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CS251-Assignment 11 Intro to ML Solved
Q.
We will code a 1-D linear regression problem. We will provide pseudo-code here, you required to transform it into a Matlab/Octave code.
Important Note: Please don’t use external library. Octave default data structures and methods should suffice.
The data files are train.csv and test.csv.
File structure:
The structure of both files are similar. Train.csv has n_train = 10^4 rows and test.csv has n_test = 10^3 rows, each row corresponding to one data point. Each row has two values separated by comma. The first value is feature and second value is label of the data point. The first row of both files may/may not contain headers.
Example-
If one row of train.txt is :
4,7
Then feature, x = 4 and label, y = 7.
Pseudo-code
Step-1
- Read files train.csv
- Create vector - X_train (dim - n_train x 1) and vector - y_train (dim - n_train x 1)
- Add a column to X_train so that its dimension becomes n_train x 2. First column of X_train should be all 1 and 2nd column is the same as before adding extra column. Example -
X_train = [
2
3
4
]
New X_train = [
1 2
1 3
1 4
]
Step-2:
Generate a 2-D vector w (dim: 2 x 1) initialised randomly with floating point numbers.
Step-3
Plot y vs x using matplotlib where x is the feature and y is the label read from the file train.csv. (A scatter plot)
Consider x’ = [1 x] (prepending 1 to x to generate 2-dimensional x-vector)
On the same figure plot the line w^T*x’ vs x. (A straight line not necessarily passing through the scatter plot)
Your figure should have a dot corresponding to each datapoint (x,y) and a straight line on the plot corresponding to w^T*x’.
Step-4
Set w_direct = (X_train^T * X_train)^(-1)* X_train^T*y_train (if M is a matrix M^(-1) is inverse of M)
X_train is the n_train x 2 matrix defined earlier and y_train is the corresponding label vector.
Plot y vs x using matplotlib where x is the feature and y is the label read from the file train.csv. (A scatter plot)
Consider x’ = [1 x] (prepending 1 to x to generate 2-dimensional x-vector)
On the same figure plot the line w_direct^T*x’ vs x. (A straight line that should ideally pass through the scatter plot)
Your figure should have a dot corresponding to each datapoint (x,y) and a straight line on the plot corresponding to w_direct^T*x’.
Step-5
w - 2-dim vector initialised earlier (step-2)
Loop: for nepoch = 1 to N (N is the number of pass through the data(~2) , play with it to
find best fit)
Loop : for j = 1 to n_train
(x,y) ← jth row of train.csv x' ← [1, x]^T
w ← w - eta*( w^T*x’ - y)*x’ (eta = 0.00000001 students can change this value)
If j%100 == 0
Then plot y vs x as earlier and use current value of w to plot
w^T*x’ vs x
Step-6
Finally redraw the plot as earlier with latest value of w from the training. (like in step 3 and 4)
Note:- Don't use test.csv for training. That is test.csv should not be used before this point. Marks will be deducted otherwise.
Step-7
- Read files test.csv
- Create vector - X_test (dim - n_test x 1) and vector - y_test (dim - n_test x 1)
- Add a column to X_test so that its dimension becomes n_test x 2. First column of X_test should be all 1 and 2nd column is the same as before adding extra column.
- Let y_pred1 = X_test*w (w is the final value after doing step 5) - Calculate root mean squared error between y_pred1 and y_test.
- Let y_pred2 = X_test*w_direct
- Calculate root mean squared error between y_pred2 and y_test.
Derivation of Updates (optional reading)
In the loop of step 5 we did the following
w ← w - eta*(w^T*x’ - y)*x’
The objective of the above step is to reduce the least squared error.
Let error = 0.5(w^T*x’ - y)^2
So, derivative w.r.t. w gives: (w^T*x’ - y)x’
We need to descend down the gradient to reach the minima, so we subtract eta times the above derivative from w to gradually reach minima. We set eta to small value because otherwise, w may overshoot the minima. W_direct can also be computed on the same line by setting derivative to zero
We will code a 1-D linear regression problem. We will provide pseudo-code here, you required to transform it into a Matlab/Octave code.
