子列表最高 sum() 的索引

Question

Given a finite list of n sub-lists:

my_list = [[2, 9999], [318, 9999], [990, 9999], [9, 9999], [7767, 9999]]

I want a one-liner concise procedure that returns the index of the sub-list with the highest sum().

If in the event 2 or more sub-lists have an equal sum(), then return the first.

My code:

for ksu in key_set_unique:
   print(sum(ksu))

Answer 1

You can use max with sum as key

index = my_list.index(max(my_list, key=sum))

Answer 2

my_list = [[2, 9999], [318, 9999], [990, 9999], [9, 9999], [7767, 9999]]

temp = []

for sub_list in my_list:
    x = sum(sub_list)
    temp.append(x)

ind_value = max(temp)

print(temp.index(ind_value))

I ran a for loop to get the sum of inherited list stored in variable my_list. And that sum is stored in the new list called temp.

Now simply i got index of maximum value stored in temp as index position will remain same so i can get the inherited list with maximum sum value. ( my_list[0] = temp[0] )

Answer 3

Since you explicitly asked for a one-liner, here it is:

>>> list(map(sum, my_list)).index(max(map(sum, my_list)))
4

This line creates a list of sums using the map function and finds a value in said list using index . The value that is searched for is the max value of the list of sums mentioned earlier.

Note that this computes the sums of my_list 's elements twice, which isn't necessary. Solution without the one-line requirement:

>>> sum_list = list(map(sum, my_list))
>>> sum_list.index(max(sum_list))
4

Answer 4

# index with element

element, index = max([(sum(item), i) for i, item in enumerate(my_list)])
print(element, index)