I finished training my Sci-Kit Learn Logistic Regression model and it is performing at 100% accuracy. However, when I went to plot the decision boundary, I got a bit confused. I am not running the typical Logistic Regression model. Instead, I have transformed the elements of the input feature matrix into nonlinear 2nd degree polynomial terms before training with the help of Sci-Kit Learn's sklearn.preprocessing.PolynomialFeatures(degree = 2) function. According to the documentation, this function transforms two features $[x,y]$ to six features $[1,x,y,x^2,xy,y^2]$. With this in mind, I wrote the general equation of the decision boundary like this: $\theta _0x^0+\theta _1x^1+\theta_2y^1+\theta_3x^2+\theta_4x^1y^1+\theta_5y^2+b=0$, while obtaining all the $\theta$ parameters from model.coef_[0][0] for $\theta_0$, model.coef_[0][1] for $\theta_1$, model.coef_[0][2] for $\theta_2$ (and so on), and the $b$ parameter from model.intercept_[0]. When I plotted this on Desmos, it came up with a weird plot nothing close to a rational decision boundary. My question: does the method model.coef_ return the parameters $\theta_0$ to $\theta_5$, IN THAT ORDER? For example, if it returns [a, b, c, d, e, f], does that mean $\theta_0$ is equal to a, $\theta_1$ is equal to b, $\theta_2$ is equal to c and so on?
I am running logistic regression on a small dataset which looks like this:    After implementing gradient descent and the cost function, I am getting a 100% accuracy in the prediction stage, However I want to be sure that everything is in order so I am trying to plot the decision boundary line which separates the two datasets.  Below I present plots showing the cost function and theta parameters. As can be seen, currently I am printing the decision boundary line incorrectly.    Extracting data  clear all; close all; clc;  alpha = 0.01; num_iters = 1000;  %% Plotting data x1 = linspace(0,3,50); mqtrue = 5; cqtrue = 30; dat1 = mqtrue*x1+5*randn(1,50);  x2 = linspace(7,10,50); dat2 = mqtrue*x2 + (cqtrue + 5*randn(1,50));  x = [x1 x2]'; % X  subplot(2,2,1); dat = [dat1 dat2]'; % Y  scatter(x1, dat1); hold on; scatter(x2, dat2, '*'); hold on; classdata = (dat&gt;40);   Computing Cost, Gradient and plotting  %  Setup the data matrix appropriately, and add ones for the intercept term [m, n] = size(x);  % Add intercept term to x and X_test x = [ones(m, 1) x];  % Initialize fitting parameters theta = zeros(n + 1, 1); %initial_theta = [0.2; 0.2];  J_history = zeros(num_iters, 1);   plot_x = [min(x(:,2))-2,  max(x(:,2))+2]  for iter = 1:num_iters  % Compute and display initial cost and gradient     [cost, grad] = logistic_costFunction(theta, x, classdata);     theta = theta - alpha * grad;     J_history(iter) = cost;      fprintf('Iteration #%d - Cost = %d... \r\n',iter, cost);       subplot(2,2,2);     hold on; grid on;     plot(iter, J_history(iter), '.r');  title(sprintf('Plot of cost against number of iterations. Cost is %g',J_history(iter)));     xlabel('Iterations')     ylabel('MSE')     drawnow      subplot(2,2,3);     grid on;     plot3(theta(1), theta(2), J_history(iter),'o')     title(sprintf('Tita0 = %g, Tita1=%g', theta(1), theta(2)))     xlabel('Tita0')     ylabel('Tita1')     zlabel('Cost')     hold on;     drawnow      subplot(2,2,1);     grid on;         % Calculate the decision boundary line     plot_y = theta(2).*plot_x + theta(1);  % &lt;--- Boundary line      % Plot, and adjust axes for better viewing     plot(plot_x, plot_y)     hold on;     drawnow  end  fprintf('Cost at initial theta (zeros): %f\n', cost); fprintf('Gradient at initial theta (zeros): \n'); fprintf(' %f \n', grad);   The above code is implementing gradient descent correctly (I think) but I am still unable to show the boundary line plot. Any suggestions would be appreciated.    logistic_costFunction.m  function [J, grad] = logistic_costFunction(theta, X, y)      % Initialize some useful values     m = length(y); % number of training examples      grad = zeros(size(theta));      h = sigmoid(X * theta);     J = -(1 / m) * sum( (y .* log(h)) + ((1 - y) .* log(1 - h)) );      for i = 1 : size(theta, 1)         grad(i) = (1 / m) * sum( (h - y) .* X(:, i) );     end  end   EDIT:  As per the below answer by @Esmailian, now I have something like this:  [m, n] = size(x);  x1_class = [ones(m, 1) x1' dat1']; x2_class = [ones(m, 1) x2' dat2'];  x = [x1_class ; x2_class]
