similar to: linear discriminant analysis

Dear R help list, I have a training dataset that looks like Table1. I have an unknown dataset that looks like Table2. I want to have a program that should search the training dataset and identify that the unknown sample belongs to which category (type1, type2 or type3) and also if the unknown does not belong to any of the categories, it should let me know. The real dataset has 600 variables and

Dear R helpers   Following is a part of R code.   data_lab <- expand.grid(c("R11", "R12", "R13"), c("R21", "R22", "R23"), c("R31", "R32", "R33"), c("R41", "R42", "R43"), c("R51", "R52", "R53"), c("R61", "R62", "R63"),

Dear all, I have a dataset that looks like this: x <- read.table(textConnection("col1 col2 3 1 2 2 4 7 8 6 5 10"), header=TRUE) I want to rewrite it as below: var1 var2 var3 var4 var5 var6 var7 var8 var9 var10 1 0 1 0 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 0 0 0 0 1 0 0

Hello, I have a large data set that has more columns than rows (sample data below). I am trying to perform a partitioning cluster analysis and then plot that using pca. I have tried using CLUSPLOT(), but that only allows for 'princomp' where I need 'prcomp' as I do not want to reduce my columns. Is there a way to edit the CLUSPLOT() code to use 'prcomp', please? #

var1 var2 var3 var4 var5 var6 var7 var8 var9 var10 gold 2 3 1 2 4 0 1 4 4 3 2 2 4 2 4 3 4 2 4 4 4 2 3 3 0 0 4 1 0 2 4 4 2 1 4 0 3 2 0 0 2 4 4 2 3 4 0 2 2 0 0 0 3 4 2 2 2 3 2 2 0 0 0 2 4 2 2 4 1 1 2 0 0 3 3 3 2 3 4 1 4 0 0 0 0 3 4 2 3 1 0 2 2 1 0 2 3 3 2 0 3 1 1 1 1 2 1 2 3 2 1

Hello, I have trouble converting a character string to a R object. Let me describe this by an example; > dim(a) [1] 270 14 > dim("a") NULL > names(a) [1] "Var1" "Var2" "Var3" "Var4" "Var5" "Var6" "Var7" "Var8" "Var9" [10] "Var10" "Var11" "Var12"

Hi, I am trying to merge data similar to the example data below > dat0 id var1 var2 var3 2 1 0 1 3 1 0 1 4 0 1 1 5 0 1 1 > dat1 id var4 var5 var6 2 1 0 1 3 1 0 1 6 0 1 1 7 0 1 1 > dat2 id

Seems like get() is what you are looking for; e.g., try dim(get("a")), names(get("a")), etc. HTH, Andy > From: Sixten Borg > > Hello, > > I have trouble converting a character string to a R object. > Let me describe this by an example; > > > dim(a) > [1] 270 14 > > dim("a") > NULL > > > names(a) > [1]

Hola.. yo lo haría de la siguiente manera ... En excel:genero una columna con la serie de fechas continuas ... con la función buscarv, agrego las variables que tienen dato a esta serie.. en base a la fecha muy largo y mecánico para mi gusto.... En R ..de igual manera genero el vector de fechas require(chron)# crear el vector continuo de fechasfch01 <- data.frame(''fch'' =

Dear all, I am trying (!!!) to generate pdfs that have 8 plots on one page: df = data.frame( day = c(1,2,3,4), var1 = c(1,2,3,4), var2 = c(100,200,300,4000), var3 = c(10,20,300,40000), var4 = c(100000,20000,30000,4000), var5 = c(10,20,30,40), var6 = c(0.001,0.002,0.003,0.004), var7 = c(123,223,123,412), var8 = c(213,123,234,435), all = as.factor(c(1,1,1,1)))

Dear all, I have a dataset that looks like id var1 var2 var4 var7 var8 1 0.0 0.1 0.3 0.9 0.0 2 0.4 0.6 0.0 0.0 0.2 3 0.0 0.0 0.0 0.8 0.7 Some columns are missed, for example, here the fourth (var3), sixth(var5) and seventh (var6) columns. I want to first determine which columns are missed in a huge dataset and then add the missed

Dear list, after fitting an lrm with the Design package (stored as "mymodel") I try running a summary, but I get the following error: dim(mydata) [1] 235 9 names(mydata) [1] "id" "VAR1" "VAR2" "VAR3" "VAR4" "VAR5" "VAR6" "VAR7" "VAR8" summary(mymodel) Error in `contrasts<-`(`*tmp*`,

I've only begun investigating R as a substitute for SPSS. I have a need to identify for each CASE the closest (or most similar) 5 other CASES (not including itself as it is automatically the closest). I have a fairly large matrix (50000 cases by 50 vars). In SPSS, I can use Correlate > Distances to generate a matrix of similarity, but only on a small sample. The entire matrix can not

Dear all, I have a dataset as below: id code1 code2 p 1 4 8 0.1 1 5 7 0.9 2 1 8 0.4 2 6 2 0.2 2 4 3 0.6 3 5 6 0.7 3 7 5 0.9 I just want to rewrite it as this (vertical to horizontal): id var1 var2 var3

Let's say I have a program that returns variables whose names may be any string within the vector NAMES=c("varA","varB","varC","varD","varE","varF"..."varZ"), but I do not ever know which ones have actually been created. So in one example output, "varA", "varC", and "varD" could exist, but

Dear R helpers Wish you all a very Happy and Prosperous New Year 2011. I have following query. country = c("US", "France", "UK", "NewZealand", "Germany", "Austria", "Italy", "Canada") Through some other R process, the result.csv file is generated as result.csv      var1   var2  var3  var4    var5    var6   var7  

I am trying to Sweave the output of calculating correlations between one variable and several others. I wanted to print a table where the odd-numbered rows contain the variable names and the even-numbered rows contain the correlations. So if VarA is correlated with all the variables in mydata.df, then it would look like var1 var2 var3 corr1 corr2 corr3 var4 var5

Hi all, I am trying to run an autologistic model using the function errorsarlm from spdep package. **I built an XY matrix extracting the two colums from matriz** coords1<-matriz[matriz$casos1==1, c(4,5)] coords1<-as.matrix(coords1) **I identify neighbours of region points** nb20<-dnearneigh(coords1,0,20,longlat=TRUE) ** I build a neighbours list with spatial weights**

I have a dataframe called "data" with 5 records (in rows) each of which has been scored on each of many variables (in columns). Five of the variables are named var1, var2, var3, var4, var5 using headers. The other variables are named using other conventions. I can create a new variable called var6 with the value 15 for each record with this code: > var6=var1+var2+var3+var4+var5

> Dear All > > I have data as follows. > > D T M L > 0.20 1 03 141 > 0.32 1 07 62 > 0.50 1 05 49 > 0.80 1 04 46 > 0.20 2 14 130 > 0.32 2 17 52 > 0.50 2 13 41 > 0.80 2 14 36 > 0.20 3 24 120 > 0.32