I have used Audrey's code to cite something of the form: \citet[p.~42]{smth}. This code is excelent if someone wants to have the parentheses affected as well. Now, I ask if it is possible to have the parentheses with another color. For example I would have preferred it to be black. A minimal script for replication and possible change is the following:   \documentclass[hidelinks,12pt]{article} \usepackage[utf8x]{inputenc} \usepackage{colortbl, tcolorbox}          \usepackage[sectionbib, sort]{natbib}  \usepackage{chapterbib} \usepackage{etoolbox}  \definecolor{darkred}{rgb}{0.5,0,0} \definecolor{darkgreen}{rgb}{0,0.5,0} \definecolor{darkblue}{rgb}{0,0,0.5}  \definecolor{dark-red}{rgb}{0.4,0.15,0.15}  \usepackage[hyperfootnotes = true]{hyperref}  \hypersetup{colorlinks,                  linkcolor=red,                                 filecolor=darkgreen,  urlcolor=darkred,  citecolor=blue}  % Audrey's code starts \makeatletter   % include postnote and \citet closing bracket in hyperlink \pretocmd{\NAT@citex}{%   \let\NAT@hyper@\NAT@hyper@citex   \def\NAT@postnote{#2}%   \setcounter{NAT@total@cites}{0}%   \setcounter{NAT@count@cites}{0}%   \forcsvlist{\stepcounter{NAT@total@cites}\@gobble}{#3}}{}{} \newcounter{NAT@total@cites} \newcounter{NAT@count@cites} \def\NAT@postnote{}  % include postnote and \citet closing bracket in hyperlink \def\NAT@hyper@citex#1{%   \stepcounter{NAT@count@cites}%   \hyper@natlinkstart{\@citeb\@extra@b@citeb}#1%   \ifnumequal{\value{NAT@count@cites}}{\value{NAT@total@cites}}     {\ifNAT@swa\else\if*\NAT@postnote*\else%      \NAT@cmt\NAT@postnote\global\def\NAT@postnote{}\fi\fi}{}%   \ifNAT@swa\else\if\relax\NAT@date\relax   \else\NAT@@close\global\let\NAT@nm\@empty\fi\fi% avoid compact citations   \hyper@natlinkend} \renewcommand\hyper@natlinkbreak[2]{#1}  % avoid extraneous postnotes, closing brackets \patchcmd{\NAT@citex}   {\ifNAT@swa\else\if*#2*\else\NAT@cmt#2\fi    \if\relax\NAT@date\relax\else\NAT@@close\fi\fi}{}{}{} \patchcmd{\NAT@citex}   {\if\relax\NAT@date\relax\NAT@def@citea\else\NAT@def@citea@close\fi}   {\if\relax\NAT@date\relax\NAT@def@citea\else\NAT@def@citea@space\fi}{}{}  \makeatother % End of Audrey's code  \bibpunct{(}{)}{;}{a}{,}{,~}     \title{My title} \author{My name}
natbib and hyperref create by default two links to the bibentry in the reference section|chapter of the document:   \begin{filecontents}{\jobname.bib} @book{author08, title = {{A Title}}, publisher = {Alpha}, year = {2008}, author = {Author, A}, address = {London} } @book{buthor90, title = {{B Title}}, publisher = {Bravo}, year = {1990}, editor = {Buthor, B}, address = {New York} } \end{filecontents}  \documentclass{article}  \usepackage{natbib} \usepackage{hyperref} \hypersetup{colorlinks,citecolor=red}  \begin{document}  \noindent citet: \citet{author08}, \citet[see][p. 20]{author08} \\ citep: \citep{author08}, \citep[see][p. 20]{author08}  \bibliographystyle{plainnat} \bibliography{\jobname}  \end{document}     In order to avoid two links to the bibliography (one for the [author] and one for the year [year], respectively), I took  to an older question and added it to my preamble.  \usepackage{etoolbox} \makeatletter  \pretocmd{\NAT@citex}{%   \let\NAT@hyper@\NAT@hyper@citex   \def\NAT@postnote{#2}%   \setcounter{NAT@total@cites}{0}%   \setcounter{NAT@count@cites}{0}%   \forcsvlist{\stepcounter{NAT@total@cites}\@gobble}{#3}}{}{} \newcounter{NAT@total@cites} \newcounter{NAT@count@cites} \def\NAT@postnote{}  % include postnote and \citet closing bracket in hyperlink \def\NAT@hyper@citex#1{%   \stepcounter{NAT@count@cites}%   \hyper@natlinkstart{\@citeb\@extra@b@citeb}#1%   \ifnumequal{\value{NAT@count@cites}}{\value{NAT@total@cites}}     {\ifNAT@swa\else\if*\NAT@postnote*\else%      \NAT@cmt\NAT@postnote\global\def\NAT@postnote{}\fi\fi}{}%   \ifNAT@swa\else\if\relax\NAT@date\relax   \else\NAT@@close\global\let\NAT@nm\@empty\fi\fi% avoid compact citations   \hyper@natlinkend} \renewcommand\hyper@natlinkbreak[2]{#1}  % avoid extraneous postnotes, closing brackets \patchcmd{\NAT@citex}   {\ifNAT@swa\else\if*#2*\else\NAT@cmt#2\fi    \if\relax\NAT@date\relax\else\NAT@@close\fi\fi}{}{}{} \patchcmd{\NAT@citex}   {\if\relax\NAT@date\relax\NAT@def@citea\else\NAT@def@citea@close\fi}   {\if\relax\NAT@date\relax\NAT@def@citea\else\NAT@def@citea@space\fi}{}{}  \makeatother     Which gets me almost where I want to be, but leaves me with two issues I would like to be solved:   while \citet includes pre- and postnotes in the link, \citep doesn't. How can \citep be changed to include postnotes? is there a way to color the brackets and commas that are typeset in black in the unmodified first example also in black in the modified second example without removing them from the link?