Markdown change: Intra-word emphasis now works There are tons of questions I could maybe post this as an answer to, but before choosing the best one, I'll just post this as an announcement question that can be linked elsewhere. The short story From now on, intra-word emphasis is possible in our Markdown version. That means you can now write  The &quot;V&quot; is for Model**View**Controller. That's un*fricking*believable.  and it renders like this:  The &quot;V&quot; is for ModelViewController. That's unfrickingbelievable.  This only works for asterisks, not for underscores. It works in posts and comments. The reasons There are two big reasons for this. For one thing, this is more or less the behavior of , which we eventually . &quot;More or less&quot; because CommonMark's emphasis behavior is still in flux, and supports a few things that we don't support yet, but at least with this change we're closer. The other reason is languages that don't even have the concept of &quot;words separated by spaces&quot;. See  and  for two examples. This change allow posts in those languages to emphasize words without having to resort to HTML tags. To use an example from the first linked post,  **北京**和**上海**是直辖市。   now renders like this:  北京和上海是直辖市。  Possible issues In the 99% cases, this should only make things better; I've only found very few posts where this change causes unintended bolding/italicizing. And even in those cases, note that the rendered versions of questions and answers are not re-rendered until a post is edited, so this will not affect anything in posts retroactively. Unlike posts, comments are rendered , so this immediately applies to all comments. In most cases this doesn't change anything or actually fixes things. Should you discover a rare case where this breaks an important comment badly, we can always edit it to fix. There's one side effect that I'd like to mention: If you find a great quicksort implementation on GitHub that has a bug that causes a crash when sorting Windows versions, and the author doesn't want to fix it, you may be inclined to say  f**k this s**t   (i.e., &quot;fork this sort&quot;). With intraword emphasis, this would however look like this:  fk this st  So you would have to escape the asterisks (one of them would actually be enough). I'm not worried about this case too much, since if you must put censor-worthy words into a post, you might as well have to jump through a little hoop. Some details This being Meta, there are probably lots of people interested in a more detailed description of the new behavior (don't feel bad if you're not one of them; just skip this section). Here are the major rules regarding intraword-emphasis:  As mentioned, it only applies to asterisks, not underscores. Underscores  this restriction was in place at all; asterisks are much less of a problem.  If a single or double asterisk is supposed to be interpreted as intra-word emphasis, it cannot have punctuation on one side; it actually must be inside a word. An example of an edge case this solves is  It's a web*-based solution. *see Wikipedia for a definition of &quot;web&quot;  where the asterisks after &quot;web&quot; could cause unintended italicizing if there was a second asterisk in the same paragraph somewhere.  For italics, the &quot;inside a word&quot; part of rule 2 is even a bit stricter. For bolding, any word or digit character is allowed; for italics, digits don't work either. That's because the asterisk is often used as a multiplication symbol, so  The game worked in 320200 and 640480 resolution.  wouldn't look right without this restriction.  