[Tutor] Getting greatest 3 numbers from list

[Manprit Singh]

Dear Sir ,

This is not related to python, but a general question, in c++ we have sets
in standard template library and iterators, and sets are basically sorted,
So doing the same problem in c++ using sets is a bad idea ?

#include<iostream>
#include<set>
#include<functional>
using namespace std;

int main()
{
    int arry[6] = {2, 4, 9, 2, 4, 1};
    set<int> second(arry, arry+6);
    auto itr = second.end();
    itr--;
    auto greatest = *itr;
    itr--;
    auto seclargest = *itr;
    itr--;
    auto thirdlargest = *itr;
    cout << greatest << " " << seclargest << " "<<thirdlargest;
    return 0;
}

Gives the right answer

My apologies for posting c++ code if it is against the rule of python
mailing list

Regards
Manprit Singh




On Mon, Feb 21, 2022 at 6:58 PM Peter Otten <__peter__ at web.de> wrote:

> On 21/02/2022 12:29, Manprit Singh wrote:
> > Dear Sir,
> >
> > My problem is to get greatest 3 numbers from list, list ls  along with
> > solution given below:
> > ls = [2, 4, 9, 6, 3, 9, 6, 1]
> > sorted(set(ls))[-3:]
> >
> > gives the right answer
> >
> > [4, 6, 9]
> >
> > Tried it with a for loop, just thinking why i am producing  a sorted
> list , for
> >
> > only getting largest 3 items
> >
> > lmax = []
> > ls = [2, 4, 9, 6, 3, 9, 6, 1]
> > for _ in range(3):
> >      mx = ls[0]
> >      for ele in ls[1:]:
> >          if ele > mx and ele not in lmax:
> >              mx = ele
> >      lmax.append(mx)
> > print(lmax)
> >
> > gives the correct answer
> >
> > [9, 6, 4]
> >
> > This problem be done with a for loop more easily ? i know it is not a
> good
> >
> > question - my apologies
> >
>
> I don't think it's a bad question. One answer is to keep the 3 largest
> values in a heap starting with the first three in the sequence.
> Then remove the smallest value and insert the current value from the
> iteration whenever it is larger than the smallest value in the heap.
>
> See the source and docs for the heapq.nlargest() function for a detailed
> explanation. If the current implementation looks too complicated start
> with an older version (before 3.5) of the module where nlargest()
> doesn't accept a 'key' parameter.
>
> https://docs.python.org/3/library/heapq.html#heapq.nlargest
> https://docs.python.org/3/library/heapq.html#theory
>
> PS: Here's an implementation that uses a sorted list to keep the largest
> n items:
>
>  >>> def nlargest(n, items):
>         items = iter(items)
>         result = sorted(islice(items, n))
>         for item in items:
>                 if item > result[0]:
>                         result[0] = item
>                         result.sort()
>         return result
>
>  >>> from itertools import islice
>  >>> from random import shuffle
>  >>> items = list(range(10))
>  >>> shuffle(items)
>  >>> items
> [1, 0, 9, 3, 4, 6, 7, 5, 2, 8]
>  >>> nlargest(3, items)
> [7, 8, 9]
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