Calculating Statistics

October 4, 2013

In today’s exercise we will do somebody’s homework:

Read a file containing integers, one integer per line. At the end of the file, write the number of integers in the file, their sum, and the mean, median, mode, minimum and maximum of the integers in the file.

Your task is to write the indicated program. When you are finished, you are welcome to read or run a suggested solution, or to post your own solution or discuss the exercise in the comments below.

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4 Responses to “Calculating Statistics”

  1. axiac said

    The median for the list of numbers in the example is 6, not 5.

  2. Paul said

    In Python.For the median I use: middle element if odd number of elements, else average of middle 2 elements.

    from __future__ import division
    
    from collections import Counter, namedtuple
    Stats = namedtuple("Stats", ['n', 'sum', 'mean', 'median', 'mode', 'min', 
                                 'max'])
    def stats(fname=None, data=None):
        """Use filename, if provided, else use data"""
        if fname:
            data = [int(line) for line in open(fname)]
        data = sorted(data)
        if not data:
            return Stats(0, None,  None,  None,  None,  None,  None)
        nr_data = len(data)
        sum_data = sum(data)
        mean_data = sum_data / nr_data
        mode_data = Counter(data).most_common(1)[0][0]
        median_data = (data[nr_data // 2] if nr_data & 1 else 
            (data[nr_data // 2] + data[nr_data // 2 + 1]) / 2)
        return Stats(nr_data, sum_data, mean_data, median_data, mode_data, data[0],
                data[-1])
    
    print stats(data=[1,2,3,4,5,7,8,2,3,6,7,1,2,3,4,9,8,6,7,3,4,6,7,4,5,4,3,2,2,5])
    # Stats(n=30, sum=133, mean=4.433333333333334, median=4.0, mode=2, min=1, max=9)
  3. Nothing special. Ruby.

    #!/usr/bin/env ruby
    
    if $PROGRAM_NAME == __FILE__
      frequency = Hash.new(0)
      median_sort = []
      File.open('whatever.txt') do |f|
        f.each do |line|
          number = line.to_i
          frequency[number] += 1
          median_sort << number
        end
      end
    
      # compute the number of integers
      size = median_sort.size
    
      # computer their sum
      sum = 0
      median_sort.each { |i| sum += i}
    
      # compute the mean
      mean = sum.to_f / size
    
      # compute the mode
      mode_pick = frequency.sort_by { |k,v| v }
      mode = mode_pick[-1][0]
    
      # compute the min and the max
      min_max = median_sort.sort
      min = min_max[0]
      max = min_max[-1]
    
      # compute the median
      half = (size.to_f - 1 )/ 2
      unless half%2 == 0
        half1 = half.floor
        half2 = half.ceil
      end
    
      median = min_max[half.to_i] if half % 2 == 0
      median = (min_max[half1] + min_max[half2]) / 2.0 unless half % 2 == 0
    
      puts "size: #{size}"
      puts "sum: #{sum}"
      puts "min: #{min}"
      puts "max: #{max}"
      puts "mean: #{mean}"
      puts "median: #{median}"
      puts "mode: #{mode}"
    end
    
  4. It’s tricky to determine the “best” approach without some knowledge of the expected data.

    Only a few dozen values, and all values are less than a thousand? Just read them all into an array of regular old integers, and do the math afterwards. The code will be simple, easy to read, and easy to maintain.

    Tens of millions of values? You might want to do something more clever, like keep a running total of count, total, min and max, and then do some hashtables for frequency (for the mode) and .. I don’t know, maybe some clever post-processing with the frequency tables for the median.

    Values could be > 4,294,967,295? Then you need to do some careful thinking. :0)

    But if you can be reasonably confident the values will be small enough (and few enough) that their sum fits comfortably in a regular old integer, and if the code is unlikely to be run more than a few dozen times on any given day, then I’d suggest the seasoned veteran programmer would take the simple approach and reduce the maintenance burden.

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