On Stack Exchange, users may gain a certain level of reputation.  What does reputation do? How can a user gain or lose reputation?   See also   in the Help Center  - Maximum amount of votes a user can use in a day and reputation requirements to vote.  - How to get an exact overview of the actions that got you your reputation.  - A full list of all privileges granted at each reputation level on all sites. How can I check if a program exists from a Bash script? Prove,by vector method,that the point of intersection of the diagonals of the trapezium lies on the line passing through the mid-points of the parallel sides.    My Attempt: Let the trapezium be $OABC$ and that the O is a origin and the position vectors of $A,B,C$ be $\vec{a},\vec{b},\vec{c}$.Then the equation of $OB$ diagonal is $\vec{r}=\vec{0}+\lambda \vec{b}................(1)$ And equation of $AC$ diagonal is $\vec{r}=\vec{a}+\mu(\vec{c}-\vec{a}).......(2)$ And the equation of the line joining the mid points of $OA$ i.e.$\frac{\vec{a}}{2}$ and $BC$ i.e. $\frac{\vec{b}+\vec{c}}{2}$ is $\vec{r}=\frac{\vec{a}}{2}+t(\frac{\vec{b}+\vec{c}}{2}-\frac{\vec{a}}{2}).........(3)$. Here $\lambda,\mu,t$ are scalars. I do not know how to solve $(1)$ and $(2)$ and put into $(3)$ to prove the desired result. Please help me.Thanks. SVM for unbalanced data I want to attempt to use Support Vector Machines (SVMs) on my dataset. Before I attempt the problem though, I was warned that SVMs dont perform well on extremely unbalanced data. In my case, I can have as much as 95-98% 0's and 2-5% 1's.   I tried to find resources which talked about using SVMs on sparse/unbalanced data, but all I could find was 'sparseSVMs' (which use a small amount of support vectors).  I was hoping someone could briefly explain:   How well SVM would be expected to do with such a dataset   Which, if any, modifications must be done to the SVM algorithm What resources/papers discuss this Add an unnmbered section to the table of Contents in Lyx How do I find this conditional expectation? Capturing screenshots on the Mac has gotten very slow recently.  It used to be nearly instantaneous but now it is taking more than 10 seconds for the image to appear on the desktop.  I seem to recall having this problem years ago. The workaround was to delete a large accumulation of log files that the os had built up. After I deleted these log files, the screen capture worked very fast again.  Problem is that I cannot remember where these log files are. Searched for quite some time and could not find the information. Does "Going Infinite" necessarily end the game? A fair coin is tossed $10$ times. What is the probability that no two   consecutive tosses are heads?   Possibilities are (dont mind the number of terms):  $H TTTTTTH$, $HTHTHTHTHTHTHT$.  But except for those,  let $y(n)$ be the number of sequences that start with $T$  $T _$, there are two options, $T$ and $H$ so,  $y(n) = y(n - 1) + x(n-1) = y(n - 1) + y(n - 2)$  Let $x(n)$ be the number of sequences that start with $H$,   $H _$ the next option is only $T$ hence,  $x(n) = y(n - 1)$   The total number of sequences $F(n)$ is:  $$F(n) = y(n) + x(n) = 2y(n - 1) + y(n - 2)$$  We are after $F(10)$,  $$F(10) = 2y(9) + y(8)$$  $$F(3) = 2y(2) + y(1) = 1 + 4 = 5$$  $$F(4) = 2y(3) + y(2) = 6 + 2 = 8$$  But I'm not quite sure where this will take me though. I have a dell laptop with 4GB of ram and 500GB of storage.I just finished installing windows 8.1 32bit version, which utilizes 3GB of ram.  Can I install and run Ubuntu to utilize that unusable portion of Ram.Then how?  Thanks in advance I've read that abelian categories can naturally be enriched in $\mathbf{Ab}.$ How does this actually work? Wikipedia defines the notion of an abelian category as follows ().     A category is abelian iff         it has a zero object,   it has all binary products and binary coproducts, and   it has all kernels and cokernels.   all monomorphisms and epimorphisms are normal.      