A function to calculate the interpolated median of a list of values.

interpolatedMedian.py

# interpolatedMedian.py

# Author: Richard Law

# Calculates the interpolated median of a list,
# which is a form of median that retains the
# important properties of a median (isn't swayed by
# unusually large or small values) but gives a better
# indication of the ditstribution of values

# See: http://aec.umich.edu/median.php

# An alternative: http://www.cad.vuw.ac.nz/wiki/index.php/Student_Feedback_Medians

import numpy as np

def interpolatedMedian(list_, alternative=False, interval=None):
	if len(list_) == 0:
		return None
	M = float(np.median(list_))
	if not alternative:
		nl = float(len([n for n in list_ if n < M]))
		ne = float(len([n for n in list_ if n == M]))
		ng = float(len([n for n in list_ if n > M]))
		if ne == 0:
			return M
		else:
			return M + (ng - nl) / (2 * ne)
	elif alternative == True and interval is not None:
		# Try the alternative method
		I = interval
		L = float(M - (interval * 0.5))
		N = float(len(list_))
		F = float(len([n for n in list_ if n == min(list_)]))
		f = float(len([n for n in list_ if n == M]))
		return L + I * (((N / 2.0) - F) / f)


# Testing
question1 = [5,5,5,5,5,5,5,5,5,4,4,4,4,4,4,4,4,4,4,2]
question2 = [5,4,4,4,4,4,4,4,4,4,4,3,3,3,3,3,3,2,1,1]
question3 = [1,1,1,1,1,1,1,2,2,2,2,2,2,2,2,3,3]
# These two lists have the same traditional median,
# but it is clear that <question1> tends higher than
# <question2>. The interpolated median recognises this.
print np.median(question1), interpolatedMedian(question1), interpolatedMedian(question1, alternative=True, interval=1)
# 4.0 4.4 4.4
print np.median(question2), interpolatedMedian(question2), interpolatedMedian(question2, alternative=True, interval=1)
# 4.0 3.6 4.3
print np.median(question3), interpolatedMedian(question3), interpolatedMedian(question3, alternative=True, interval=1)
# 2.0 1.7 1.7