similar to: help with a regression

Hello, I have a big data frame where consecutive time dates and corresponding observed values for each subject (ID) are on a line. I want to compute the linear slope for each subject. I would like to use apply but I do not know how to express the corresponding function. An example using a loop follows # # create dummy data set There are missing values a <- c(1,2,3,4, 1,1,1,1, 2,2,3,3,

I am using R 2.8 and the latest versions of RMySQL on a Windows XP 64 bit machine. I was wondering if someone could help me figure out how to use dbWriteTable without inserting \r into my table. Consider the following code snippet, which is run after I connect to my database. myDFOut = dbReadTable(conn, "myDF") print(myDFOut) myDFIn = data.frame(x=paste("x", 1:5, sep =

Hi Dear all; I would like to create a plot for regression coefficients with each independent variable (x) along the side and the phenotypes (y) across the top (as given below). For each data point, direction and magnitude of effect could be color and significance could be the size of the circle? Is this possible? I would greatly be appreciated your help. Thanks, Greg y1 y2 y3 y4 y5 y6 x1

Hi all, I have a data frame with "clustered" rows as follows: Cu1 x1 y1 z1 ... Cu1 x2 y2 z2 ... Cu1 x3 y3 z3 ... # end of first cluster Cu1 Cu2 x4 y4 z4 ... Cu2 x5 y5 z5 Cu2 ... # end of second cluster Cu2 Cu3 ... ... "cluster"-size is 3 in the example above (rows making up a cluster are always consecutive). Is there any faster way to sample n clusters (with

Dear list, I have problem that I cannot solve and would like to ask your opinion. I tried to ask a few days ago already but got no answer and all my attempts to solve it by myself since then failed. Sorry for repeated posting! Here the problem broken down a bit. My problem basically is, that I want to use the elements of a character vector as names for objects and by recalling only the

Hi, All I get following question: A data format like following: [Day time x y] Jan1 18:56:24 x1 y1 Jan1 18:56:25 x2 y2 Jan1 18:56:27 x3 y3 Jan1 18:56:28 x4 y4 Jan1 18:56:31 x5 y5 ..................... ..................... what I wanna do is to partion the time interval by unit of 5 seconds. and pick x,y corresponding to the last time within that interval. for the example above, suppose

Dear R-users; Previously I posted a question about the problem of rank deficiency in summary.manova. As somebody suggested, I'm attaching a small part of the data set. #*************************************************** "test" <- structure(.Data = list(structure(.Data = c(rep(1,3),rep(2,18),rep(3,10)), levels = c("1", "2", "3"), class =

Dear help-list members, I am a student at Durham University (UK) conducting a PhD on spatial representation in baboons. Currently, I'm analysing the effect of sampling interval on home range calculations. I have followed the baboons for 234 days in the field, each day is represented by about 1000 waypoints (x,y coordinates) recorded at irregular time intervals. Consecutive waypoints in

Dear members, I want to create 8 graphs and write it into one page using mfrow=c(4,2). How to make all graphs (including the titles, legends, line types) to be scale down (resized proportionally). As an illustration, below is the code: pdf("testmfrow.pdf") par(mfrow=c(4,2)) x<-seq(1:10) y1<-rnorm(10) y2<-rnorm(10,mean=2,sd=1) y3<-rnorm(10,mean=3,sd=1)

Fwiw, encoding magnitude in color is generally a bad idea. Using area(*not* radius) is also not great, but maybe it will work for you. See here for some explanation: https://www.amazon.com/Visual-Display-Quantitative-Information/dp/0961392142/ref=sr_1_1?s=books&ie=UTF8&qid=1518092778&sr=1-1&keywords=Tufte Bert On Feb 7, 2018 11:13 PM, "greg holly" <mak.hholly at

I would like to conduct an extra sum-of -squares test that compares a full MANOVA model (with all 1st order interactions) to a reduced model (no interactions) to determine if I can drop all interactions at the same time. This is analagous to an extra sum-of-squares F-test in ANOVA, but instead using MANOVA. Is there a command in R that does this? If not, is there a command that calculates

Dear R-users I have 5 dependent variables (y1 to y5) and one independent variable (x) and 3 conditioning variables (m, n, and 0). Each of the conditioning variables has 2 levels. I created 2*4 panel plots. xyplot(y1+y2+y3+y4+y5 ~ x | m*n*o,layout = c(4,2)) I would like to reorder the 8 panels. I tried to use index.cond (e.g., index.cond = list(c(1,3,2,4,5,7,6,8)) but it didn't work out.

