python：当最小元素数大于1时列表中的第二个最小值

Question

the following code returns the second smallest value only when the elements in the list are different from each other, but in this case, since '1' appears twice, the second smallest printed number is still '1'. How to fix this issue?

l=[1,2,1,3,4,5,6,2,7]
print(sorted(l)[1])

Answer 1

This is not the most efficient method, but much quicker than sorting a set:

l=[1,2,1,3,4,5,6,2,7]
min1 = min(l)
min2 = min(i for i in l if i > min1)

Answer 2

print(sorted(list(set(l)))[1])

你必须摆脱重复

Answer 3

You can use a set to get only the unique elements of your list. Then you sort the set and get the element on index 1.

l = [1, 2, 1, 3, 4, 5, 6, 2, 7]
sorted(set(l))[1]
>> 2

Answer 4

l=[1,2,1,3,4,5,6,2,7]
min = 100
sub_min = 100
for i in range(len(l)):
    if l[i] < min:
        min = l[i]
    if l[i] < sub_min and l[i] > min:
        sub_min = l[i]
# print(min, sub_min)

This might seem crude, but if you don't want to use the 'sorted' method and need a linear runtime, you can use this.