search for: var7

...nde hay discontinuidades. No me interesa llenar los vacíos > en las variables, sólo las fechas y dejar las variables en blanco ¿es > posible hacerlo en R?. Mi serie de tiempo es del tipo (con fecha en formato > dd-mm-yyyy hh:mm): > > 01-01-2011 00:00 var1 var2 var3 var4 var5 var6 var7 var8 > 01-01-2011 00:01 var1 var2 var3 var4 var5 var6 var7 var8 > 01-01-2011 00:02 var1 var2 var3 var4 var5 var6 var7 var8 > 01-01-2011 00:03 var1 var2 var3 var4 var5 var6 var7 var8 > 01-01-2011 00:04 var1 var2 var3 var4 var5 var6 var7 var8 > 01-01-2011 00:07 var1 var2 var3 var4 var5...

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))) pdf("test1.pdf", width=20, heigh=27, paper="a4") print(plot(groupedData(var1 ~ day | all, data = df), main = "var1", xlab="", ylab=""), split=c(1,1,2,4), more=TRUE...

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...

...r3 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 var7 var8 var9 2 1 0 1 5 1 0 1 6 0 1 1 8 0 1 1 Basically what I'd like to do is combine these variables on id and create one large data frame that looks like the following. > dat3 id var1 var2...

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 1 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 1...

...itting 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*`, value = "contr.treatment") : contrasts can be applied only to factors with 2 or more levels but, VAR1 is my dependent > nlevels(factor(VAR1)) [1] 2 VAR2 is continuous > summary(psa) Min. 1st Qu. M...

...prob[[7]]  <- c(0.44,0.07,0.49) range_prob[[8]]  <- c(0.54,0.06,0.40) range_prob[[9]]  <- c(0.26,0.62,0.12) range_prob[[10]] <- c(0.65,0.19,0.16)   pdf <- expand.grid(range_prob)   data_lab$probs <- apply(pdf, 1, prod)   joint_probs = xtabs(probs ~ Var1 + Var2+Var3+Var4+Var5+Var6+Var7+Var8+Var9+Var10, data = data_lab)   write.csv(data.frame(joint_probs), 'joint_probs.csv', row.names = FALSE) ONS = read.csv('joint_probs.csv')   Names = NULL   for (i in 1:length(joint_probs))  {        Names[i] = paste(ONS$Var1[i], ONS$Var2[i], ONS$Var3[i], ONS$Var4[i], ONS$Var5[...

...data with repeated measures (measurements are equally spaced in time, subjects are measured several times a year). I need to allow slope and intercept vary. SAS codes are: proc mixed data = survey method=reml; class subject var1 var3 var2 time; model score = var2 score_base var4 var5 var3 var6 var7 var1 time/ noint solution; random intercept timecontinious / subject=subject type=un g gcorr v vcorr; run; Thank you a lot! [[alternative HTML version deleted]]

...LRM executes fine for the following commands: mod1 <- lrm(death ~ score, x=T, y=T, data = env1) mod1<- lrm(death ~ factor(score)+ factor(var1)+factor(var2)+factor(var3)+ factor(var4)+factor(var5)+factor(var6)+ factor(var7), x=T, y=T, data = env1) I do not understand why I get the error. I have tried to use glm and it works perfectly fine. -- View this message in context: http://r.789695.n4.nabble.com/Unable-to-fit-model-using-lrm-fit-tp4631525.html Sent from the R help mailing list archive at Nabble.com.

> 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

...y. 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   var8 1      25     45    29    92     108     105     65     56 2      80    132    83    38      38      11     47     74 3     135     11    74    56      74      74     74     29                    I need the country names to be column heads i.e. I need an output like > result_new     US ...

...### require(randomForest) set.seed(17) ## Number of points x <- 10 df <- rbind( data.frame(var1=runif(x, 10, 50), var2=runif(x, 2, 7), var3=runif(x, 0.2, 0.35), var4=runif(x, 1, 2), var5=runif(x, 5, 8), var6=runif(x, 1, 2), var7=runif(x, 5, 8), cls=factor("CLASS-2") ) , data.frame(var1=runif(x, 10, 50), var2=runif(x, -3, 3), var3=runif(x, 0.1, 0.25), var4=runif(x, 1, 2), var5=runif(x, 5, 8), var6=runif(x, 1, 2), var7=runif(x...

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

...25 50 51 46 18 20 16 var2 4 5 4 9 8 9 10 9 10 var3 7 7 7 12 12 12 9 6 6 var4 4 5 4 10 12 9 2 2 2 var5 4 5 4 10 9 10 3 2 3 var6 5 4 5 2 3 2 1 3 5 var7 5 4 5 7 7 7 3 3 3 var8 3 4 3 10 10 8 4 2 4 var9 3 4 3 2 2 2 2 2 2 var10 3 3 3 4 4 4 3 1 2 Table 2 unknown var1 23 var2 4 var3 7 var4 4 var5 4 var6 6 var7 5 var8...

...25 50 51 46 18 20 16 var2 4 5 4 9 8 9 10 9 10 var3 7 7 7 12 12 12 9 6 6 var4 4 5 4 10 12 9 2 2 2 var5 4 5 4 10 9 10 3 2 3 var6 5 4 5 2 3 2 1 3 5 var7 5 4 5 7 7 7 3 3 3 var8 3 4 3 10 10 8 4 2 4 var9 3 4 3 2 2 2 2 2 2 var10 3 3 3 4 4 4 3 1 2 Table 2 unknown var1 23 var2 4 var3 7 var4 4 var5 4 var6 6 var7 5 var8...

...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? # sample of my data PRVID,VAR1,VAR2,VAR3,VAR4,VAR5,VAR6,VAR7,VAR8,VAR9,VAR10,VAR11 PRV1,0,54463,53049,62847,75060,184925,0,0,0,0,0 PRV2,0,2100,76,131274,0,0,0,0,0,0,18 PRV3,967,0,0,0,0,0,0,0,0,3634,0 PRV4,817,18344,3274,9264,1862,0,0,141,0,0,0 PRV5,0,0,0,0,0,0,29044,0,0,0,0 PRV6,59,6924,825,3008,377,926,0,0,10156,0,5555 PRV7,11,24902,36040,47223,20086,0,0,74...

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...

...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 var4 var5 var6 var7 var8 1 0 0 0 0.1 0.9 0 0.9 0.1 2 0.4 0.2 0.6 0.6 0 0.2 0 0.4 3 0 0 0 0 0.8 0.7 0.9 0 For the third subject, there are two values being equal to 5 in code1 and code2, but different values in p: 0.7 a...

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" "Var13" "Var14" > names("a") NULL I realise that the character string lacks both a dimension and any column names; my question is how to make R understand that I l...

...to use xgboost to model and predict count data. The predictions are however not as expected as shown below. # sponge count data in library(spm) library(spm) data(sponge) data(sponge.grid) names(sponge) [1] "easting" "northing" "sponge" "tpi3" "var7" "entro7" "bs34" "bs11" names(sponge.grid) [1] "easting" "northing" "tpi3" "var7" "entro7" "bs34" "bs11" range(sponge[, c(3)]) [1] 1 39 # count sample data # the e...