After reviewing the code below, you will see that sections 1 thru 3 merely prepare the incoming data to be in the right format for the least squares steps in section 4, which is merely 4 lines of code. Is there yet another way to derive a least squares solution? We’ll use python again, and even though the code is similar, it is a bit different. Both of these files are in the repo. It’s hours long, but worth the investment. Data Scientist, PhD multi-physics engineer, and python loving geek living in the United States. Every step involves two rows: one of these rows is being used to act on the other row of these two rows. How does that help us? We’ll cover more on training and testing techniques further in future posts also. We then operate on the remaining rows, the ones without fd in them, as follows: We do this for columns from left to right in both the A and B matrices. If you’ve never been through the linear algebra proofs for what’s coming below, think of this at a very high level. a \footnotesize{Mx3} matrix can only be multiplied on a \footnotesize{3xN} matrix or vector, where the \footnotesize{M ~ and ~ N} could be any dimensions, and the result of the multiplication would yield a matrix with dimensions of \footnotesize{MxN}). Realize that we went through all that just to show why we could get away with multiplying both sides of the lower left equation in equations 3.2 by \footnotesize{\bold{X_2^T}}, like we just did above in the lower equation of equations 3.9, to change the not equal in equations 3.2 to an equal sign? IF you want more, I refer you to my favorite teacher (Sal Kahn), and his coverage on these linear algebra topics HERE at Khan Academy. The documentation for numpy.linalg.solve (that’s the linear algebra solver of numpy) is HERE. It’s a worthy study though. Computes the “exact” solution, x, of the well-determined, i.e., full rank, linear matrix equation ax = b. Computes the “exact” solution, x, of the well-determined, i.e., full rank, linear matrix equation ax = b. Parameters We have not yet covered encoding text data, but please feel free to explore the two functions included in the text block below that does that encoding very simply. A detailed overview with numbers will be performed soon. Once a diagonal element becomes 1 and all other elements in-column with it are 0’s, that diagonal element is a pivot-position, and that column is a pivot-column. That is we want find a model that passes through the data with the least of the squares of the errors. Now, let’s consider something realistic. This is a conceptual overview. If we used the nth column, we’d create a linear dependency (colinearity), and then our columns for the encoded variables would not be orthogonal as discussed in the previous post. Thus, if we transform the left side of equation 3.8 into the null space using \footnotesize{\bold{X_2^T}}, we can set the result equal to the zero vector (we transform into the null space), which is represented by equation 3.9. Let’s use equation 3.7 on the right side of equation 3.6. Understanding this will be very important to discussions in upcoming posts when all the dimensions are not necessarily independent, and then we need to find ways to constructively eliminate input columns that are not independent from one of more of the other columns. Yes, \footnotesize{\bold{Y_2}} is outside the column space of \footnotesize{\bold{X_2}}, BUT