Hello, I am new to R. I would like others to explain to me how to add
absolute values inside the individual stacked bars in a consistent way
using the basic R plotting function (R base). I tried to plot a stacked bar
graph using R base but the values appear in an inconsistent/illogical way
in such
By considering this reproducible example
#start code
library(lattice)
dotplot(variety ~ yield | site, data = barley,
layout = c(1,6),
index.cond= function(x,y){median(x)},
panel = function(x,y,...) {
panel.dotplot(x,y,...)
median.values <- median(x)
Hello,
I did computation of the model like in the paper.
And now I have RMSE from my estimated model and RMSE from the paper.
The question is: How to do the test for two RMSEs?
Thank you.
[[alternative HTML version deleted]]
__
R-help@r-projec
You are good!
Many thanks
Alex
From: Dennis Murphy
Cc: R help
Sent: Saturday, February 16, 2013 3:11 AM
Subject: Re: [R] Plot a Matrix as an Image with ggplot
Hi:
See if the following works for you:
library(reshape2)
library(ggplot2)
tdm <- melt(testDat
On 02/18/2013 06:05 PM, nay-nancy Laiser wrote:
Hello, I am new to R. I would like others to explain to me how to add
absolute values inside the individual stacked bars in a consistent way
using the basic R plotting function (R base). I tried to plot a stacked bar
graph using R base but the value
Hi
I'm having difficulty working out how to get what I think is the appropriate
partitioning of variability in a repeated measures setup.
I have G=5 treatment-groups, each containing n=6 subjects, and a response is
measured on each subject on t=4 occasions.
I think the anova degrees of free
On 02/18/2013 09:51 PM, Jim Lemon wrote:
...
In fact, it is easier to label the bars like this:
barlabels(xpos,mydata,cex=0.7)
(I forgot that I had solved the stacked bar problem)
Jim
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailm
sorry, I've managed to find a (partial) solution to my problem (even not much
elegant);
now what is left out is the proper tweaking of labels...
any better solutions?
dotplot(variety ~ yield | site, data = barley,
layout = c(1,6),
index.cond= function(x,y){median(x)},
pan
sorry again, there is something more to consider...
in case I need to plot x axis in logarithmic scale the above solution is not
working:
so now my further question is how to do that when I set
scales=list(x=list(log=TRUE))
I've been trying to simplify my original example to the bare bone (for s
Hi all,
Firstly, it is not a homework. I am working for a hotel booking company in
Thailand but I don't want to explain a complex equation and concept here so
I keep it simple and closely related to what I am trying to solve.I apology
if my question is not clear enough.
I am new to R and previous
Dear All
I am kind of stuck up with a code a part of which seems to be causing a
problem, or at least I think so. May be the community can help me. Its
simple but I suppose I am missing something.
I generate a data matrix X, say of order n*p, where n represents
independent row-vectors and p co
Hi
I'm having difficulty working out how to get what I think is the
appropriate partitioning of variability in a repeated measures setup.
I have G=5 treatment-groups, each containing n=6 subjects, and a response
is measured on each subject on t=4 occasions.
I think the anova degrees of freedom
Dear R experts,
I am trying to arrange multiple plots, creating one graph for each
size1 factor variable in my data frame, and each plot has the median
price on the y-axis and the size2 on the x-axis grouped by clarity:
library(ggplot2)
df <- data.frame(price=matrix(sample(1:1000, 100, replace =
Can anyone point me to a comparision of the various packages avaliable to do
Finite Mixture Models or Latent Class Models in R. As ever my googling only
produce papers on specific packages or implementations. I am looking of a
considered comparision that criticlaly compares them. Faili
Hi! I'm a recent convert from Stata, so forgive my ignorance.
In Stata, I can write foreach loops (example below)
foreach var of varlist p1-p14 {
foreach y of varlist p15-p269 {
reg `var' `y'
}
}
It's looping p1-p15, p1-p16, p1-p269, p2-p15, p2-p16,... p2-p269,...
variable pai
Hi! I'm a recent convert from Stata, so forgive my ignorance.
