Question

def print_most_numbers_occurrences(numbers_str):
    number_list = numbers_str.split() 
    for i in number_list:
        i=max(number_list,key=number_list.count)
    print(i)


def test_print_most_numbers_occurrences():
     print_most_numbers_occurrences('2 3 40 1 5 4 3 3 9  9')
     print_most_numbers_occurrences('9 30 3 9 3 1 4')
     print_most_numbers_occurrences('19 30 13 4 9 3 1 4')

def main():
    test_print_most_numbers_occurrences()


main()

输出：

3
9
4

我希望'9 30 3 9 3 1 4'得到所有最重复的数字：9和3出现两次，因此不仅要报告两次出现9

输出如下：

Answer 1

首先：当您使用for时，您不需要max - 循环。它已经为你做到了。

第二：如果你想拥有多个值，那么max并不是一个好的选择。对于这类计数操作，您应该使用collections.Counter（它还可以避免多次计算出现次数）。

from collections import Counter

def print_most_numbers_occurrences(numbers_str):
    number_list = numbers_str.split() 
    # count the occurrences
    cnts = Counter(number_list)
    # Get the maximum count
    maximum_cnt = max(cnts.values())
    # print all values that have the "maximum" count
    print(*[val for val, cnt in cnts.items() if cnt == maximum_cnt])

输入的测试打印出来：

3
9 3
4

如果您更喜欢简单的循环而不是理解（或者您使用不带print函数的python-2.x），您也可以使用：

def print_most_numbers_occurrences(numbers_str):
    number_list = numbers_str.split() 
    cnts = Counter(number_list)
    maximum_cnt = max(cnts.values())
    for value, cnt in cnts.items():
        if cnt == maximum_cnt:
            print(value)

给出：

Answer 2

from collections import Counter

a = ('2 3 40 1 5 4 3 3 9  9','9 30 3 9 3 1 4','19 30 13 4 9 3 1 4')

for x in a:
    x = x.split()
    b = Counter(x)
    z = max(b.values())
    for k in b.keys():
        if b[k] == z:
            print(k)

输出：

如果您想要其他输出

from collections import Counter

a = ('2 3 40 1 5 4 3 3 9  9','9 30 3 9 3 1 4','19 30 13 4 9 3 1 4')

k = []
for x in a:
    x = x.split()
    b = Counter(x)
    z = max(b.values())
    k.append([f for f in b.keys() if b[f] == z])

print(k)

输出

[['3'], ['9', '3'], ['4']]

使用功能

from collections import Counter

def maxmy(sequences):
    k = []
    for x in sequences:
        x = x.split()
        b = Counter(x)
        z = max(b.values())
        k.append([f for f in b.keys() if b[f] == z])
    return k

maxmy(('2 3 40 1 5 4 3 3 9  9','9 30 3 9 3 1 4','19 30 13 4 9 3 1 4'))

输出

[['3'], ['9', '3'], ['4']]

如何在列表中找到大多数重复元素？

2 个答案: