Petr, your code does not create a native Excel file, and it is misleading to
name it with an xls extension.
---
Jeff NewmillerThe . . Go Live...
DCN:Basics: ##.#. ##.#.
Have a look at this example
h1 <- cut(rnorm(1000, 4), breaks = seq(0, 8, by = 2))
h2 <- cut(rnorm(1000, 3), breaks = seq(0, 8, by = 2))
h3 <- cut(rnorm(1000, 5), breaks = seq(0, 8, by = 2))
dataset <- data.frame(Set = rep(c("A", "B", "C"), each = 100), value = c(h1,
h2, h3))
dataset$fvalue <- fac
>
> Hi, Dear all,
>
> Could you please tell me how to select specified column in dataset and
how
> to do
my.data[, x]
my.data[, "aa"]
my.data$aa
> the concentration-time profiles in R?
What is that?
Besides you could help youself a lot reading an intro to R which I believe
is in doc direct
On 05/07/2012 02:57 AM, nagy edina wrote:
I would like to ask some help for the following object, because I don’t know which command in
R-project should I use or how to start to solve it. I attached the file if it is needed.Any help or advice
is appreciated. FWHM (Full Width at Half Maximum) at
Hi
>
> Petr, your code does not create a native Excel file, and it is
misleading
> to name it with an xls extension.
Yes, you are right. It saves tab delimited file without row names which
can be directly opened by Excel by double clicking. At least in my comps.
I agree that it is not "propp
In vim, first move to the top line of the block.
Then press Shift+V (i.e. upper-case V); this line will then be
highlighted.
Then move down (down-arrow key) to the bottom line of the block;
the whole block will then be highlighted.
At this stage enter
:s/^/# /
(The "g" in Don's sequence is not n
On 05/07/2012 02:31 PM, Manish Gupta wrote:
HI,
Below is third example.
# here we want the full scale from zero to one
color.legend(2,6,4,6.4,legend=c("100% guys","100% girls"),
rect.col=color.scale(seq(0,1,by=0.25),c(0.2,1),c(0.2,0.4),c(1,0.4)))
par(mar=c(5,4,4,2))
# use barp to display a mult
Hi everybody!
I would like to plot a barplot, but, unfortunately, when I change the
y-axis limits
the bars do not start at 0 any more but get negative:
#
# Data (just a short example):
a<-c(1.61, 2.1)
b<-c(1.5, 1.9)
c<-c(1.85, 2.2)
d<-c(1.63, 2.3)
Dear all,
I am using glmnet (Coxnet) for building a Cox Model and
to make actual prediction, i.e. to estimate the survival function S(t,Xn) for a
new subject Xn. If I am not mistaken, glmnet (coxnet) returns beta, beta*X and
exp(beta*X), which on its own cannot generate S(t,Xn). We miss baseline
At 18:03 05/05/2012, Jin Choi wrote:
Dear users of metafor,
I am working on a meta-analysis using the metafor package. I have a
excel csv database that I am working with. I am interested in pooling
the effect measures for a particular subgroup (European women) in this
csv database. I am conducti
On 07-05-2012, at 09:55, David Studer wrote:
> Hi everybody!
>
> I would like to plot a barplot, but, unfortunately, when I change the
> y-axis limits
> the bars do not start at 0 any more but get negative:
>
> #
> # Data (just a short example):
>
Michael just provided a good suggestion, using the subset argument to make sure
that you are really using the same data in both analyses.
However, I would not expect the results to be exactly the same anyway. Remember
that these are random/mixed-effects models you are using. So, when you use the
Esteemed UseRs,
This must be embarrassingly trivial to achieve with e.g., melt() and
cast(): deduplicating records ("pw.X" in example) for a given set of
responses ("cond.Y" in example).
