similar to: get mean from cdf

My statistics teacher wrote a textbook and included .rdata files for the problem set data. I downloaded these and opened one of them up. Then I used the ls() command, and it gave me this: > ls() [1] "ex02-12" However, when i try to use the object, it interprets it as ex02 - 12, and I get the following error: > ex02-12 Error: object 'ex02' not found I don't know why

There seems to be a latent flaw in the definition of struct SEXPREC in Rinternals.h, which likely doesn't cause problems now, but could if the relative sizes of data types changes. The SEXPREC structure contains a union that includes a primsxp, symsxp, etc, but not a vecsxp. However, in allocVector in memory.c, zero-length vectors are allocated using allocSExpNonCons, which appears to

Dear List, i would like to mimic the behaviour or the following indexing with a do.call construct to be able to supply the arguments to `[` as a list: test = matrix[1:4,2] result = test[2,] My try, however, did not work: result = do.call(`[`,list(test,2,NULL)) result = do.call(`[`,list(test,2,)) result = do.call(`[`,list(test,2,'')) How can I use the do.call in that way with

Dear colleagues, I have a data set that looks roughly like this; mydat<-data.frame(state=c(rep("Alabama", 5), rep("Delaware", 5), rep("California", 5)), news=runif(15, min=0, max=8), cum.news=rep(0, 15)) For each state, I'd like to cumulatively sum the value of "news" and make that put that value in cum.news. I'm trying as follows but I get

From SVN revision 57032, with configuration R is now configured for i686-pc-linux-gnu Source directory: . Installation directory: /usr/local C compiler: gcc -std=gnu99 -g -O2 Fortran 77 compiler: gfortran -g -O2 C++ compiler: g++ -g -O2 Fortran 90/95 compiler: gfortran -g -O2 Obj-C compiler: Interfaces supported: X11

Full_Name: Murray H Smith Version: 0.63.3 OS: Windows NT Submission from: (NULL) (130.216.5.57) The ppois function is displaced by -0.5. Try: > ppois(-0.5,1) [1] 0.3678794 > ppois(-0.51,1) [1] 0 > ppois(0,1) [1] 0.3678794 and > par(mfrow=c(2,1)) > x<-seq(-1,5,0.01) > plot(x,ppois(x,1),type="s",ylab="F(x)",main="Poisson CDF?") >

I have the following results for an ANOVA comparing two nested models. I wasn't sure how I am supposed to report this result in the area of psychology. Specifically, am I supposed to report the DF's or just the F ratio? I could manually calculate the degrees of freedoms, but there must be a reason why R does not give this information, i.e. those are not conventionally used in the

Hi, I've data from an incomplete fatorial design. One level of a factor doesn't has the levels of the other. When I use lm(), the summary() return NA for that non estimable parameters. Ok, I understant it. But I use contrast::contrast(), gmodels::estimable(), multcomp::glht() and all these fail when model has NA estimates. This is becouse vcov() and coef() has different dimensions. Is

Dear all, I am trying to aggregate a table (divided in two lists here), but get a memory error. Here is the code I'm running : sessionInfo() print(paste("memory.limit() ", memory.limit())) print(paste("memory.size() ", memory.size())) print(paste("memory.size(TRUE) ", memory.size(TRUE))) print(paste("size listX ", object.size(listX)))

Hi, All, How can I get the indices of the minimum elements in a matrix without using a loop? For example, if the matrix is 4 5 2 2 8 9 5 2 3 Then I want to output (1,3), (2,1), (3,2). Thanks, Annie [[alternative HTML version deleted]]

Hi, sorry for repeating the question but this is kind of important to me and i don't know whom should i ask. So as noted before when I do a parameter fit to the beta distr i get: fitdist(vectNorm,"beta"); Fitting of the distribution ' beta ' by maximum likelihood Parameters: estimate Std. Error shape1 2.148779 0.1458042 shape2 810.067515 61.8608126 Warning

Hi, R users, I am trying to use glm to do logistic regression. I know generally when I have two covariates, say x1 and x2, then I do fit <- glm(y~x1+x2,famliy='binomial') But now my covariates form a n*p matrix, say x, so actually each column is a covariate. So I think I should do fit <- glm(y~x,family='binomial') Then I need to predict new data. How should I write the

Hi All, I have a 2-dim density defined on 0<x<1, 0<y<1, x<y. I know the exact formula of the density. How can I visualize it? What plot functions can I use? Thanks, Annie [[alternative HTML version deleted]]

Why does R think these numbers ***are*** equal? In a somewhat bizarre set of circumstances I calculated x0 <- 0.03580067 x1 <- 0.03474075 y0 <- 0.4918823 y1 <- 0.4474461 dx <- x1 - x0 dy <- y1 - y0 xx <- (x0 + x1)/2 yy <- (y0 + y1)/2 chk <- yy*dx - xx*dy + x0*dy - y0*dx If you think about it ***very*** carefully ( :-) )

dear all, i have a quite simple question, i want to fill up a Matrix like done in the following function, but the performance is very bad for large dimensions is there a way to do this like with apply or something similar? makeMatrix <- function(a, b,dim) { X=matrix(0,ncol=dim,nrow=dim) for (i in c(1:dim)){ for (j in c(1:dim)) { if (i==j) {X[i,j]<-a} else { X[i,j]<- exp((

Le 04/12/2018 ? 11:27, I?aki Ucar a ?crit?: > On Tue, 4 Dec 2018 at 11:12, <qweytr1 at mail.ustc.edu.cn> wrote: >> function ppois is a function calculate the CDF of Poisson distribution, it should generate a non-decreasing result, but what I got is: >> >>> any(diff(ppois(0:19,lambda=0.9))<0) >> [1] TRUE >> >> Actually, >> >>>

Hi, all. I'm new to R. I've been a SAS programmer for 20 years. I seem to be having trouble with the most basic task - bringing a table in an *.RData file into a data.frame object. Here's how I created the *.RData file. library(RODBC) db <- odbcConnect("*******") df <- sqlQuery( db , "select * from schema.table where year(someDate)=2006" ) save(

function ppois is a function calculate the CDF of Poisson distribution, it should generate a non-decreasing result, but what I got is: > any(diff(ppois(0:19,lambda=0.9))<0) [1] TRUE Actually, > ppois(19,lambda=0.9)<ppois(18,lambda=0.9) [1] TRUE Which could not be TRUE. Code is tested in both R 3.5.1 and Microsoft R Open 3.5.1. _

Hello, I was wondering if someone knows the formula used by the function lm to compute the t-values. I am trying to implement a linear regression myself. Assuming that I have K variables, and N observations, the formula I am using is: For the k-th variable, t-value= b_k/sigma_k With b_k is the coefficient for the k-th variable, and sigma_k =(t(x) x )^(-1) _kk is its standard deviation.

Dear R-helpers, In the following example I compute ret and returns the SAME way. In ret I use compute returns for EACH column and in returns I do it for the whole data frame. Could someone please tell me why I see a lagged result,by which I mean ret and returns are different by one lag. getSymbols("GOOG",src="yahoo") ret<-apply(GOOG,2,function(x) diff(log(x)) / lag(x,1) )