如何找到数组的最大数量？

Question

Say I have this array:

Letters = ["A","B","C","D"]
           #A           #B     #C   #D
numbers = [ [1,4,3],[2,4,5,5],[3],[2,3] ]

I would like to find which of these arrays is the longest and want an output like this:

B has the most values

I attempted something like this

Letters = ["A","B","C","D"]
               #A       #B     #C   #D
numbers = [ [1,4,3],[2,4,5,5],[3],[2,3] ]
length = 0
maximum = 0
for i in numbers:
    for x in i:
        lenght = len(numbers)
        if length > maximum:
            maximum = length
print(maximum)

Answer 1

You could zip both lists and use the max built-in function with a custom key to find the tuple with the longest list:

from operator import itemgetter
s = max(zip(Letters, numbers), key= lambda x: len(itemgetter(1)(x)))[0]

print(s, 'has the most values')
# B has the most values

Answer 2

1. Since you only need the lists' lengths and their positions, convert numbers to a list of lengths, where the positions of each length correspond to the positions of each sublist:

    numbers_lengths = list(map(len, numbers)) # using `list` is not necessary # numbers_lengths == [3, 4, 1, 2] 

2. You want to associate each element of letters with a length, which means you need a different data structure:

    Map = dict(zip(letters, numbers_lengths)) # Map == {'A': 3, 'B': 4, 'C': 1, 'D': 2} 

3. Just find the key whose value is the greatest:

    result = max([(length, letter) for letter, length in Map.items()]) # result == (4, 'B') 

   This works because you can compare tuples: ((2, 'a') > (0, 'b')) is True .

   3.1 Another approach:

    result = max(Map.items(), key=lambda pair: pair[1]) # result == ('B', 4) 

4. So, the answer is result[1] == 'B'