What are the pros and cons of logistic regression and SVM (support vector machines)?
Can someone please give me some intuition as to when to choose either SVM or LR? I want to understand the intuition behind what is the difference between the optimization criteria of learning the hyperplane of the two, where the respective aims are as follows:     SVM: Try to maximize the margin between the closest support vectors LR: Maximize the posterior class probability    Let's consider the linear feature space for both SVM and LR.  Some differences I know of already:   SVM is deterministic (but we can use Platts model for probability score) while LR is probabilistic. For the kernel space, SVM is faster (stores just support vectors)
The new 17.10 only has 64 bit ISOs. However 32 bit is still supported, there just are no live ISOs for it.  I have a 32 bit machine with WIFI, and I need to install from a USB disk. I tried the "mini.iso" but it doesn't have WIFI support.  What options do I have?  (PS I don't want another distro, and I can't use ethernet for installation, must be WIFI). Unfair 3 Sided Die Question Say you have an unfair 3-sided die.  Side 1 will be rolled with probability $P_1.$ Side 2 will be rolled with probability $P_2.$ Side 3 will be rolled with probability $P_3.$  Find a formula for the probability that the 8th 3 will be rolled on the 10th roll.  Here is how I started:  $$ \sum_{i=8}^{10}P(T_8 = i) $$  Im just having trouble with that expansion? Show that a positive operator on a complex Hilbert space is self-adjoint Let $(\mathcal{H}, (\cdot, \cdot))$ be a complex Hilbert space, and $A :  \mathcal{H} \to \mathcal{H}$ a positive, bounded operator ($A$ being positive means $(Ax,x) \ge 0$ for all $x \in \mathcal{H}$). Prove that $A$ is self-adjoint. That is, prove that $(Ax,y) = (x, Ay)$ for all $x,y \in \mathcal{H}$. Here's what I have so far. Because $A$ is positive we have $\mathbb{R} \ni (Ax,x) = \overline{(x,Ax)} = (x,Ax)$, all $x \in \mathcal{H}$. Next, I have seen some hints that tell me to apply the polarization identity: $$(x,y) = \frac{1}{4}((\lVert x+y \rVert^2 + \lVert x-y \rVert^2) - i(\lVert x + iy \rVert^2 - \lVert x - iy \rVert^2)),$$ where of course the norm is defined by $\lVert \cdot \rVert^2 = (\cdot, \cdot)$. So my guess is that I need to start with the expressions: $$(Ax,y) = \frac{1}{4}((\lVert Ax+y \rVert^2 + \lVert Ax-y \rVert^2) - i(\lVert Ax + iy \rVert^2 - \lVert Ax - iy \rVert^2)),$$ $$(x,Ay) = \frac{1}{4}((\lVert x+Ay \rVert^2 + \lVert x-Ay \rVert^2) - i(\lVert x + iAy \rVert^2 - \lVert x - iAy \rVert^2)),$$ and somehow show they are equal. But here is where I have gotten stuck. Hints or solutions are greatly appreciated. Connecting subplots Is there a way to draw a sort of connection between subplots in TiKz/pgfplot, like in this ? I'm looking to achieve something like this:    But I would need to draw the "x 2000" line between the plots.  EDIT: So right now I have something like this:    Coming from my main.tex:  \input{aipcheck}  \documentclass[final]{aipproc}  \layoutstyle{8x11single} \usepackage[utf8]{inputenc} \usepackage{algorithm2e} \usepackage{color} \usepackage[usenames,dvipsnames]{xcolor} \usepackage{cleveref} \usepackage{pgfplots}  \pgfplotsset{ tick label style={font=\tiny}, %axis numbers label style={font=\small}, %axis labels legend style={font=\tiny} }  \begin{document}  (...)  \begin{figure}     \centering     \newlength\figureheight      \newlength\figurewidth      \setlength\figureheight{5cm}      \setlength\figurewidth{5cm}     \input{figs/_ddmp2d.tikz} \label{fig2D} \end{figure}   with _ddmp2d.tikz, originally created from matlab2tikz, but already modified, containing:  % This file was created by matlab2tikz v0.3.3. % Copyright (c) 2008--2013, Nico Schlömer &lt;nico.schloemer@gmail.com&gt; % All rights reserved. %  % The latest updates can be retrieved from %   http://www.mathworks.com/matlabcentral/fileexchange/22022-matlab2tikz % where you can also make suggestions and rate matlab2tikz. %  %  %   \begin{tikzpicture}  \begin{axis}[% width=\figurewidth, height=\figureheight, scale only axis, xmin=0, xmax=15, xlabel={Number of iterations}, ymin=0, ymax=16000000000000, name=plot1, after end axis/.append code={\coordinate (a) at (axis description cs:1,0);}]  \addplot [ thick, color=red, dash pattern=on 1pt off 3pt on 3pt off 3pt, forget plot ] table[row sep=crcr]{ 1 15841000000000\\ (...) 