Hi Yves lavaan looks like a very nice package. From the tutorial introduction I see you create path diagrams for some of the models you describe. How did you do this? I don't see a function for this in the package. I know there is a path.diagram function in the sem package that uses dot to draw the diagram, but I've always found the layouts from dot somewhat strange for path diagrams

R 2.2 on windows XP I have a dataset with multiple columns. Some of the columns represent independent variables, some represent dependent variables. I would like to run the same analyses on a fixed set of independent variables, changing only the dependent variable, e.g. y1-y2=x1+x2+x3 y3-y4=x1+x2+x3 y5-y6=x1+x2+x3, etc. I know I can write a function to perform the analyses, however in order to

Hi After melt you can change levels of your factor variable. Again with the toy example. > levels(temp$variable) [1] "y1" "y2" "y3" "y4" > levels(temp$variable) <- levels(temp$variable)[c(2,4,1,3)] > levels(temp$variable) [1] "y2" "y4" "y1" "y3" > And you will get graphs with this new levels ordering.

Hello, I think the following qualify as a bug given: x<-c(1,2,3,4,2,4,2,2,4,2,2,2,4,3,2,1) z<-embed(x,4) z [,1] [,2] [,3] [,4] [1,] 4 3 2 1 [2,] 2 4 3 2 [3,] 4 2 4 3 [4,] 2 4 2 4 [5,] 2 2 4 2 [6,] 4 2 2 4 [7,] 2 4 2 2 [8,] 2 2 4 2 [9,] 2 2 2 4 [10,] 4 2 2 2

Hello, I think the following qualify as a bug given: x<-c(1,2,3,4,2,4,2,2,4,2,2,2,4,3,2,1) z<-embed(x,4) z [,1] [,2] [,3] [,4] [1,] 4 3 2 1 [2,] 2 4 3 2 [3,] 4 2 4 3 [4,] 2 4 2 4 [5,] 2 2 4 2 [6,] 4 2 2 4 [7,] 2 4 2 2 [8,] 2 2 4 2 [9,] 2 2 2 4 [10,] 4 2 2 2

Hi Petr; Thanks for your reply. It is much appreciated. A small example is given below for 4 independent and 4 dependent variables only. The values given are regression coefficients.I have looked ggplot documents before writing to you. Unfortunately, I could not figure out as my experience in ggplot is ignorable Regards. Greg y1 y2 y3 y4 x1 -0.19 0.40 -0.06 0.13 x2 0.45 -0.75 -8.67 -0.46 x3

Hello, I am trying to modify some functions in an existing package. I can get a function by just typing the name in the console, but how can I get all the functions in a package? the ./R/ subdirectory has files xx.rdb and xx.rdx which I can not decompress with 7-zip or with unzip, unrar etc. The help files are compressed with gzip in the ./man/ folder and can be unzipped easily. Thanks for your

Dear all, I am quite new to R; facing certain problems: Say, I have a text file( named as "try"): Year C1 C2 C3 C4 C5 C6 Y1 3.5 13.8 9.5 6.8 0.4 24.2 Y2 3.8 13.9 9.9 7.6 0.7 12.8 Y3 4.5 14.5 14.2 9.2 0.6 14.5 Y4 5.9 16.2 24.6 12.7 0.2 24.3 Y5 7.2 20.4 40.6 18.2 0.8 28.2 Y6 5.9 18.6 37.4 14.5 0.3 36.9 Y7 8.0 16.1 88.6 24.1 0.1 34.6 Y8 13.6 21.1 56.3 19.0 0.7 33.3 I wish to import the