In Stata, I can write foreach loops (example below)
foreach var of varlist p1-p14 {
foreach y of varlist p15-p269 {
reg `var' `y'
}
}
It's looping p1-p15, p1-p16, p1-p269, p2-p15, p2-p16,... p2-p269,...
variable pai
On 18-02-2013, at 10:34, Prakasit Singkateera wrote:
> Hi all,
>
> Firstly, it is not a homework. I am working for a hotel booking company in
> Thailand but I don't want to explain a complex equation and concept here so
> I keep it simple and closely related to what I am trying to solve.I apolo
Hi,
You are making it more complicated than it needs to be. You already
provided the data.frame in the ggplot call, so you don't need to
specify it in the aes calls. The various factor() and as.factor()
calls are also unnecessary. So stripping away this extra stuff your
plot looks like
ggplot(dat
Hello All,
Have recently started learning Sweave and Knitr. Am now trying to learn
odfWeave as well. Things went pretty smoothly with Sweave and Knitr but I'm
having some trouble with odfWeave.
My understanding was that odfWeave should work in pretty much the same way as
Sweave. With odfWeave,
On the same help page, it says "'RNGkind' is a more friendly interface to query
or set the kind of
RNG in use."
Try
RNGkind()
to see the currently selected RNG. The same function with kind= specified will
also change to another
S Ellison
> -Original Message-
> From: r-help-boun
Readers,
How to solve this unicode input error please?
> postscript("~/tmp/test.eps",width=100/25.4,height=100/25.4,horizontal=FALSE,onefile=TRUE,paper="special")
> testx<-seq(1:5)
> testy<-seq(1:5)
> plot(testy~testx)
> mtext('text (O₂)\n more text',side=3,line=1)
Warning messages:
1: In mtext("
Hi All
I am getting the above mentioned error when I run the code below. I don't
know why because I have implemented the function and I'm calling the
function with a parameter. I'm obviously missing the plot ... Can someone
perhaps point out the error of my ways?
Error:
> out<-ode(y=init, times=t
1. Please do not multiple post.
2. One approach would be mixed effects models. See the nlme and lme4 packages.
3. But there are a host of others (including, perhaps, just ?aov.).
4. This is a statistical question, and off topic here. Post on the
r-sig-mixed-effects models list instead or s stati
Hello,
I did computation of the model like in the paper.
And now I have RMSE from my estimated model and RMSE from the paper.
The question is: How to do the test for two RMSEs?
Thank you.
2013/2/18 ghjf fghjk
> Hello,
>
> I did computation of the model like in the paper.
> And now I have RMSE
Le lundi 18 février 2013 à 13:48 +0100, Jamora, Nelissa a écrit :
> Hi! I'm a recent convert from Stata, so forgive my ignorance.
>
>
>
> In Stata, I can write foreach loops (example below)
>
>
>
> foreach var of varlist p1-p14 {
>
> foreach y of varlist p15-p269 {
>
> reg `var' `y'
>
That's not a reproducible example. There is no sessionInfo() and you
omitted code (where did 'fp' come from?).
It works fine for me (see sessionInfo below) using the code in ?odfWeave.
As for the file paths: you can point to different paths for the files
(although don't change the working directo
On 18/02/2013 14:15, e-letter wrote:
Readers,
How to solve this unicode input error please?
On what system, in what locale? And where is Unicode mentioned?
If you intended a subscript 2, use plotmath. That is not a character in
the standard Postscript fonts, nor is in the encoding you selec
On 18-02-2013, at 15:19, Jannetta Steyn wrote:
> Hi All
>
> I am getting the above mentioned error when I run the code below. I don't
> know why because I have implemented the function and I'm calling the
> function with a parameter. I'm obviously missing the plot ... Can someone
> perhaps poin
Hello,
I did computation of the model like in the paper.
And now I have RMSE from my estimated model and RMSE from the paper.
The question is: How to do the test for two RMSEs? e.g. like Null
hypothesis: RMSE1-RMSE2=0
I have the paper and I am doing computation exactly like it is described in
the
Hi, Can someone help me with the following? I currently have the following
plot: plot(c(0,14,0,14),c(0,5,0,5),ann = FALSE, xaxs="i", yaxs="i",bty="L" )
polygon(c(0,18,18), c(0,6,0), col='light grey', border = NA)
polygon(c(0,18,18), c(0,2,0), col='grey', border = NA) However, I would like to
Dear colleagues,
This is to inform you that Version 1.0 of the R package MVN has been released
on CRAN yesterday. This package contain several multivariate normality tests
such as Henze-Zirkler, Mardia and Royston.