Hopefully the runnable example shows clearly what i have and what i'm
trying to convert it to. But i'm ju
you could try aggregate(), e.g.,
my.df <- data.frame(pathway = c(rep("pw.A", 2), rep("pw.B", 3),
rep("pw.C", 1)),
cond.one = c(0.5, NA, 0.4, NA, NA, NA),
cond.two = c(NA, 0.6, NA, 0.9, NA, 0.2),
cond.three = c(NA, NA, NA, NA, 0.1, NA))
Hi
I wold vote aggregate
> aggregate(my.df[,-1], list(pathway=my.df$pathway), mean, na.rm=T)
pathway cond.one cond.two cond.three
1pw.A 0.5 0.6NaN
2pw.B 0.4 0.90.1
3pw.C NaN 0.2NaN
>
Regards
Petr
>
> Esteemed UseRs,
>
> Thi
I making an xyplot and the y label is too long and needs to be in two rows, but
when I brake it there is a huge gap between the last text string and the
expression, and I can't get rid of it. Any ideas?
Data:
structure(list(Temp = c(8L, 8L, 8L, 8L, 8L, 8L, 12L, 12L, 12L,
12L, 12L, 12L), CO2 = c
Dimitris, Petra,
Thank you! aggregate() is my lesson for today, not melt() | cast()
Really appreciate the super fast help,
Karl
On 07/05/12 12:09, Dimitris Rizopoulos wrote:
you could try aggregate(), e.g.,
my.df <- data.frame(pathway = c(rep("pw.A", 2), rep("pw.B", 3),
rep("pw.C", 1)),
cond
What you are doing is correct. However because it is not informed about
the penalty parameters used in creating the fit, the coxph command is not
able to correctly reconstruct the variance matrix; this is the source of all
of the ³singular² messages. The confidence bands for the survival curves
Hello,
I'm using fArma package to estimate the value of Hurst exponent using R/S
method. However, for a certain set of data I get H ~ 1.8. How do I
interpret this?
Following are the output that I get for this set:
> mean(data[,2])
[1] 400.5433
> sd(data[,2])
[1] 1139.786
>
> rsFit(data[,2], leve
Hello List,
I have some plots with the wireframe() function, and I'd like to
display them in a single jpeg file. I know that par(mfrow=c(x,y)) will
divide my display window in x rows and y columns, and although this
works with plot(), it looks like it's not working with wireframe.
here's my code:
using reshape:
library(reshape)
m <- melt(my.df, id.var="pathway", na.rm=T)
cast(m, pathway~variable, sum, fill=NA)
Jan
On 05/07/2012 12:30 PM, Karl Brand wrote:
Dimitris, Petra,
Thank you! aggregate() is my lesson for today, not melt() | cast()
Really appreciate the super fast help,
Karl
Thanks Tal for answering,
Anyway I still have no idea on why the binomial GLM is missing the
relationship between the response variable and the explanatory variable
"cohort".
Is there anyone who might help me to understand this?
--
View this message in context:
http://r.789695.n4.nabble.com/Bi
And if the y-axis starts with 0: Instead of ylim=c(1,2), you take ylim=c(0,2).
Thus barplot(x[,1], ylim=c(0,2)) Hopefully it is now correct. Kind
regards,Esra Atescelik
> Date: Mon, 7 May 2012 09:55:06 +0200
> From: stude...@gmail.com
> To: r-help@r-project.org
> Subject: [R] y-axis-problem (b
Thanks. It does work now.
I also get another problem when I use naivebayes and prediction in myapplication
There was anerror message:
Error in table(predict(nb.obj, test.data[, subset, drop = FALSE]), test.data[, :
all arguments must have the same length
The length of test.data[,subset
Hi everyone,
I have a file in which the dates are subscribed as for instance: 20101020.
This is 20th Octobre 2010.
My problem is that R won't except this as a date, since there is no sign to
seperate the Year, Month and Day
and that it will only see it as an origin, which it is not.
Does anyone k
Dear All,
I have a codes which calculates the result of Ripley's K function of my
data. I want to repeat this process 999 times. However, i am getting an
error when i use the "for i in" function. Is there any way to repeat this
analysis 999 times. Here are the codes i used ;
data4 <- matrix(c(s
Dear All
I would really appreciate some help with a script which a colleague wrote for
me, but I am having problems running (and have not been able to contact my
colleague).
The script is designed to compare the area of suitable habitat in binary
projections of a large number of species c
Hi there,
I am trying to interface c++ code in R and make a package. With R CMD SHLIB
the dll was created, but when I try R CMD check, I am getting 'undefined
reference to..' linkage error messages.
The relevant c++ source from conf-infomap.cpp:
#include "conf-infomap.h"
#include "R.h" // R fu
I have a decent sized matrix (36 x 11,000) that I have preformed a PCA on
with prcomp(), but due to the large number of variables I can't plot the
result with biplot(). How else can I plot the PCA output?
I tried posting this before, but got no responses so I'm trying again.
Surely this is a commo
By the way, my "for" function is below, I can't find the mistake
rand.max.t<- function(n){
f<-rep(NA,n)
for (i in 1:n) {
reassign[i]<-matrix(c(sample(id),data1),203,3)
new.data<-reassign[,1]
random.cas=reassign[new.data==0,2:3]
random.con=reassign[new.data==1,2:3]
f<- list(x=random.cas[,1],y=ra
Thank you both so much for your help! I ended up using
bquote(expression(...)), and it's working perfectly!
On Sat, May 5, 2012 at 1:05 PM, David Winsemius wrote:
>
> On May 5, 2012, at 2:42 PM, Josh Browning wrote:
>
> Hello useRs!
>>
>> So, I have a random question. I'm trying to build a cha
Hi Collin,
Look in the package 'pwr' for 'pwr.r.test'.
A.K.
- Original Message -
From: Collin Lynch
To: r-help@r-project.org
Cc:
Sent: Monday, May 7, 2012 1:44 AM
Subject: [R] Statistical power of correlations.
My apologies for the statistical naivete of my question but...
Is there a
Dear R users,
I am working with panel data and I want the difference of a variable with
its t+1 value.
For example, I have a data frame as below.
> a <- data.frame(c(rep(2,5), rep(3,5)), c(2005:2009, 2004:2008),
c(NA,10,34,23,12, 23,45, NA, 45, NA))
> colnames(a) <- c("firm","year","var")
I want
That depends on what you want to plot there. Basically, you could just use
plot() with pcaResult$x. You might need to define which PCs you want to plot
there though.
pcaResult<-prcomp(iris[,1:4])
plot(pcaResult$x) # gives the first 2 PCs
plot(pcaResult$x[,2:3]) #gives the second vs the 3rd PC
o
On 07.05.2012 10:24, BrittD wrote:
Hi everyone,
I have a file in which the dates are subscribed as for instance: 20101020.
This is 20th Octobre 2010.
strptime("20101020", format="%Y%m%d")
seems to work for me...
UWe Ligges
My problem is that R won't except this as a date, since there is
To add: If thats not it, maybe you could be a bit more specific about what you
consider the "result", and how you want it visualized.
Am 07.05.2012 um 15:24 schrieb Jessica Streicher:
> That depends on what you want to plot there. Basically, you could just use
> plot() with pcaResult$x. You mig
Christian, is that 36 samples x 11K variables? Sounds like it. Is this
spectroscopic data?
In any case, the scores are in the list element $x as follows:
answer <- prcomp(your matrix)
answer$x contains the scores, so if you want to plot the 1st 2 pcs, you could do
plot(answer$x[,1], answer$x
Hello,
I have been using 'lda' package for topic modeling and wish to predict
new topics using a fitted model by giving new dataset as input. While
doing this, I came across the function predictive.distribution(). But
I am not able to apply it on a new dataset as the input parameter asks
for docume
Hi all,
One of my variables looks like this:
.7_-.3_-.2_.9
And this is a character variable. I made this by combining four different
number like .7, -.3, -.2, and .9 using paste function.
Now, I want to go back to original format from this one combined character
variable. For instance, I want to
Apologies for cross-posting
Members of this mailing list may be interested in the following book:
Zero Inflated Models and Generalized Linear Mixed Models with R.
Zuur, Saveliev, Ieno. (2012)
This book is only available from:
http://www.highstat.com/book4.htm
Kind regards,
Alain Zuur
--
D
Hi YN,
I use strsplit for this:
x <- ".7_-.3_-.2_.9"
> strsplit(x, split = "_")
[[1]]
[1] ".7" "-.3" "-.2" ".9"
> strsplit(x, split = "_")[[1]][3]
[1] "-.2"
Best,
Ista
On Mon, May 7, 2012 at 9:54 AM, YN Kim wrote:
> Hi all,
>
> One of my variables looks like this:
>
> .7_-.3_-.2_.9
>
> And
Biplot, depending on what parameters you give it, scales the data in a certain
way.
