0]]
minScore = scores[best_scores[-1]]
average =  sum(scores.values())/len(scores.values())

edited Jul 12 '17 at 18:03

answered Jul 12 '17 at 17:57

efirvida

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score 0 · Answer 5 · answered Jul 12 '17 at 17:58

0

sorted(scores.items(), cmp=lambda x, y: x[1] - y[1])

answered Jul 12 '17 at 17:58

AGN Gazer

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define a function for max score from a dictionary in python

5 Answers5