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I am still learning Python and I have been trying to sort this problem for hours but somehow cannot get it working: Suppose we have a list of tuples containing names and grades:

[("Mark", [5,4,1,2,4,3]), ("John", [1,2,3,1,2,1]), ("Patrick", [1,3,5,1,2,4])]

And we want to get this output:

[("John", [1,2,3,1,2,1]), ("Patrick", [1,3,5,1,2,4]), ("Mark", [5,4,1,2,4,3])]

, where the original list is is sorted in ascending order of students' marks score (suppose we get marks score by discarding the highest mark and adding all remaining marks together). I.e. John has marks 1,2,3,1,2,1, so we discard mark 3 and add 1+2+1+2+1 together, which gives us 7.

This task wouldnt be problem at all, if I wasnt required to get that output with all their original marks. I wrote this function:

def sortedFinalRes(input): finalRes = 0 sortedRes = [] b = [] for i in range(0, len(input)): total = 0 a = sorted(input[i][1], key = lambda x: x) a.pop() sortedRes += input[i][0], a total = sum(sortedRes[i*2+1]) b += (sortedRes[i*2], total) c = sorted(b, key = lambda x: str(x)) print(sortedRes) print(b)

, where I basically at first sort marks of each student so that I can then discard the highest one by using pop() and then I sum all remaining ones, which gives me marks score. Then I sort it according to marks score, but the output is required to be sorted, but with original marks, not with the mark score. Any ideas appreciated! Cheers

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3 Answers 3

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We can define a scoring function like this:

def score(x): sum(x) - max(x)

This computes your so-called "marks score" by summing the scores and throwing out the max score. We'll then use it as a key for sorting, making sure to give it the second element of the tuple:

In [4]: sorted(d, key=lambda x: score(x[1])) Out[4]: [('John', [1, 2, 3, 1, 2, 1]), ('Patrick', [1, 3, 5, 1, 2, 4]), ('Mark', [5, 4, 1, 2, 4, 3])]
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4 Comments

great answer and snuck it in just before me :P
Why the double lambdas? Lambdas should never be assigned to variables, it misses the point. How about sorted(d, key=lambda x: sum(x) - max(x)) instead?
@tdelaney - sorted(d, key=lambda x: sum(x[1]) - max(x[1])) to be exact. x will be the tuples in the list. But good idea overall.
@JoranBeasley - oops, I should have looked more closely!
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def sort_key(item): name,scores = item return sum(scores)-max(scores) print sorted(students,key=sort_key)

you could easily do it with a single lambda as shown below ... I just did this for added verbosity

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you can refuse from max value by sorted(x[1])[:-1] that first sort the list and slice from beginning to highest:

>>> sorted (l,key=lambda x : sum(sorted(x[1])[:-1])) [('John', [1, 2, 3, 1, 2, 1]), ('Patrick', [1, 3, 5, 1, 2, 4]), ('Mark', [5, 4, 1, 2, 4, 3])]

answers benchmark :

>>> s="""l=[("Mark", [5,4,1,2,4,3]), ("John", [1,2,3,1,2,1]), ("Patrick", [1,3,5,1,2,4])] ... sorted (l,key=lambda x : sum(sorted(x[1])[:-1]))""" >>> timeit.timeit(stmt=s, number=100000) 0.407818078994751 >>> s="""l=[("Mark", [5,4,1,2,4,3]), ("John", [1,2,3,1,2,1]), ("Patrick", [1,3,5,1,2,4])] ... sorted(l, key=lambda x: sum(x[1]) - max(x[1]))""" >>> timeit.timeit(stmt=s, number=100000) 0.3033828830718994 >>> s="""l=[("Mark", [5,4,1,2,4,3]), ("John", [1,2,3,1,2,1]), ("Patrick", [1,3,5,1,2,4])] ... def sort_key(item): ... name,scores = item ... return sum(scores)-max(scores) ... sorted(l,key=sort_key)""" >>> timeit.timeit(stmt=s, number=100000) 0.2999439334869385

As you can see there no much difference between all the answers , any way Joran Beasley's answer is the best !

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13 Comments

This is actually nlogn whereas the other solution is n
@1_CR how you think that using max inside sorted has n order ?
Actually, if n is the size of the outer list, and m is the size of the inner, this solution is O(n m log m + n log n), whereas solution with max is O(nm + n log n), I believe.
@PadraicCunningham actually i mean as its key ! but any way im sure that order is not n
@PadraicCunningham The key function sum(x) - max(x) takes time O(m), where m is the size of the score list. We have to evaluate the key function for each of the n entries, giving O(mn). Then we need to sort n keys, adding the O(n log n).
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