search for: timepoint

Hi, I'm having difficulty importing my textfile that looks something like this: #begin text file Timepoint 1 ObjectNumber Volume SurfaceArea 1 5.3 9.7 2 4.9 8.3 3 5.0 9.1 4 3.5 7.8 Timepoint 2 ObjectNumber Volume SurfaceArea 1 5.1 9.0 2...

I have the following dataset: 96 plots 12 varieties 2 time points The experiment is arranged as follows: A single plot has two varieties tested on it. With respect to time points, plots come in 3 kinds: (1) varietyA, timepoint#1 vs. variety B, timepoint#1 (2) varietyA timepoint #2 vs. varietyB timepoint #2 (3) varietyA timepoint #1 vs. variety A timepoint#2 - there are 36 of each kind of within timepoint comparison and 24 between timepoint comparisons. so it isn't a fully connected design. Plots and varieties...

I am attempting to model data with the following variables: timepoint - n=48, monthly over 4 years hospital - n=3 opsn1 - no of outcomes total.patients skillmixpc - skill mix percentage nurse.hours.per.day Aims To determine if skillmix affects rate (i.e. no.of.outcomes/total.patients). To determine if nurse.hours.per.day affects rate. To determine if r...

Hi Folks, I have repeated measures for data on association time (under 2 acoustic condtions) in male and female frogs as they grow to adulthood (6 timepoints). Thus, two within-subject variables (Acoustic Condition: 2 levels, Timepoint: 6 levels) and one between-subject variable (Sex:male or female). I am pretty sure my distributions depart from normality but I would first like to simply run a RM anova on the data. My problem is that when I do this I...

...trix" (Intercept) (Intercept) 8.519916 attr(,"sc") scale 3.215072 > VarCorr(fm)[[1]][1] [1] 8.519916 > VarCorr(fm)[[2]][1] Error in VarCorr(fm)[[2]] : subscript out of bounds ########################################################## set.seed(500) n.timepoints <- 4 n.subj.per.tx <- 20 sd.d <- 5; sd.p <- 2; sd.res <- 1.3 drug <- factor(rep(c("D", "P"), each = n.timepoints, times = n.subj.per.tx)) drug.baseline <- rep( c(0,5), each=n.timepoints, times=n.subj.per.tx ) #Patient <- rep(1:(n.subj.per.tx*2), each = n...

...attached base packages: [1] grid stats graphics grDevices utils datasets methods base other attached packages: [1] effects_1.0-12 lattice_0.17-8 loaded via a namespace (and not attached): [1] Matrix_0.999375-11 lme4_0.999375-24 nlme_3.1-89 tools_2.7. set.seed(500) n.timepoints <- 4 n.subj.per.tx <- 20 sd.d <- 5; sd.p <- 2; sd.res <- 1.3 drug <- factor(rep(c("D", "P"), each = n.timepoints, times = n.subj.per.tx)) drug.baseline <- rep( c(0,5), each=n.timepoints, times=n.subj.per.tx ) Patient <- rep(1:(n.subj.per.tx*2), each =...

Full_Name: Dav Clark Version: 2.0.1 OS: OS X 10.3 Submission from: (NULL) (128.122.87.35) When the timepoints that reshape uses (in direction="long") are negative or fractional, the time label is assigned incorrectly. It is easier to give an example than to describe the problem abstractly: Assume you have a data.frame header with values related to peri-stimulus time like this: "HRF -5&qu...

Hello, I have a number of files output1.dat, output2.dat, ... , output20.dat, each of which monitors several variables over a fixed number of timepoints. From this I want to create a data frame which contains the mean value between all files, for each timepoint and each variable. The code below works, but it seems like I should be able to do the second part without a for loop. I played with sapply(myList, mean), but that seems to take the m...

...on it. if variety A on plot #1 has treatment T1 applied to it, then variety B on plot #1 has treatment T2 applied to it. across the whole experiment variety A is exposed to treatment T1 the same number of times as treatment T2. with respect to time points, plots come in 3 kinds. (1) varietyA, timepoint#1 vs. variety B, timepoint#1 (2) varietyA timepoint #2 vs. varietyB timepoint #2, and (3) varietyA timepoint #1 vs. variety A timepoint#2 plots and varieties are random samples from a population of plots and varieties, so they are random effects. The technical treatment and timepoints are fix...

dear list, i am using a multifactorial design with two treatments (factor A: drugs, three levels; factor B: theraphy, two levels) and a time factor (three levels, different timepoint). hypothetically, i measured the same subjects for all treatements and timepoints, so its a repeated measurement design. now i ran an anova in R and also some Tukey post-hoc tests using glht. but what i am actually interested in is to perform conditional contrasts, f.i. A1 for the 1st timepoint vs....

dear list, i am running a within-design ANOVA with 4 factors (4,4,2 and 2 levels each). the last one is a time factor comprising two different treatment timepoints. i fit a mixed-effects model using lme and apply the anova function to the outcome. according to this analysis, there are highly significant main effect on the first and the time factor. i then checked the boxplots for the two 4-level factors for each timepoint separately: there is a difference or...

