Hi,
x1<-read.table(text="
Year AP EU LA NA total
Jun2012 2.32 2.26 5.38 13.74 23.70
Jul2012 2.46 2.21 5.33 12.94 22.94
Aug2012 2.69 2.24 5.28 13.32 23.54
Sep2012 2.62 2.28 5.14 12.99 23.06
Oct2012 2.61 2.27 5.31 12.59 22.80
Nov2012 2.55 2.18 5.08 12.56 22.39
",sep="",stringsAsFactors=FALS
Hi everyone, I am having trouble with creating a loop to subtract arrays.
In R, this is what I have done:
> Vobsr <- read.csv("Observed_Flow.csv", header = TRUE, sep =",") # see data
> below
> Vsimr <- read.csv("1000Samples_Vsim.csv", header = TRUE, sep =",") # see
> data below
> Vobsr <- as.matr
Hi, I have data with binomial response variable (survival) and 2 categorical
independent variables (site and treatment) (see below). I have run a binomial
GLM and found that both IVs and the interaction are significant. Now I want to
do a post-hoc test for all pairwise comparisons to see whi
On Nov 18, 2012, at 21:32 , Thomas Lumley wrote:
> On Fri, Nov 16, 2012 at 4:48 PM, frespider wrote:
>
>> Hi,
>>
>> I am fitting a weighted least square regression and trying to compute
>> SSE,SST and SSReg but I am not getting SST = SSReg + SSE and I dont know
>> what I am coding wrong. Can y
Hi
maybe
?rbind
Petr
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of Mat
> Sent: Friday, November 16, 2012 3:42 PM
> To: r-help@r-project.org
> Subject: [R] Sum Column in data.frame for Excel-Export
>
> Hello together,
>
Hi
Values ve or su came as factors during the input. Probably some mismatch in
decimal point, nonumeric value somewhere, thousands separator or something else.
At least str(mat) can help you (and us) what is the problem.
Regards
Petr
> -Original Message-
> From: r-help-boun...@r-projec
Hi
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of Jeff Newmiller
> Sent: Saturday, November 17, 2012 3:21 PM
> To: Partha Sinha; r-help@r-project.org
> Subject: Re: [R] Basic Stats question
>
> If you read the posting guide
> I'd like to resurrect this issue: is the "varwidth" equivalent (boxplot
> box-width scaling according to number of data points) emulatable in the
> 0.9.* versions of ggplot2? Width still doesn't seem capable of accepting a
> vector with length > 1 ...
No, and it's not currently on the to do list
On Nov 18, 2012, at 7:05 PM, Tom Roche wrote:
https://stat.ethz.ch/pipermail/r-help/2012-November/329479.html
Hopefully [sufficiently] "small, self-contained example":
mailquotes omitted from "start example" to "end example" to
ease rerunning the following code:
# start example
library(res
On Nov 18, 2012, at 6:34 PM, Eiko Fried wrote:
When I run this script on 9 variables, it works without problems.
Z <-
data
[,c
("s1_1234_m
","s2_1234_m
","s3_1234_m
","s4_1234_m
","s5_1234_m","s6_1234_m","s7_1234_m","s8_1234_m","s9_1234_m"
)]
However, when I run the script on 9 different vari
On Mon, Nov 19, 2012 at 10:31 AM, R. Michael Weylandt <
michael.weyla...@gmail.com> wrote:
> On Sunday, November 18, 2012, Rlotus wrote:
>
> > Gives a mistake that "probability" is not found
> >
> > x<- runif(1,0,1)
> > y<-x^5+x^8+(cos(x))^3
> > yrandom<- runif(10,0,3)
> > if (y > area=probability
https://stat.ethz.ch/pipermail/r-help/2012-November/329479.html
>> Hopefully [sufficiently] "small, self-contained example":
mailquotes omitted from "start example" to "end example" to
ease rerunning the following code:
# start example
library(reshape2)
library(lattice)
lon=11
lat=7
lev=5
len=l
When I run this script on 9 variables, it works without problems.
Z <-
data[,c("s1_1234_m","s2_1234_m","s3_1234_m","s4_1234_m","s5_1234_m","s6_1234_m","s7_1234_m","s8_1234_m","s9_1234_m"
)]
However, when I run the script on 9 different variables, it does not work:
Z <-
data[,c("d_s1_m","d_s2_m","
Sorry, got the answer (ticktype):
persp(x, y, z1,box = TRUE, axes = TRUE,ticktype="detailed")
Thanks!
D.
