Hi:
df <- read.table(textConnection("
+ xloc yloc yield
+ 1 101295
+ 2 111081
+ 3 121120
+ 4 121110"), header = TRUE)
with(df, xtabs(yield ~ yloc + xloc))
xloc
yloc 10 11 12
10 0 81 0
11 0 0 30
12 95 0 0
HTH,
Dennis
On Tue, Feb 22, 2011 at
Hi Dennis,
Thanks for your quick response and sorry for not being clear. That
helped, but I need an actual matrix of e.g., 12 x 12 and those functions
give me a matrix with only the "filled" locations. I need a 12 by 12
matrix with sums (0 if there's not data and the actual sum where there
is
Hi:
> df$xloc <- factor(df$xloc, levels = 1:12)
> df$yloc <- factor(df$yloc, levels = 1:12)
> with(df, xtabs(yield ~ xloc + yloc))
yloc
xloc 1 2 3 4 5 6 7 8 9 10 11 12
1 0 0 0 0 0 0 0 0 0 0 0 0
2 0 0 0 0 0 0 0 0 0 0 0 0
3 0 0 0 0 0 0 0 0 0 0
Dear list,
this works fine:
x <- split(iris, iris$Species)
x1 <- lapply(x, function(L) transform(L, g = L[,1:4] * 3))
but I would like to multiply each Species with another factor:
setosa by 2, versicolor by 3 and virginica by 4. I've tried mapply but
without success.
Any thoughts? Thanks fo
Hi Albert,
Did u get the answer for this problem? I am connecting to a mysql database
through RJDBC and i am facing the same problem?
Regards,
Raji
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Hi Ista, David,
your answers indeed helped me to solve the problem!
So thanks to you and have a good day!
Stephan
--- On Wed, 2/23/11, David Winsemius wrote:
> From: David Winsemius
> Subject: Re: [R] Plot Stepped Line chart with multiple lines
> To: "Ista Zahn"
> Cc: "Techni X" , r-help@r-p
Hi,
I am trying to make a palaeoenvironmental transfer function using the R
package rioja that predicts the water-table (measured as depth to the water
table) of an area given the testate amoebae that are found there. I've
carried out weighted averaging of the data and am trying to produce a grap
Hello,
I am having trouple installing doSMP and revoIPC from:
https://r-forge.r-project.org/R/?group_id=950
My Linux system:
Linux 2.6.35-25-generic #44-Ubuntu SMP Fri Jan 21 17:40:44 UTC 2011
x86_64 GNU/Linux
Ubuntu 10.10 \n \l
I get this reply:
costas@ELEPHANT:~/Downloads$ sudo R CMD INSTA
Dear All,
I fear I am badly misunderstanding something fundamental about the
mclust package.
Please considered the dataset pasted at the end of the email (you can
also have a look at
http://dl.dropbox.com/u/5685598/dataedges.csv
).
Now, I would like to use the info on columns 1,2 and 4 to
I was wondering if anyone has produced any "R" code for the Bonnet-Price
95% confidence interval for the difference between two medians?
Refs:
Price, R.M. & Bonnet, D.G. 2002. Distribution-free confidence intervals
for difference and ratio of medians. J. Statist. Comput. Simul. 72(2),
119-124.
Hi,
I am having problems carrying out a mle for 3 parameters in a non-homogenous
poisson process.
I am trying to use the optim function to minimise the -ve log-likelihood.
When I use assumed values of my three parameters (20,1,1) the -ve
log-likelihood function returns a value of 1309122 but I
Hello all,
as the combination DA and R is rather new to me I would like to know:
are there packages that implement a fuzzy version of Discriminant
Analysis?
Thanks!
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PLE
Hello Everyone,
I am using R ver. 2.10.1 (2009-12-14). I installed lattice today via
install.packages.
I am doing this minimal example:
M = read.csv("/path/to/csv")
cloud(x ~ y * z, data = M)
this correctly gives me a scatterplot of the csv data which has 3 columns.
Every column contains 1800
Dear R users group
I have a couple of questions:
1. I'm fitting a linear mixed effects model using lme (package nlme) with
gaussian errors (using exponential correlation structure at the moment): Is it
possible to extract predictions for the gaussian process?
exp1 <- corExp(value=c(7,(40/300))
On Feb 22, 2011, at 11:34 PM, clc wrote:
Hi i am doing an environmental research
The equation is as follow:
gam(y1 ~ x1 + s(x2) + s(x3) + s(x4), family = gaussian, fit = true)
I would like to obtain the beta coefficient and 95CI of x4 (or
s(x4)), what
should I do?
You should re-think y
Thanks a lot. In many papers, they reported the changes and 95% ci in the
interquartile range, how can I obtain such figure using gam? Thanks!
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Dear R List
Could I ask again my question about where the size of a plot should be
specified (in par or pdf?). I still did not figure out, and any help
would be much appreciated!!
Thanks a lot!
