Pfthb

If a have a list within a another list that looks like this...

[['Harry',9,1],['Harry',17,1],['Jake',4,1], ['Dave',9,2],['Sam',17,2],['Sam',4,2]]

How can I add the middle element together so so for 'Harry' for example, it shows up as ['Harry', 26] and also for Python to look at the group number (3rd element) and output the winner only (the one with the highest score which is the middle element). So for each group, there needs to be one winner. So the final output shows:

[['Harry', 26],['Sam',21]]

THIS QUESTION IS NOT A DUPLICATE: It has a third element as well which I am stuck about

The similar question gave me an answer of:

grouped_scores = 
for name, score, group_number in players_info:
 if name not in grouped_scores:
 grouped_scores[name] = score
 grouped_scores[group_number] = group_number 
 else:
 grouped_scores[name] += score

But that only adds the scores up, it doesn't take out the winner from each group. Please help.

I had thought doing something like this, but I'm not sure exactly what to do...

grouped_scores = 
for name, score, group_number in players_info:
 if name not in grouped_scores:
 grouped_scores[name] = score
 else:
 grouped_scores[name] += score
 for group in group_number:
 if grouped_scores[group_number] = group_number:
 [don't know what to do here]

Use itertools.groupby, and collections.defaultdict:

l=[['Harry',9,1],['Harry',17,1],['Jake',4,1], ['Dave',9,2],['Sam',17,2],['Sam',4,2]]
from itertools import groupby
from collections import defaultdict
l2=[list(y) for x,y in groupby(l,key=lambda x: x[-1])]
l3=
for x in l2:
 d=defaultdict(int)
 for x,y,z in x:
 d[x]+=y
 l3.append(max(list(map(list,dict(d).items())),key=lambda x: x[-1]))

Now:

print(l3)

Is:

[['Harry', 26], ['Sam', 21]]

you can try to use Counter and it's method most_common

Return a list of the n most common elements and their counts from the
most common to the least

from collections import Counter
l = [['Harry',9,1],['Harry',17,1],['Jake',4,1], ['Dave',9,2],['Sam',17,2],['Sam',4,2]]
step = 3
results = list()
for x in range(0, len(l), step):
 cnt = Counter()
 for y in l[x: x+step]:
 cnt.update(y[0]: y[1])
 results.append(cnt.most_common(1)[0])

will give you:

print(results)
[('Harry', 26), ('Sam', 21)]

I would aggregate the data first with a defaultdict.

>>> from collections import defaultdict
>>> 
>>> combined = defaultdict(lambda: defaultdict(int))
>>> data = [['Harry',9,1],['Harry',17,1],['Jake',4,1], ['Dave',9,2],['Sam',17,2],['Sam',4,2]]
>>> 
>>> for name, score, group in data:
...: combined[group][name] += score
...: 
>>> combined
>>> 
defaultdict(<function __main__.<lambda>()>,
 1: defaultdict(int, 'Harry': 26, 'Jake': 4),
 2: defaultdict(int, 'Dave': 9, 'Sam': 21))

Then apply max to each value in that dict.

>>> from operator import itemgetter
>>> [list(max(v.items(), key=itemgetter(1))) for v in combined.values()]
>>> [['Harry', 26], ['Sam', 21]]

use itertools.groupby and then take the middle value from the grouped element and then append it to a list passed on the maximum condition

import itertools
l=[['Harry',9,1],['Harry',17,1],['Jake',4,1], ['Dave',9,2],['Sam',17,2],['Sam',4,2]]
maxlist=
maxmiddleindexvalue=0
for key,value in itertools.groupby(l,key=lambda x:x[0]):
 s=0
 m=0
 for element in value:
 s+=element[1]
 m=max(m,element[1])
 if(m==maxmiddleindexvalue):
 maxlist.append([(key,s)])

 if(m>maxmiddleindexvalue):
 maxlist=[(key,s)]
 maxmiddleindexvalue=m

print(maxlist)

OUTPUT

[('Harry', 26), [('Sam', 21)]]