HZ.test: The Henze-Zirkler (HZ) test is based on a non-negative functional
dist
Hi Milan,
Thanks for responding to my question. I'm actually not interested in LM, it
was more for example.
You are right, I'm trying an enormous set of model runs. My Var1 n=14; Var2
n=255 ==> 3570!
But first, I need be able to set up 2 variables in each model run. Those 2
variables are dif
Hi guys, I am a biologist and an R newbie, and I'm learning how to create a
simple population model.
So, I have a population matrix ("pop")of 30 age classes of female (1:4 are
non-breeders, 5:30 are breeders) which will be modelled for 100 years.
> pop <- matrix(0,30,100)
I then populate this ma
I have an irregular time series, which cannot easily be converted to a ts
object (long stretches with NA).
I have calculated the climatological mean and st.dev, but need help on how
to calculate the anomalies from the climatology and my original data set.
Below is an example, where I have indi
Dear Ista,
Thank you! It works perfectly!
-fra
2013/2/18 Ista Zahn :
> Hi,
>
> You are making it more complicated than it needs to be. You already
> provided the data.frame in the ggplot call, so you don't need to
> specify it in the aes calls. The various factor() and as.factor()
> calls are als
melswed slu.se> writes:
>
> I understand. I want to specify that drug is only a fixed factor and family
> should be the only random factor. So maybe, my R code is wrong If I
> specify random=~1|drug/family it is only because I wanted to specify that
> family is nested within drug.
Then yo
Le lundi 18 février 2013 à 17:09 +0100, Jamora, Nelissa a écrit :
> Hi Milan,
>
> Thanks for responding to my question. I'm actually not interested in LM, it
> was more for example.
>
> You are right, I'm trying an enormous set of model runs. My Var1 n=14; Var2
> n=255 ==> 3570!
> But first,
Hi Max,
Sorry I didn't provide sufficient information. Below is my sessionInfo with all
code included.
Thanks,
Paul
sessionInfo()
R version 2.13.1 (2011-07-08)
Platform: x86_64-pc-mingw32/x64 (64-bit)
locale:
[1] LC_COLLATE=English_United States.1252
[2] LC_CTYPE=English_United States.125
Currently your code does not seem to make any sense
===comments in line
John Kane
Kingston ON Canada
> -Original Message-
> From: s...@st-andrews.ac.uk
> Sent: Mon, 18 Feb 2013 16:09:49 +
> To: r-help@r-project.org
> Subject: [R] help with population matrix
>
> Hi guys, I am a b
If you have the same set of predictors for multiple outcomes, you can
speed up the process considerably by taking advantage of the fact that
the design matrix is the same for multiple outcomes. As an example:
set.seed(10)
y <- matrix(rnorm(1 * 14), ncol = 14)
x <- matrix(rnorm(1 * 2), nco
Please don't repeat yourself.
Please read the Posting Guide. Don't post here using HTML email.
This is a statistics question, not an R question, so it is off topic here.
---
Jeff NewmillerThe .
See if this helps at all:
http://cran.r-project.org/web/packages/popbio/popbio.pdf
-Roy M.
On Feb 18, 2013, at 8:49 AM, John Kane wrote:
> Currently your code does not seem to make any sense
> ===comments in line
>
> John Kane
> Kingston ON Canada
>
>
>> -Original Message-
>> Fro
Hello,
Try the following.
Month <- format(Dat$open_date, "%m")
anom <- Dat$Dry_w - ave(Dat$Dry_w, Month, FUN = function(x) mean(x,
na.rm = TRUE))
st.anom <- Dat$PP_int - ave(Dat$PP_int, Month, FUN = function(x) mean(x,
na.rm = TRUE))
Hope this helps,
Rui Barradas
Em 18-02-2013 14:03, He
It will be better if you can give us a complete example of what you want to
do. You start off with a typo in your for() command using "y" but then
switching to "t" in the code. You also say the rows "all have different
survival probabilities" but it is not clear where that fits into the picture
(b/
Hi Duncan,
Thank you for quick reply.
I am not sure that your solution solves the problem. If I use
useOuterStrips(dotplot(count ~ subject|risk*treatment,df))
the order of subjects and panel y-limits in each panel are
A, B, BB, CCC, DD, .
