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# simonoff/linearregression.rb

Forked from scottsbaldwin/linearregression.rb
Created Nov 18, 2013
 # Adapted from a C# example here: # http://stackoverflow.com/questions/43224/how-do-i-calculate-a-trendline-for-a-graph # And thanks to John Esser for helping figure out how to # calculate the targets to stabilize a negative slope! class LinearRegression attr_accessor :slope, :intercept # Pass in an array of values to get the regression on def initialize y_values @y_values = y_values @size = @y_values.size @x_values = (1..@size).to_a #initialize everything to 0 sum_x = 0 sum_y = 0 sum_xx = 0 sum_xy = 0 # calculate the sums @y_values.zip(@x_values).each do |y, x| sum_xy += x*y sum_xx += x*x sum_x += x sum_y += y end # calculate the slope @slope = 1.0 * ((@size * sum_xy) - (sum_x * sum_y)) / ((@size * sum_xx) - (sum_x * sum_x)) @intercept = 1.0 * (sum_y - (@slope * sum_x)) / @size end # Get the y-axis values of the trend line def trend return @x_values.map { |x| predict(x) } end # Get the Y value for any given X value # from y = mx + b, or # y = slope * x + intercept def predict( x ) y = @slope * x + @intercept end # Get the "next" value if the sequence # was continued one more element def next predict (@size + 1) end # Determine the target needed to stabilize # the slope back to 0, assuming slope is # negative. The target returned will be # the number needed to have stabilized slope # inclusive of all values in the set def stabilize_over_all target = stabilize @y_values target end # Determine the target needed to stabilize # the slope back to 0, assuming slope is # negative. The target returned will be # the number needed to have stabilized slope # over a set of values of size n (including # the target value def stabilize_over_n_values(n) values = [] values = @y_values.last(n-1) if n - 1 >= 0 target = stabilize values target end private def stabilize(values) n = values.length sum = 0 target = 0 values.each_with_index do |value, i| sum += (n - (2 * i)) * value end target = 1.0 * sum / n if n > 0 target end end
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