15 25160000000\\ }; \addplot [ thick, color=black, dashed, forget plot ] table[row sep=crcr]{ 1 15841000000000\\ (...) 15 52501000000\\ }; \addplot [ thick, color=blue, dotted, forget plot ] table[row sep=crcr]{ 1 15841000000000\\ (...) 15 5324200000\\ }; \addplot [ thick, color=green, solid, forget plot ] table[row sep=crcr]{ 1 15841000000000\\ (...) 15 551920000\\ }; \addplot [ thick, color=green, solid, forget plot ] table[row sep=crcr]{ 3 0\\ 3 16000000000000\\ };  \addplot [ thick, color=blue, dotted, forget plot ] table[row sep=crcr]{ 4 0\\ 4 16000000000000\\ }; \addplot [ thick, color=black, dashed, forget plot ] table[row sep=crcr]{ 5 0\\ 5 16000000000000\\ }; \addplot [ thick, color=red, dash pattern=on 1pt off 3pt on 3pt off 3pt, forget plot ] table[row sep=crcr]{ 8 0\\ 8 16000000000000\\ }; \end{axis}  \begin{scope}[xshift=5cm] \begin{axis}[% width=0.35\figurewidth, height=\figureheight, scale only axis, xmin=995, xmax=1000, xtick={997,998,999,1000}, xticklabels={$997$,$998$,$999$,$1000$}, ymin=0, ymax=8000000000, at=(plot1.right of south east), anchor=left of south west, yticklabel pos=right, axis x discontinuity=crunch, after end axis/.append code={\coordinate (b) at (axis description cs:0,1);} ] \addplot [ thick, color=red, dash pattern=on 1pt off 3pt on 3pt off 3pt ] table[row sep=crcr]{ 995 6093100000\\ 996 6093100000\\ 997 6093100000\\ 998 6093100000\\ 999 6093100000\\ 1000 6093100000\\ };  \addplot [ thick, color=black, dashed ] table[row sep=crcr]{ 995 1577100000\\ 996 1598000000\\ 997 1598500000\\ 998 1628600000\\ 999 1631000000\\ 1000 1595900000\\ };  \addplot [ thick, color=blue, dotted ] table[row sep=crcr]{ 995 2592000000\\ 996 2591900000\\ 997 2591700000\\ 998 2591600000\\ 999 2591500000\\ 1000 2591400000\\ };  \addplot [ thick, color=green, solid ] table[row sep=crcr]{ 995 391810000\\ 996 391810000\\ 997 391810000\\ 998 391810000\\ 999 391810000\\ 1000 391810000\\ };  \end{axis} \draw[gray] (a) -- (b) node[pos=0.5,sloped,fill=white] {$\times$2000}; \end{scope}  \end{tikzpicture}% I just upgraded my texlive distribution and can no longer insert space in table rows using \\[x pt] syntax when colortbl is loaded.  It seems that recent changes to array.sty might be the problem as this is the only package to change recently.  Can anyone suggest a workaround?  MWE  \listfiles \documentclass{article} \usepackage{colortbl} \begin{document} \begin{tabular}{l}   \hline   No space below with \texttt{colortbl.sty} loaded. \\[10mm]   \hline \end{tabular} \end{document}   File versions in use:   article.cls    2014/09/29 v1.4h Standard LaTeX document class   size10.clo    2014/09/29 v1.4h Standard LaTeX file (size option) colortbl.sty    2012/02/13 v1.0a Color table columns (DPC)    array.sty    2018/04/07 v2.4g Tabular extension package (FMi)    color.sty    2016/07/10 v1.1e Standard LaTeX Color (DPC)    color.cfg    2016/01/02 v1.6 sample color configuration   pdftex.def    2018/01/08 v1.0l Graphics/color driver for pdftex Refused entry to UK, not sure what to do I'm a writer and theater/film director who has been traveling nearly all my life without incident. In February, I decided to pull up stakes in NYC and move to London. I am a freelancer and I'm working with three different organizations in NYC so I wasn't working illegally in the UK. I was, however, seeing if there might be some opportunities in London, ie. meeting with Artistic Directors of various theatre companies to see if they might be interested in producing my work.   