However, the order of subjects and y-limits which
Hello R'users,
I've tried to find a package or some code about Uplift modeling within R (or
Netlift modeling, Incremental or Differential response modeling) but failed.Â
If you have any clue of source about "Uplift modeling with R" on the web, i
would appreciate to share it with you.
Thank's
Boris:
If I understand you correctly, you wish to set panel limits by row. I
know of no "slick" way of doing this (others may), as what is in a row
can change depending on layout and how you determine the scale for the
rows' contents may depend on the application context.
So I would do it manuall
Search! Google on "uplift modeling in R" or similar. I got:
http://blog.data-miners.com/2009/12/differential-response-or-uplift.html
There is undoubtedly more.
-- Bert
On Mon, Feb 18, 2013 at 9:38 AM, wrote:
> Hello R'users,
>
> I've tried to find a package or some code about Uplift modeling
Hi Vera,
Not sure I understand your question.
Your statement
"In my lista I can´t merge rows to have the group, because the idea is
for each file count frequencies of mm, when b<0.01. after that I
want a graph like the graph in attach."
files<-paste("MSMS_",23,"PepInfo.txt",sep="")
read.dat
On 02/16/2013 05:00 AM, r-help-requ...@r-project.org wrote:
Then I perform cox regression as follows
m2_1<-coxph(Surv(X_t0,X_t, vlsupp) ~ nvp + as.factor(cd4pccat) +
as.factor(vlcat) + as.factor(agecat) + as.factor(whostage) +
as.factor(hfacat) + as.factor(wfacat) + as.factor(wfhcat) +
as.fac
Hi,
I am not able to open your graph. I am using linux.
Also, the codes in the function are not reproducible
directT <- direct[grepl("^t", direct)]
directC <- direct[grepl("^c", direct)]
It takes double the time to know what is going on.
dir()
#[1] "a1" "a2" "a3" "b1" "b2" "c1"
direct<- list
Some (quite a few!) years ago I wrote myself a wee function called
compInt() ("compound interest") to do --- I think --- just what you require.
I have attached the code for this function and a help file for it.
If anyone else wants this code, and if the attachments don't get through
the list,
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
> On Behalf Of Mauricio Zambrano-Bigiarini
> Sent: Monday, February 18, 2013 1:33 AM
> To: r-help@r-project.org
> Subject: [R] Random number generator used in 'runif'
>
> Dear list,
>
> For the
Rolf,
Your attachments got through.
But where is the function newt(…)
Berend
On 18-02-2013, at 21:25, Rolf Turner wrote:
>
> Some (quite a few!) years ago I wrote myself a wee function called
> compInt() ("compound interest") to do --- I think --- just what you require.
> I have attached the
Hello,
I'm trying to create an Excel chart using the package RDCOMClient. I am not
able to title the excel chart and axis using RDCOMClient "ChartTitle" and
"Axis"?
Any suggestions or references is appreciated.
Thanks,
uak.
[[alternative HTML version deleted]]
___
I have 12 binary (raster) files
https://echange-fichiers.inra.fr/get?k=k3M2jatJyHy65Cs99G4 .
I would like to calculate the moving average for the 12 values for each
pixel in the 12 files.
For a simple vector we can get a moving average by using this :
x <- c(1,2,3,NA
monicamir88 wrote
> Hello!
> I have a doubt with the R software. I have this function:
>
> results <- function(bCODBOD, BOD, VSS, COD, sBOD, sCOD, TSS, TKNpa, T,
> NH3Ne, DO, Q, TKN, MLSS, NO3Ne, RAS, tanoxic1, tanoxic2, rbCOD, SDNR1,
> SDNR2, tdanoxic, tanaerobic, IRp, P) {
> bCOD <- bCODBOD*BOD
On Feb 18, 2013, at 12:28 PM, Jonsson wrote:
> I have 12 binary (raster) files
> https://echange-fichiers.inra.fr/get?k=k3M2jatJyHy65Cs99G4 .
> I would like to calculate the moving average for the 12 values for each
> pixel in the 12 files.
>
> For a simple vector we can get a moving average
Dear all,
I tried running the following syntax but it keeps running for about 4 hours
and then i got the following errors:
Error in if (is.na(f0$objective)) { : argument is of length zero
In addition: Warning message:
In is.na(f0$objective) :
is.na() applied to non-(list or vector) of type 'NUL
You're kidding, right?!
