Question

我堆了一会儿。我试过调试，但我找不到解决方案。我试图按升序计算数字的出现次数例如，我有这些输入：2 5 6 5 4 3 23 43 2 我得到的输出 -2发生2次 -3发生一次 -4次发生一次 -5次发生2次 -6发生一次 -43发生一次＃为什么会发生这种情况？ 23次出现

def main（）：

while True:
    try:
        numbers = input("Enter integers between 1 and 100:-")
        #my_list= [int(x) for x in input().split(' ')]
        my_list = list(map(int, numbers.strip().split(' ')))

        break
    except ValueError:
        print ("Oops ! enter an integer... ")
count = {}
for x in set(my_list):
    count[x] = my_list.count(x)
for key, value in count.items():
    if value == 1:
        print('{} occurs one time'.format(key))
    else:
        print('{} occurs {} times'.format(key, value))

main（）

Answer 1

输出正确：

输入：

Country

输出：

2 5 6 5 4 3 23 43 2

必须按键对字典进行排序才能获得所需的输出：

Enter integers between 1 and 100:-2 5 6 5 4 3 23 43 2
2 occurs 2 times
3 occurs one time
4 occurs one time
5 occurs 2 times
6 occurs one time
43 occurs one time
23 occurs one time

输出以下

：

import collections
while True:
    try:
        numbers = input("Enter integers between 1 and 100:-")
        #my_list= [int(x) for x in input().split(' ')]
        my_list = list(map(int, numbers.strip().split(' ')))

        break
    except ValueError:
        print ("Oops ! enter an integer... ")
count = {}

for x in set(my_list):
    count[x] = my_list.count(x)
count = collections.OrderedDict(sorted(count.items()))
for key, value in count.items():
    if value == 1:
        print('{} occurs one time'.format(key))
    else:
        print('{} occurs {} times'.format(key, value))

计算数字的出现次数

1 个答案: