similar to: strip function

Dear all, I am working with a csv file. Some data of the file are not valid and they are marked with a star '*'. For example : *789. I have attached with this email a example file (test.txt) that looks like the data I have to work with. I see 2 possibilities ..thast I cannot manage anyway in R: 1-first & easiest solution: Read the data with read.csv in R, and define as na strings

Dear all, I have a list of precipitation record and a list of time I would like to sum them up per hour, or per day. Does such a function exist ? example: time<-c("2000-10-03 14:00:00","2000-10-03 14:10:00","2000-10-03 14:20:00","2000-10-03 15:30:00","2000-10-03 16:40:00","2000-10-03 16:50:00","2000-10-03

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Hi, I would like to fit some experimental points by a exponential function. I ignore the parameters of this exponential and what I would like is to ask R to calculate the best fitting curve an the associated parameters (as the linear model function (lm) does for linear models). Is it possible ? Do anyone have an idea about how to do that ? Thanks by advance Jessica Gervais [[alternative

Hi, I am trying to fit experimental points by exponemtial curve my data are stored into a matrix data the first column is the geographical point (a number = data[,1] ) ( I would like to plot several graphes at the same time) the second column is the time of measurement (x in the plot) the third column is a speed (y in the plot) if we assume the point are folowing this exponential behaviour

Hi, I would like to know how to extract the arguments from a list : For example, I have a list of charchacter x x<- c("Bentazone","Atrazine","Epoxiconazol","Metolachlor","Epoxiconazol","Atrazine desethyl","Fenpropimorph","Epoxiconazol","Metolachlor","Simazine","Atrazine

Hi, Being a R-newbie I am wondering how to calculate a correlation coefficient (preferably with an associated p-value) for data like: > d[,1] [1] 25.5 25.3 25.1 NA 23.3 21.5 23.8 23.2 24.2 22.7 27.6 24.2 ... > d[,2] [1] 0.0 11.1 0.0 NA 0.0 10.1 10.6 9.5 0.0 57.9 0.0 0.0 ... Apparently corr(d) from the boot-library fails with NAs in the data, also cor.test cannot cope with a

This is FAQ 7.22. Lattice functions produce graphic objects, which are not displayed by default. If you print your graph, you should be fine. Also, take a look at the documentation for panel.xyplot. Using type = c("p", "r") should make things simpler. Regards, Matt Wiener -----Original Message----- From: r-help-bounces at stat.math.ethz.ch [mailto:r-help-bounces at

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, 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) -

Dear all, I am working with a data file which is the record of precipitation measurement normaly done every 10 minutes. I would like to check if there are missing times in my data file. Is there a function existing able to check for that in R ? Thanks by advance, Jessica

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, 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, 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

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

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

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

Good day! I'm creating HDL IP CORE (for using in FPGA) for theora encoder (now only I-frames). I don't undestand one moment. Now i develop such stages: 1. From RBG(byer) to YCbCr converter 2. DCT processing (8x8 pixels blocks) 3. Quantizator of DCT coeff. 4. Zig-Zag of quantized DCT coeff. and now i have uresolved last stage of compression - how i must send 8x8 blocks to huffman

Hello, I have different results from these two softwares for a simple binomial GLM problem. >From Genmod in SAS: LogLikelihood=-4.75, coeff(intercept)=-3.59, coeff(x)=0.95 >From glm in R: LogLikelihood=-0.94, coeff(intercept)=-3.99, coeff(x)=1.36 Is there anyone tell me what I did wrong? Here are the code and results, 1) SAS Genmod: % r: # of failure % k: size of a risk set data

Dear all, I have some additionale question concerning the spgrass6 package. * When you set a region in GRASS, does the readGRASS6 function in R only load data contained in the zoomed region or the whole map ? * When you have a MASK map in grass, does the readGRASS6 function in R only load data contained inside the MASK area ? Could this be the problem ? Thanks, Jessica