ML | Mathematical explanation of standard deviation and R-squared error

RMSE: root mean square error — it is a measure of how well the regression line fits the data points. RMSE can also be interpreted as the standard deviation of the residual.
Consider these points: (1, 1), (2, 2), (2, 3), (3, 6).
We split the above data points into the 1st list.
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  x =  [1, 2, 2, 3]  y =  [1, 2, 3, 6]

Code: Regression Plot

import matplotlib.pyplot as plt

import math

# dots
plt. plot (x, y)

# x-axis title

plt.xlabel ( `x - axis` )

# Y-axis name

plt.ylabel ( `y - axis` )

# giving a title to my graphic

plt.title ( `Regression Graph` )

# plot show function
plt.show ()

Code: Average

       # in the next step we will find the equation of the line of best fit  
  # we will use the slope of a linear algebra point to find the equation of the regression line  
  # the shape of the slope is represented by y = mx + c  
  # where m means slope (change in y) / (change in x)  
  # c is a constant, it represents where the line will cross the Y-axis  
  # Slope m can be formulated like this:  
  "" " 
     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 x mean and y_mean

Output:

 Value of X mean 2.0 value of Y mean 3.0

Code: linear equation

# below is the process of finding a linear equation in mathematical terms
# our line`s slope is 2.5
# evaluate c to find the equation

m = 2.5

c = y_mean - m * x_mean

print ( `Intercept` , c)

Output:

 Intercept -2.0

Code: Medium square error

Our regression line equation looks like this:
# y_pred = 2.5x-2.0
# we name the string y_pred
# insert a regression line plot

   from   sklearn.metrics   import   mean_squared_error 
   # y_pred for our exhaustive data points as shown below  
   
   y   =   [  1  ,   2  ,   3  ,   6  ] 
   y_pred   =   [  0.5  ,   3  ,   3  ,   5.5  ]

# sklearn root mean square

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

# let`s see how the mean square is calculated mathematically
# let`s introduce a term called residuals
# remainder - it is basically the distance from the data point to the regression line
# the residuals are indicated by the red line in the graph below
# RMS and residuals are calculated as shown below
# we have 4 data points
"" "
r = 1, ri = yi-y_pred
y_pred is mx + c
ri = yi- (mx + c)
eg x = 1, we have the y value as 1
we want to evaluate exactly what our model 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 ) # on top of the calculation we have residual values residuals = [ 0.5 , - 1 , 0 ,. 5 ] # now calculate the root mean square error # N = 4 tons data points N = 4 rmse = math.sqrt ((r1 * * 2 + r2 * * 2 + r3 * * 2 + r4 * * 2 ) / N) print ( `Root Mean square error using maths` , rmse) # RMS value actually calculated using math # both RMSEs are calculated the same

Output:

Root Mean square error using maths 0.6123724356957945

R-squared error or coefficient of determination
Error R2 answers the following question.
How much y changes with change in x. Basically, the percentage change in y when changing from 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 the total deviation of Y from the mean # the change in y is calculated 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 computing y_var, we compute the distance # between data points y and mean y # so answer to our question,% of total deviation # of x is denoted as below: r_squared = 1 - (SE_line / SE_mean)    # [SE_line / SE_mean] - & gt; tells us what% variation # not described by the regression line # 1- (SE_li ne / SE_mean) - & gt; gives us the exact meaning # how much% y varies with x print ( `Rsquared error` , r_squared)

Output:

Rsquared error 0.8928571428571429

Code: R-Squared Error with sklear

from sklearn.metrics import r2_score


Error # r2 calculated by sklearn is similar
# to our mathematically calculated error r2
# compute r2 error with sklearn
r2 _score (y, y_pred)

Output:
0.8928571428571429

ML | Mathematical explanation of standard deviation and R-squared error

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