Important Note: Please don’t use external library. Octave default data structures and methods should suffice.
The data files are train.csv and test.csv.
File structure:
The structure of both files are similar. Train.csv has n_train = 10^4 rows and test.csv has n_test = 10^3 rows, each row corresponding to one data point. Each row has two values separated by comma. The first value is feature and second value is label of the data point. The first row of both files may/may not contain headers.
Example-
If one row of train.txt is :
4,7
Then feature, x = 4 and label, y = 7.
Pseudo-code
Step-1
- Read files train.csv
- Create vector - X_train (dim - n_train x 1) and vector - y_train (dim - n_train x 1)
- Add a column to X_train so that its dimension becomes n_train x 2. First column of X_train should be all 1 and 2nd column is the same as before adding extra column. Example -
X_train = [
2
3
4
]
New X_train = [
1 2
1 3
1 4
]
Step-2:
Generate a 2-D vector w (dim: 2 x 1) initialised randomly with floating point numbers.
Step-3
Plot y vs x using matplotlib where x is the feature and y is the label read from the file train.csv. (A scatter plot)
Consider x’ = [1 x] (prepending 1 to x to generate 2-dimensional x-vector)
On the same figure plot the line w^T*x’ vs x. (A straight line not necessarily passing through the scatter plot)
Your figure should have a dot corresponding to each datapoint (x,y) and a straight line on the plot corresponding to w^T*x’.
Step-4
Set w_direct = (X_train^T * X_train)^(-1)* X_train^T*y_train (if M is a matrix M^(-1) is inverse of M)
X_train is the n_train x 2 matrix defined earlier and y_train is the corresponding label vector.
Plot y vs x using matplotlib where x is the feature and y is the label read from the file train.csv. (A scatter plot)
Consider x’ = [1 x] (prepending 1 to x to generate 2-dimensional x-vector)
On the same figure plot the line w_direct^T*x’ vs x. (A straight line that should ideally pass through the scatter plot)
Your figure should have a dot corresponding to each datapoint (x,y) and a straight line on the plot corresponding to w_direct^T*x’.
Step-5
w - 2-dim vector initialised earlier (step-2)
Loop: for nepoch = 1 to N (N is the number of pass through the data(~2) , play with it to
find best fit)
Loop : for j = 1 to n_train
(x,y) ← jth row of train.csv x' ← [1, x]^T
w ← w - eta*( w^T*x’ - y)*x’ (eta = 0.00000001 students can change this value)
If j%100 == 0
Then plot y vs x as earlier and use current value of w to plot
w^T*x’ vs x
Step-6
Finally redraw the plot as earlier with latest value of w from the training. (like in step 3 and 4)
Note:- Don't use test.csv for training. That is test.csv should not be used before this point. Marks will be deducted otherwise.
Step-7
- Read files test.csv
- Create vector - X_test (dim - n_test x 1) and vector - y_test (dim - n_test x 1)
- Add a column to X_test so that its dimension becomes n_test x 2. First column of X_test should be all 1 and 2nd column is the same as before adding extra column.
- Let y_pred1 = X_test*w (w is the final value after doing step 5) - Calculate root mean squared error between y_pred1 and y_test.
- Let y_pred2 = X_test*w_direct
- Calculate root mean squared error between y_pred2 and y_test.
Derivation of Updates (optional reading)
In the loop of step 5 we did the following
w ← w - eta*(w^T*x’ - y)*x’
The objective of the above step is to reduce the least squared error.
Let error = 0.5(w^T*x’ - y)^2
So, derivative w.r.t. w gives: (w^T*x’ - y)x’
We need to descend down the gradient to reach the minima, so we subtract eta times the above derivative from w to gradually reach minima. We set eta to small value because otherwise, w may overshoot the minima. W_direct can also be computed on the same line by setting derivative to zero
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