Matthieu
Le 22. 02. 11 13:53, Matthieu Stigler a écrit :
Hi
I want to have a rectangular plot o
Dear list,
i would like to build an own package using:
R CMD INSTALL --html --build --auto-zip -l folder spectral.methods
it seems to succeed as I get the following messages:
* installing *source* package 'spectral.methods' ...
** R
** data
Warning: empty 'data' directory
** preparing package f
On Feb 23, 2011, at 4:11 AM, Patrick Hausmann wrote:
Dear list,
this works fine:
x <- split(iris, iris$Species)
x1 <- lapply(x, function(L) transform(L, g = L[,1:4] * 3))
but I would like to multiply each Species with another factor:
You will probably confuse yourself by not adopting R ter
On Feb 23, 2011, at 7:32 AM, clc wrote:
Thanks a lot. In many papers, they reported the changes and 95% ci
in the
interquartile range, how can I obtain such figure using gam? Thanks!
I suppose I could be wrong (not being an avid user of GAMs), but
reporting a 95% CI for the interquartil
It should be specified in pdf() as you did. If you try
pdf("try.pdf", height=2, width=2)
par(pin=c(0.5, 0.3),mai=rep(0.1,4), omi=rep(0.01,4), mar=c(5,4,4,2))
plot(runif(100))
dev.off()
I think the problem will become apparent.
best,
Ista
On Wed, Feb 23, 2011 at 12:39 PM, Matthieu Stigler
wr
Dear Ista
Thanks a lot for your help! However, when I spedify the right size I
wish (the same as specified in par() ) I get an error:
pdf("try.pdf", height=0.5, width=0.3)
par(pin=c(0.5, 0.3),mai=rep(0.1,4), omi=rep(0.01,4), mar=c(5,4,4,2))
plot(runif(100))
dev.off()
I don't see why my m
On Wed, 23 Feb 2011, Jannis wrote:
Dear list,
i would like to build an own package using:
R CMD INSTALL --html --build --auto-zip -l folder spectral.methods
it seems to succeed as I get the following messages:
* installing *source* package 'spectral.methods' ...
** R
** data
Warning: empty '
Thanks for your reply. I was puzzled as well with the message that a zipped
file was (said to be) created, but I could not find any zipped file. Running
the command only creates an unpacked folder with the package name at the
specified path.
Jannis
--- Prof Brian Ripley schrieb am Mi, 23.2.2
I am using getSymbols function from quantmod package to get price data from
internet.
Currently I have:
my.ticker <- "IBM"
getSymbols(my.ticker,src="google")
This creates an xts object named my.ticker which contains historical price
data for IBM.
How can I call and manipulating this xts object
Dear R-help,
my aim is to fit a constrained 2 component gamma mixture using mix
within package mixdist. The constraints being fixing both of the means
and both of the sigmas. Thus the only unconstrained parameter I wish to
fit is pi the proportions of the mixtures.
This is the code I used:
PP
On Feb 23, 2011, at 8:32 AM, Matthieu Stigler wrote:
Dear Ista
Thanks a lot for your help! However, when I spedify the right size
I wish (the same as specified in par() ) I get an error:
pdf("try.pdf", height=0.5, width=0.3)
par(pin=c(0.5, 0.3),mai=rep(0.1,4), omi=rep(0.01,4), mar=c(5,4,
Hello buddies.
I'm trying to use the "granger.test" function wich is included in MSBVAR
package, but I'm also using the package "var" in order to develop vector
autorregresion analysis.
Everything is well behaved...but when I'm trying to use the "irf" function
when I have both libraries open ther
Sorry, it seems as this was my mistake. I did not set the PATH environment
correctly. Now it works perfectly! As the status messages told me that a zip
file was created I thought that rtools was integrated correctly.
Thanks for your help
Jannis
--- Prof Brian Ripley schrieb am Mi, 23.2.2011:
On Feb 23, 2011, at 8:55 AM, Kushan Thakkar wrote:
I am using getSymbols function from quantmod package to get price
data from
internet.
Currently I have:
my.ticker <- "IBM"
getSymbols(my.ticker,src="google")
This creates an xts object named my.ticker which contains
historical price
da
Hello, list:
I'm not sure about where to send this question. I have several repeatability
calculations, together with their 95% confidence intervals, and I would like to
plot them, in a way similar to error bars. I was wondering if there is any
specific function to do this, or any method
@Scott:
I can't just use
colnames(IBM) <- c("open","high","low","close","vol")
I have nearly 100 tickers stored in an vector and I am looping through them.
Its not a good idea to hardcode all of them.
@David:
Column names is just an example. I have a long script that takes in an xts
object and
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
On Feb 23, 2011; 03:32pm Matthieu Stigler wrote:
>>> I want to have a rectangular plot of size 0.5*0.3 inches. I am having
>>> surprisingly a difficult time to do it...
<...snip...>
>>> If I specifz this size in pdf(), I get an error...
>>>
>>> pdf("try.pdf", height=0.3, width=0.5)
>>>
>>> p
In one of the papers...
