similar to: How can I get "predict.lm" results with manual calculations ? (a floating point problem)

find the position of the first value who equals certain number in a vector: Say a=c(0,0,0,0,0.2, 0.2, 0.4,0.4,0.5) i wish to return the index value in a for which the value in the vector is equal to 0.4 for the first time. in this case, it is 7. [[alternative HTML version deleted]]

Hello I've stumbled upon a problem for inversion of a matrix with large values, and I haven't found a solution yet... I wondered if someone could give a hand. (It is about automatic optimisation of a calibration process, which involves the inverse of the information matrix) code: ********************* > macht=0.8698965 > coeff=1.106836*10^(-8) >

Hello:I am trying to match the value 0.3 in the sequence seq(.2,.3). I get > 0.3 %in% seq(from=.2,to=.3) [1] FALSE Yet > 0.3 %in% c(.2,.3) [1] TRUE For arbitrary sequences, this "invisible .3" has been problematic. What is the best way to work around this? Thank you. Dan [[alternative HTML version deleted]]

With floating point numbers I'm seeing 'cut' putting values in the wrong bands. An example below places 0.3 in (0.3,0.6] i.e. 0.3 > 0.3. > x = 1:5*.1 > x [1] 0.1 0.2 0.3 0.4 0.5 > cut(x, br=c(0,.3,.6)) [1] (0,0.3] (0,0.3] (0.3,0.6] (0.3,0.6] (0.3,0.6] Levels: (0,0.3] (0.3,0.6] I'm sure this is probably the same issue documented in the FAQ (7.31 Why doesn't R

Hi, I am not sure if this is just me using R (R-2.3.1 and R-2.4.0) in the wrong way or if there is a more serious bug. I was having problems getting some calculations to add up so I ran the following tests: > (2.34567 - 2.00000) == 0.34567 <------- should be true [1] FALSE > (2.23-2.00) == 0.23 <------- should be true [1] FALSE > 4-2==2 [1] TRUE > (4-2)==2 [1] TRUE >

Perhaps somebody could clarify for me if the following is a floating point matter or otherwise, and how am I to correct for it? > floor(100*.1) [1] 10 > 100*(1.0-.9) [1] 10 > floor(100*(1-0.9)) [1] 9 Thanks! Michael _______________________________________________________ Michael Folkes Salmon Stock Assessment Canadian Dept. of Fisheries & Oceans Pacific Biological Station

Hello, I can`t figure out how can increase the velocity of the fitting data by nls. I have a long data .csv I want to read evry time the first colunm to the other colunm and analisy with thata tools setwd("C:/dati") a<-read.table("Normalizzazione.csv", sep=",", dec=".", header=F) for (i in 1:dim(a[[2]]]) { #preparazione dati da analizzare

Hi All, Another nls related problem (for background, I'm migrating a complicated modelling package from S-plus to R). Below I've reduced this to the minimum necessary to demonstrate my problem (I think); the real situation is more complicated. Two similar selfStart functions, ssA and ssB. The 'initial' function for ssB modifies its arguments a little and then calls getInital

Hello, I was trying to build some functions which I would like to integrate over an interval using the function 'integrate' from the 'stats' package. As an example, please consider the function h(u)=sin(pi*u) + sqrt(2)*sin(pi*2*u) + sqrt(3)*sin(pi*3*u) + 2*sin(pi*4*u) Two alternative ways to 'build' this function are as in f and g below: coeff<-sqrt(1:4)

Hello, from Verzani, simpleR (pdf), p. 80, I created the following function to test the coefficient of lm() against an arbitrary value. coeff.test <- function(lm.result, var, coeffname, value) { # null hypothesis: coeff = value # alternative hypothesis: coeff != value es <- resid(lm.result) coeff <- (coefficients(lm.result))[[coeffname]] # degrees of freedom = length(var) -

I am interested in testing two similar nls models to determine if the lines are statistically different when fitted with two different data sets; one corn, another soybean. I know I can do this in linear models by testing for interactions. See Introductory Statistics with R by Dallgaard p212-218 for an example. I have two different data sets I am comparing to lai. ci.re should have very

I've been playing around with contrasts in lm by specifying the contrasts argument. So, I want to specify a specific contrast to be tested Say: > y _ rnorm(100) > x _ cut(rnorm(100, mean=y, sd=0.25),c(-3,-1.5,0,1.5,3)) > reg _ lm(y ~ x, contrasts=list(x=c(1,0,0,-1))) > coef(reg)[2] x1 -1.814101 I was surprised to see that I get a different estimate for the

The following series adds initial vector support for PowerPC64. On POWER9, flac --best is about 3.3x faster. Amitay Isaacs (2): Add m4 macro to check for C __attribute__ features Check if compiler supports target attribute on ppc64 Anton Blanchard (5): configure.ac: Remove SPE detection code configure.ac: Add VSX enable/disable configure.ac: Fix FLAC__CPU_PPC on little endian, and add

Dear list, I'd like to use path.analysis in the package agricolae in batch format on many files, retrieving the path coefficients for each run and appending them to a table. I don't see any posts in the help files about this package or the path.analysis package. I've tried creating an object out of the call to path.analysis, but no matter what I try, the function automatically prints

Hi R users I am trying to run a non linear parameter optimization using the function optim() and I have problems regarding memory allocation. My data are in a dataframe with 9 columns. There are 656100 rows. >head(org_results) comb.id p H1 H2 Range Rep no.steps dist aver.hab.amount 1 1 0.1 0 0 1 100 0 0.2528321

Hi, I've a list containing parameters (intercepts & coefficients) of 12 regressions fitted > coeff [[1]] (Intercept) anno -427017.1740 217.0588 [[2]] (Intercept) anno -39625.82146 21.78025 ..... [[12]] (Intercept) anno 257605.0343 -129.7646 I want create a data frame with two columns (intercept and anno)using data in these list. Any help

Hi all, This is probably going to come off as unnecessary (and show my ignorance) but I am trying to understand the parameter estimates I am getting from R when doing an ANCOVA. Basically, I am accustomed to the estimate for the categorical variable being equivalent to the respective cell means minus the grand mean. I know is the case in JMP - all other estimates from these data match the

Hi all, I have a problem to estimate Std. Error and t-value by ?polr? in library Mass. They result from the summary of a polr object. I can obtain them working in the R environment with the following statements: temp <- polr(formula = formula1, data = data1) coeff <- summary(temp), but when the above statements are enclosed in a function, summary reports the following error:

Dear all, I am trying to calculate the fitted values using a ridge model (lm.ridge(), MASS library). Since the predict() does not work for lm.ridge object, I want to get the fitted_value from the coefficients information. The following are the codes I use: fit = lm.ridge(myY~myX,lambda=lamb,scales=F,coef=T) coeff = fit$coef However, it seems that "coeff" (or "fit$coef") is

Hi, I would like to asign titles into strip of a panel of curves designed thanks to a xyplot function (lattice package) I ve tried the "strip" function, ... here is my code I would like to write in the strip of each panel conditionning varialbe (here, it is data[,1]) xyplot (data[,3] ~ data[,2] | data[,1] , strip.default (which.given=1, which.panel = c(1:15), var.name= c