如何在 python 的列表中找到第二小的数字

Question

I wrote this code to find the second lowest number in a list, but I'm not getting the correct result. Can you show me what I did wrong?

def second_lowest_number(list):
    
    the_minimum = min(list)
    new = []
    for i in range(len(t)):
        if list[i] > the_minimum:
            new.append(list[i])

    return min(new)

Answer 1

print(sorted(the_list)[1])

I would think would work and would not be any slower than any other solution

generally just keep it simple

if it has duplicates just use a set

 print(sorted(set(the_list))[1]) 

note that the first solution will fail if there is only one item in the list, and the second solution will fail if all items in the list are the same value

Answer 2

One problem with your function is that there is no variable t so the for loop would not run. Also, "list" is an in-built python keyword so avoid using that as the parameter name. Here is the working version of your implementation:

def seconde_lowest_number(arr):
    
    the_minimum = min(arr)
    new = []
    for i in range(len(arr)):
        if arr[i] > the_minimum:
            new.append(arr[i])

    return min(new)

although I would recommend just sorting the list and returning the 2nd index like so:

def second_lowest_number(arr):
    arr.sort()
    return arr[1]

Answer 3

Actually, you have a typo in your for loop.
It needs to be like this

for i in range(len(list)):

But, I would like to say something to you.
1. Don't use built-in function name as your variable/list name. You can use newlist instead.

2 . It's better to avoid using range(len(listname)) .
You can try this instead

for i in list:

Answer 4

The answer depends on what your input list looks like, if your list contains unique numbers (no repetition) you can do something like below

numbers = [3, 6, 1, 7, 8, 2]


def second_lowest_number(arr):
    arr.sort()
    return arr[1]

However, if your list contains duplicated numbers, you could try something like this

numbers = [3, 3, 1, 1, 5, 6]


def second_lowest_number(arr):
    arr.sort()
    for i in arr:
        if i > min(arr):
            return i

Answer 5

If you need it in linear time, I think this might work:

def second_lowest_number(list):
    the_minimum = list[0]
    second_min = the_minimum
    flag = False
    for i in range(len(list)):
       if list[i] < the_minimum:
           second_min = the_minimum
           the_minimum = list[i]
       elif the_minimum == second_min and list[i] > the_minimum and flag == False:
           second_min = list[i]
           flag = True
       elif list[i] < second_min and list[i] != the_minimum:
           second_min = list[i]
    return second_min
print(second_lowest_number([-2,3,5,2,2,0,5,1,2,3,5,-1]))

Please don't use this as a final answer, I'm just trying to brainstorm and aid as best as I can. If someone found cases where this code fails please share with me.