search for: ldply

I've examining a number of linear regression models on a large dataset following the basic ideas presented here http://www.r-bloggers.com/r-calculating-all-possible-linear-regression-models-for-a-given-set-of-predictors/ Calculating all possible linear regressions . I run into a problem with ldply when I have a formula that includes no intercept. Here's a simple test to show what happens. # data and two linear model regressions xy <- data.frame(cbind(x=(0:10),y=2*x + 0.2*rnorm(11))) models <- as.list(c('y ~ x', 'y ~ -1 + x')) models <- lapply(models, function(x)...

hey, folks, I have two very simple questions. I am not quite sure if I can do this using R. so, I am analyzing a large data frame with thousands of variables. For example: Dependent variables: d1, d2, d3 (i.e., there are three dependent variables) Independent variables: s1, s2, s3, ......s1000 (i.e., there are 1000 independent variables) now I want to run simple linear regression analyses of

Hi all, I have the next problem: I have a matrix with size 8,000,000x18. My personal computer...blocks...so I have cut my original file into 100 different file. I have written a function that should be run on each of this file. So imagine I need to read data from q1 to q100 file data<-read.table("q1.txt",sep="") and each time I read 1 file execute my personal function

..., 0.72, 0.67, 0.73, 0.76, 0.65, 0.68, 0.78, 0.73, 0.71) test <- data.frame(individual, day, condition) ind.id <- unique(test$individual) ind.list <- lapply(1:length(ind.id), function(i){ subset(test, test$individual==ind.id[i])}) lms <- lapply(ind.list, lm, formula=condition~day) ldply(lms, function(x) x$coefficients) here I can display coefficients, here I need to write code for r2.squared I tried with following script summary(ldply(lms, function(x) )$r.squared, $p value) but it did not work. can any one help me? thanks Kristi [[alternative HTML version dele...

...n variables back into a dataframe. mydf <- data.frame(x1=rnorm(100), x2=rnorm(100), x3=rnorm(100)) mydf$fac<-factor(sample((0:2),replace=T,100)) mydf$y<- mydf$x1+0.01+mydf$x2*3-mydf$x3*19+rnorm(100) dlply(mydf,.(fac),function(df) lm(y~x1+x2+x3,data=df))->dl here I?d like to use ldply(dl,coef(summary)) or something similar but I cant figure it out... Best, M

...of JSON lists mapping lat/long route points between locations using CloudMade's API. post_url is the URL of the HTTP request for (n in json_dir) { i = i + 1 if (typeof(json_dir[[i]]) != "NULL") { if (i == 1) { dat_add <- ldply(json_dir[[i]], function(x) t(data.frame(x)), .progress = "text") names(dat_add) <- c("lat", "lon") json_path <- list(dat_add) } else { dat_add <- ldply(json_dir[[i]], function(x) t(dat...

...t column in each datframe contains a vector of "Interval" class after I have produced this column using "lubridate" package. I needed to convert my list of dataframes to be in a single dataframe for further analysis. I did this using the following syntax : SingleDataframe <- ldply (MylistofDFs,data.frame) The problem is that after this conversion, the the column with time intervals has been converted into numeric type with number of seconds during the time interval instead of the interval itself as follows : The column Before conversion looks like this: TimeInterval (clas...

...???????? m <- as.numeric(x$y) ????????????? x1 <- m <- as.numeric(x$x1) ????????????? x2 <- m <- as.numeric(x$x2) ????????????? summary(glm(m~1+x1+x2, family=Gamma),dispersion=NULL) ?????????????? }) #Save the results of the estimated parameters str(FGLM,no.list = TRUE) SFGLMC<- ldply(FGLM, function(x) x$coefficients) SFGLMD<- ldply(FGLM, function(x) x$dispersion) GLM <- cbind(SFGLMC,SFGLMD)

for example : I have files with the name "ma01.dat","ma02.dat","ma03.dat","ma04.dat",I want to read the data in these files into one data.frame flnm<-paste("obs",101:114,"_err.dat",sep="") newdata<-read.table(flnm,skip=2) data<-(flnm,skip=2) but the data only contains data from the flnm[1] I also tried as below : for

...n each datframe contains a vector of "Interval" class after I have produced this column using "lubridate" package. I needed to convert my list of dataframes to be in a single dataframe for further analysis. I did this using the following syntax : > > SingleDataframe <- ldply (MylistofDFs,data.frame) > > The problem is that after this conversion, the the column with time intervals has been converted into numeric type with number of seconds during the time interval instead of the interval itself as follows : > > The column Before conversion looks like this:...

Hi there, I ran the following code: vols=read.csv(file="C:/Documents and Settings/Hugh/My Documents/PhD/Swaption vols.csv" , header=TRUE, sep=",") X<-ts(vols[,2]) #X dcOU<-function(x,t,x0,theta,log=FALSE){ Ex<-theta[1]/theta[2]+(x0-theta[1]/theta[2])*exp(-theta[2]*t) Vx<-theta[3]^2*(1-exp(-2*theta[2]*t))/(2*theta[2]) dnorm(x,mean=Ex,sd=sqrt(Vx),log=log) }

Hi I would like to shorten mod1 <- nls(ColName2 ~ ColName1, data = table, ...) mod2 <- nls(ColName3 ~ ColName1, data = table, ...) mod3 <- nls(ColName4 ~ ColName1, data = table, ...) ... is there something like cols = c(ColName2,ColName3,ColName4,...) for i in ... mod[i-1] <- nls(ColName[i] ~ ColName1, data = table, ...) I am looking forward to help Christof

I need to run binomial tests (binom.test) on a large set of data, stored in a table - 600 tests in total. The values of x are stored in a column, as are the values of n. The data for each test are on a separate row. For example: X N 11 19 9 26 13 21 13 27 18 30 It is a two-tailed test, and P in all cases is 0.5. My question is: Is there a quicker way of running these tests without having to

Hi group, I imported 16 data frames using the function "list.files" temp <- list.files(path="...........") myfiles = lapply(temp, read.table,sep = "") Now I have 16 data set imported in R window. I want to combine them by row and tried some thing like (Here I am considering only 20 columns) for(i in 1:16){ data<- cbind(myfiles[[i]][,1:20]) } but it

Dear All Sorry for this simple question, I could not solve it by spending days. My data looks like this: # data set.seed(1234) clvar <- c( rep(1, 10), rep(2, 10), rep(3, 10), rep(4, 10)) # I have 100 level for this factor var; yvar <- rnorm(40, 10,6); var1 <- rnorm(40, 10,4); var2 <- rnorm(40, 10,4); var3 <- rnorm(40, 5, 2); var4 <- rnorm(40, 10, 3); var5 <- rnorm(40, 15,

Dear List, I'm trying to use different R packages for my Teaching Assistantship classes. And I cam out to an (apparently) very simple problem. I would like to retrieve the vertices coordinate of a SpatialPolygon data. I know this is stored in the "coords" slot, but I can't get access to it! I tried to coerce the SpatialPolygon into a data.frame but it doesn't work.

Folks, I am trying to get a loop to run which increments the object name as part of the loop. Here "fit1" "fit2" "fit3" and "fit4" are linear regression models that I have created. > for (ii in c(1:4)){ + SSE[ii]=rbind(anova(fit[ii])$"Sum Sq") + dfe[ii]=rbind(summary(fit[ii])$df) + } Error in anova(fit[ii]) : object 'fit' not found

Greetings all, I am getting this error that is driving me nuts... (not a long trip, haha) I have a set of files and in these files I want to calculate ttests on rows 'compareA' and 'compareB' (these will change over time there I want a variable here). Also these files are in many different directories so I want a way filter out the junk... Anyway I don't believe that this is

I have a list of dataframes i.e. each list element is a dataframe with three columns and differing number of rows. The third column takes on only two values. I wish to split the list into two sublists based on the value of the third column of the list element.  Second issue with lists as well. I would like to reduce each of the sublist based on the range of the second column, i.e. if the range of

...or (j in 1:m) resultDF4[(((j-1)*nr) + 1):(j*nr), ] <- mylist[[j]] ## This is a bit error prone for my taste. If the data frames have ## different numbers of rows, some major code surgery will be needed. ## ## Dennis Murphy pointed out the plyr package, by Hadley Wickham. ## Dennis said " ldply() in the plyr package. The following is the same ## idea as do.call(rbind, l), only faster." library("plyr") resultDF5 <- ldply(mylist, rbind) all.equal(resultDF, resultDF5) ## Plyr author Hadley Wickham followed up with "I think all you want here is rbind.fill:" r...