It later explains that an abelian category can naturally be enriched in $\mathbf{Ab},$ as a result of the first three axioms above.  How does this actually work? How can I determine the sudo password? What is the default password of the sudo? I visit some forums, it tells that the default password is password, but it doesn't work, how can I determine the password? What if table columns cannot fit on webpage? If I've 12 - 14 columns in a table to show on a page but there's not enough space to show all columns (let's say, there's only enough space to show 7 columns of the table), what is the solution for this?  I can make the page scroll horizontally but it's ugly and the last resort. Shrinking the font size of the table data only has rooms for 1  - 2 more columns at best. Does $\{f_ng_n\}\to fg$ uniformly? Show that:  If $U$ is a principal ultrafilter, then the canonical inmersion $j$ is an isomorphism between $\mathcal A$ and $Ult_{U}\mathcal A$ Jutifying: irrational + rational = irrational I am not sure how to show this. This is what I tried.   Proposition: If x is rational number and y is an irrational number then x+y is an irrational number.  Logically speaking this is  P=x is rational  Q= Y is irrational  R= x+y is irrational  $P\wedge Q \rightarrow R$  Explanation: What I did  $x=\frac{p}{q}$ P and Q integer  Y cannot be written as the a rational by definition  So therefore  $x+y=\frac{p}{q}+y$  Will not be a rational number as there cannot be a common denominator. Since upgrading to 13.10, I can't log in to unity desktop. Light dm works correctly, but attempting to log in tries to start the session then drops back to light. I've already dropped to terminal (ctrl+alt+f2) and done this:  sudo apt-get update  sudo apt-get install --reinstall ubuntu-desktop  sudo apt-get install unity   Logging in as a guest session also fails. Logging in to other window managers works with varying degrees of success.  Note: I have Cinnamon 2.0 installed from PPA.  I'm using a 2 monitor setup.  Also of note is that the session prior to my upgrade to 13.10 the background of unity failed to display at all, instead showing what was there in the screen buffer from the previous frame. The entire OS worked correctly otherwise though, so I just ignored it for the session. No other upgrades or even updates were done prior to this occurring.  My upgrade path to 13.10 was basically this: Install 13.04 alongside Windows 7, use ubuntu as a glorified web browser for a while, get updates (in preparation for 13.10), install 13.10.  I also used Unity Tweak Tool to change some aspects of unity, particularly auto-hide.  Any help or ideas would be appreciated, as I'm typing this on my phone :( JUnit @Test expected annotation not working How does someone get to cast more than 40 votes in a day? And by someone, I mean that evil feline WesleyDavid:        That screenshot was taken just before 4pm PST, i.e. just before 0000 UTC.  That shows KittehHead's votes for the month, which is the same as for the day since it's March 1st.  Just to be sure it wasn't that SE had the time wrong, I tried upvoting something and got the expected "Daily Vote Limit Reached, Please Vote Again in 6 minutes."  So how did Kitteh^H^H^H^H^H^HWesley get 42 votes in one day?  ADDED: It's getting worse... Now it's March 2nd, and Wesley/Cat has 3 extra votes:        Now I finally got an extra vote (take that evil kitteh!), but it hasn't helped me understand why: I just need to check the reasoning in my proof is correct, I think it is valid although I'm not totally convinced because I can't follow the logic; does proving that $x$ is an integer prove that $ac|bd$?  Theorem: Let $a$, $b$, $c$, $d$ be integers. If $a|b$ and $c|d$, then $ac|bd$.  Proof: $aj=b$ and $ck=d$ for integers $j$, $k$. Then, $ac|bd$ implies $acx=ajck$ and thus $x=jk$ for some $x$. Since the product $jk$ is an integer, $x$ is an integer and thus $acx=ajck$ and thus $ac|bd$.