See
http://stat.ethz.ch/R-manual/R-patched/library/stats/html/biplot.princomp.html
scale
The variables are scaled by lambda ^ scale and the observations are scaled by
lambda ^ (1-scale) where lambda are the sin
On May 7, 2012, at 6:20 AM, Beatriz De Francisco wrote:
I making an xyplot and the y label is too long and needs to be in
two rows, but when I brake it there is a huge gap between the last
text string and the expression, and I can't get rid of it. Any ideas?
My first idea would be that y
And i always forget the question..
I haven't understood biplots a 100%, but from what i gleaned this scaling is
done so it looks better/is easier to read, while the scaling retains certain
properties of the biplot (something about projecting).
If you want to use the data for anything else, i wo
On May 7, 2012, at 7:15 AM, Simone Gabbriellini wrote:
Hello List,
I have some plots with the wireframe() function, and I'd like to
display them in a single jpeg file. I know that par(mfrow=c(x,y)) will
divide my display window in x rows and y columns, and although this
works with plot(), it l
Thanks a lot for pointing me to that!
Best,
Simone
2012/5/7 David Winsemius :
>
> On May 7, 2012, at 7:15 AM, Simone Gabbriellini wrote:
>
>> Hello List,
>>
>> I have some plots with the wireframe() function, and I'd like to
>> display them in a single jpeg file. I know that par(mfrow=c(x,y)) wil
I don't know the answer, Jessica gave some insight.
I avoid the biplot at all costs, because IMHO it violates one of the tenets of
good graphic design: It has two entirely different scales on axes. These are
maximally confusing to the end-user. So I never use it.
If it is gene expression dat
I don't see anything that looks like it should throw an error, but I
haven't tested your code without "data1" and "id" -- you might look at
? replicate() though -- it's designed for these sorts of things. E.g.,
replicate(100, mean(rexp(50)))
gets me a hundred draws of the mean of 50 random expone
On May 7, 2012, at 6:28 AM, efulas wrote:
By the way, my "for" function is below, I can't find the mistake
rand.max.t<- function(n){
f<-rep(NA,n)
for (i in 1:n) {
reassign[i]<-matrix(c(sample(id),data1),203,3)
new.data<-reassign[,1]
random.cas=reassign[new.data==0,2:3]
random.con=reassign[ne
Hi Bryan,
Many thanks for the replies.
The data is gene expression data for 36 samples over 11k genes.
I see that I can plot PC1 vs PC2 by using $x, but compared to biplot() I
can see that the range of values are different. For example, if I use
plot() the PC1 scale ranges from -150 to 150 wher
In RStudio select the lines to be commented (or uncommented) and press
Ctrl+/ or select comment/uncomment on the Edit menu tab
--
David L Carlson
Associate Professor of Anthropology
Texas A&M University
College Station, TX 77843-4352
> -Original Me
Hi Jessica,
THanks for pointing that out. The scaling in biplot() doesn't seem to make
sense to me, however. The default value for scale=1 therefore lambda ^
(1-scale) -> lambda ^ 0 which is 1 regardless of what lambda is. Which
can't be right?
Anyway, I won't worry about it anymore as you and B
Hi Jessica,
Yes, that does help. It confirms my digging around in the prcomp object.
I was plotting $x, but wasn't sure whether this was appropriate. Mainly
because the data ranges are different in $x than when plotted by biplot()
- as I mentioned my reply to Bryan. Do you know if this difference
Hi Bryan,
On 07/05/2012 15:33, "Bryan Hanson" wrote:
>I don't know the answer, Jessica gave some insight.
>
>I avoid the biplot at all costs, because IMHO it violates one of the
>tenets of good graphic design: It has two entirely different scales on
>axes. These are maximally confusing to the
I tried the subset argument as Michael suggested, which led to the
same results. The results between meta-regression and subgroup
analyses were only slightly different as Wolfgang had suggested. I
also believe that these minor differences must be arising from the use
of random effects.
Thank you v
On Mon, May 7, 2012 at 3:21 PM, Apoorva Gupta wrote:
> Dear R users,
> I am working with panel data and I want the difference of a variable with
> its t+1 value.
