Python - Find second smallest number

Question

I found this code on this site to find the second largest number:

def second_largest(numbers):
    m1, m2 = None, None
    for x in numbers:
        if x >         


        

Answer 1

I am writing the code which is using recursion to find the second smallest element in a list.

def small(l):
 small.counter+=1;
 min=l[0];

 emp=[]

 for i in range(len(l)):
    if l[i]<min:
        min=l[i]

 for i in range(len(l)):
    if min==l[i]:
     emp.append(i)

 if small.counter==2:
    print "The Second smallest element is:"+str(min)
 else:
   for j in range(0,len(emp)):

     l.remove(min)

   small(l)
small.counter = 0

list=[-1-1-1-1-1-1-1-1-1,1,1,1,1,1]
small(list)

You can test it with various input integers.

Answer 2

a = [6,5,4,4,2,1,10,1,2,48]
s = set(a) # used to convert any of the list/tuple to the distinct element and sorted sequence of elements
# Note: above statement will convert list into sets 
print sorted(s)[1] 

Answer 3

l = [41,9000,123,1337]

# second smallest
sorted(l)[1]
123


# second biggest
sorted(l)[-2]
1337

Answer 4

Yes, except that code relies on a small quirk (that raises an exception in Python 3): the fact that None compares as smaller than a number.

Another value that works is float("-inf"), which is a number that is smaller than any other number.

If you use that instead of None, and just change -inf to +inf and > to <, there's no reason it wouldn't work.

Edit: another possibility would be to simply write -x in all the comparisons on x, e.g. do if -x <= m1: et cetera.