查找数组中第二大项无法按预期工作

Question

我找到第二个最小数字的程序按预期工作，但是当运行代码找到第二大数字时，第二个最大输出不正确。我以类似的方式完成了第二个功能，但翻转了大于号。修复代码的最佳方法是什么？

def second_smallest():
    num_list = [1,5,6,44,5,7,2,444,3]
    smallest_num = num_list[0]
    second_smallest_num = num_list[1]
    for i in num_list:
        if i < smallest_num:
            smallest_num = i
        elif smallest_num < i < second_smallest_num:
            second_smallest_num = i
    print("Smallest Number:", smallest_num)
    print("Second smallest number:", second_smallest_num)

def second_largest():
    num_list = [1,5,6,44,5,7,2,444,3]
    largest_num = num_list[0]
    second_largest_num = num_list[1]
    for i in num_list:
        if i > largest_num:
            largest_num = i
        elif second_largest_num < i < largest_num:
            second_largest_num = i
    print("Largest:", largest_num)
    print("Second largest:", second_largest_num)

Answer 1

分配新的最高编号时，需要将第二个最高编号分配给旧的最高编号。

def second_smallest():
     num_list = [1,5,6,44,5,7,2,444,3]
     smallest_num = num_list[0]
     second_smallest_num = num_list[1]
     for i in num_list:
         if i < smallest_num:
            smallest_num = i
         elif smallest_num < i < second_smallest_num:
         second_smallest_num = i
     print("Smallest Number:", smallest_num)
     print("Second smallest number:", second_smallest_num)

    def second_largest():
    num_list = [1,5,6,44,5,7,2,444,3]
    largest_num = num_list[0]
    second_largest_num = num_list[1]
    for i in num_list:
        if i > largest_num:
            second_largest_num = largest_num
            largest_num = i
        elif second_largest_num < i < largest_num:
            second_largest_num = i
    print("Largest:", largest_num)
    print("Second largest:", second_largest_num)

    second_smallest()
    second_largest()

Answer 2

您没有设置变量来保持最大值。如果您创建一个tmp变量来保存最大值并将其分配给第二大值，它将起作用

def second_largest():
num_list = [1,5,6,44,5,7,2,444,3]
largest_num = num_list[0]
second_largest_num = num_list[1]
for i in num_list:
    if i > largest_num:
        tmp = largest_num
        largest_num = i
        second_largest_num = tmp
    elif second_largest_num < i < largest_num:
        second_largest_num = i
print("Largest:", largest_num)
print("Second largest:", second_largest_num)

此外，如果从数组的索引2开始迭代作为第一个和第二个已经分配将更好。

Answer 3

在顶部，您声明的地方

    num_list = [1,5,6,44,5,7,2,444,3]
    smallest_num = num_list[0]
    second_smallest_num = num_list[1]

你是否意识到你正在访问列表中的第0个和第一个元素，即

smallest_num = 1
second_smallest_num = 5

顺便说一句，第0个元素（即= 1）恰好是该列表中的最小元素，这就是为什么它适用于其中一个函数但不适用于另一个函数的原因。

应该做些什么

def second_largest():
num_list = [1,5,6,44,5,7,2,444,3]
largest_num = max(num_list)
### SOME FUNCTION THAT TAKES IN THAT LIST, REMOVES THE LARGEST NUM ELEMENT, AND THEN FINDS AND RETURNS THE NEXT LARGEST ELEMENT - I'll leave this to you ##
for i in num_list:
    if i > largest_num:
        largest_num = i
    elif second_largest_num < i < largest_num:
        second_largest_num = i
print("Largest:", largest_num)
print("Second largest:", second_largest_num)

Answer 4

为什么不使用heapq？

>>> import heapq
>>> heapq.nlargest(2, num)
[10, 8]
>>> heapq.nsmallest(2, num)
[1, 2]
>>>

有关https://docs.python.org/2/library/heapq.html的更多信息

Answer 5

这里最简单的方法就是使用python builtins max()和min()。请注意，我们正在制作传入列表的副本，以便原始的num_list不会受到remove()的影响：

num_list = [1,5,6,44,5,7,2,444,3]

def second_smallest(list):
    temp_num_list = list.copy()
    temp_num_list.remove(min(temp_num_list))
    smallest_num = min(list)
    second_smallest_num = min(temp_num_list)
    return smallest_num, second_smallest_num

def second_largest(list):
    temp_num_list = list.copy()
    temp_num_list.remove(max(temp_num_list))
    largest_num = max(list)
    second_largest_num = max(temp_num_list)
    return largest_num, second_largest_num

largest_num, second_largest_num = second_largest(num_list)
smallest_num, second_smallest_num = second_smallest(num_list)

print("Largest:", largest_num)
print("Second largest:", second_largest_num)

print("Smallest:", smallest_num)
print("Second smallest:", second_smallest_num)

输出：

Largest: 444
Second largest: 44
Smallest: 1
Second smallest: 2

没有必要进行任何比较。我们只是找到max和min，然后从副本中删除它们，然后从剩余部分中取出max和min。

请注意，我已从函数中抽象出num_list，但将它们留在函数中会起作用。此方法允许您在代码中重用具有不同列表的函数。

或者，正如在另一个答案中所提到的，从@andercruzbr，你可以导入heapq模块，并简单地做：

import heapq

num_list = [1,5,6,44,5,7,2,444,3]

def smallest_nums(list):
    return heapq.nsmallest(2, num_list)

def largest_nums(list):
    return heapq.nlargest(2, list)

print("Largest:", largest_nums(num_list)[0])
print("Second largest:", largest_nums(num_list)[1])

print("Smallest:", smallest_nums(num_list)[0])
print("Second smallest:", smallest_nums(num_list)[1])

输出：

Largest: 444
Second largest: 44
Smallest: 1
Second smallest: 2