similar to: Running regression (lm, lrm) 100+ times and saving the results as matrix

Hi R-expert, If I have this daily rainfall data, how do call a particular day? Year,Month,Day,Amount 1900,12,22,1.3 1900,12,23,0 1900,12,24,0 1900,12,25,0 1900,12,26,0 1900,12,27,0 1900,12,28,0 1900,12,29,4.8 1900,12,30,0.3 1900,12,31,0.5 1901,1,1,0 1901,1,2,3 1901,1,3,0 1901,1,4,0.5 1901,1,5,0 1901,1,6,0 ... I used to use julian.date in S-Plus. Thank you so much for your kind attention and help.

Plotting with 2 x axis? One axis inside another, for example salary within state, 1-50 | 50 ? 100 | 100+ | 1- 50 | 50 -100 | 100+ | ? repeated bins for salary AL ! AR ?? more states The values are all stored with a single data frame. I have tried different things with the axis function and done many

Hello I'm using logistic regression from the Design library (lrm), then fastbw to undertake a backward selection and create a reduced model, before trying to make predictions against an independent set of data using predict.lrm with the reduced model. I wouldn't normally use this method, but I'm contrasting the results with an AIC/MMI approach. The script contains: # Determine full

Hi, May be this helps: funcName<- function(df1, x){ ?whatCol=df1[[x]] ?print("Got it") ?print(whatCol) ?} ? funcName(df,"ColA") #[1] "Got it" #[1] 1 2 3 4 5 ? funcName(df,"ColB") #[1] "Got it" #[1] A B C D E #Levels: A B C D E A.K. >I am trying to extract the 2nd column from a dataframe using a function called funcName. Note this is an

> On Sep 14, 2017, at 12:30 AM, Bonnett, Laura <L.J.Bonnett at liverpool.ac.uk> wrote: > > Dear all, > > I am using the publically available GustoW dataset. The exact version I am using is available here: https://drive.google.com/open?id=0B4oZ2TQA0PAoUm85UzBFNjZ0Ulk > > I would like to produce a nomogram for 5 covariates - AGE, HYP, KILLIP, HRT and ANT. I have

Hi R-helpers I have a dataframe DF (lets say with the variables, y, x1, x2, x3, ..., clust) containing relatively many NAs. When I fit an ordinal regression model with the function lrm from the Design library: model.lrm <- lrm(y ~ x1 + x2, data=DF, x=TRUE, y=TRUE) it will by default delete missing values in the variables y, x1, x2. Based on model.lrm, I want to apply the robust covariance

Dear R helpers, >From the lrm( ) model used for binary logistic regression, we used the L.R. model value (or the G2 value, likelihood-ratio chi-squared statistic) to evaluate the goodness-of-fit of the models. The model with the lowest G2 value consequently, has the best performance and the highest accuracy. However our model includes rsc() functions to account for non-linearity. We

Given this data frame (a simplified, essential reproducible example) A<-c(8,7,10,1,5) A_flag<-c(10,0,1,0,2) B<-c(5,6,2,1,0) B_flag<-c(12,9,0,5,0) mydf<-data.frame(A, A_flag, B, B_flag) # this is my initial df mydf I want to get to this final situation i<-which(mydf$A_flag==0) mydf$A[i]<-NA ii<-which(mydf$B_flag==0) mydf$B[ii]<-NA

Fixed 'maxiter' in the help file. Thanks. Please give the original source of that dataset. That dataset is a tiny sample of GUSTO-I and not large enough to fit this model very reliably. A nomogram using the full dataset (not publicly available to my knowledge) is already available in http://biostat.mc.vanderbilt.edu/tmp/bbr.pdf Use lrm, not lrm.fit for this. Adding maxit=20 will

Hi, I have to build a logistic regression model on a data set that I have. I have three input variables (x1, x2, x3) and one output variable (y). The syntax of lrm function looks like this lrm(formula, data, subset, na.action=na.delete, method="lrm.fit", model=FALSE, x=FALSE, y=FALSE, linear.predictors=TRUE, se.fit=FALSE, penalty=0, penalty.matrix, tol=1e-7,

Hi R, I am getting this error while trying to use 'lrm' function with nine independent variables: > res = lrm(y1994~WC08301+WC08376+WC08316+WC08311+WC01001+WC08221+WC08106+WC0810 1+WC08231,data=y) singular information matrix in lrm.fit (rank= 8 ). Offending variable(s): WC08101 WC08221 Error in j:(j + params[i] - 1) : NA/NaN argument Now, if I take choose only four

Hi, I am running a logistic regression model using lrm library and I get the following error when I run the command: mod1 <- lrm(death ~ factor(score), x=T, y=T, data = env1) Unable to fit model using ?lrm.fit? where score is a numeric variable from 0 to 6. LRM executes fine for the following commands: mod1 <- lrm(death ~ score, x=T, y=T, data = env1) mod1<- lrm(death ~

Hi all, Can anyone explain why the following use of the subset() function produces a different outcome than the use of the "[" extractor? The subset() function as used in density(subset(mydf, ht >= 150.0 & wt <= 150.0, select = c(age))) appears to me from documentation to be equivalent to density(mydf[mydf$ht >= 150.0 & mydf$wt <= 150.0, "age"])

Hi List, I don't understand why 'lrm' doesn't recognize the '~.' formula. I'm pretty sure it was working before. Please see below: I'm using R2.3.0, WinXP, Design 2.0-12 thanks, ...Tao > dat <- data.frame(y=factor(rep(1:2,each=50)), x1=rnorm(100), x2=rnorm(100), x3=rnorm(100)) > lrm(y~., data=dat, x=T, y=T) Error in terms.formula(formula, specials =

Hi, I've come across a strange error when using the lrm.fit function and the subsequent predict function. The model is created very quickly and can be verified by printing it on the console. Everything looks good. (In fact, the performance measures are rather nice.) Then, I want to use the model to predict some values. I get the following error: "fit was not created by a Design

Hello! I have a data frame with dates. I need to create a new "month" that starts on the 20th of each month - because I'll need to aggregate my data later by that "shifted" month. I wrote the code below and it works. However, I was wondering if there is some ready-made function in some package - that makes it easier/more elegant? Thanks a lot! # Example data:

Dear all, given I have data in a data.frame which indicate the number of people in a specific year at a specific age: n <- 10 mydf <- data.frame(yr=sample(1:10, size=n, replace=FALSE), age=sample(1:12, size=n, replace=FALSE), no=sample(1:10, size=n, replace=FALSE)) Now I would like to make a matrix with (in this simple example) 10 columns (for the

Hi, I've come across a strange error when using the lrm.fit function and the subsequent predict function. The model is created very quickly and can be verified by printing it on the console. Everything looks good. (In fact, the performance measures are rather nice.) Then, I want to use the model to predict some values. I get the following error: "fit was not created by a Design

What causes the error report: logical(0) to arise in the rms function lrm? Here's my data: But both the dependent and the independent variable seem fine... > str(AABB) 'data.frame': 1176425 obs. of 9 variables: $ sex : int 1 1 0 1 1 0 0 0 0 0 ... $ faint : int 0 0 0 0 0 0 0 0 0 0 ... Here's the simplified model and error AABB$model1 < lrm (faint ~ sex)

Dear all, I wanted to make a two-way-table of two variables with a counting variable stored in another column of a dataframe. In version 1.9.1, the behavior is as expected as shown in the simplified example code. > sex <- rep(c("F", "M"), 5) > income <- c(rep("low", 5), rep("high", 5)) > count <- 1:10 > mydf <-