>
> Could you tell me if such a function exists in the plm package?
>
Perhaps diff() or lag(). See the plm vignette.
Liviu
_
Hello,
Try
x <- 20102010
as.Date(as.character(x), format="%Y%d%m")
[1] "2010-10-20"
as.POSIXct(as.character(x), format="%Y%d%m")
[1] "2010-10-20 BST"
Note that you must pass x as a character vector.
If not, the date functions will see it as the number of days since an origin
such as 1970-01-01
Hello,
efulas wrote
>
> By the way, my "for" function is below, I can't find the mistake
>
>
> rand.max.t<- function(n){
> f<-rep(NA,n)
>
> for (i in 1:n) {
> reassign[i]<-matrix(c(sample(id),data1),203,3)
> new.data<-reassign[,1]
> random.cas=reassign[new.data==0,2:3]
> random.con=reassign[
Oops,
Sorry, wrong number.
x <- 20101020
as.Date(as.character(x), format="%Y%m%d")
as.POSIXct(as.character(x), format="%Y%m%d")
Rui Barradas
--
View this message in context:
http://r.789695.n4.nabble.com/Dates-in-R-tp4614266p4614756.html
Sent from the R help mailing list archive at Nabble.co
Hello,
YN wrote
>
> Hi all,
>
> One of my variables looks like this:
>
> .7_-.3_-.2_.9
>
> And this is a character variable. I made this by combining four different
> number like .7, -.3, -.2, and .9 using paste function.
> Now, I want to go back to original format from this one combined char
Hi David,
I am sorry for cross posting. I will keep this in mind next time.
Regarding details, my doubt is regarding the documentation in the
package where, it speaks about applying the function to held-out words
but the input parameters don't take new datasets, rather they use the
output generated
Thank you for replies. I sort out the problem by defining the reassign
matrix.
Best Wishes,
efulas
--
View this message in context:
http://r.789695.n4.nabble.com/Repeating-tp4614371p4615072.html
Sent from the R help mailing list archive at Nabble.com.
___
Great thanks Peter!
Collin.
On Mon, 7 May 2012, peter dalgaard wrote:
>
> On May 7, 2012, at 07:44 , Collin Lynch wrote:
>
> > My apologies for the statistical naivete of my question but...
> >
> > Is there an established method or calulating the statistical power of a
> > correlation te
Thank you Arun!
Collin.
On Mon, 7 May 2012, arun wrote:
> Hi Collin,
Look in the package 'pwr' for 'pwr.r.test'.
A.K.
- Original Message -
From: Collin Lynch
To: r-help@r-project.org
Cc:
Sent: Monday, May 7, 2012 1:44 AM
Subject: [R] Statistical power of correlations.
My apo
Hello
I need to use a coxme model with my data (survival analysis with
right-censoring and hierarchical nesting), but I cant find a way to get
predicted values from a new data table (or even from the original one).
Has anyone had this problem before? I cant find anything about that
anywhere.
Tha
Hey all who have responded to this post. I am a newbie to ANOVA analysis in
R, and let me tell you- resources for us learners are scant, horrible,
unclear, imprecise.. in other words.. the worst ever. So advice like "go
look it up" in your "classical" textbook or on google is not helpful at all.
I
Hi All,
Sorry for posting the same question again. I was not sure if the message
was sent initially since it was my first post the forum.
Can the MNP package available in R be used to analyze panel data as well?
*i.e., *if there are 3 observed discrete choices for three time periods for
the same
Hi all,
I can't find the error in the binomial GLM I have done. I want to use that
because there are more than one explanatory variables (all categorical) and
a binary response variable.
This is how my data set looks like:
> str(data)
'data.frame': 1004 obs. of 5 variables:
$ site : int 0 0
1. As this is a statistical, rather than an R issue, you would do
better posting on a statistical help site like stats.stackexchange.com
(although some generous soul here may respond).
2. You would also probably do better consulting with a local
statistical resource if available, as it is difficul
Hello. I'm a newbie here.
