Some implementations of decision trees (eg cran/tree) can split on categorical variables where the split separates the variable into 2 groups:  &gt; library("tree") &gt; df &lt;- read.csv('https://forge.scilab.org/index.php/p/rdataset/source/file/master/csv/datasets/Titanic.csv')  &gt; tree(Survived ~ Class, data=df, weights=Freq, minsize=10) node), split, n, deviance, yval, (yprob)       * denotes terminal node  1) root 2201 2769 No ( 0.6770 0.3230 )     2) Class: 3rd,Crew 1591 1772 No ( 0.7549 0.2451 ) *   3) Class: 1st,2nd 610  844 Yes ( 0.4738 0.5262 ) *   For a numerical variable, splits are calculated by trying all possible splits between the minimum and maximum value of the variable.  How are the splits enumerated for a categorical variable? Do you have to try all combinations of values? (this would seem to be computationally crazy)  Or is there a heuristic you use to limit the number of combinations tried?
I am curious about the practical implementation of a binary split in a decision tree - as it relates to levels of a categorical predictor $X{j}$.  Specifically, I often will utilize some sort of sampling scheme (e.g. bagging, oversampling etc) when building a predictive model using a decision tree - in order to improve its predictive accuracy and stability. During these sampling routines, it is possible for a categorical variable to be presented to a tree fitting algorithm with less than the complete level set.  Say a variable X takes on levels {A,B,C,D,E}. In a sample, maybe only levels {A,B,C,D} are present. Then, when the resulting tree is used for prediction, the full set may be present.  Continuing from this example, say a tree splits on X and sends {A,B} to the left and {C,D} to the right. I would expect the logic of the binary split to then say, when faced with new data: "If X has value A or B, send to the left, otherwise, send this case to the right". What seems to happen in some implementations is "if X has value A or B, send to the left, if X has value C or D send to the right". When this case takes on value E, the algorithm breaks down.   What is the "right" way for a binary split to be handled? It seems the much more robust way is implemented often, but not always (see Rpart below).  Here are a couple examples:  Rpart fails, the others are ok.  #test trees and missing values  summary(solder) table(solder$PadType)  # create train and validation set.seed(12345) t_rows&lt;-sample(1:nrow(solder),size=360, replace=FALSE) train_solder&lt;-solder[t_rows,] val_solder&lt;-solder[-t_rows,]  #look at PadType table(train_solder$PadType) table(val_solder$PadType) #set a bunch to missing levels(train_solder$PadType)[train_solder$PadType %in% c('L8','L9','W4','W9')] &lt;- 'MISSING'   #Fit several trees, may have to play with the parameters to get them to split on the variable  ####RPART mod_rpart&lt;-rpart(Solder~PadType,data=train_solder) predict(mod_rpart,val_solder) #Error in model.frame.default(Terms, newdata, na.action = na.action, xlev = attr(object,  :  #factor 'PadType' has new level(s) D6, L6, L7, L8, L9, W4  ####TREE mod_tree&lt;-tree(Solder~PadType,data=train_solder,split="gini") predict(mod_tree,val_solder) #works fine  ####ctree mod_ctree&lt;-ctree(Solder~PadType,data=train_solder,control = ctree_control(mincriterion = 0.05)) predict(mod_ctree,val_solder) #works fine
Identical creatures entering the battlefield If I have two cards called  in hand and I play my first, Sin Collector (A). I use its triggered ability.   "When Sin Collector enters the battlefield, target opponent reveals his or her hand. You choose an instant or sorcery card from it and exile that card."  Now, I play Sin Collector (B), I use B's triggered ability.  But now my question. Does Sin Collector (A)'s ability trigger again as well? 'Sin Collector' entered the battlefield again, right? I would like to know how to make this effect of merged circle selection. Here are images to illustrate:     Basically I'm looking for this effect:    How the merge effect of the circles can be achieved ? I didn't found any explanation concerning this effect. I know that to project those texture I can develop a decal system but I don't know how to create the merging effect.  If possible, I'm looking for purely shaders solution. I'm using leaflet to display a map.  I have a control section that I'd like to be open by default, but I can't find a way to do it through the . Is there a way to do this? Compute approximately the probability that more than 50 of the observations of the random sample are less than 3. Let $X_1,X_2,...,X_n$ be a random sample of size $n = 72$ from a distribution with probability density function $f(x) =\begin{cases} 1/x^2,&amp; 1 &lt; x &lt; infinity\\0,&amp;\text{otherwise}\end{cases}$ Compute approximately the probability that more than $50$ of the ob- servations of the random sample are less than $3$. It's easy to check that for any natural $n$ $$\frac{n+1}{n}=\cfrac{1}{2-\cfrac{n+2}{n+1}}.$$  Now,   $$1=\frac{1}{2-1}=\frac{1}{2-\cfrac{1}{2-1}}=\frac{1}{2-\cfrac{1}{2-\cfrac{1}{2-1}}}=\cfrac{1}{2-\cfrac{1}{2-\frac{1}{2-\frac{1}{2-1}}}}=\ldots =\cfrac{1}{2-\cfrac{1}{2-\frac{1}{2-\frac{1}{2-\frac{1}{2-\dots}}}}},$$  $$2=\cfrac{1}{2-\cfrac{3}{2}}=\cfrac{1}{2-\cfrac{1}{2-\cfrac{4}{3}}}=\cfrac{1}{2-\cfrac{1}{2-\frac{1}{2-\frac{5}{4}}}}=\cfrac{1}{2-\cfrac{1}{2-\cfrac{1}{2-\frac{1}{2-\frac{6}{5}}}}}=\ldots =\cfrac{1}{2-\frac{1}{2-\frac{1}{2-\frac{1}{2-\frac{1}{2-\ldots}}}}}.$$  Since the right hand sides are the same, hence $1=2$. Can we deploy PROFILES alone from one sandbox to another? Split-like environment inside cases environment I'd like to have a cases environment with some cases that are too long to fit on one line. I tried this:  \begin{cases}   \begin{split}     long expression \\ second line of long expression   \end{split} &amp; condition \\   ...   but I get the warning     Package amsmath Warning: Cannot use 'split' here;   trying to recover with 'aligned' on input line 201.   