there is a projection of \footnotesize{\bold{Y_2}} back onto the column space of \footnotesize{\bold{X_2}} is simply \footnotesize{\bold{X_2 W_2^*}}. Let’s start with the function that finds the coefficients for a linear least squares fit. There are multiple ways to solve such a system, such as Elimination of Variables, Cramer's Rule, Row Reduction Technique, and the Matrix Solution. Instead of a b in each equation, we will replace those with x_{10} ~ w_0, x_{20} ~ w_0, and x_{30} ~ w_0. Let’s look at the 3D output for this toy example in figure 3 below, which uses fake and well balanced output data for easy visualization of the least squares fitting concept. In the first code block, we are not importing our pure python tools. We’ll use python again, and even though the code is similar, it is a bit differ… This blog’s work of exploring how to make the tools ourselves IS insightful for sure, BUT it also makes one appreciate all of those great open source machine learning tools out there for Python (and spark, and there’s ones for R of course, too). Applying Polynomial Features to Least Squares Regression using Pure Python without Numpy or Scipy, \tag{1.3} x=0, \,\,\,\,\, F = k \cdot 0 + F_b \\ x=1, \,\,\,\,\, F = k \cdot 1 + F_b \\ x=2, \,\,\,\,\, F = k \cdot 2 + F_b, \tag{1.5} E=\sum_{i=1}^N \lparen y_i - \hat y_i \rparen ^ 2, \tag{1.6} E=\sum_{i=1}^N \lparen y_i - \lparen mx_i+b \rparen \rparen ^ 2, \tag{1.7} a= \lparen y_i - \lparen mx_i+b \rparen \rparen ^ 2, \tag{1.8} \frac{\partial E}{\partial a} = 2 \sum_{i=1}^N \lparen y_i - \lparen mx_i+b \rparen \rparen, \tag{1.9} \frac{\partial a}{\partial m} = -x_i, \tag{1.10} \frac{\partial E}{\partial m} = \frac{\partial E}{\partial a} \frac{\partial a}{\partial m} = 2 \sum_{i=1}^N \lparen y_i - \lparen mx_i+b \rparen \rparen \lparen -x_i \rparen), \tag{1.11} \frac{\partial a}{\partial b} = -1, \tag{1.12} \frac{\partial E}{\partial b} = \frac{\partial E}{\partial a} \frac{\partial a}{\partial b} = 2 \sum_{i=1}^N \lparen y_i - \lparen mx_i+b \rparen \rparen \lparen -1 \rparen), 0 = 2 \sum_{i=1}^N \lparen y_i - \lparen mx_i+b \rparen \rparen \lparen -x_i \rparen), 0 = \sum_{i=1}^N \lparen -y_i x_i + m x_i^2 + b x_i \rparen), 0 = \sum_{i=1}^N -y_i x_i + \sum_{i=1}^N m x_i^2 + \sum_{i=1}^N b x_i, \tag{1.13} \sum_{i=1}^N y_i x_i = \sum_{i=1}^N m x_i^2 + \sum_{i=1}^N b x_i, 0 = 2 \sum_{i=1}^N \lparen -y_i + \lparen mx_i+b \rparen \rparen, 0 = \sum_{i=1}^N -y_i + m \sum_{i=1}^N x_i + b \sum_{i=1} 1, \tag{1.14} \sum_{i=1}^N y_i = m \sum_{i=1}^N x_i + N b, T = \sum_{i=1}^N x_i^2, \,\,\, U = \sum_{i=1}^N x_i, \,\,\, V = \sum_{i=1}^N y_i x_i, \,\,\, W = \sum_{i=1}^N y_i, \begin{alignedat} ~&mTU + bU^2 &= &~VU \\ -&mTU - bNT &= &-WT \\ \hline \\ &b \lparen U^2 - NT \rparen &= &~VU - WT \end{alignedat}, \begin{alignedat} ~&mNT + bUN &= &~VN \\ -&mU^2 - bUN &= &-WU \\ \hline \\ &m \lparen TN - U^2 \rparen &= &~VN - WU \end{alignedat}, \tag{1.18} m = \frac{-1}{-1} \frac {VN - WU} {TN - U^2} = \frac {WU - VN} {U^2 - TN}, \tag{1.19} m = \dfrac{\sum\limits_{i=1}^N x_i \sum\limits_{i=1}^N y_i - N \sum\limits_{i=1}^N x_i y_i}{ \lparen \sum\limits_{i=1}^N x_i \rparen ^2 - N \sum\limits_{i=1}^N x_i^2 }, \tag{1.20} b = \dfrac{\sum\limits_{i=1}^N x_i y_i \sum\limits_{i=1}^N x_i - N \sum\limits_{i=1}^N y_i \sum\limits_{i=1}^N x_i^2 }{ \lparen \sum\limits_{i=1}^N x_i \rparen ^2 - N \sum\limits_{i=1}^N x_i^2 }, \overline{x} = \frac{1}{N} \sum_{i=1}^N x_i, \,\,\,\,\,\,\, \overline{xy} = \frac{1}{N} \sum_{i=1}^N x_i y_i, \tag{1.21} m = \frac{N^2 \overline{x} ~ \overline{y} - N^2 \overline{xy} } {N^2 \overline{x}^2 - N^2 \overline{x^2} } = \frac{\overline{x} ~ \overline{y} - \overline{xy} } {\overline{x}^2 - \overline{x^2} }, \tag{1.22} b = \frac{\overline{xy} ~ \overline{x} - \overline{y} ~ \overline{x^2} } {\overline{x}^2 - \overline{x^2} }, \tag{Equations 2.1} f_1 = x_{11} ~ w_1 + x_{12} ~ w_2 + b \\ f_2 = x_{21} ~ w_1 + x_{22} ~ w_2 + b \\ f_3 = x_{31} ~ w_1 + x_{32} ~ w_2 + b \\ f_4 = x_{41} ~ w_1 + x_{42} ~ w_2 + b, \tag{Equations 2.2} f_1 = x_{10} ~ w_0 + x_{11} ~ w_1 + x_{12} ~ w_2 \\ f_2 = x_{20} ~ w_0 + x_{21} ~ w_1 + x_{22} ~ w_2 \\ f_3 = x_{30} ~ w_0 + x_{31} ~ w_1 + x_{32} ~ w_2 \\ f_4 = x_{40} ~ w_0 + x_{41} ~ w_1 + x_{42} ~ w_2, \tag{2.3} \bold{F = X W} \,\,\, or \,\,\, \bold{Y = X W}, \tag{2.4} E=\sum_{i=1}^N \lparen y_i - \hat y_i \rparen ^ 2 = \sum_{i=1}^N \lparen y_i - x_i ~ \bold{W} \rparen ^ 2, \tag{Equations 2.5} \frac{\partial E}{\partial w_j} = 2 \sum_{i=1}^N \lparen y_i - x_i \bold{W} \rparen \lparen -x_{ij} \rparen = 2 \sum_{i=1}^N \lparen f_i - x_i \bold{W} \rparen \lparen -x_{ij} \rparen \\ ~ \\ or~using~just~w_1~for~example \\ ~ \\ \begin{alignedat}{1} \frac{\partial E}{\partial w_1} &= 2 \lparen f_1 - \lparen x_{10} ~ w_0 + x_{11} ~ w_1 + x_{12} ~ w_2 \rparen \rparen x_{11} \\ &+ 2 \lparen f_2 - \lparen x_{20} ~ w_0 + x_{21} ~ w_1 + x_{22} ~ w_2 \rparen \rparen x_{21} \\ &+ 2 \lparen f_3 - \lparen x_{30} ~ w_0 + x_{31} ~ w_1 + x_{32} ~ w_2 \rparen \rparen x_{31} \\ &+ 2 \lparen f_4 - \lparen x_{40} ~ w_0 + x_{41} ~ w_1 + x_{42} ~ w_2 \rparen \rparen x_{41} \end{alignedat}, \tag{2.6} 0 = 2 \sum_{i=1}^N \lparen y_i - x_i \bold{W} \rparen \lparen -x_{ij} \rparen, \,\,\,\,\, \sum_{i=1}^N y_i x_{ij} = \sum_{i=1}^N x_i \bold{W} x_{ij} \\ ~ \\ or~using~just~w_1~for~example \\ ~ \\ f_1 x_{11} + f_2 x_{21} + f_3 x_{31} + f_4 x_{41} \\ = \left( x_{10} ~ w_0 + x_{11} ~ w_1 + x_{12} ~ w_2 \right) x_{11} \\ + \left( x_{20} ~ w_0 + x_{21} ~ w_1 + x_{22} ~ w_2 \right) x_{21} \\ + \left( x_{30} ~ w_0 + x_{31} ~ w_1 + x_{32} ~ w_2 \right) x_{31} \\ + \left( x_{40} ~ w_0 + x_{41} ~ w_1 + x_{42} ~ w_2 \right) x_{41} \\ ~ \\ the~above~in~matrix~form~is \\ ~ \\ \bold{ X_j^T Y = X_j^T F = X_j^T X W}, \tag{2.7b} \bold{ \left(X^T X \right) W = \left(X^T Y \right)}, \tag{3.1a}m_1 x_1 + b_1 = y_1\\m_1 x_2 + b_1 = y_2, \tag{3.1b} \begin{bmatrix}x_1 & 1 \\ x_2 & 1 \end{bmatrix} \begin{bmatrix}m_1 \\ b_1 \end{bmatrix} = \begin{bmatrix}y_1 \\ y_2 \end{bmatrix}, \tag{3.1c} \bold{X_1} = \begin{bmatrix}x_1 & 1 \\ x_2 & 1 \end{bmatrix}, \,\,\, \bold{W_1} = \begin{bmatrix}m_1 \\ b_1 \end{bmatrix}, \,\,\, \bold{Y_1} = \begin{bmatrix}y_1 \\ y_2 \end{bmatrix}, \tag{3.1d} \bold{X_1 W_1 = Y_1}, \,\,\, where~ \bold{Y_1} \isin \bold{X_{1~ column~space}}, \tag{3.2a}m_2 x_1 + b_2 = y_1 \\ m_2 x_2 + b_2 = y_2 \\ m_2 x_3 + b_2 = y_3 \\ m_2 x_4 + b_2 = y_4, \tag{3.1b} \begin{bmatrix}x_1 & 1 \\ x_2 & 1 \\ x_3 & 1 \\ x_4 & 1 \end{bmatrix} \begin{bmatrix}m_2 \\ b_2 \end{bmatrix} = \begin{bmatrix}y_1 \\ y_2 \\ y_3 \\ y_4 \end{bmatrix}, \tag{3.2c} \bold{X_2} = \begin{bmatrix}x_1 & 1 \\ x_2 & 1 \\ x_3 & 1 \\ x_4 & 1 \end{bmatrix}, \,\,\, \bold{W_2} = \begin{bmatrix}m_2 \\ b_2 \end{bmatrix}, \,\,\, \bold{Y_2} = \begin{bmatrix}y_1 \\ y_2 \\ y_3 \\ y_4 \end{bmatrix}, \tag{3.2d} \bold{X_2 W_2 = Y_2}, \,\,\, where~ \bold{Y_2} \notin \bold{X_{2~ column~space}}, \tag{3.4} \bold{X_2 W_2^* = proj_{C_s (X_2)}( Y_2 )}, \tag{3.5} \bold{X_2 W_2^* - Y_2 = proj_{C_s (X_2)} (Y_2) - Y_2}, \tag{3.6} \bold{X_2 W_2^* - Y_2 \isin C_s (X_2) ^{\perp} }, \tag{3.7} \bold{C_s (A) ^{\perp} = N(A^T) }, \tag{3.8} \bold{X_2 W_2^* - Y_2 \isin N (X_2^T) }, \tag{3.9} \bold{X_2^T X_2 W_2^* - X_2^T Y_2 = 0} \\ ~ \\ \bold{X_2^T X_2 W_2^* = X_2^T Y_2 }, BASIC Linear Algebra Tools in Pure Python without Numpy or Scipy, Find the Determinant of a Matrix with Pure Python without Numpy or Scipy, Simple Matrix Inversion in Pure Python without Numpy or Scipy, Solving a System of Equations in Pure Python without Numpy or Scipy, Gradient Descent Using Pure Python without Numpy or Scipy, Clustering using Pure Python without Numpy or Scipy, Least Squares with Polynomial Features Fit using Pure Python without Numpy or Scipy, Single Input Linear Regression Using Calculus, Multiple Input Linear Regression Using Calculus, Multiple Input Linear Regression Using Linear Algebraic Principles. We now do similar operations to find m. Let’s multiply equation 1.15 by N and equation 1.16 by U and subtract the later from the former as shown next. As always, I encourage you to try to do as much of this on your own, but peek as much as you want for help. Here, due to the oversampling that we have done to compensate for errors in our data (we’d of course like to collect many more data points that this), there is no solution for a \footnotesize{\bold{W_2}} that will yield exactly \footnotesize{\bold{Y_2}}, and therefore \footnotesize{\bold{Y_2}} is not in the column space of \footnotesize{\bold{X_2}}. These steps are essentially identical to the steps presented in the matrix inversion post. Now let’s use those shorthanded methods above to simplify equations 1.19 and 1.20 down to equations 1.21 and 1.22. This is good news! As we perform those same steps on B, B will become the values of X. Let’s cover the differences. Also, we know that numpy or scipy or sklearn modules could be used, but we want to see how to solve for X in a system of equations without using any of them, because this post, like most posts on this site, is about understanding the principles from math to complete code. To understand and gain insights. Each column has a diagonal element in it, of course, and these are shown as the S_{kj} diagonal elements. I hope that the above was enlightening. How to do gradient descent in python without numpy or scipy. B has been renamed to B_M, and the elements of B have been renamed to b_m, and the M and m stand for morphed, because with each step, we are changing (morphing) the values of B. Recall that the equation of a line is simply: where \hat y is a prediction, m is the slope (ratio of the rise over the run), x is our single input variable, and b is the value crossed on the y-axis when x is zero. Let’s start with single input linear regression. the code below is stored in the repo as System_of_Eqns_WITH_Numpy-Scipy.py. Now, let’s arrange equations 3.1a into matrix and vector formats. In case you weren’t aware, when we multiply one matrix on another, this transforms the right matrix into the space of the left matrix. We’re only using it here to include 1’s in the last column of the inputs for the same reasons as explained recently above.

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