Given 3 numbers, I need to find which number lies between the two others.
ie,given 3,5,2 I need 3 to be returned.
I tried to implement this by going thru all three and using if else conditions to check if each is between the other two.But this seems a naive way to do this.Is there a better way?
Put them in a list, sort them, pick the middle one.
sorted((a, b, c))[1] in that regard.>>> x = [1,3,2] >>> sorted(x)[len(x) // 2] 2 The fastest obvious way for three numbers
def mean3(a, b, c): if a <= b <= c or c <= b <= a: return b elif b <= a <= c or c <= a <= b: return a else: return c 
mean3(3,5,2) will give 2 instead of 3 ?What you want is the median. You can use this code below for any number of numbers:
import numpy numbers = [3,5,2] median = numpy.median(numbers) for a custom solution you can visit this page.
You could do
numbers = [3, 5, 2] sorted(numbers)[1] 
1This is a O(n) implementation of the median using cumulative distributions. It's faster than sorting, because sorting is O(ln(n)*n).
def median(data): frequency_distribution = {i:0 for i in data} for x in data: frequency_distribution[x] =+ 1 cumulative_sum = 0 for i in data: cumulative_sum += frequency_distribution[i] if (cumulative_sum > int(len(data)*0.5)): return i Check this (Suppose list already sorted):
def median(list): ceil_half_len = math.ceil((len(list)-1)/2) # get the ceil middle element's index floor_half_len = math.floor((len(list)-1)/2) # get the floor middle element 's index' return (list[ceil_half_len] + list[floor_half_len]) / 2 
TypeError: list indices must be integers, not float. You need to wrap math.ceil and math.floor with int()If it's just for 3 numbers. First find the max and the min numbers in list. Remove them, the remainder of list is the medium. If you don't want to use the "if-else" or "sorted()" that is.
def mid(a,b,c): num = [a,b,c] small, big = min(num), max(num) num.remove(small) num.remove(big) return num[0] # return the remainder Here is my attempt using a more pythonic version
def median(a): sorted_a = sorted(a) if len(a) % 2 == 0: median = sum(sorted_a[(len(a)//2)-1:(len(a)//2)+1])/2. else: median = sorted_a[(len(a)-1)//2] >>> x = [64630, 11735, 14216, 99233, 14470, 4978, 73429, 38120, 51135, 67060] >>> median(x) >>> 44627.5 >>> y = [1, 2, 3] >>> median(y) >>> 2 If you wish to avoid sorting, you can do:
def find_median(x): return sum(x) - max(x) - min(x) 
[1,1,2]?