similar to: result numeric(0) when using variable1[which(variable2="max(variable2)"]

Hello, I have a question concerning ?for loops? on multiple columns. I made 91 columns with results (all made together with a for loop) and I want to us lm to fit the model. I want to compare the results of all these calculated columns (91) with one column with observed values. I use the function lm to fit the model and calculate r.squared. I manage to do this for each column separately: For

Dear all, I have a question concerning manipulating data of several columns of a dataframe at the same time. I manage to do it for one column (with the use of the specific name for this column). In each columns, I have 60 values. But I should reorganize the values (because I created this as an output before and I want to compare it with an other dataset). I want that the value on row 2 becomes

I estimate two competing simple regression models, A and B where the LHS is the same in both cases but the predictor is different ( I handle the intercept issue based on other postings I have seen ). I estimate the two models on a weekly basis over 24 weeks. So, I end up with 24 RSquaredAs and 24 RsquaredBs, so essentally 2 time series of Rsquareds. This doesn't have to be necessarily thought

Hi all, I'm struggling with nls. How do you know if your model is significant? For a lm, you get a p-value, but you don't get it for a nls. Is there a way to calculate it? For a lm I use this: a<-summary(lm(model ~obs)) f.stat<-a$fstatistic p.value<-1-pf(f.stat["value"],f.stat["numdf"],f.stat["dendf"]) Is there something similar for a nls?

Hi, I'm using glm function to do logistic regression and now I want to know if it exists a kind of R-squared with this function in order to check the model. Thank you.

Hi, I have a question about repeating something for several columns. I calculated a lot of values for different objects. The values are normally calculated step-wise for every day for each object. With this obtained data frame, I make further calculations. Here I have to calculate some things and I have to take into account the years because I want a number for each year. I do these first for one

Hi all, Like a lot of people I noticed that I get different results when I use nls in R compared to the exponential fit in excel. A bit annoying because often the R^2 is higher in excel but when I'm reading the different topics on this forum I kind of understand that using R is better than excel? (I don't really understand how the difference occurs, but I understand that there is a

I calculate an anova test in the following way: expdata<-read.table("/home/dorien/UA/meta-music/optimuse/optimuse1-build-desktop/results/results_processedCP", header=TRUE)

Hello, I'm trying to annotate my plots nicely and am running into trouble. This example contains two problems: a) the 'text()' arguments do not show the conditional behavior I'm trying to give them. I try to test for the intercept of my regression and reformat the output accordingly ('+ intercept' in the >= 0 case and '- sqrt(intercept^2)' in the other case),

hello mailing list! i still consider myself an R beginner, so please bear with me if my questions seems strange. i'm in the field of biology, and have done consecutive hydraulic conductivity measurements in three parallels ("Sample"), resulting in three sets of conductivity values ("PLC" for percent loss of conductivity, relative to 100%) at multiple pressures

Hi all, I have a basic question concerning graphs in R. I?m using the par() function and I?m working with multiple figures by row (mfrow) but my the hight of my figures become compressed. I have 4 rows and 2 columns (because I want to plot 8 histograms (freq = FALSE ) on it. I know I can adapt my margins with for example ?oma? and ?mai? but I don?t know how to choose the size of the figure? I

Greetings, I have been exploring the use of the caret package to conduct some plsda modeling. Previously, I have come across methods that result in a R2 and Q2 for the model. Using the 'iris' data set, I wanted to see if I could accomplish this with the caret package. I use the following code: library(caret) data(iris) #needed to convert to numeric in order to do regression #I

Hi all, I have a question concerning using several conditions in an ifelse function used as the function in apply. I want to create a new value with the function ifelse ? object which can be coerced to logical mode ?test[n,] >1 & test[n-1,]==0? With n I mean the row. I don?t know how I could do this without a loop. I want to avoid the usage of loops and was thinking about apply. This

Hi Mark, I don't know wether you recived a sufficient reply or not, so here are my comments to your problem. Supressing the constant term in a regression model will probably lead to a violation of the classical assumptions for this model. From the OLS normal equations (in matrix notation) (1) (X'X)b=X'y and the definition of the OLS residuals (2) e = y-Xb you get - by

Hi all, I was wondering if there is a function kind of similar that splits a dataframe, applies a function to each row and returns in a data frame. I know ddply but this one isn?t useful in this situation. I have a dataframe with values for each day (rows) for different objects (columns). I have values for several years. Now, I want to do calculations on only the data of that year. With the

Hello, I'm with a conceptual doubt regarding Rsquared of both lm() and postResample(library caret). I've got a multiple regression linear model (lets say mlr) with anR² value of 67.52%. Then I use this model pro make predictions with predict() function using the same data as input , that is, use the generated model to predict the value associated with data that I used as input. Next, if

Hi, I know that if i want to see the SPLUS or R-source code of a function,i should give the command without the brackets.For example:  rsquared.lmRobMM function(x) {  str0 <- "Initial S-estimate"  str1 <- "Final M-estimate"  if(x$est == "final") {   z <- x$r.squared   attr(z, "info") <- str1  }  if(x$est == "initial") {   z <-

Hello, I'm doing some linear regression: >lm<-lm(osas~alp,data) >summary(lm) However, the Rsquared in the output of summary() is not the same as the "standard" Rsquared calculated by spreadsheets, and outlined in statistical guidebooks, being SSR/SSTO. The output says "multiple Rsquared", but it is no multiple regression... What's the difference? Thanks,

All, Happy World Cup and Wimbledon. This morning finds me with the first of my many daily questions. I am running a ks.test on residuals obtained from a regression model. I use this code: > ks.test(Year5.lm$residuals,pnorm) and obtain this output One-sample Kolmogorov-Smirnov test data: Year5.lm$residuals D = 0.7196, p-value < 2.2e-16 alternative hypothesis: two.sided Warning

Dear R-helpers, I am developing a Mixed-Effects model for a study of immunoassays using 'lme4' library. The study design is as follows: 10 samples were run using 7 different immunoassays, 3 times each, in duplicates. Data attached. I have developed the following model: c.lme <- lmer(Result~SPL + (SPL|Assay/Run) -1, data=data) This model has excellent predictions - the Rsquared of