On Sun, Nov 18, 2012 at 8:20 PM, Dimitri Liakhovitski <
dimitri.liakhovit...@gmail.com> wrote:
> Dear R-ers,
>
> I have 3 vectors - x, y, and z and want to plot a surface (z as the 3rd
> di
On Nov 18, 2012, at 12:42 PM, Tom Roche wrote:
Thanks to the lattice gurus on this list, and having reference to the
excellent open-access Sarkar 2008
ISBN 978-0-387-75968-5
e-ISBN 978-0-387-75969-2
http://dx.doi.org/10.1007/978-0-387-75969-2
I now know how to label lattice panels by variabl
1. This is not really an R question. I suggest you post to a
statistical list like stats.stackexchange.com instead.
2. I also suggest you consult a local statistician. You have a time
series with a categorical response. That's not so simple as you think
(and folks on stackexchange may provide you
Hello David,
Many thanks - this does exactly what I want and it lets me see whether the
clusters make sense in terms of the patetrn of values & where they join a
cluster.
Regards
Bob
> Something like this?
>
>> split(FS1, hcli8)
> $`1`
>X1 X2 X3 X4 X5 X6 X7 X8 X9 X10 X11 X12
> 1 1 1 0
Try x1$`NA` instead of x1$NA.
HTH
Quoting eric :
I inserted na.strings='' and that seemed to work except for a problem with a
plot statement
plot(x1$NA,type='l',ylab='M kg/ y ',xlab='')
Error: unexpected numeric constant in "plot(x1$NA"
tail(x1)
AP EU LANA total
Jun 20
In many situations the interactions indicated by additive risk models are
false in the sense that they merely reflect restrictions so that risk are in
[0,1]. I tend to prefer models that have no restrictions on the parameters.
In fact, I think that a test of goodness of fit of a model can be made
I inserted na.strings='' and that seemed to work except for a problem with a
plot statement
plot(x1$NA,type='l',ylab='M kg/ y ',xlab='')
Error: unexpected numeric constant in "plot(x1$NA"
> tail(x1)
AP EU LANA total
Jun 2012 2.32 2.26 5.38 13.74 23.70
Jul 2012 2.46 2.21 5.33 1
Hi Thomas,
Can you please edit my code in way it works. I appericated your help. I need to
calculate the R2 in the Z scale and it is not make sense to me.
or Can I have some documentation to read?
Thanks
Date: Sun, 18 Nov 2012 12:35:03 -0800
From: ml-node+s789695n4649969...@n4.nabble.com
To
You tried setting shorter variable names, but the two lines of your data set
that you gave us still have the original names, eg.
Telephone.lines..per.100.people.
--
David L Carlson
Associate Professor of Anthropology
Texas A&M University
College Station,
Most likely the value you are using in the 'hist' function is a
factor. It would help if you included an 'str' of the object you are
using.
On Sun, Nov 18, 2012 at 4:45 PM, 9man wrote:
> Hello all,
>
> I hope someone of you can help me out, I have searched other posts as well
> but I can't find
You can do the following to remove the directory names from the list:
setdiff(list.files(full.names = TRUE), list.dirs(recursive = FALSE))
On Sun, Nov 18, 2012 at 12:20 PM, Jannis wrote:
> Dear R developers,
>
>
> as far as I understand the manual of list.files(), there is only a way to
> excl
Something like this?
> split(FS1, hcli8)
$`1`
X1 X2 X3 X4 X5 X6 X7 X8 X9 X10 X11 X12
1 1 1 0 1 0 0 1 1 0 1 1 1
3 1 0 1 0 0 1 1 0 0 1 0 1
4 1 1 0 0 0 0 1 1 1 1 1 1
7 0 1 0 1 0 0 1 1 0 1 0 1
9 1 1 1 1 0 1 1 0 1 1 1 0
Hello all,
I hope someone of you can help me out, I have searched other posts as well
but I can't find any solution to the problem I'm dealing with.
I want to make a histogram from the data Telephone Lines
MDGdataset <-read.csv("MDG_dataset_2010.csv", header=T)
MDGdatasetAdapted <- subset(MDGdat
Hi,
Try this:
If you wanted to replace "<" and ">" from NA:
dat1<-read.table(text="
635 LA 201207 557329
636 LA 201208 683771
637 LA 201209 613851
638 LA 201210 764217
639 LA 201211 212897
782 200701 875634
783 200702 614856
784 200703 521520
785 200704 1406400
",sep=
use na.strings = ''
> x <- read.table(text = "635 LA 201207 557329
+ 636 LA 201208 683771
+ 637 LA 201209 613851
+ 638 LA 201210 764217
+ 639 LA 201211 212897
+ 782 NA 200701 875634
+ 783 NA 200702 614856
+ 784 NA 200703 521520
+ 785 NA 200704 1406400", as.is
On Sunday, November 18, 2012, Rlotus wrote:
> I am a beginner in programmin in general and R specifically.
