在Python中使用循环找到第二小的数字 [英] Finding the second smallest number using loops in python
问题描述
我想知道如何从def函数的用户输入列表中找到第二小的数字。另外,使用任何排序函数,导入模块和min()和max()函数的 WITHOUT ,我如何通过使用循环和关系运算符来查找数字?
I was wondering how to find the second smallest number from a user-input list with def functions. Also, WITHOUT using any sorting functions, imported modules, and min() and max() functions, how would I find the numbers by using just loops and relational operators?
这是我的下面的代码(目前为止我只找到最小的数字......):
Here's my following code (I only have finding the smallest number so far...):
def second_smallest():
smallest = second_smallest[0]
for i in second_smallest[1:]:
if smallest > i:
smallest = i
return smallest
显示以下测试的示例:
Examples of following tests are shown:
print(second_smallest([5, 7, 2, 1, 3]))
2
print(second_smallest([100, 51, 31, 5, 10]))
10
感谢!
推荐答案
>>> def second_smallest(lst):
... first = second = float("inf")
... for num in lst:
... if num < first:
... second, first = first, num
... elif first < num < second:
... second = num
... return second
请注意,对于含有 len(lst)的列表,这返回
,因为列表中没有第二项。 float('inf')
(初始值) = 1
Please note that this returns float('inf')
(initial value) for lists with len(lst) <= 1
as there is no second item in the list.
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