A minor addition that I made that's only tangentially related to intra-word emphasis is that a sequence of more than three underscores or more than three asterisks will no longer take part in any emphasis. This shouldn't really cause any problems, but fixes some (e.g. the underline in the second blockquote in , while not really visible, is a weird combination of bold and italic underscore characters; after this change, it would just be (more) underscore characters). This behavior also matches CommonMark. Check whether $\sum _{m=1}^{\infty }\sum _{n=1}^{\infty }\frac{1}{\left(m+n\right)^2}$ converges or NOT? Problem with packages subfiles and import.sty after using TexLive Utility for updating Everything worked fine before using TexLive for updating my packages three days ago. Now it is not possible anymore to use the package subfiles   Here are the MWE  The minimum Preamble.tex file  \documentclass{amsart} \usepackage{subfiles}   The MWE.tex file  \documentclass[MWE.tex]{subfiles} \begin{document} MWE \end{document}   When I compile the MWE file I get on the console  This is pdfTeX, Version 3.14159265-2.6-1.40.20 (TeX Live 2019) (preloaded &gt;format=pdflatex)  restricted \write18 enabled. entering extended mode  (./MWE.tex  LaTeX2e &lt;2020-02-02&gt; patch level 5  L3 programming layer &lt;2020-02-25&gt; (/usr/local/texlive/2019/texmf-dist/tex/latex/subfiles/subfiles.cls  Document Class: subfiles 2020/02/14 v16 Multi-file projects (class)  Preamble taken from file `MWE.tex'  (/usr/local/texlive/2019/texmf-dist/tex/latex/tools/verbatim.sty)  (/usr/local/texlive/2019/texmf-dist/tex/latex/import/import.sty)  ! LaTeX Error: File `import.sty' not found.  Type X to quit or &lt;RETURN&gt; to proceed, or enter new name. (Default extension: sty)  Enter file name:   I checked and the import.sty file is in the right directory, and everything worked fine before. Do you have any idea about what happened (apart from the fact that I should not upgrade stuff before the official releases of TexLive I suppose  ...)? Is an integer a sum of two rational squares iff it is a sum of two integer squares? Let $a\in \mathbb Z$. Is it true that $a$ is a sum of two squares of rational numbers if and only if it is a sum of two squares of integers?  I came to face this problem while dealing with quaternion algebras. Allow questions to be saved as drafts prior to posting If I have an i.i.d. random variables $X_i$ from a uniform distribution on $[0,1]$.   How would I find scaling sequences $a_n$, $b_n$ such that $a_n(M_n − b_n)$ converges in distribution to a non-trivial limit function $G$ for  $ Y_i = X_i,$ and $M_n = \max(Y_1,...,Y_n) $ Can I rent a camper in the US/Canada with a European B license? We are considering a holiday in North America with a camper. I have a European driver's license (B) which allows me to drive personal cars up to 3500 kg, although I can imagine that an average camper easily weighs more.  Does my license qualify me to drive a camper? If not, can I take an American/Canadian driver's test which would allow me to drive a camper? Acount suspension reason is only shown on the main site profile not meta Group of order 15 is abelian How do I prove group of order 15 is abelian?  Is there any general strategy to prove that a group of particular order(composite order) is abelian? I was recently told (  ):     You cannot delete a class in production directly. You will need delete   the class from your sandbox and then deploy the deletions to your   production org.   When you deploy from sandbox to production, the   missing classes will come up in red and you can choose to deploy these   deletions to Production.   However, I'd like some clarification because finding information on how to do this is like pulling teeth, unfortunately.  Am I supposed to:   Delete the Trigger in Sandbox Make a deployment package in Sandbox Add the Trigger which is going to be listed in RED now Deploy (and that deletes it from Production)?   I am not seeing deleted Triggers in Red in Sandbox.  Are they NOT listed under "Triggers" for deployment, but elsewhere?   Frustrated = true.   Does this ONLY apply to Classes??  UPDATE:  I just tried again, with a CLASS, and it still doesn't show up at all as a Deployable package (red or not) after I delete it in the Sandbox.  UPDATE 2:  I was just able to delete a class, that I know was written via IDE from a third party, in both Sandbox and Production.  