>
> I would like to generate a set of random numbers in a normal distribution
> but to limit the decimal places in these numbers to only 2.
>
> I have been using x1 <- runif(1,0,1) to generat
Please stop posting and re-posting the same questions.
Michael
On Sunday, November 18, 2012, Haris Rhrlp wrote:
> Dear R users,
>
>
> I want to check matrices when i change the order of the rows or/and the
> order of the columns or/and the combination of them
>
> i will give an example what i wa
On Sunday, November 18, 2012, Rlotus wrote:
> Gives a mistake that "probability" is not found
>
> x<- runif(1,0,1)
> y<-x^5+x^8+(cos(x))^3
> yrandom<- runif(10,0,3)
> if (y area=probability*3
>
>
You're ignoring the warning that's also given. Read ?ifelse and be
enlightened
Cheers
Michael
>
>
I am a beginner in programmin in general and R specifically.
I would like to generate a set of random numbers in a normal distribution
but to limit the decimal places in these numbers to only 2.
I have been using x1 <- runif(1,0,1) to generate my numbers.
Can I add something to it to enable me t
I am reading some data into R from an Excel spreadsheet using read.csv. Some
of the original data that comes into column 1 from the spreadsheet is text
that says NA. The NA stands for north america. As it comes in, R converts
the NA over to .
What is the cleanest way to change the values to so
Hi everyone, I am having trouble using my own data in the Nash-Sutcliffe
efficiency (NSE) function.
In R, this is what I have done:
Vobsr <- read.csv("Observed_Flow.csv", header = TRUE, sep =",") # see data
below
Vsimr <- read.csv("1000Samples_Vsim.csv", header = TRUE, sep =",") # see
data below
Gives a mistake that "probability" is not found
x<- runif(1,0,1)
y<-x^5+x^8+(cos(x))^3
yrandom<- runif(10,0,3)
if (yhttp://r.789695.n4.nabble.com/How-to-fix-it-tp4649955.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.
I have a rather basic set of data. It is simply a variable that can be 0, 1 or
2 and its value over a series of time t0 - t9 like so:
y: 1 1 2 0 1 2 2 1 2
1
x: t0 t1 t2 t3 t4 t5 t6 t7 t8
David,
Many thanks, I'm sure this will be helpful. What would also be
helpful is if I can extract each cluster and examine id by variable,
within the respective cluster. I could index the variables for each
cluster and run such an analysis but thre must be a more efficient
way of doing this
https://stat.ethz.ch/pipermail/r-help/2012-November/329450.html
>> I [can now] prepare my netCDF data specifying gas concentrations over
>> a 3D space (dimensions longitude, latitude, and (vertical) level),
>> so as to call lattice::levelplot() like
>> levelplot(conc ~ lon * lat | lev, data=data.
Dear R users,
I want to check matrices when i change the order of the rows or/and the order
of the columns or/and the combination of them
i will give an example what i want
1 -1 1 1 1 1 1 1
-1 -1 -1 -1 -1 -1 -1 -1
1 1 1 1 1 -1 1 1
these 2 matrices are iden
If you just want a summary of the mean for each variable in each
cluster, this will get you there:
> set.seed=42
> FS1 <- data.frame(matrix(sample(c(0, 1), 12*63, replace=TRUE),
nrow=63,
+ ncol=12))
> dmat <- dist(FS1, method="binary")
> cl.test <- hclust(dmat, method="average")
> plot(cl.test, h
Thanks to the lattice gurus on this list, and having reference to the
excellent open-access Sarkar 2008
ISBN 978-0-387-75968-5
e-ISBN 978-0-387-75969-2
http://dx.doi.org/10.1007/978-0-387-75969-2
I now know how to label lattice panels by variable value: see thread
starting @
https://stat.ethz.c
On Fri, Nov 16, 2012 at 4:48 PM, frespider wrote:
> Hi,
>
> I am fitting a weighted least square regression and trying to compute
> SSE,SST and SSReg but I am not getting SST = SSReg + SSE and I dont know
> what I am coding wrong. Can you help please?
>
For a start, you need to replace your mu
HI,
Not sure if I understand it clearly.
