Find the second last number from a list-CodePudding

It was working properly,but while tried with

my_nums = [16,5,7,9,12] it's not working.

def get_second_largest(nums):
    largest = nums[0]
    second_largest = nums[0]
    for i in range(1,len(nums)):
        if nums[i] > largest:
            second_largest = largest
            largest = nums[i]
        elif nums[i] > second_largest:
            second_largest = nums[i]
    return second_largest

my_nums = [16,5,7,9,12]
second_largest = get_second_largest(my_nums)
print("Second highest number is : ",second_largest)

CodePudding user response：

my_nums = [16, 5, 7, 9, 12]
my_nums.remove(max(set(my_nums))) #removing the maximum element  
print(max(my_nums)) #printing the second largest element

One more method

CodePudding user response：

Sort the list (default ascending) and return the penultimate element from the output of the sort.

def get_second_largest(nums):
    return None if len(nums) < 2 else sorted(nums)[-2]

CodePudding user response：

There is an easier way to do this with sort:

my_nums = [16, 5, 7, 9, 12]
my_nums.sort(reverse=True)  # Descending order
print(my_nums[1])  # Output: 12

If you want to cope without sorting, as someone commented above you need to set the initial value of the variables outside of the list scope. -1 will suffice with your example, but a better way to do it is setting the starting value so that there is nothing smaller: float('-inf').

CodePudding user response：

Your only mistake is to set both largest and second_largest to nums[0], which in the example happens to be the largest number available - so it will never be replaced.

As a side note, why don't you iterate directly on the list items instead of an index?

So your code may become like this:

def second_largest(nums):
    largest = second_largest = 0 #or -math.inf if you want to account for any possible value
    for n in nums:
        if n > largest:
               second_largest = largest
               largest = n
        elif n > second_largest:
               second_largest = n
    return second_largest

Finally, this could be generalized to find the k-th largest number (though sorting would probably be a better strategy at this point):

def kth_largest(nums, k):
    largest_nums = [0 for x in range(k)]
    for n in nums:
        for i,large in enumerate(largest_nums):
            if n > large:
                largest_nums.insert(i,n)
                del largest_nums[-1]
                break
    return largest_nums[k-1]

CodePudding user response：

You can use np.sort() to sort the array and access the second last element.

It looks something like this:

my_nums_sorted = np.sort(np.array(my_nums))  # Convert to np.array and sort  
second_largest = my_nums_sorted[-2]