In April, I was invited to a conference in San Diego and then a film festival in Switzerland. I didn't think it would be a problem if I went and then bummed around in France for a few days since I hadn't broken any laws and had left the UK well under my six month leave to stay. Well, when I tried to return to London, I was denied entry at Luton. The border control officer said I didn't have enough funds for a two month stay and put a black cross over a stamp in my passport. They put me in a detention room until the next flight back to Lyon, which was the following morning. I don't know anyone in Lyon so I ended up going to a friend's place in Paris.   After a few days of not knowing what to do, a friend suggested that perhaps I should try going back to London on the Eurostar since border control is in Paris. I wondered if I ought to go to the British Embassy but was deterred by something on their website that says that a visitors visa could take up to three weeks (I can't sit around here in Paris for three weeks) and anyway, the decision ultimately is with the border control, so I decided to take a chance on the Eurostar. I bought myself a return ticket in two weeks to Paris just so I had something to show.   Well, the UK border control denied me entry after giving me a third degree. They accused me of lying and seeking work in the UK. They also insisted that the amount of money I had was not enough to live on for two weeks. So I'm back at my friend's place in Paris, wondering what to do.   Is it possible for me to ever get back to the UK? I had to cancel one meeting and I'm wondering if I have to cancel going to the Sheffield DocFest. (Two of my projects were accepted to programs there.)   Should I apply for a visa from the British Embassy? Will this really take three weeks? When I started to say something about obtaining a visa, the border control officer cut me off and said, "Why are you going to the Embassy? This has nothing to do with the Embassy. We make the decisions." How to choose a predictive model after k-fold cross-validation? Basic question, how do I resume interrupted / failed downloads in Firefox? reversing plot axis for Plot, LogPlot, LogLogPlot $\varepsilon$-number countability without choice Let $\alpha\mapsto\varepsilon_\alpha$ be the enumeration of the $\varepsilon$-numbers--that is, those $\alpha$ such that $\omega^\alpha=\alpha$--by the ordinals.  If we know that countable unions of countable sets are countable, (slightly weaker than countable choice), then we can show fairly easily that $\varepsilon_\alpha$ is countable iff $\alpha$ is countable.  I think I saw a result at some point that $\varepsilon_0$ is countable without relying on any choice principle (though I can't recall where I saw that). Is this true? More generally, for which $\alpha$ can we conclude that $\varepsilon_\alpha$ is countable in ZF?  Update: I have been able to prove (in ZF) that if $\varepsilon_\alpha$ is countable, then $|\alpha|&lt;\aleph_1$, so that's one direction. I'm still a bit uncertain about the answers I've got so far. In Ubuntu, I want to scroll up while a command is running in my terminal window and generates new output (the command that is running is rspec). I can scroll up, but each time rspec generates a new line of console output, it throws me back down to the bottom. Is there a way to scroll all the way up? I'm using terminator on Linux Mint/Ubuntu 12.04 LTS. I found a possible workaround.  says that with stty ixon  you can turn on &quot;XON/XOFF flow control for terminal output. Being able to pause output by typing ^S and resume it by typing ^Q [...]