Как найти второй максимальный элемент массива python

задан 6 окт 2019 в 6:26

Андрей КрузликАндрей Крузлик

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Можно написать функцию:

def find_maxes(array, count):
    # копируем список чтобы не изменить старую
    copied_array = array.copy()
    maximums = []
    if count > len(copied_array):
        raise ValueError('Количество не может превышать длину списка')
    for _ in range(count):
        max_val = max(copied_array)  # получаем максимальное значение
        copied_array.remove(copied_array)  # удаляем его из списка
        maximums.append(max_val)  # добавляем в наш ожидаемый результат
    return maximums

или же можно поступить хитро

def find_maxes(array, count):
    if count > len(array):
        raise ValueError('Количество не может превышать длину списка')
    sorted_array = sorted(array)  # отсортировать список
    # Забрать последние элементы из спика так как они будут максимальными
    return sorted_array[len(array)-count: len(array)]

ответ дан 6 окт 2019 в 6:44

FeToRFeToR

12 бронзовых знака

You could use the heapq module:

>>> el = [20,67,3,2.6,7,74,2.8,90.8,52.8,4,3,2,5,7]
>>> import heapq
>>> heapq.nlargest(2, el)
[90.8, 74]

And go from there…

Since @OscarLopez and I have different opinions on what the second largest means, I’ll post the code according to my interpretation and in line with the first algorithm provided by the questioner.

def second_largest(numbers):
    count = 0
    m1 = m2 = float('-inf')
    for x in numbers:
        count += 1
        if x > m2:
            if x >= m1:
                m1, m2 = x, m1            
            else:
                m2 = x
    return m2 if count >= 2 else None

(Note: Negative infinity is used here instead of None since None has different sorting behavior in Python 2 and 3 – see Python – Find second smallest number; a check for the number of elements in numbers makes sure that negative infinity won’t be returned when the actual answer is undefined.)

If the maximum occurs multiple times, it may be the second largest as well. Another thing about this approach is that it works correctly if there are less than two elements; then there is no second largest.

Running the same tests:

second_largest([20,67,3,2.6,7,74,2.8,90.8,52.8,4,3,2,5,7])
=> 74
second_largest([1,1,1,1,1,2])
=> 1
second_largest([2,2,2,2,2,1])
=> 2
second_largest([10,7,10])
=> 10
second_largest([1,1,1,1,1,1])
=> 1
second_largest([1])
=> None
second_largest([])
=> None

Update

I restructured the conditionals to drastically improve performance; almost by a 100% in my testing on random numbers. The reason for this is that in the original version, the elif was always evaluated in the likely event that the next number is not the largest in the list. In other words, for practically every number in the list, two comparisons were made, whereas one comparison mostly suffices – if the number is not larger than the second largest, it’s not larger than the largest either.

answered Apr 25, 2013 at 23:15

You could always use sorted

>>> sorted(numbers)[-2]
74

answered Apr 25, 2013 at 22:21

Try the solution below, it’s O(n) and it will store and return the second greatest number in the second variable. UPDATE: I’ve adjusted the code to work with Python 3, because now arithmetic comparisons against None are invalid.

Notice that if all elements in numbers are equal, or if numbers is empty or if it contains a single element, the variable second will end up with a value of None – this is correct, as in those cases there isn’t a “second greatest” element.

Beware: this finds the “second maximum” value, if there’s more than one value that is “first maximum”, they will all be treated as the same maximum – in my definition, in a list such as this: [10, 7, 10] the correct answer is 7.

def second_largest(numbers):
    minimum = float('-inf')
    first, second = minimum, minimum
    for n in numbers:
        if n > first:
            first, second = n, first
        elif first > n > second:
            second = n
    return second if second != minimum else None

Here are some tests:

second_largest([20, 67, 3, 2.6, 7, 74, 2.8, 90.8, 52.8, 4, 3, 2, 5, 7])
=> 74
second_largest([1, 1, 1, 1, 1, 2])
=> 1
second_largest([2, 2, 2, 2, 2, 1])
=> 1
second_largest([10, 7, 10])
=> 7
second_largest( [1, 3, 10, 16])
=> 10
second_largest([1, 1, 1, 1, 1, 1])
=> None
second_largest([1])
=> None
second_largest([])
=> None

answered Apr 25, 2013 at 22:51

You can find the 2nd largest by any of the following ways:

Option 1:

numbers = set(numbers)
numbers.remove(max(numbers))
max(numbers)

Option 2:

sorted(set(numbers))[-2]

answered Jul 12, 2018 at 16:12

The quickselect algorithm, O(n) cousin to quicksort, will do what you want. Quickselect has average performance O(n). Worst case performance is O(n^2) just like quicksort but that’s rare, and modifications to quickselect reduce the worst case performance to O(n).

