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ML | Mathematical explanation of RMSE and R-squared error

  • Last Updated : 02 Aug, 2019

RMSE: Root Mean Square Error is the measure of how well a regression line fits the data points. RMSE can also be construed as Standard Deviation in the residuals.
Consider the given data points: (1, 1), (2, 2), (2, 3), (3, 6).
Lets break the above data points into 1-d lists.
Input :

x = [1, 2, 2, 3]
y = [1, 2, 3, 6]

Code : Regression Graph




import matplotlib.pyplot as plt 
import math
  
# plotting the points  
plt.plot(x, y) 
  
# naming the x axis 
plt.xlabel('x - axis'
  
# naming the y axis 
plt.ylabel('y - axis'
  
# giving a title to my graph 
plt.title('Regression Graph'
  
# function to show the plot 
plt.show() 



Code: Mean Calculation




# in the next step we will find the equation of the best fit line
# we will use Linear algebra's Point slope form to find regression line equation
# point-slope form is represented by y = mx + c
# where m is slope means (change in y) / (change in x)
# c is constant, it represents at which point line will intercept y-axis
# slope m can be formulated as below:
'''
   n
m =? (xi - Xmean) (yi - Ymean)/?(xi - Xmean)^2
 i = 1
'''
# calculate Xmean and Ymean
ct = len(x)
sum_x = 0
sum_y = 0
  
for i in x:
    sum_x = sum_x + i
x_mean = sum_x / ct
print('Value of X mean', x_mean)
  
for i in y:
    sum_y = sum_y + i
y_mean = sum_y / ct
print('value of Y mean', y_mean)
  
# we have the values of x mean and y_mean

Output :

Value of X mean 2.0
value of Y mean 3.0

Code : Line Equation






# below is the process of finding line equation in mathematical terms
# slope of our line is 2.5
# calculate c to find out the equation
  
m = 2.5
c = y_mean - m * x_mean
print('Intercept', c)

Output :

Intercept -2.0

Code : Mean Squared Error




# equation of our Regression line comes out to be as below:
# y_pred = 2.5x-2.0
# we call the line y_pred
# paste regression line graph
from sklearn.metrics import mean_squared_error 
# y_pred for our exusting data points is as below
  
y =[1, 2, 3, 6]
y_pred =[0.5, 3, 3, 5.5]




# root mean square calculated by sklearn package
mse = math.sqrt(mean_squared_error(y, y_pred))
print('Root mean square error', mse)

Output :

Root mean square error 0.6123724356957945

Code : RMSE Calculation




# lets check how the Root mean square is calculated mathematically
# lets introduce a term called residuals
# residual are basically the distance of data point from the regression line
# residuals are denoted by red marked line in below graph
# root mean square and residuals are calculated as below
# we have 4 data points 
'''
r = 1, ri = yi-y_pred
y_pred is mx + c
ri = yi-(mx + c) 
e.g. x = 1, we have value of y as 1
we want to evaluate what exactly our model has predicted for x = 1
(1, 1)r1 = 1, x = 2 
'''
# y_pred1 = 1-(2.5 * 1-2.0)= 0.5
r1 = 1-(2.5 * 1-2.0)
  
#(2, 2) r2 = 2, x = 2 
# y_pred2 = 2-(2.5 * 2-2.0)=-1
r2 = 2-(2.5 * 2-2.0)
  
#(2, 3) r3 = 3, x = 2 
# y_pred3 = 3-(2.5 * 2-2.0)= 0
r3 = 3-(2.5 * 2-2.0)
  
#(3, 6) r4 = 4, x = 3 
# y_pred4 = 6-(2.5 * 3-2.0)=.5
r4 = 6-(2.5 * 3-2.0)
  
# from above calculation we have values of residuals
residuals =[0.5, -1, 0, .5]
  
# now calculate root mean square error
# N = 4 data points
N = 4
rmse = math.sqrt((r1**2 + r2**2 + r3**2 + r4**2)/N)
print('Root Mean square error using maths', rmse)
  
# root mean square actually calculated using mathematics
# both of RMSE calculated are same


Output :

Root Mean square error using maths 0.6123724356957945

R-squared Error or Coefficient of Determination
R2 error answers the below question.
How much y varies with variation in x.Basically the % variation of y on variation with x

Code : R-Squared Error




# SEline =(y1-(mx1 + b)**2 + y2-(mx2 + b)**2...+yn-(mxn + b)**2)
# SE_line =(1-(2.5 * 1+(-2))**2 + (2-(2.5 * 2+(-2))**2) +(3-(2.5*(2)+(-2))**2) + (6-(2.5*(3)+(-2))**2))
  
val1 =(1-(2.5 * 1+(-2)))**2
val2 =(2-(2.5 * 2+(-2)))**2
val3 =(3-(2.5 * 2+(-2)))**2
val4 =(6-(2.5 * 3+(-2)))**2
SE_line = val1 + val2 + val3 + val4
print('val', val1, val2, val3, val4)
  
# next to calculate total variation in Y from mean value
# variation in y is calcualted as
# y_var =(y1-ymean)**2+(y2-ymean)**2...+(yn-ymean)2
  
y =[1, 2, 3, 6]
  
y_var =(1-3)**2+(2-3)**2+(3-3)**2+(6-3)**2
SE_mean = y_var
  
# by calculating y_var we are calculating the distance
# between y data points and mean value of y
# so answer to our question, % of the total variation
# of wrt x is denoted as below:
r_squared = 1-(SE_line / SE_mean)
  
# [SE_line / SE_mean] -->tells us the what % of variation
# in y is not described by regression line
# 1-(SE_line / SE_mean) --> gives us the exact value of 
# how much % y varies with variation in x
print('Rsquared error', r_squared)

Output :

Rsquared error 0.8928571428571429

Code : R-Squared Error with sklearn




from sklearn.metrics import r2_score
  
# r2 error calculated by sklearn is similar 
# to ours mathematically calculated r2 error
# calculate r2 error using sklearn
r2_score(y, y_pred)

Output :

0.8928571428571429

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