Statistics with Python

Question

Write a program to determine the mean, median, and standard deviation of a list of numbers
Count the duplicate numbers in the code, listing any that occur more than once with the number of times it occurs.

My implementation:

import math
def mean(input):
 return sum(input) / float(len(input))
def median(input):
 sorted_input = sorted(input)
 length = len(input)
 if length < 1:
 return None
 if length % 2 == 0:
 return (sorted_input[length / 2] + sorted_input[length / 2 - 1]) / float(2)
 else:
 return sorted_input[length / 2]
def duplicate_counts(input):
 distinct_input = set(input)
 for distinct in set(input):
 count = input.count(distinct)
 if (count > 1):
 print("{} occurs {} times".format(distinct, count))
def standard_deviation(input):
 return math.sqrt(mean([(x - mean(input)) ** 2 for x in input]))
def main():
 #sample run/test
 the_list = [3, 6, 7, 2, 8, 9, 2, 3, 7, 4, 5, 9, 2, 1, 6, 9, 6]
 print("The mean is: {}".format(mean(the_list)))
 print("The median is: {}".format(median(the_list)))
 print("The standard deviation is: {}".format(standard_deviation(the_list)))
 duplicate_counts(the_list)
if __name__ == "__main__":
 main()

Sample output:

The mean is: 5.23529411765 
The median is: 6 
The standard deviation is: 2.64640514805 
2 occurs 3 times 
3 occurs 2 times 
6 occurs 3 times 
7 occurs 2 times 
9 occurs 3 times 

Simple exercise I decided to do with Python for practice and to test/try all the excellent feedback I received on my previous two questions. There may be built in functions for these, and feel free to mention them, but for the sake of practice I implemented them this way.

• SuperBiasedMan: Are you using Python2 or 3?

Answer 1

• A variable name input is confusing. After all, input is a Python built-in.

• standard_deviation calls mean too many times (add a debug printout to mean to see). Better do

  def standard_deviation(input):
   input_mean = mean(input)
   return math.sqrt(mean([(x - input_mean) ** 2 for x in input]))
  

• Median can be calculated in \$ O(\log n)\$. Sorting a list in \$O(n \log n)\$ just for a median is an overkill.

• SirPython: I may be missing something (I probably am) but, in your second tip, it seems to me that all you are doing is storing one of the mean calls in a variable. Was tip a tip for increasing readability?
• Legato: Just to clarify, for calculating the median, you're suggesting I create a better performing sorting algorithm?
• vnp: @Legato Not sorting, but equal partitioning. To find a median you don't need partitions to be sorted.
• 301_Moved_Permanently: @SirPython the storage into a variable seems mainly to avoid computing the mean for each element in the input list, thus saving on CPU.
• SirPython: @MathiasEttinger Ah, that makes perfect sense! Thank you for clearing that up.

Answer 2

In mean, you're wrapping the result of len in a float. I'm guessing this is because of Python 2's odd division that will always return an int and not a float.

>>> 12/5
2

In Python 3.x this is no longer an issue as it will return proper float values, and you can get this in Python 2 as well, by using from __future__ import division. from __future__ is a way of importing new additions in Python 3 that were added to 2 but not in a way that overrides the old functionality. By importing this, division returns floats in Python 2 as well:

>>> from __future__ import division
>>> 12/5
2.4

Also I notice you use float later on the literal value 2. But you could just use 2.0 and get the same effect even if you don't do the import.

>>> 12/5.0
2.4

Should median really return None for input of length 0? Shouldn't that raise ValueError rather than silently return a non result?

Also you should indent by 4 spaces rather than 2. That's the commonly accepted Python style, and since it affects flow people could easily misread your code this way.