See ?traceback for browsing the call stack after an error.
Also ?browser, ?debug, and ?recover for other debugging options.
-- Bert
On Mon, Feb 18, 2013 at 1:28 PM, Aya Anas wrote:
> Dear all,
>
> I tried running the following syntax but it keeps running for about 4 ho
Following up on what Bert said, also put
options(error=utils::recover)
at the start of the script so that on errors you get the traceback
printed out and the you can also use the browser to look at the values
of the objects at the point of the error. This is debugging 101.
On Mon, Feb 18, 2013
Hi Elisa,
I think your question is regarding "why the third column (Frequency) is absent??
aggregate(row.names(dat1)~class,dat1,function(x) x)[1] #Here, the first two
class are not there, because it was not in the dataset. (0.10.
# class
#1 0.2
To: "smartpink...@yahoo.com"
Sent: Monday,
On 02/19/2013 01:20 AM, Elisabeth wever wrote:
Hi, Can someone help me with the following? I currently have the following plot:
plot(c(0,14,0,14),c(0,5,0,5),ann = FALSE, xaxs="i", yaxs="i",bty="L" )
polygon(c(0,18,18), c(0,6,0), col='light grey', border = NA)
polygon(c(0,18,18), c(0,2,0), c
simonj16 wrote
> Consider an urn that contains 10 tickets, labelled: 1,1,1,1,2,5,5,10,10,10
>
> I want to draw with replacement n=40 tickets. I am interested in the sum,
> Y, of the 40 ticket values that I draw
>
> Write an R function named urn.model that simulates this experiement. What
> I have
On 02/19/2013 09:50 AM, Jim Lemon wrote:
Hi Elisabeth,
For a quick hack, try this:
segments(rep(0,100),seq(-4.5,0,length.out=200),rep(14,200),
seq(0,4.5,length.out=200),
col=rgb(seq(0.8,0.5,length.out=200),seq(0.8,0.5,length.out=200),
seq(0.8,0.5,length.out=200)),lwd=2)
and even easier:
libra
Hello,
Or with base R only,
smooth <- colorRampPalette(c('white', 'darkgrey'))
segments(rep(0,100),seq(-4.5,0,length.out=200),rep(14,200),
seq(0,4.5,length.out=200), col = smooth(200), lwd = 2)
Hope this helps,
Rui Barradas
Em 18-02-2013 23:30, Jim Lemon escreveu:
On 02/19/2013 09:50 AM, J
What version of odfWeave and XML?
On Feb 18, 2013, at 11:49 AM, Paul Miller wrote:
> Hi Max,
>
> Sorry I didn't provide sufficient information. Below is my sessionInfo with
> all code included.
>
> Thanks,
>
> Paul
>
> sessionInfo()
> R version 2.13.1 (2011-07-08)
> Platform: x86_64-pc-
hello, all.
in my previous research, i have always used existing data. i am trying
something new as an exploratory exercise and have never create my own variable
form scratch.
essentially, i am creating a variable for party affiliation.
here is an example.
var =party.
levels= democrat, repu
Hi ,
Are there any multivariate GARCH package (e.g., BEKK) in R?
I do see a few "projects" constructing multivariate GARCH package, but
wonder if there is anyone that is ready for use.
Thanks,
Miao
[[alternative HTML version deleted]]
_
On Feb 18, 2013, at 5:16 PM, jpm miao wrote:
> Hi ,
>
> Are there any multivariate GARCH package (e.g., BEKK) in R?
>
> I do see a few "projects" constructing multivariate GARCH package, but
> wonder if there is anyone that is ready for use.
What efforts have you made at searching? In par
Hi Nicole,
On Mon, Feb 18, 2013 at 7:01 PM, Nicole Ford wrote:
> hello, all.
>
> in my previous research, i have always used existing data. i am trying
> something new as an exploratory exercise and have never create my own
> variable form scratch.
>
> essentially, i am creating a variable for
On 02/19/2013 09:44 AM, Berend Hasselman wrote:
Rolf,
Your attachments got through.
But where is the function newt(…)?