The idea of quickselect is to use the same pivot, lower, higher idea of quicksort, but to then ignore the lower part and to further order just the higher part.

answered Jun 13, 2015 at 18:36

Why to complicate the scenario? Its very simple and straight forward

1. Convert list to set – removes duplicates
2. Convert set to list again – which gives list in ascending order

Here is a code

mlist = [2, 3, 6, 6, 5]
mlist = list(set(mlist))
print mlist[-2]

answered Nov 7, 2016 at 7:36

This is one of the Simple Way

def find_second_largest(arr):
    first, second = float('-inf'), float('-inf')

    for number in arr:
        if number > first:
            second = first
            first = number
        elif second < number < first:
            second = number

    return second

answered Oct 19, 2021 at 11:58

kpiekpie

9,4705 gold badges28 silver badges49 bronze badges

O(n): Time Complexity of a loop is considered as O(n) if the loop variables is incremented / decremented by a constant amount. For example following functions have O(n) time complexity.

 // Here c is a positive integer constant   
   for (int i = 1; i <= n; i += c) {  
        // some O(1) expressions
   }

To find the second largest number i used the below method to find the largest number first and then search the list if thats in there or not

x = [1,2,3]
A = list(map(int, x))
y = max(A)
k1 = list()
for values in range(len(A)):
if y !=A[values]:
    k.append(A[values])

z = max(k1)
print z

answered Aug 2, 2016 at 6:27

MikeMike

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BilyBily

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Using reduce from functools should be a linear-time functional-style alternative:

from functools import reduce

def update_largest_two(largest_two, x):
    m1, m2 = largest_two
    return (m1, m2) if m2 >= x else (m1, x) if m1 >= x else (x, m1)


def second_largest(numbers):
    if len(numbers) < 2:
        return None
    largest_two = sorted(numbers[:2], reverse=True)
    rest = numbers[2:]
    m1, m2 = reduce(update_largest_two, rest, largest_two)
    return m2

… or in a very concise style:

from functools import reduce

def second_largest(n):
    update_largest_two = lambda a, x: a if a[1] >= x else (a[0], x) if a[0] >= x else (x, a[0])

    return None if len(n) < 2 else (reduce(update_largest_two, n[2:], sorted(n[:2], reverse=True)))[1]

answered Aug 17, 2022 at 13:11

>>> l = [19, 1, 2, 3, 4, 20, 20]
>>> sorted(set(l))[-2]
19

answered Jul 21, 2015 at 22:52

This can be done in [N + log(N) – 2] time, which is slightly better than the loose upper bound of 2N (which can be thought of O(N) too).

The trick is to use binary recursive calls and “tennis tournament” algorithm. The winner (the largest number) will emerge after all the ‘matches’ (takes N-1 time), but if we record the ‘players’ of all the matches, and among them, group all the players that the winner has beaten, the second largest number will be the largest number in this group, i.e. the ‘losers’ group.

The size of this ‘losers’ group is log(N), and again, we can revoke the binary recursive calls to find the largest among the losers, which will take [log(N) – 1] time. Actually, we can just linearly scan the losers group to get the answer too, the time budget is the same.

Below is a sample python code:

def largest(L):
    global paris
    if len(L) == 1:
        return L[0]
    else:
        left = largest(L[:len(L)//2])
        right = largest(L[len(L)//2:])
        pairs.append((left, right))
        return max(left, right)

def second_largest(L):
    global pairs
    biggest = largest(L)
    second_L = [min(item) for item in pairs if biggest in item]

    return biggest, largest(second_L)  



if __name__ == "__main__":
    pairs = []
    # test array
    L = [2,-2,10,5,4,3,1,2,90,-98,53,45,23,56,432]    

    if len(L) == 0:
        first, second = None, None
    elif len(L) == 1:
        first, second = L[0], None
    else:
        first, second = second_largest(L)

    print('The largest number is: ' + str(first))
    print('The 2nd largest number is: ' + str(second))

answered Sep 29, 2016 at 5:47

Objective: To find the second largest number from input.