So now I'm wondering if it has to do with where it was originally deployed from (if from IDE, IDE can delete it - it from Sandbox, you have to jump through hoops).  :-( Illustrator Vector Graphics Appearing Pixellated Message appearing when a user was removed.     It seems I lost 1765 reputation points on Stack Overflow yesterday. The reason given in my history is "removed", with the description of "User was removed ()". I don't understand what that means. What caused this? How to tell whether a field is required or not in Apex by describe fields What are the proposed theoretical explanations for the EmDrive? My actual WhatsApp back total size is 474Mb but when I go to I cloud back up it is showing 10GB, how is it possible? I mean I have deleted all media files from my WhatsApp and in WhatsApp application the total space is 474MB and also in iCloud storage it is 474MB, but when I entered further in to Backups and then choose my phone, it shows total backup size is 14GB and WhatsApp itself is taking 10GB. Can someone help with this? Travelling on Business visa to UK for tourism purposes I already have a 6 month valid UK business visa received prior to April 24 2015. I was supposed to travel to the UK in April, but my trip was cancelled on medical grounds.  I am now planning to visit the UK in July for tourism. Would it be fine for me to do so? Will I be allowed to enter the UK? Should I carry the original invite letter and medical certificate to prove why I didn't go the first time? Can a university retract a degree for actions taken after the degree is awarded? I read that it is a best practice to double quote all of your variable expansions in Bash. I also read that one cannot use a shell glob (wildcard(*)) right after a double quoted variable expansion. These situations conflict when one wants use the value of a regex pattern, expanded out of a variable.   The reason I would sometimes want to combine regex with a wildcard is to keep my regex minimal, neater, in my personal taste.  My particular problem  I downloaded phpmyadmin into my document root directory and unzipped it, but I fail to rename it with mv by a regex pattern I put in a variable, and available when expanding its variable. Here's the exact trace:  userName@compName:/var/www/html# ll total 11336 drwxr-xr-x  3 root root     4096 Feb 14 07:04 ./ drwxr-xr-x  3 root root     4096 Feb 14 06:56 ../ -rw-r--r--  1 root root      612 Feb 14 06:57 index.nginx-debian.html drwxr-xr-x 12 root root     4096 Dec 23 08:50 phpMyAdmin-4.7.7-all-languages/ -rw-r--r--  1 root root 11589684 Dec 23 14:08 phpMyAdmin-latest-all-languages.zip userName@compName:/var/www/html# echo $pma [pP][hH][pP][mM][yY][aA][dD][mM][iI][nN] userName@compName:/var/www/html# mv "$pma"*/ phpmyadmin/ mv: cannot stat '[pP][hH][pP][mM][yY][aA][dD][mM][iI][nN]*/': No such file or directory   If I'll unquote the variable expansion to ${pma} I would indeed be able to combine the variable expansion and the regex, as in ${pma}*, but it is important to me to follow the best practice without exceptions, if I can.  My question  How could I keep the variable expansion double quoted, but still using the extracted value with a wildcard? I sometimes get files from my clients that have the wrong file extension. For example, the name is image.jpg but the file is actually a TIFF image. In many cases I can clarify it by opening the file in a text editor, looking at the first few bytes, then deducing which file type it is.  This works for me with JPEG, TIFF, GIF and PDF files. However there are many more file types out there.  Is it possible to automate identification of the correct file type by analyzing the containing data? Proving that orthogonal vectors are linearly independent I am trying to prove that $a_1$, $a_2$, $a_3$ are linearly independent.  I am asked to use vector product and prove that if $c_{1}a_{1} + c_{2}a_{2} + c_{3}a_{3} = 0$ then $c_1 = c_2 = c_3 = 0$  I am completely stuck on where to go with this problem. I would think that linearly independent then the null space of the space can only be equal to $0$. Would this prove it?