Sort tuple by averages

Question

I have a tuple which currently looks like the following:

[['Hussain', 7, 8, 0], ['Max', 3, 4, 3], ['Alexandra', 10, 9, 8]]

I want to be able to sort this tuple so that it prints out the tuple in order from highest to lowest averages. This is what I expect as the outcome:

['Alexandra', 9],['Hussain', 5],['Max', 3.3333333333]

I would appreciate any help given.

TIA

Oh al right, then what differentiates a list and a tuple? - I've totally forgotten. — Hussain, Apr 10 '15 at 22:33
Check this post out to learn about the differences: http://stackoverflow.com/questions/626759/whats-the-difference-between-list-and-tuples — Julien Spronck, Apr 10 '15 at 22:35

Padraic Cunningham · Accepted Answer · 2015-04-10T23:28:17.990

3

from statistics import mean
from operator import itemgetter
l = [['Hussain', 7, 8, 0], ['Max', 3, 4, 3], ['Alexandra', 10, 9, 8]]

print(sorted(([ele[0],mean(ele[1:])] for ele in l),key=itemgetter(1),reverse=True))

[['Alexandra', 9.0], ['Hussain', 5.0], ['Max', 3.3333333333333335]]

If you are not using python >= 3.4 you can just do the average calculation manually.

print(sorted(([ele[0],sum((ele[1:])) /  len(l) - 1] for ele in l),key=itemgetter(1),reverse=True))

ele[0],mean(ele[1:]) takes the name and the remaining elements which are the scores, we then sort using itemgetter(1) which is the score/second element as the key and setting reverse=True to go from high to low.

I would also use itertools.islice to get the slice avoiding building new lists with normal slicing:

from itertools import islice
print(sorted(([ele[0],mean(islice(ele,1,None))] for ele in l),key=itemgetter(1),reverse=True))

Or without using mean:

from itertools import islice
print(sorted(([ele[0],sum(islice(ele,1,None)) / len(l) - 1] for ele in l),key=itemgetter(1),reverse=True))

edited Apr 10 '15 at 23:28

answered Apr 10 '15 at 22:25

Padraic Cunningham

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Thanks for the quick response, I really do appreciate it :) – Hussain Apr 10 '15 at 22:25
Is "ele" a python syntax? or is it something you have used as like an index. – Hussain Apr 10 '15 at 22:34
@Hussain, it is just a variable name, you could call it anything you want – Padraic Cunningham Apr 10 '15 at 22:36
Thanks just wanted to clarify – Hussain Apr 10 '15 at 22:36
I understand the use of `itertools.islice` but which parts are you splitting here? – Hussain Apr 10 '15 at 22:58
@Hussain, instead of creating slices which normal slicing does islice returns an islice object that is basically an iterator so more memory efficient. Not an issue in your code but on larger datasets it may well be. – Padraic Cunningham Apr 10 '15 at 23:22

score 1 · Answer 2 · answered Apr 10 '15 at 22:31

1

s = [['Hussain', 7, 8, 0], ['Max', 3, 4, 3], ['Alexandra', 10, 9, 8]]
result = sorted([[e[0], sum(e[1:])/len(e[1:])] for e in s], key=lambda x: x[1], reverse=True)

answered Apr 10 '15 at 22:31

Jaakko

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Sorted needs an iterable. I'm using list comprehension to give sorted the first parameter: [[e[0], sum(e[1:])/len(e[1:])] for e in s] = > [['Hussain', 5.0], ['Max', 3.3333333333333335], ['Alexandra', 9.0]] – Jaakko Apr 10 '15 at 22:48
I used timeit to see how our functions compare. With number=10000, your answer gets: 1.5141523819997929 Mine gets: 0.08450542999980826 – Jaakko Apr 10 '15 at 22:58
Your first answer: from statistics import mean from operator import itemgetter sorted(([ele[0],mean(ele[1:])] for ele in l),key=itemgetter(1),reverse=True) Maybe the imports slow it down. Anyway, what's the problem with my answer? It's fast, one line, clear and list comprehensions are great. – Jaakko Apr 10 '15 at 23:16
Dude, why did you delete your comments? – Jaakko Apr 10 '15 at 23:22

Sort tuple by averages

2 Answers2