Rearranging a function for lists to work backwards

Admin 1
Viewed 44
price_list = [["apple", 1], ["sugar", 5], ["mango", 3],["coffee", 9], ["trail mix", 6]]

def most_expensive_item(list):
    result = max(price_list, key=lambda x: x[1])
    return result[0]

I have the code above, but I have run into a bit of a problem. Lets say coffee and sugar are both $9, I want the function to return the item nearest to the end so it would be coffee, like this:

price_list = [["apple", 1], ["sugar", 9], ["mango", 3],["coffee", 9], ["trail mix", 6]]

most_expensive(price_list)
>>> coffee

I feel like theres an easy solution but I'm not seeing it, any help would be appreciated.

nested-lists python
2 Answers

So you may easily achieve that by reversing the array

def most_expensive(li):
    result = max(li, key=lambda x: x[1])
    return result[0]

price_list = [["apple", 1], ["sugar", 9], ["mango", 3],["coffee", 9], ["trail mix", 6]]

most_expensive(price_list[::-1])

Or just use the conventional loop:

def most_expensive(list):
    maxi = price_list[-1]
    for i in range(len(price_list)-1, 0, -1):
        if price_list[i][1] > maxi[1]:
            maxi = price_list[i]
    return maxi[0]

most_expensive(price_list)
user_146318850

max() tiebreaks by picking the first element in the sequence that is involved in the tie. If you want the last element, reverse the list and put that into max() instead:

most_expensive(price_list[::-1])
user_177698150
Related