In my script (I name it readData.R), I wrote the followings:
readData <-function(){ med = read.csv("medicalData.csv");}
Then I tested the script by 'Source R Code' then on the command I typed
'readData()' then I typed 'med' to check if the variable contains the medical
R is a functional language so, by default, assignments (and other
things) within function scope doesn't have global effects. This is
generally considered a _very good thing_ in language design. You'd
perhaps prefer something like:
readData <- function() {
read.csv("medialData.csv")
}
med <- r
Hi Suhaila,
You don't need to make a function: your script should just contain:
med <- read.csv("medicalData.csv")
If you do want to make a function, then you need to assign the
resulting value to something, eg:
med <- readData()
but there's no reason to do that. Values that are assigned within
Chris,
I think that you really need to quantify what you mean by correlation.
Things to consider would depend on what the matrices represent--are they
the estimates of the same set of N geographic points, are they traces of
the same line, are they traces of the same polygon outline? If either
Hello,
I have an experimental design where I would like to use separate ranking
events to predict an independent ranking event. I have been using function
clmm in the ordinal library but now realize that I am violating one of the
assumptions. I have included a subset of the data.
I am looking a
On 07-05-2012, at 19:41, Suhaila Haji Mohd Hussin wrote:
>
> Hello. I'm a newbie here.
> In my script (I name it readData.R), I wrote the followings:
> readData <-function(){med = read.csv("medicalData.csv");}
> Then I tested the script by 'Source R Code' then on the command I typed
> '
Perhaps I haven't explained it that well as I would have liked to.
To me this was an R issue because I didn't understand why the binomial GLM
is getting these results and I believed this was something due to the way I
am implementing it in R, not to the binomial GLM itself.
If I was wrong and this
Thank Sarah!
> Date: Mon, 7 May 2012 14:06:31 -0400
> Subject: Re: [R] Problem in executing R-script
> From: sarah.gos...@gmail.com
> To: bell_beaut...@hotmail.com
> CC: r-help@r-project.org
>
> Hi Suhaila,
>
> You don't need to make a function: your script should just contain:
> med <- read.cs
Hello.
I'm trying to compute median for a filtered column based on other column but
there was something wrong. I'll show how I did step by step.
Here's the data:
a b c class
1 12 0 90 A-B2 3 9711 A-B3 78 NA123 A-C4
NA NA12A-C5
Please use dput() to give us your data (eg dput(data) ) rather than
simply pasting it in.
Sarah
On Mon, May 7, 2012 at 2:52 PM, Suhaila Haji Mohd Hussin
wrote:
>
> Hello.
> I'm trying to compute median for a filtered column based on other column but
> there was something wrong. I'll show how I
I might be silly but if I was going to type in dput() then how should I send
the data over here?
Instead, I've just uploaded the image online, you can access it via the link
below.
http://i1165.photobucket.com/albums/q585/halfpirate/data.jpg
> Date: Mon, 7 May 2012 14:55:24 -0400
> Subject: R
Hi Lincoln,
Some thoughts:
1) Did you intend to use "cohort" as a factor and not as a numeric? (at
least that is what it looks like in your output)
2) Is there a strong correlation between "cohort" and the
other explanatory variables you are trying in your model?
Contact
Detail
Thank you so much!
Suhaila.
> Date: Mon, 7 May 2012 15:08:47 -0400
> Subject: Re: [R] Problem with Median
> From: jholt...@gmail.com
> To: bell_beaut...@hotmail.com
>
> Your problem is that a.AC is a dataframe:
>
>
> > x <- read.table(text = " a b c class
> + 1 12 0 90
On 07/05/2012 3:05 PM, Suhaila Haji Mohd Hussin wrote:
I might be silly but if I was going to type in dput() then how should I send
the data over here?
Cut and paste. For example, if I have a dataframe named x and type
dput(x), I see
structure(list(a = 1:10), .Names = "a", row.names = c(N
Afternoon-
I am trying to subtract a matrix, basically a vector of 28 values, each by
the same number to account for differences in regression fitting. I am
taking the 1x28 and minus it by the mean value of the matrix, each number.
The result I receive is a 1X29 matrix. Does anyone know why the r
On May 7, 2012, at 3:26 PM, meredith wrote:
Afternoon-
I am trying to subtract a matrix, basically a vector of 28 values,
each by
the same number to account for differences in regression fitting. I am
taking the 1x28 and minus it by the mean value of the matrix, each
number.
