search for: mdata

...<- function(formula, family, data){ ret <- glm(formula, family=binomial, data=data) return(ret) } myfacpred <- function(object, newdata) { ret <- as.factor(ifelse(predict.glm(object, newdata, type='response') < 0.5, 0, 1)) return(ret) } # logerrorest takes four arguments # mdata is a data frame holding the data to be modeled # form is the output of the is.formula function # rvar is the response variable as a string, for example 'birdx' # fvar is the fold variable, for example 'recordyear' logerrorest <- function(mdata, form, rvar, fvar) { require(Hmisc)...

...<- function(dat=v) { test<-nlsList(result~a+(b-a)/(1+(conc/(c+z*cdiff))^d) |rep,start=dat$init,data=dat$mixeddat) return(nlme(test,random=b~1)) } Here, v is an appropriate data frame in the main workspace. Inserting two lines in nlme.nlsList will fix it: Replacing line 21 from bottom "mData<-eval(mData) " by " mData <- eval(mData,parent.frame())" Inserting thisCall[["data"]]<-mData on line 4 from bottom before val <- do.call("nlme.formula", thisCall) Doing this makes nlme.nlsList recognize the input data of the regression. But is...

....data, form, 'basp', 'recordyear') Error in order(na.last, decreasing, ...) : Argument 1 is not a vector My questions are, why doesn't this work, and how do I fix it. I'm using the formula function to create the formula that I'm sending to my function. And the mdata is a data.frame. I'm assumed that if I passed the column names as strings (response variable - rvar, fold variable - fvar) this would work. Apparently however it doesn't. Lastly, since I don't have access to MASS and there are apparently many examples of doing this kind of thing in MAS...

...st whether the effect is different for animals and plants/whether "group" (animal/plant) influences the effect size, but am having trouble interpreting the results I get. I've read previous posts about QM in metafor, but I'm still a bit confused. I've dummy-coded the factors: MData$k.animal<-ifelse(MData$Type=="Animal",1,0)MData$k.plant<-ifelse(MData$Type=="Plant", 1,0) I've then run the model with the following script: MixModel<-rma(z, v.z., mods=cbind(k.animal, k.plant), intercept=FALSE, data=MData, subset=(Response=="Tolerance"...

...# months months <- c("Jan", "Feb", "Mar", "Apr", "May", "Jun", "Jul", "Aug", "Sep", "Oct", "Nov", "Dec"); n <- length(months); m <- 10; mdata <- data.frame(x = runif(m*n), y = runif(m*n), z = rep(months, each = m)); h <- xyplot(y ~ x | z, data = mdata); plot(h); #------------------------------------------------------------- The output of the xyplot() is sorted by the alphabetical order of months (First Apr, then Aug so on). I need...

...r: Error in `row.names<-.data.frame`(`*tmp*`, value = c("NA.NA", "NA.NA", : duplicate 'row.names' are not allowed In addition: Warning message: non-unique value when setting 'row.names': ‘NA.NA’ My code: > flevel1$Q120<-as.factor(flevel1$Q120) > MData<-mlogit.data(flevel1, varying=NULL,choice="Q120", shape="wide") > LogitModel<-mlogit(flevel1$Q120~1|flevel1$Q92, data=MData) MData[1:10,] Q56 Q92 Q102 Q119 Q120 1.1 10 7 10 10 FALSE 1.10 10 7 10 10 TRUE 1.2 10 7 10 10 FALSE 1.3 10 7...

*I have the following function:* /plot_mi_time = function(mdata, miname) { mdata2 = mdata[row.names(hakat) == miname, ] print(mdata2) xcoords <- c(1,1,2,2,3,3,4,4,5,5,6,6) plot(c(xcoords), mdata2, xaxt="n", ylab="Expression", xlab="Time(h)", , main=miname) axis(1, at=xcoords, labels=c("a","a",...

Hi everyone, I have some data about a market research which I want to arrange in one plot for easy viewing, the data looks something like: Product Color StoreA StoreB StoreC StoreD Price ProdA R NA 4.33 2 4.33 35 G NA 4.33 2 4.33 35 B

Hi, I have the following function that i use to plot graphs. plot_mi_time = function(mdata, miname) { mdata2 = mdata[row.names(mir_test3) == miname, ] # print(mdata2) xcoords <- c(1,1,2,2,3,3) plot(xcoords, mdata2, xaxt="n", ylab="Expression", xlab="Time(h)", , main=miname) axis(1, at=xcoords, labels=c("a","a","b&q...

