similar to: error on predict

With ggplot2, I can plot the glm stat_smooth for binomial data when the response is binary or a two-level factor as follows: data("Donner", package="vcdExtra") ggplot(Donner, aes(age, survived)) + geom_point(position = position_jitter(height = 0.02, width = 0)) + stat_smooth(method = "glm", family = binomial, formula = y ~ x, alpha = 0.2, size=2) But how can I

Estimados compañeros hace unos días instale la versión última de R (3.4.1) en windows 10 y hoy me tiene loco el siguiente error cuando ejecuto predict para un árbol de clasificación-regresión. El problema da incluso con el ejemplo del comando predict library(tree) data(shuttle, package="MASS") shuttle.tr <- tree(use ~ ., shuttle, subset=1:253, mindev=1e-6,

Hello, i have a xls file. I will read it in r, what library-command i use for this?? any ideas?? Thanks Michele [[alternative HTML version deleted]]

Dear R Wizards, After much frustration and days of confusion I have finally broken down and am asking for help, which I don’t like doing, but I just can’t figure this one out on my own. I’ve conducted a laboratory experiment testing the effects of temperature and salinity on whether or not a biological event will occur (Go or NoGo). I’ve coded the factors temperature and salinity as factors for

Dear all, I have built glm models based on presences/absences and a number of predictor maps and would like to compute habitat suitability based on the modelled coefficients. I thought this is pretty straight forward and wanted to use predict() and supply the new data in a data frame, with one column for each predictor. However, I do get an error msg warning me that the number of rows for

Gracias Carlos miraré el fix y en su defecto intentaré mandar un mensaje al autor. De todas las maneras espero que alguien lo pruebe en windows 10 Saludos, Juan El 24/09/2017 a las 20:32, Carlos Ortega escribió: > Hola Juan Antonio, > > Acabo de instalar "tree" en Mac OSX y no aparece este error... > > > shuttle1 <- shuttle[254:256, ]  # 3 missing cases >

I'm trying to solve a homework problem using R. The problem gives a list of cricket chirps per second and corresponding temperature, and asks to give the equation for the linear model and then predict the temperature to produce 18 chirps per second. So far, I have: > # Homework 11.2.1 and 11.3.3 > chirps <- scan() 1: 20 2: 16 3: 19.8 4: 18.4 5: 17.1 6: 15.5 7: 14.7 8: 17.1 9: 15.4

Dear all Please help me with correct syntax of predict.nlme. I would like to predict from nlme object for new data. I used predict(fit.nlme6, data=newdata) but I have always got fitted values, no matter how I changed newdata. I have > summary(fit.nlme6) Nonlinear mixed-effects model fit by maximum likelihood Model: konverze ~ SSfpl(tepl, A, B, xmid, scal) Data: limity.gr AIC

Un ejemplo con un modelo más simple: He especificado este modelo: >formula(m2.pile) ger ~ tem + pot + time Si hago predict me da: >predict(m2.pile,newdata=data.frame(tem=25,pot=0,time=3),type="response") 0.08243262 Extraigo los coeficientes: > coef(m2.pile) (Intercept) tem pot time -1.89521331 -0.02303313 4.74499714 0.02043222 Ahora calculo la

I maintain the Devore5 package for R. This package provides the data sets from Jay Devore's text "Probability and Statistics for Engineering and the Sciences (5th ed)". I am having difficulty reproducing some logistic regression results from the textbook. Perhaps this is because I am not using the glm function correctly. The data from Example 13.5 (page 559 for those with a copy

Hi all I have run into a case where I don't understand why predict.lrm and predict.glm don't yield the same results. My data look like this: set.seed(1) library(Design); ilogit <- function(x) { 1/(1+exp(-x)) } ORDER <- factor(sample(c("mc-sc", "sc-mc"), 403, TRUE)) CONJ <- factor(sample(c("als", "bevor", "nachdem",

Hi, I have run into another problem using offsets, this time with the predict function, where there seems to be a contradiction again between the behavior and the help page. On the man page for predict.lm, it says Offsets specified by offset in the fit by lm will not be included in predictions, whereas those specified by an offset term in the formula will be. While it indicates nothings about

I've recently encountered an issue when trying to use the predict.glm function. I've gotten into the habit of using the dataframe$variablename method of specifying terms in my model statements. I thought this unambiguous notation would be acceptable in all situations but it seems models written this way are not accepted by the predict function. Perhaps others have encountered this

Hi, I split a data set into two partitions (80 and 42), use the first as the training set in glm and the second as testing set in glm predict. But when I call glm.predict, I get the warning message:  Warning message: 'newdata' had 42 rows but variable(s) found have 80 rows  ---------------------  s = sample(1:122)

Estimad en s errer en s He hecho este modelo glm m1.pile<-glm(ger~tem+pot+time+I(tem^2)+I(tem^2):pot ,family="binomial" ,data=long.PILE ) Que nos da la probabilidad de germinación de una semilla en función de tem (Temperatura), pot (Humedad del suelo) y time (Tiempo que la semilla pasa en esas condiciones). Ahora quiero, para diferentes tem, pot

Hi, See below I reply your message for <https://stat.ethz.ch/pipermail/r-help/2008-April/160966.html>[R] predict.glm & newdata posted on Fri Apr 4 21:02:24 CEST 2008 You say it ##works fine but it does not: if you look at the length of yhat2, you will find 100 and not 200 as expected. In fact predict(reg1, data=x2) gives the same results as predict(reg1). So I am still looking for

I am fittling a spline to a variable in a regression model, I am then using the predict.glm funtion to make some predictions. When I use bs to fit the spline I don't have any problems using the predict.glm function however when I use ns I get the following error: Error in model.frame(formula, rownames, variables, varnames, extras, extranames, : variable lengths differ (found for

Hi guys... I have problem with this excersise... Consider the pressure data frame again. (a) Plot pressure against temperature, and use the following command to pass a curve through these data: > curve((0.168 + 0.007*x)?(20/3), from=0, to=400, add=TRUE) (b) Now, apply the power transformation y3/20 to the pressure data values. Plot these transformed values against temperature. Is a linear

Dear all, the help file for predict.gbm states that "The predictions from gbm do not include the offset term. The user may add the value of the offset to the predicted value if desired." I am just not sure how exactly, especially for a Poisson model, where I believe the offset is multiplicative ? For example: library(MASS) fit1 <- glm(Claims ~ District + Group + Age +

I imitated predict.glm, my thing worked, now I need to revise. It would help me very much if someone would explain predict.glm line 28, which says object$na.action <- NULL # kill this for predict.lm calls I want to know 1) why does it set the object$na.action to NULL 2) what does the comment after mean? Maybe I need a pass by value lesson too, because I can't see how changing that