Are you looking for something like this:
res1<-list()
fun1<-function(x,res){
for(i in 1:ncol(x)){
res[[i]]<-list()
res[[i]]<-combn(1:ncol(x),i)
}
res}
fun1(m1,res1)
[[1]]
[,1] [,2] [,3] [,4] [,5]
[1,] 1 2 3 4 5
[[2]]
[,1]
OK -- to answer my own question below, here's how to compute them on
the fly, which may be a better answer than what I gave in my earlier
reply to Tom Roche:
x <- 1:10; y <- runif(10); f <- factor(rep(1:2,5))
xyplot(y~x|f,
strip=function(which.panel,factor.levels,...){
lab <- level
I looked at strip.custom()'s code and now get it, too. Here's the
version for that:
xyplot(y~x|f, strip=strip.custom(factor.levels=c("a","b"),style=1))
Both of these require precomputing the factor.levels vector of
character (or expression) labels. I see no obvious way to calculate
them on the f
Ken:
Aha! Thank you.
I was thrown by the phrase in the Help file, "currently being drawn,"
which I parsed to the levels rather the conditioning variable. As I
suspected, user error!
Still not sure why strip.custom fails.
-- Bert
On Sun, Nov 18, 2012 at 10:25 AM, Ken Knoblauch wrote:
> Bert,
On 16.11.2012 02:39, sarahw wrote:
Did you ever get a response to this or resolve this yourself?
You sent a message to the mailing list, but apparently not to the
original poster.
Uwe Ligges
Many thanks!
--
View this message in context:
http://r.789695.n4.nabble.com/additive-inter
On 16.11.2012 16:05, Catarina Maia wrote:
Hello,
I am using the mda package and in particular the fda routine to classify in
term of gear a set of 20 trips.
I preformed a flexible discriminant analysis (FDA) using a set of 151
trips.
FDAT1 <- fda(as.factor(gear) ~ . , data =matrizR)
A t
On 15.11.2012 10:52, Gabor Grothendieck wrote:
On Wed, Nov 14, 2012 at 12:05 PM, Murat Tasan wrote:
hi all - i've seen versions of this question before, but none seem to get
directly at my solving my (probably very simple) issue:
i simply want to annotate the tick marks on an axis with (supe
Bert,
A little bit of experimentation and I got this:
library(lattice)
x <- 1:10
y <- runif(1:10)
f <- factor(rep(1:2,5))
xyplot(y ~ x | f,
strip = function(which.panel, factor.levels,...){
lab <- c("a","b") #[which.panel]
strip.default(factor.levels = lab,
On 17.11.2012 00:41, nmend wrote:
Hi there,
I was wondering if anyone could explain how you should set tol in the prcomp
function.
Using help(prcomp) explains it as "a value indicating the magnitude below
which components should be omitted. (Components are omitted if their
standard deviations
Dear R users,
i want to put the results of a list to a for loop.
i will give an example
m1<-matrix(rep(1,15),ncol=5)
ind.sgn <- lapply(1:ncol(m1), combn, x = ncol(m1))
ind.sgn
[[1]]
[,1] [,2] [,3] [,4] [,5]
[1,] 1 2 3 4 5
[[2]]
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
On 18.11.2012 18:37, Rlotus wrote:
I have function y=x^5+x^8+ (cos(x))^3;
I need to get random x;
random y;
and after that determine whether y below or above the given function.
I tried a lot of version. but my code even doesnt workplz help with this
expressions ;( I am a begiiner in R
Which message are you referring to? Please quote.
And please read the posting guide to the mailing list that asks you to
do that!
On 18.11.2012 18:55, Rlotus wrote:
I mean i need to put random x in equation in order to determine value y.
i used runifbut it doesn work...
y= (runif(0,1))^
On 18.11.2012 13:02, nany23 wrote:
Good morning,
I have a sampled probability density function and I have to fill in with
colour a fraction of the total area beneath the curve (e.g. the 50% of the
total area beneath the curve). I would like to know which is the simplest
way to do this.
See ?
On 16.11.2012 16:59, Rui Barradas wrote:
Hello,
I believe it is, but see package XLConnect. The vignette is very
helpfull, with lots of examples.
... and many other ways. Please see the manual "R Data Import/Excel"
that ships with R and has a whole section devoted to Excel.
Best,
Uwe Ligg
Hello,
I believe that's a question for r-devel, but good point. It's
docummented that in non-recursive calls to list.files subdirectory names
are always included. (With a typo, "There always are" instead of "They
always are".)
Rui Barradas
Em 18-11-2012 17:20, Jannis escreveu:
Dear R develo
I mean i need to put random x in equation in order to determine value y.
i used runifbut it doesn work...
y= (runif(0,1))^5+(runif(0,1))^8+((cos runif(0,1)))^3; (<=it doesnt work
properly)
now i have value y
need to determine whether y is above or below my function.
so after that y1=r
On 18.11.2012 14:25, eliza botto wrote:
Dear UseRs,i want to find centroid of clusters, which i generated by hclust. Is
there a way doing that? i took mean to elements in each cluster to get centroid
but i am not sure if i am right?