&quot; This it not totally what I wanted though, without having to pause the output (and maybe wondering when forgetting to turn it on again), I thought there might be a simple modifier key like shft-alt with mouse wheel scroll-up. Are there better solutions? I'm trying to include a tabu inside a new definition of an environment but it fails if the postamble (enddef) of the new environment includes tabu's contents, so cells. With tabular or tabular* works. The error is:   ! Missing number, treated as zero. \tabu@0.H5   The MWE:  \documentclass{article} \usepackage{tabu} \newenvironment{testt} {\begin{tabu} to \textwidth{cccc}  11 &amp; 12 &amp; 13 &amp; 14 \\  21 &amp; 22 &amp; 23 &amp; 24 \\} {  41 &amp; 42 &amp; 43 &amp; 44 \end{tabu} }  \begin{document}  \begin{testt}  31 &amp; 32 &amp; 33 &amp; 34 \\ \end{testt} \end{document}   How to solve this? How to check a particular directory is mounted on the Linux machine. For instance there is a folder named test, I want to check if it is mounted or not. I was viewing some old posts on Stack Overflow (two-year-old posts) and found that due to "fast technology trend changing," some of them are no longer valid. Stack Overflow (and the Stack Exchange sites generally) should have some sort of criteria to tag questions as Obsolete, or a vote-based system to mark them as such.  I am not in favour of closing obsolete questions. I'm looking for a way to insert continuous colour legend for a raster layer.  I managed to create discrete symbols with values along the bar, but I would like to have a continuous bar with some values on the bar, similar to the old plugin "One-Band Raster", that it doesn't work with the QGIS version higher than 1.7  I tried to set in properties->style, where I selected the  Render Type "Singleband pseudocolor" and see the "generate new color ramp" box.  I tried both "Continuous" or "Equal interval". However, in the Composer the legend always is displayed with discrete symbols.  Is there any way that can make QGIS Composer to display a nice continuous band for a raster layer? If you are falling, can you Misty Step to the ground to avoid damage? If I fall off a cliff - say, 1000 feet - can I wait (readied action?) until I am near enough to the ground (my understanding is you fall 500 ft/round) then  safely to the ground? Or, does the momentum from the fall carry over through the teleport and I still splatter across the rocks? Salesforce to Xero integration using Named Credentials by OAuth 2.0 I know that I have to use some tweaking tool and then a extension for this. But all extensions that are suggested for this(Workspace grid, Workspace matrix), aren't available in the software center. Why might one also use a blank constructor? I was reading some Java recently and came across something (an idiom?) new to me: in the program, classes with multiple constructors would also always include a blank constructor. For example:   public class Genotype {   private boolean bits[];   private int rating;   private int length;   private Random random;    public Genotype() {              //  &lt;= THIS is the bandit, this one right here     random = new Random();   }    /* creates a Random genetoype */   public Genotype(int length, Random r) {     random = r;     this.length = length;     bits = new boolean[length];      for(int i=0;i&lt;length;i++) {         bits[i] =random.nextBoolean();     }   }    /* copy constructor */   public Genotype(Genotype g,Random r) {     random = r;     bits = new boolean[g.length];     rating = g.rating;     length = g.length;      for(int i=0;i&lt;length;i++) {         bits[i] = g.bits[i];     }    } }   The first constructor doesn't seem to be a "real" constructor, it seems as though in every case one of the other constructors will be used. So why is that constructor defined at all? Is there anyway in a flex application to figure out (in runtime) when the running swf was compiled?  I'd like to compare this with the latest file version of the swf on the server and detect if there's a newer version on the server and if so force the user to reload their browser to get the new version.   I need this to also work with swfs from the cache.