...e between-subjects factor (SEX) belongs in the Error Term of the aov or not? And if it does belong, where in the Error Term does it go? The 3 possible scenarios are listed below: e.g., 1. Omit Sex from the Error Term: >My.aov = aov(Locomotor.Response~(Age*AcousticCond*Sex) + Error (Subject/(Timepoint*Acx.Cond)), data=locomotor.tab) note: Placing SEX outside the double paretheses of the Error Term has the same statistical outcome effect as omitting it all together from the Error Term (as shown above in #1). 2. Include SEX inside the Error Term (inside Double parentheses): >My.aov = ao...

library(ggplot2) a<- read.table("data", header=T) b = na.omit(a) ggplot(data=b) + geom_line(aes(x=timepoint, y=value,group=sample, colour= factor(sample))) +? geom_point(aes(x=timepoint, y=value, group=s ample)) + facet_wrap(~bio, scales = "free",ncol = 5) +theme_bw() + opts(legend.direction = "horizontal",??? legend.position = "top",???? legend.background = theme_blank()) +...

Hi, i do have a dataframe representing data from a repeated experiment. PID is a subject identifier, Time are timepoints in an experiment which was repeated twice. For each subject and all three timepoints there are 2 sets of four values. df <- data.frame(PID = c(rep("A", 12), rep("B", 12), rep("C", 12)), Time = rep(c(0, 0, 0, 0, 30, 30, 30, 30, 60, 60, 60, 60),...

...llow 6 8 8 7 7 7 6 6 6 7 7 7 6 6 6 6 6 5 red 15 15 15 15 15 15 14 13 12 11 12 10 9 8 7 6 8 9 pink 11 11 11 11 11 10 12 11 13 14 14 15 15 14 14 17 17 17 blue 17 15 15 16 17 17 17 17 18 18 18 19 20 20 20 21 22 21 the column names are timepoints 450 in total. When I run the heatmap code: x <- as.matrix(o4) ramp <- colorRamp(c("yellow","green","blue")) cv<-rgb( ramp(seq(0, 1, length = 83)), max = 255) heatmap(x, col = cv, Colv=NA, Rowv=NA,xaxt="n", yaxt="n", scale="column&q...

...g a difficult time finding some guidelines about this (perhaps I am not searching the correct terms or references). My KM figures contain typically two curves that I am comparing using the logrank test. Inspecting the ?survfit.object yields the std.err field that gives the standard error for each timepoint on the curve. Is it recommended that I just name the timepoint at which the standard error exceeds 0.1 in the figure legend? For example, "The standard error exceeds 10% at time points beyond 394 days." I have seen this strategy in other publications. What is your approach? Thanks fo...

Hi, I cannot find a 'vectorized' solution to this 'for loop' kind of problem. Do you see a vectorized, fast-running solution? Objective: Take the value of X at each timepoint and calculate the corresponding value of Y. Leading 0's and all 1's for X are assigned to Y; otherwise Y is incremented by the number of 0's adjacent to the last 1. The frequency and distribution of X vary widely and may have ~100 repeated 0's or 1's in a vector of 10k timepoi...

I have a data frame with two variables that are factors. One is actually a TRUE/FALSE factor, and I have coded it as 1/0, a continuous variable, but I could turn it back into a factor. The second is an ordered factor and consists of five timepoints. There are several continuous variables as well. Now I want to fit a linear model to my data, using lm (or another R procedure if recommended). Question: should I use polynomial contrasts? My timepoints are very far from being evenly spaced, so ordinary R contrasts seem more natural. But I'm...

System: R 2.3.1 on a Windows XP computer. I am validating several cancer prognostic models that have been published with a large independent dataset. Some of the models report a probability of survival at a specified timepoint, usually at 5 and 10 years. Others report only the linear predictor of the Cox model. I have used Harrell's c index for censored data (rcorr.cens) as a measure of discrimination and have constructed smoothed calibration plots. I would like to include some measures of overall model performanc...

Dear all R users, I want to realize 800 000 ANOVAS and to store Sum of Squares of the effects. Here is an extract of my table data Product attribute subject rep t1 t2 t3 … t101 P1 A1 S1 R1 1 0 0 … 1 I want to realize 1 ANOVA per timepoint and per attribute, there are 101 timepoints and 8 attributes so I want to realize 808 ANOVAS. This will be an ANOVA with two factors : Here is one example: Aov(t1~Subject*Product,data[data$attribute==”A1”,]) I want to store for each ANOVA SSprod,SSsujet,SSerreur,SSinter and SStotal. In fact I want...