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# 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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