Input : 5
2 3 6 6 5

Output: 5

*n = int(raw_input())
arr = map(int, raw_input().split())
print sorted(list(set(arr)))[-2]*

Brian SanchezBrian Sanchez

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answered Nov 5, 2020 at 13:46

You can also try this:

>>> list=[20, 20, 19, 4, 3, 2, 1,100,200,100]
>>> sorted(set(list), key=int, reverse=True)[1]
100

answered Oct 24, 2016 at 5:54

A simple way :

n=int(input())
arr = set(map(int, input().split()))
arr.remove(max(arr))
print (max(arr))

answered Dec 18, 2017 at 15:41

use defalut sort() method to get second largest number in the list.
sort is in built method you do not need to import module for this.

lis = [11,52,63,85,14]
lis.sort()
print(lis[len(lis)-2])

answered Apr 23, 2018 at 12:20

if __name__ == '__main__':
    n = int(input())
    arr = list(map(float, input().split()))
    high = max(arr)
    secondhigh = min(arr)
    for x in arr:
        if x < high and x > secondhigh:
            secondhigh = x
    print(secondhigh)

The above code is when we are setting the elements value in the list
as per user requirements. And below code is as per the question asked

#list
numbers = [20, 67, 3 ,2.6, 7, 74, 2.8, 90.8, 52.8, 4, 3, 2, 5, 7]
#find the highest element in the list
high = max(numbers)
#find the lowest element in the list
secondhigh = min(numbers)
for x in numbers:
    '''
    find the second highest element in the list,
    it works even when there are duplicates highest element in the list.
    It runs through the entire list finding the next lowest element
    which is less then highest element but greater than lowest element in
    the list set initially. And assign that value to secondhigh variable, so 
    now this variable will have next lowest element in the list. And by end 
    of loop it will have the second highest element in the list  
    '''
    if (x<high and x>secondhigh):
        secondhigh=x
print(secondhigh)

answered Oct 2, 2019 at 8:38

Max out the value by comparing each one to the max_item. In the first if, every time the value of max_item changes it gives its previous value to second_max. To tightly couple the two second if ensures the boundary

def secondmax(self, list):
    max_item = list[0]
    second_max = list[1]
    for item in list:
        if item > max_item:
            second_max =  max_item
            max_item = item
        if max_item < second_max:
            max_item = second_max
    return second_max

Most of previous answers are correct but here is another way !

Our strategy is to create a loop with two variables first_highest and second_highest. We loop through the numbers and if our current_value is greater than the first_highest then we set second_highest to be the same as first_highest and then the second_highest to be the current number. If our current number is greater than second_highest then we set second_highest to the same as current number

#!/usr/bin/env python3
import sys
def find_second_highest(numbers):

    min_integer = -sys.maxsize -1
    first_highest= second_highest = min_integer
    for current_number in numbers:
        if current_number == first_highest and min_integer != second_highest:
            first_highest=current_number
        elif current_number > first_highest:
            second_highest = first_highest
            first_highest = current_number
        elif current_number > second_highest:
            second_highest = current_number
    return second_highest

print(find_second_highest([80,90,100]))
print(find_second_highest([80,80]))
print(find_second_highest([2,3,6,6,5]))

answered Nov 12, 2018 at 6:42

Below code will find the max and the second max numbers without the use of max function. I assume that the input will be numeric and the numbers are separated by single space.

myList = input().split()
myList = list(map(eval,myList))


m1 = myList[0]
m2 = myList[0]

for x in myList:
    if x > m1:
        m2 = m1
        m1 = x
    elif x > m2:
        m2 = x

print ('Max Number: ',m1)
print ('2nd Max Number: ',m2)

answered Feb 6, 2021 at 17:22

Here I tried to come up with an answer.
2nd(Second) maximum element in a list using single loop and without using any inbuilt function.

def secondLargest(lst):
    mx = 0
    num = 0
    sec = 0
    for i in lst:
        if i > mx:
            sec = mx
            mx = i
        else:
            if i > num and num >= sec:
                sec = i
            num = i
    return sec

answered Apr 22, 2021 at 11:27

We can use 2 loop to compare and find the second largest number from list rather than removing max number from list:

def second_largest(list1):
   second_max = list1[0]
   max_nu = max(list1)

   for i in range(len(list1) -1):
       for j in range(1,len(list1)):
          if list1[i] > list1[j] and list1[i] < max_nu :
              second_max = list1[i]
          elif list1[i] < list1[j] and list1[j] < max_nu:
              second_max = list1[j]
   return second_max

l = [2, 4, 5, 6, 8, 7, 21, 20]
print(second_largest(l))

answered Nov 17, 2021 at 8:09