How can I change the output? (Python, lists, dictionaries help)

Question

Possible Duplicate:
changing the output

This is the code:

def voting_borda(args):
    results = {}
    for sublist in args:
        for i in range(0, 3):
            if sublist[i] in results:
                results[sublist[i]] += 3-i
            else:
                results[sublist[i]] = 3-i

    winner = max(results, key=results.get)
    return winner, results

print(voting_borda(
    ['GREEN','NDP', 'LIBERAL', 'CPC'],
    ['GREEN','CPC','LIBERAL','NDP'],
    ['LIBERAL','NDP', 'CPC', 'GREEN']
))

The output produced is

"('GREEN', {'LIBERAL': 5, 'NDP': 4, 'GREEN': 6, 'CPC': 3})"

I don't want the party names in the output (liberal, ndp, green and cpc) I just need the values, How can I edit the code to achieve that?

edit:

the error message i got after testing the above code (with: >>>voting_borda([['NDP', 'CPC', 'GREEN', 'LIBERAL'],['NDP', 'CPC', 'LIBERAL', 'GREEN'],['NDP', 'CPC', 'GREEN', 'LIBERAL']])

Traceback (most recent call last): File "", line 1, in voting_borda([['NDP', 'CPC', 'GREEN', 'LIBERAL'],['NDP', 'CPC', 'LIBERAL', 'GREEN'],['NDP', 'CPC', 'GREEN', 'LIBERAL']]) File "C:\Users\mycomp\Desktop\work\voting_systems.py", line 144, in voting_borda winner = max(results, key=results.get) NameError: global name 'results' is not defined

Thank you so much! I am able to run my module efficiently, however when when I try a test case, it gives me error saying "global name 'results' is not defined'' how may I fix this? thanks in advance — عائشة نديم, Dec 02 '12 at 04:44
Look at my post, made a change. Is this what you are doing ? — asheeshr, Dec 02 '12 at 04:46
Yes I added "return winner, list(results.values())" instead of "return winner, results" and kept the rest of the code same — عائشة نديم, Dec 02 '12 at 04:49
Post you code in the question with error message, below the original question. — asheeshr, Dec 02 '12 at 05:04
Your classmates already [asked this question](http://stackoverflow.com/questions/13639268/changing-the-output) and [here as well](http://stackoverflow.com/questions/13542458/how-to-give-points-for-each-indices-of-list) — Burhan Khalid, Dec 02 '12 at 05:54

asheeshr · Accepted Answer · 2012-12-02T04:45:53.603

1

For Python 2.7 :

return winner, [value for value in results.values()])

For Python 3.x :

return winner, list(results.values())

edited Dec 02 '12 at 04:45

answered Dec 02 '12 at 03:58

asheeshr

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Why not just `return winner, results.values()` ? (or `list(results.values())` in 3.x ? – Jon Clements Dec 02 '12 at 04:00
@JonClements results.values() returns a dict_values() object which is iterable but doesnt look clean on printing – asheeshr Dec 02 '12 at 04:03
I mentioned both 2.x and 3.x options... – Jon Clements Dec 02 '12 at 04:08
@JonClements Mentioned where ? – asheeshr Dec 02 '12 at 04:13
(or list(results.values()) in 3.x ? – Jon Clements 13 mins ago – Jon Clements Dec 02 '12 at 04:14

score 0 · Answer 2 · edited May 23 '17 at 12:18

Very old fashioned Python:

#!/usr/bin/env python
# -*- coding: utf-8 -*-

myResults=(['GREEN','NDP', 'LIBERAL', 'CPC'],
    ['GREEN','CPC','LIBERAL','NDP'],
    ['LIBERAL','NDP', 'CPC', 'GREEN'])

def count(results):
  counter = dict()
  for resultList in results:
    for result in resultList:
      if not(result in counter):
        counter[result] = 1
      else:
        counter[result] += 1
  print "counter (before): %s" % counter
  return counter.values()


if __name__ == "__main__":
  print "%s" % count(myResults)

If you're using Python >= 2.7, check the "collections.Counter" (as explained in this question)

How can I change the output? (Python, lists, dictionaries help)

2 Answers2