ans  = [1, 1, 0, 0, 1, 1]
all_ans = {ans.count(ans[i]): ans[i] for i in range(len(ans))}
print(all_ans)

all_ans={4: 1, 2: 0}
max_key = max(all_ans.keys())

4

print(all_ans[max_key])

1

def most_frequent(List):

    counter = 0

    num = List[0]

 

    for i in List:

        curr_frequency = List.count(i)

        if(curr_frequency> counter):

            counter = curr_frequency

            num = i


    return num


List = [2, 1, 2, 2, 1, 3]

print(most_frequent(List))

>>> li  = ['goose', 'duck', 'duck']

>>> def foo(li):
         st = set(li)
         mx = -1
         for each in st:
             temp = li.count(each):
             if mx < temp:
                 mx = temp 
                 h = each 
         return h

>>> foo(li)
'duck'

我这样做使用scipy统计模块和lambda:

import scipy.stats
lst = [1,2,3,4,5,6,7,5]
most_freq_val = lambda x: scipy.stats.mode(x)[0][0]
print(most_freq_val(lst))

结果:

 most_freq_val = 5

简单的一行代码:

def most_common(lst):
    return max(set(lst), key=lst.count)

# use Decorate, Sort, Undecorate to solve the problem

def most_common(iterable):
    # Make a list with tuples: (item, index)
    # The index will be used later to break ties for most common item.
    lst = [(x, i) for i, x in enumerate(iterable)]
    lst.sort()

    # lst_final will also be a list of tuples: (count, index, item)
    # Sorting on this list will find us the most common item, and the index
    # will break ties so the one listed first wins.  Count is negative so
    # largest count will have lowest value and sort first.
    lst_final = []

    # Get an iterator for our new list...
    itr = iter(lst)

    # ...and pop the first tuple off.  Setup current state vars for loop.
    count = 1
    tup = next(itr)
    x_cur, i_cur = tup

    # Loop over sorted list of tuples, counting occurrences of item.
    for tup in itr:
        # Same item again?
        if x_cur == tup[0]:
            # Yes, same item; increment count
            count += 1
        else:
            # No, new item, so write previous current item to lst_final...
            t = (-count, i_cur, x_cur)
            lst_final.append(t)
            # ...and reset current state vars for loop.
            x_cur, i_cur = tup
            count = 1

    # Write final item after loop ends
    t = (-count, i_cur, x_cur)
    lst_final.append(t)

    lst_final.sort()
    answer = lst_final[0][2]

    return answer

print most_common(['x', 'e', 'a', 'e', 'a', 'e', 'e']) # prints 'e'
print most_common(['goose', 'duck', 'duck', 'goose']) # prints 'goose'