def outlier(*args):
    outlist=[]
    def median(args1):
        if(len(args1)%2==1):
            return list(sorted(args1))[int((len(args1)/2)-0.5)]
        else:
            return (list(sorted(args1))[int(len(args1)/2)] list(sorted(args1))[int(len(args1)/2)-1])/2
    def fmax(args2):
        sortargs=sorted(args2)
        return sortargs[-1]
    def fmin(args3):
        sortargs=sorted(args3)
        return sortargs[0]
    q1=median(list(range(fmin(args),floor(median(args)) 1)))
    q3=median(list(range(floor(median(args)),fmax(args) 1)))
    for i in args:
        if(i<(q1-1.5*(q3-q1)) or i>(q3 1.5*(q3-q1)*(q3-q1))):
            outlist.append(i)
    return outlist

print(outlier(1,2,3,4,5,6,7,8,9,10,100000000))

I have tried to get the outlier values of a list in Python , but everytime I try it ,it returns an empty list or throws an error.

CodePudding user response：

If the list returns empty, the reason is that neither parts of your if condition is met and so nothing is appended to the list:

if(i<(q1-1.5*(q3-q1)) or i>(q3 1.5*(q3-q1)*(q3-q1))):   # never met

If you are not adverse to useing bigger module, you could calculate the quartiles using numpy, see

• How to calculate 1st and 3rd quartiles?

or use this answer that gives you a function for manual calculation:

BTW:

• sorted() returns a list so list(sorted(..)) is redundant
• while smaller functions are nice, multiple times sorting the same data is not efficient - sort it once - to get the min/max of a list use:
• all of your calculations need sorted input - you may as well sort it once on top and use the sorted list as inputs further down

You can also get the min and max in one sorting:

def minmax(data):
    if len(data) < 2: 
        raise ValueError("Must be iterable of len 2 or more")
    srt = sorted(data)

    return srt[0], srt[-1]

instead of

def fmax(args2):
    sortargs=sorted(args2)
    return sortargs[-1]

def fmin(args3):
    sortargs=sorted(args3)
    return sortargs[0]