A --- dang! It's another item from my "personal miscellany package"
which I'm so used to having around that I forgot that other people don't
have it.
Attached. Along
On 02/19/2013 11:01 AM, Nicole Ford wrote:
hello, all.
in my previous research, i have always used existing data. i am trying
something new as an exploratory exercise and have never create my own variable
form scratch.
essentially, i am creating a variable for party affiliation.
here is an
Bert,
Thank you very much for your reply. Your suggestions to set manually y-limits
of panels in lattice plot certainly puts me on the right track.
It works really well with xyplot():
hf <- data.frame(
  x=seq(1,20,by=1),
  y=seq(0.1,2,by=0.1),
  risk=gl(2,10,labels=c("low","high")),
Pete: Please don't do people's homework for them!
At most, give them a ***hint*** only.
cheers,
Rolf Turner
On 02/19/2013 12:01 PM, Pete Brecknock wrote:
simonj16 wrote
Consider an urn that contains 10 tickets, labelled: 1,1,1,1,2,5,5,10,10,10
I want to draw with replacement n=
hello ,Rusers:
I have a dataset for example
totdata=c(8.020 7.412 8.110 8.270)
plot(totdata[1:4],type='l',pch=1,lty=1,col='blue',lwd=1)
and the data could be plotted as the attachment figure.
now I want to exchange its x-axis and y-axis in the figure .How could I
modify the plot command ?
than
Hi Boris
Just a different take on it, quick glimpse of it
library(lattice)
library(latticeExtra)
z= unique(df$subject)
df$nsub <- sapply(df$subject, pmatch, z)
useOuterStrips(xyplot(count ~
nsub|treatment*risk, df, type = "h", lwd = 2,
scales = list(x = list(at = 1:6, labels = z
The o
On 02/19/2013 02:17 PM, Jie Tang wrote:
hello ,Rusers:
I have a dataset for example
totdata=c(8.020 7.412 8.110 8.270)
plot(totdata[1:4],type='l',pch=1,lty=1,col='blue',lwd=1)
and the data could be plotted as the attachment figure.
now I want to exchange its x-axis and y-axis in the figure .
Hi
plot(totdata,seq_along(totdata),type='l',pch=1,lty=1,col='blue',lwd=1)
A.K.
- Original Message -
From: Jie Tang
To: r-help@r-project.org
Cc:
Sent: Monday, February 18, 2013 10:17 PM
Subject: [R] how to change the plot from X-axis to Y-axis
hello ,Rusers:
I have a dataset for ex
Hi all
I have a set of data whose scatter plot shows a very nice power
relationship. My problem is when I fit a Power Trend Line in an Excel
spreadsheet, I get the model y= 44.23x^2.06 with an R square value of 0.72.
Now, if I input the same data into R and use
model< -nls(y~ a*x^b , trace=TRUE, d
Thanks. I tried this code below, why does expanded dataset 'res1' has m1=3
and n1=3
, dataset 'd3' doesn't have m1=3, n1=3.
d3<-structure(list(m1 = c(2, 3, 2), n1 = c(2, 2, 3), cterm1_P0L = c(0.9025,
0.857375, 0.9025), cterm1_P1L = c(0.64, 0.512, 0.64), cterm1_P0H =
c(0.9025,
0.9025, 0.857375), c
Have you plotted the data and the lines to see how they compare? (see
fortune(193)).
Is there error around the line in the data? The nls function is known to
not work well when there is no error around the line. Also check and make
sure that the 2 methods are fitting the same model.
You might
To paraphrase Bill Venables (see fortune(217)):
Simple. Excel must be broken. Have you reported it to them?
(The difference in this case is that it is probable that Excel *is* broken.
It usually is.)
cheers,
Rolf Turner
On 02/19/2013 05:49 PM, David Gwenzi wrote:
Hi all
I
Excel definitely does not use nonlinear least squares fitting for power curve
fitting. It uses linear LS fitting of the logs of x and y. There should be no
surprise in the OP's observation.
---
Jeff Newmiller
Hi Berend,
Thank you for your information. However, I am still not clear on the Rpmt
function please elaborate more about
- What is the purpose of Rpmt function? Why do you subtract
pmt(xrate,x,xpv) from xpmt? / What is the meaning of xpmt ?
- And what is the purpose of doing Rpmt(xpv/xpmt/2,xra
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