Yes, that is a common definition for centroid.
Uwe Ligge
Ken:
I would have thought so, too. However:
x <- 1:10; y <- runif(1:10); f <- factor(rep(1:2,5))
xyplot(y~x|f, strip= function(which.panel,factor.levels,...){
lab <- c("a","b")[which.panel]
strip.default(factor.levels=lab,which.panel=which.panel,style=1,...)} )
produces label "a" fo
Good morning,
I have a sampled probability density function and I have to fill in with
colour a fraction of the total area beneath the curve (e.g. the 50% of the
total area beneath the curve). I would like to know which is the simplest
way to do this.
Thank you for answering
Best regards
Nany23
I have function y=x^5+x^8+ (cos(x))^3;
I need to get random x;
random y;
and after that determine whether y below or above the given function.
I tried a lot of version. but my code even doesnt workplz help with this
expressions ;( I am a begiiner in R
--
View this message in context:
http:
Dear R developers,
as far as I understand the manual of list.files(), there is only a way
to exclude directories from the returned vector if you use list.files
recursively. In non recursive mode, there seems to be no way of
excluding directories (the include dirs argument does not seem to hav
Michael, this comment doesn't seem appropriate to the question, since the
sample data is a ragged array that requires the addition of NAs to fit into a
wide format.
---
Jeff NewmillerThe .
The reason I was working with a correlation matrix is because I wanted
to calculate polychoric correlations first before submitting it to the
actual sem command. I was unsure whether R would use polychoric
correlations when indicating which of the variables are ordered. A final
question on the sam
Tom Roche pobox.com> writes:
>
>
> As described @
<<< clipped >>>
>
> However I will need to before-and-after compare this to
the results of a
> reboxing, or 3D regridding, of this data, so I would
prefer instead to
> label each panel in the lattice with the _value_ of
the level (an
> atm
At 08:56 17/11/2012, Kemi Racheal wrote:
Dear list member,
I have the following data example
ke <- data.frame(patid=c(1,1,1,2,3,3),a=c(1,2,2,1,1,2))
I want to add another variable b, such that the max of 'a' by id is returned
i.e data ke becomes
ke <- data.frame(patid=c(1,1,1,2,3,3),a=c(1,2,2,1
AFAICS, you will have to pre-compute the values and create a factor
with levels these values as character strings. Then use this factor as
the conditioning variable
?sprintf
?formatC
?round
and the like may be useful to format the numerical values suitably for
your labels.
Cheers,
Bert
On Sun,
Dear Yves
Many thanks for your quick reply, I really appreciate it!
The reason I was working with a correlation matrix is because I wanted to
calculate polychoric correlations first before submitting it to the actual
sem command. I was unsure whether R would use polychoric correlations when
indic
As described @
https://stat.ethz.ch/pipermail/r-help/2012-November/329442.html
I now know how to prepare my netCDF data specifying gas concentrations
over a 3D space (dimensions longitude, latitude, and (vertical) level),
so as to call lattice::levelplot() like
levelplot(conc ~ lon * lat | lev,
Dear UseRs,i want to find centroid of clusters, which i generated by hclust. Is
there a way doing that? i took mean to elements in each cluster to get centroid
but i am not sure if i am right?
thanks in advanceeliza
[[alternative HTML version deleted]]
Hello,
I'd like to resurrect this issue: is the "varwidth" equivalent (boxplot
box-width scaling according to number of data points) emulatable in the
0.9.* versions of ggplot2? Width still doesn't seem capable of accepting a
vector with length > 1 ...
Thank you for your input.
Sincerely, Joh
I am new to R as well, it sounds like you would want to look at clustering,
perhaps k-means clustering.
Brian
On Nov 18, 2012, at 12:19 AM, avadhoot velankar
wrote:
> I am working on morphometry of hairs and want to see if selected variables
> are giving significantly distinct groups.
>
> I
Hello,
I used the following code to perform a cluster analysis on a
dataframe consisting of 12 variables (coded as 1,0) and 63 cases.
FS1 <- read.csv("D://Arsontest2.csv",header=T,row.names=1)
str(FS1)
dmat <- dist(FS1, method="binary")
cl.test <- hclust (dist(FS1, method ="binary"), "av
>
Hello,
I'd like to resurrect this issue: is the "varwidth" equivalent (boxplot
box-width scaling according to number of data points) emulatable in the
0.9.* versions of ggplot2? Width still doesn't seem capable of accepting a
vector with length > 1 ...
Thank you for your input.
Sincerely, J
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