It automatically replaces the split with an aligned environment, which looks like this:    Is there any way to get these long expressions to behave like they were in split or multline environments, with the top line flush left and the next line indented a little? Any idea what kind of tool should I but to fix the loose left crank on my bike? I found this  but not sure if it's gonna suit. Or is it enough to buy only nut driver? How do I find the package that provides a file? Padding a number in a filename to a fixed length I have a directory of files such as:  file.1.png file.2.png … file.10.png … file.1000.png file.1001.png …   How can I pad the number to 4 digits? It should end up like this:  file.0001.png file.0002.png … file.0010.png … file.1000.png file.1001.png … I am using Sql Server 2008 on Small Business Server 2003; client is using WinXP.   I added a user to my Active Directory Security Group; Why can't this user immediately access the database? It seems there is a delay before the user is recognized in Sql Server.   I am using AD Security Groups for permissions expressly so that I don't need to add individual users in Sql Server. So effectively I don't need to do anything but add the user to the AD Security Group in order to grant access.  But for some reason Sql Server doesn't immediately recognize the addition. I've seen this a number of times. I add the user to the group but that user can't access data until the next day. It seems that it doesn't query the Active Directory in real time. Can you confirm that is so?  What do I need to do so that Sql Server "refreshes" the list of users from Active Directory? Log form of variables and descriptive measures Because of larger values of variables, I did a log- transformation in my dataset. Now I want to give a descriptive table regarding my variables like mean, max, min, median, skewness, kurtosis. Can I give standard measures like arithmetic mean or should I give geometric mean? In terms of skewness or kurtosis, do I need a transformation? What is exactly the density of a black hole and how can it be calculated? Why didn't the Seons change? In the book Elantris, the Reod was caused by a giant earthquake opening a chasm which changed the shape of all the Aons. The Seons, which are basically living Aons, were presumably unchanged, since nobody noticed the change, which should have been obvious on the simpler Aons. Alternative definition of hyperbolic cosine without relying on exponential function Ordinary trigonometric functions are defined independently of exponential function, and then shown to be related to it by Euler's formula.   Can one define hyperbolic cosine so that the formula $$\cosh{x}=\dfrac{e^x+e^{-x}}{2}$$ becomes something to be proven? I have a table story_category in my database with corrupt entries. The next query returns the corrupt entries:  SELECT *  FROM  story_category  WHERE category_id NOT IN (     SELECT DISTINCT category.id      FROM category INNER JOIN         story_category ON category_id=category.id);   I tried to delete them executing:  DELETE FROM story_category  WHERE category_id NOT IN (     SELECT DISTINCT category.id      FROM category        INNER JOIN story_category ON category_id=category.id);   But I get the next error:      #1093 - You can't specify target table 'story_category' for update in FROM clause    How can I overcome this? I'm aware there are been several similar questions, and some great answers with hints on how to prove this, for example  and .  But I've never really seen a detailed proof of the absolute convergence of the Riemann zeta function in the half-plane $\Re(s)&gt; 1$.  I hope there's interest on a detailed proof of this for reference. Here is my attempt. Please, fill in any detail that might be missing, or point out any mistake!  On $\Re(s)=\sigma&gt; 1$, we have  $$\sum_{n=1}^{\infty}\bigg|\frac{1}{n^s}\bigg|=\sum_{n=1}^{\infty}\frac{1}{n^\sigma}$$  and  $$\frac{1}{n^\sigma}\leq\frac{1}{n^{1+\epsilon}}$$  for any $\epsilon &gt;0$, so by the direct comparison test, if   $$\sum_{n=1}^{\infty}\frac{1}{n^{1+\epsilon}}$$  converges absolutely, then so do $\sum_{n=1}^{\infty}|1/n^s|$ and $\zeta(s)$.  By the integral test, the convergence of $\sum_{n=1}^{\infty}1/n^{1+\epsilon}$ is equivalente to the finitude of the integral  $$\int_1^\infty \frac{1}{n^{1+\epsilon}}=\bigg[-\frac{1}{\epsilon x^\epsilon}\bigg]=\frac{1}{\epsilon}$$  which holds for any $\epsilon &lt; \infty$.  Any correction or improvement is welcomed! Thanks in advance. What character / attack / stage combination can get me a combo of 100? For the first two challenge panels, I just used my golden hammers on the "get a combo of x," but now I'm seeing on challenge panel 3, the game expects you to get a combo of 100! I can barely get past a combo of 9 without the counter resetting. What character / stage combination can get you a combo of 100? Are there multiple, so I can more easily get the "get a total combo of 400 across all characters" challenge?  Additionally, what exactly constitutes a combo? Sometimes my combos are cut short and sometimes they're longer, so I must not understand how they're tracked. How can I change the icon of an application in the Unity launcher? I'm trying to change the icon of a specific application (Pidgin) in my Ubuntu 11.10 Unity launcher. I have tried searching in the options and in the Compiz settings but could locate no option for changing individual icons.  I can change the theme but that's not what I want.   I am willing to edit configuration files to get this change. It doesn't have to be a UI solution. Is there a way to "auto login" in PuTTY with a password? Is there a way to configure a password for a stored session in ?  I know there is the capability to specify an "auto-login username" (under Connection/Data), but is there a way to do the same with the password?