..., absent2001 = c(11.6, 12.8, 10.1, 9.2, 9.4, 8.1, 8.7, 8.5, 8.6, 8.6, 7.8, 7.5, 8.5, 8.5, 8, 6.3, 7.5, 7.3, 7.3, 8.5, 7.9, 8.5, 9, 6.2, 6.5, 5.1)), .Names = c("group", "absent2011", "absent2001"), class = "data.frame", row.names = c(NA, -26L)) mdata <- structure(list(group = structure(c(7L, 10L, 24L, 19L, 12L, 3L, 14L, 22L, 6L, 13L, 8L, 11L, 5L, 17L, 21L, 20L, 23L, 2L, 1L, 4L, 25L, 9L, 15L, 26L, 16L, 18L), .Label = c(" Accommodation and food services", "Agriculture ", " Business, building & other support se...

Hi, Sorry for my ignorance, but how can I see if a function is 'compiled code' or plain R? E.g. the daisy function from the cluster package.

...alues to a small value, close to zero { min.val = min(col[col > 0]) col[col < 0] = (min.val / 10) col # Column index })) I think this is how to start, but the NA replacement part doesn't work... newMatrix = as.matrix(apply(oldMatrix, 2, function(col) { min.val = min(mData, na.rm = T) # Find the smallest value in the dataset col[col == NA] = (min.val / 10) # Doesn't work... col # Column index } Does any of you have any suggestions? Best regards, Joel _________________________________________________________________ Hitta kär...

...// Allocate our own buffers (1 channel, 16 bits per sample, thus 16 bits per frame, thus 2 bytes per frame). // Practice learns the buffers used contain 512 frames, if this changes it will be fixed in processAudio. tempBuffer.mNumberChannels = 1; tempBuffer.mDataByteSize = FRAME_SIZE * 2; tempBuffer.mData = malloc( FRAME_SIZE * 2 ); XThrowIfError(AudioSessionSetActive(true), "couldn't set audio session active\n"); // Initialise status = AudioUnitInitialize(rioUnit);...

Dear all, I need some help on plotting multiple boxplots on one figure. I have three matrix A, B and C. Each of them is a 1000 by 10 matrix. The 10 columns of all three matrix correspond to the 10 values of the same parameter, say k=1, ..., 10. I want to make a plot where x axis represents different values of k. For each k value, I want to plot three boxplots, one on top of another. For

Dear All Can you please do me favor and let me know how can I compare two files but not in line-by-line basis on my CentOS server ? I mean say row#1 in file1 has the same data as say row#5 in file2 , but the comm compares them in line-by-line basis that is not intended . It seems that the diff cannot do the job as well . Thank you in advance -------------- next part -------------- An HTML

...y, I know that glusterfs creates link files (size zero and sticky bit set) when the new filename is hashed to a different brick. However, these link files are only visible to the glusterfsd, not to the users. I also noted that now the "link files" have an additional attibute (glusterfs.mdata) and previously link files stored the "real" brick name in "glusterfs.dht.linkto". Therefore, I have a two of questions: 1) Is this a bug, or is it intentionally that glusterfs 10 clients don't work with glusterfs 9 servers? 2) If it's not a bug, how does the attrib...

I am using the following function to plot columns from a CSV-file: plot_mi_time = function(mdata, miname) { mdata2 = mdata[row.names(mir_test) == miname, ] # print(mdata2) plot(c(1:3), mdata2, xaxt="n", ylab="Expression", xlab="Time(h)", , main=miname, pch=16) axis(1, at=c(1,2,3),labels=c("a","b","c")) } No I have a CSV f...

...atching with replacement (the "M=1" option). # Estimating the treatment effect on the treated (the "estimand" option defaults to ATT). rr <- Match(Y=Y, Tr=Tr, X=X, M=1);* And here's where the 'problem' occurs: *summary(rr) # gives an estimate of 2153.3 mean(rr$mdata$Y[rr$index.treated])-mean(rr$mdata$Y[rr$index.control]) # gives an estimate of 1083.848 * Notice that when I simply subtract the means from one another, I get a different estimate (1083.848) than when the algorithm outputs (2153.3). It seems that the obvious answer is that I'm not computing th...

Hi, I want to know which distance is using in the function kmeans and if we can change this distance. Indeed, in the function pam, we can put a distance matrix in parameter (by the line "pam<-pam(dist(matrixdata),k=7)" ) but we can't do it in the function kmeans, we have to put the matrix of data directly ... Thanks in advance, Nicolas BOUGET

Hi I was trying to import a dataset from access database into R My code looks like below >library(RODBC) >setwd("//STATESFPSOP1/docs/CRMPO/DEPT/Retail PD") >channel<-odbcConnectAccess("MortgagePD.mdb") >Mdata<-sqlFetch(channel, "MortgagePD") But I got error messages saying Error: cannot allocate vector of size 27.9 Mb In addition: Warning messages: 1: In names(value) <- make.unique(colnames) : Reached total allocation of 1535Mb: see help(memory.size) ….. Then I tried to increase...