search for: heights

How does one create a vector whose contents is the list of variables in a dataframe pertaining to a particular pattern? This is so simple but I cannot find a straightforward answer. I want to be able to pass the contents of that list to a "for" loop. So let us assume that one has a dataframe whose name is Data. And let us assume one had the height of a group of people measured at

I see the new kernel available in the plus repository... is there some way to view what the differences are between that and the mainline kernel?

Hi, I'm scratching my head as to why I can't use the subset() command to remove one line of data from a data frame. There is just one row (out of 45840) that I'd like to remove and it can be identified using.... > dim(raw.all.clean) [1] 45840 10 > subset(raw.all.clean, Height.1 == 0 & Height.2 == 0) Sample.Name Well SNP Allele.1 Allele.2 Size.1 Size.2

...intuitively, to me at least, the command x[-integer(0)] should leave x untouched. However the actual output under R2.4.0 is integer(0). A slightly more involved example demonstrates why I think this behaviour is back to front. First we define a data frame, in this case some people, with their heights. peoples.heights = data.frame(names = c("Alice", "Bob", "Carol"), heights = c(1.67, 1.85, 175)) To make sure the heights are sensible, we define a filter out impossibly tall people. dubious.records = which(peoples.heights$heights > 2.5) #3 peoples.height...

If these things bother you, could you please supply fixes? For some reason you omitted the part of the help page dealing with one of these (which is as documented), and I also think you are confusing `plot' and `device' sizes. The latter will be the displayed device surface, I believe. Nothing on the help page suggested that the fine details of the R read-only parameters will be as you

Sorry for the confusion, here is the edited question. The data= Stand_Height (attached) is recorded from 12/1/2009 to 12/31/2015 (25 observations) and the other dataset (leafbiom) is recorded from 10/7/2009 to 12/29/2016 (daily observations). I want to use the 25 observations of stand height to predict the daily stand height from 10/7/2009 to 12/29/2016. The daily stand height will be multiplied

Estimados usuarios de R: Soy nuevo en R, le ruego disculpa por mi *Ignorancia*. Estoy corriendo la siguiente instruccion, ggsave(paste('img/',plotName,'.png', sep='') y resulta el error: ggsave(paste('img/',plotName,'.png', sep='') Saving 7.32 x 4.87 in image Error in grDevices::png(..., width = width, height = height, res = dpi, :

############################################################################### Before reporting 4 problems with windows(rescale=) I want to congrat on R1.2 and to thank r-developers for quickly adding the rescale workaround to the windows version. Happy New Year Jens Oehlschlaegel ###############################################################################

Dieter Menne pointed out that the (small) xml attachment didn't make it. Here is an in-line version (see end of message). Let's hope it works this time. I'm struggling with interpreting XML files created by ADODB as data.frames and I'm looking for advice. Note: This xlm contains a result set which comes from a rectangular data array. I've been trying to play with

Hello, In the following I tried 3 versions of an example in R help. Only the two first predict command work. After : library(splines) require(stats) 1) fm1 <- lm(weight ~ bs(height, df = 5), data = women) ht1 <- seq(57, 73, len = 200) ph1 <- predict(fm1, data.frame(height=ht1)) # OK plot(women, xlab = "Height (in)", ylab = "Weight (lb)") lines(ht1, ph1) 2)

Hi folks, I have 25 stand height observations over 7 years period and daily leafbiomass data during this period. I want to use the 25 plant height observations as inputs and predict the daily stand height during the 7 years. SH=matrix(data=NA , nrow = 2641, ncol = 1) for (i in 1:2641) { SH<- predict(lm(height~Date, data=Stand_Height)); dl=leafbiom$Date[i-1]; de=leafbiom$Date[i];

hi everyone, wondering if you could help me with a novice problem. I have a data frame called subjects with a height and weight variable and want to calculate a bmi variable from the two. i have tried: attach(subjects) bmi <- (weight)/((height/100)^2) but it comes up with the error: Warning messages: 1: In Ops.factor(height, 100) : / not meaningful for factors 2: In Ops.factor((weight),

libswfdec/swfdec_image.c | 204 ++++++++++++++++++++++++++--------------------- libswfdec/swfdec_image.h | 11 +- 2 files changed, 123 insertions(+), 92 deletions(-) New commits: commit 54d034a21fa13a6afb4ad788a82b240002def2cf Author: Benjamin Otte <otte at gnome.org> Date: Sun Nov 11 00:21:54 2007 +0100 implement reading from PNG files diff --git a/libswfdec/swfdec_image.c

I have a sample of lung capacities from a population measured against height. I need to know the 95% CI of the lung capacity of a person of average height. I have fitted a regression line. How do I get a minimum and maximum values of the 95% CI? My thinking was that this has something to do with covariance, but how? My other thinking was that I could derive the 0.975 (sqrt 0.95) CI for the

Dear all, I'm trying to get a series of lattice levelplots to appear in viewports in a particular way but struggling to exert fine control over their appearence. There are two conditions: (a) I only want the levelplot to appear (I don't want axes, colour key, etc) in the viewport and (b) I want the levelplot to expand to the maximum allowable space in the viewport while observing

Once again, you are over-writing your variable. This time, you are overwriting the entirety of Stand_Height with the timeseries of height. Perhaps you should spend some time with one of the good introductory R resources out there, and think a bit more about your procedure. Sarah On Fri, Jun 30, 2017 at 11:23 AM, Ahmed Attia <ahmedatia80 at gmail.com> wrote: > Sorry for the confusion,

There are a bunch of things wrong here, although without a reproducible example I can't really fix most of them. - You're overwriting SH within the loop. - You're running the regression 2641 times, even though the result never changes. - You're never predicting from your linear model using the other data not in the regression. - Leaf biomass data is never used for anything. I

A secure macro It’s like static but well optimized. Any ideas how I can implement it universally? I thought about a secure application to help with compiling and security. It’s like sudo but more secure and can chain. I don’t know much about pipes and streams, but I’m pretty sure I can write it into llvm with enough time and help. Another thing, and I got trolled off gcc’s mailing list for this,

Dear R-users, I am using R version 2.10.1 under windows. In a barplot, I want to mark one of the bars with a special border color. For example: barplot(c(3, 7, 11), border = c(NA, "red", NA)) But how to do this when the bars are stacked? for example: barplot(matrix(1:6, ncol=3)) # border of second bar (i.e. the one with total height = 7) should be red again, I try: barplot(matrix(1:6,

Hello, Despite Sergio''s excellent documentation on prototype, I’m still struggling to get started. I hope somebody is able to help me a with a little example to get me going. Please see html code below – I have this little navigation bar and I like to add some events/functions to the buttons. Let’s say, onmouseover, I like to change e.g. the background-color of the button and