The result I
*
Disregard message please
Afternoon-
I am trying to subtract a matrix, basically a vector of 28 values, each by
the same number to account for differences in regression fitting. I am
taking the 1x28 and minus it by the mean value of the matrix, each number.
The result I receive is a 1X29 matri
I'm sorry but where is 28 coming from? It looks to me like you have a vector
of length 34 and the result is of length 34. Oh, I see, the 28 and 29 are the
indices for the first number on the line not the total length.
John Kane
Kingston ON Canada
> -Original Message-
> From: mmbal..
I'm sorry but where is 28 coming from? It looks to me like you have a vector
of length 34 and the result is of length 34. Oh, I see, the 28 and 29 are the
indices for the first number on the line not the total length.
John Kane
Kingston ON Canada
> -Original Message-
> From: mmbal..
R has functions for computing kappa, fleiss's kappa, etc., but can it compute
Gwet's AC1?
Thanks,
Matt.
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Hello,
meredith wrote
>
> Afternoon-
> I am trying to subtract a matrix, basically a vector of 28 values, each
> by the same number to account for differences in regression fitting. I am
> taking the 1x28 and minus it by the mean value of the matrix, each number.
> The result I receive is a 1X
Hello,
I am using SSfol in nlme to fit some data for the change of N concentration
(N) in plant tissue over time (gdd). The model works nicely for 2 out of 3
treatments, so I would really like to use it, but it consistently has a bad
fit for my third treatment. I am pasting the figure for the t
On May 7, 2012, at 19:39 , Bert Gunter wrote:
> 1. As this is a statistical, rather than an R issue, you would do
> better posting on a statistical help site like stats.stackexchange.com
> (although some generous soul here may respond).
>
> 2. You would also probably do better consulting with a
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Hello.
Now the median is solved but then I'm still figuring out how to put the updated
column back to replace the original column of the whole data. I'll show you
what I meant:
Continuing from the previous commands you guys helped out I continued as
followed:
Original Data: http://i1165.
Dear R people.
I´m facing a big problem.
I need to create a matrix with 10.000 columns and 750.000 rows.
matrix<- as.data.frame(matrix(data=0L, nrow=75, ncol=1)
as you can see, the data frame has huge dimesions. I was able to find out
about thr "L" in data, this way I´m telling that my data
See ?sparseMatrix from package Matrix.
Eloi
On 12-05-07 02:23 PM, Lucas wrote:
> Dear R people.
> I´m facing a big problem.
> I need to create a matrix with 10.000 columns and 750.000 rows.
> matrix<- as.data.frame(matrix(data=0L, nrow=75, ncol=1)
> as you can see, the data frame has huge
Hello everyone,
I am trying to add the following text (in proper notation) to a
graphic using expression().
X-bar (with a subscript of cv) = XX.
Note: Ideally "cv" would be a subscript, but it doesn't have to be.
I have the following code:
> text(625,.012,expression(bar(X)cv = 552.01))
Error
Hi Dan,
If I understood correctly, the following will do:
plot(0, main = expression(bar(X)^{cv}))
HTH,
Jorge.-
On Mon, May 7, 2012 at 5:39 PM, Dan Abner <> wrote:
> Hello everyone,
>
> I am trying to add the following text (in proper notation) to a
> graphic using expression().
>
> X-bar (wit
The code below should have been
plot(0, main = expression(bar(X)[cv]))
Apologies for the noise.
Jorge.-
On Mon, May 7, 2012 at 5:44 PM, Jorge I Velez <> wrote:
> Hi Dan,
>
> If I understood correctly, the following will do:
>
> plot(0, main = expression(bar(X)^{cv}))
>
> HTH,
> Jorge.-
>
>
>
>
Hello,
I have a data frame which looks like
> head(d)
a1 a2n j col
1 1 1 88341002 11 #E7E7E7
2 1 2 25094882 11 #E7E7E7
3 1 3 16916246 11 #E7E7E7
4 1 4 14289229 11 #E7E7E7
5 1 5 11945929 11 #E7E7E7
6 1 6 8401235 11 #E7E7E7
The values in 'j' run from 1 to 11. I would li
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