We developed core models with a generalized additive Poisson regression
allowing for over-dispersion in the model (Wood, 2006). For each mortality
outcome, variations in seasonality, trends, mean temperature, and mean
humidity of current and previous days (lag 0–1) were fi
this works:
colnames(IBM) <- c("open","high","low","close","vol")
scott
On Wednesday, February 23, 2011 at 7:55 AM, Kushan Thakkar wrote:
> colnames(my.ticker) <- c("open","high","low","close","vol")
[[alternative HTML version deleted]]
__
R
dear R-utents,
does anybody of you knows if there are some specific function, already stored
in some libraries, which allow me to calculate straightforward
McFadden-R^2,,pseudo-R^2, p-hat (the ML estimator for p) and the goodness of
fit R^2p?
thanks in advance!
michi
[[al
Thanks Matthew that worked great. What a great forum this is: I am learning a
lot!
PS. I am now running both solutions on two similar computers. Let's see
which is fastest.
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Se
Hi every one,
I am using the package 'drc' to model root elongation using dose
response data. I don't know which model I should use. Though I don't
know which model I should use, I tried the following codes given
below. But it produced the error messages.Can any one tell me the code
in 'drc' packa
I've been wondering what L means in the R computing context, and was
wondering if someone could point me to a reference where I could read about
it, or tell me what it's called so that I can search for it myself. (L by
itself is a little too general for a search term).
I encounter it in strange p
Hi Iker,
matlines function to plot 956 % confidence intervals after using
predict function.
barplot function for errors bars.
hope this help
M
Regards
Le 23/02/11 16:00, Iker Vaquero Alba a écrit :
> Hello, list:
>
> I'm not sure about where to send this question. I have several
> re
On Wed, 23 Feb 2011, Jannis wrote:
Sorry, it seems as this was my mistake. I did not set the PATH environment
correctly. Now it works perfectly! As the status messages told me that a zip
file was created I thought that rtools was integrated correctly.
I don't see a mention of your R version.
Gene, it's described in ?NumericConstants
HTH, Erik
Gene Leynes wrote:
I've been wondering what L means in the R computing context, and was
wondering if someone could point me to a reference where I could read about
it, or tell me what it's called so that I can search for it myself. (L by
itse
If you are looping through and have the names of the files as characters,
say in ticker.names <- c("IBM", ...), you may be able to do this using:
for(nam in ticker.names)
{
[snip...]
eval(parse(text = paste('colnames(', nam, ') <- c(
"open","high","low","close","vol")', sep = '')))
}
HTH
On 02/23/2011 05:08 PM, Gene Leynes wrote:
I've been wondering what L means in the R computing context, and was
wondering if someone could point me to a reference where I could read about
it, or tell me what it's called so that I can search for it myself. (L by
itself is a little too general for
On Wed, Feb 23, 2011 at 8:55 AM, Kushan Thakkar wrote:
> I am using getSymbols function from quantmod package to get price data from
> internet.
>
> Currently I have:
>
> my.ticker <- "IBM"
> getSymbols(my.ticker,src="google")
>
> This creates an xts object named my.ticker which contains historic
Hi Gene,
It means 'Literal integer'.
So 1L is a proper integer 1, and 0L is a proper integer 0.
Hope it helps,
Tsjerk
On Wed, Feb 23, 2011 at 5:08 PM, Gene Leynes wrote:
> I've been wondering what L means in the R computing context, and was
> wondering if someone could point me to a reference
Thanks Mark!
But how does the resulting plot look like on your machine? For me, it is
terrible... I tried then removing the mai par (to 0.01), and then got
huge circles... It looks like there are many more parameters, I changed
then cex, got better but still many adjustments... seems a nightma
On Feb 23, 2011, at 10:23 AM, Kushan Thakkar wrote:
@Scott:
I can't just use
colnames(IBM) <- c("open","high","low","close","vol")
Well, you can.
I have nearly 100 tickers stored in an vector and I am looping
through them. Its not a good idea to hardcode all of them.
Perhaps putting t
Isn't the point here that they used smooths for the other terms but a linear
effect (or several) for (lagged) visibility. So you do have a beta in this
case, and the gam summary would tell you what it is estimated to be and what
the associated standard error is. The 'coef' and 'vcov' functions a
The notation '1L' mean to interprete the data as an 'integer'.
> str(1)
num 1
> str(1L)
int 1
>
> str(0xaa)
num 170
> str(0xaaL)
int 170
On Wed, Feb 23, 2011 at 11:08 AM, Gene Leynes wrote:
> I've been wondering what L means in the R computing context, and was
> wondering if someone could
I expected, that I will get the same prediction, if I multiply the
weights for all classes with a constant factor, but I got different
results. Please look for the following code.
> library(e1071)
> data(Glass, package = "mlbench")
> index <- 1:nrow(Glass)
> testindex <- sample(index, trunc(lengt
Matthieu,
A couple of points that you may be misunderstanding:
1.
your 'mar' setting overwrites the 'mai' setting;
use one or the other.
2.
pdf(width=0.5, height=0.3) sets the *device* region
dimensions, not the *plot* region. The thing to
remember is that "device contains figure contains plot"
I called svm and predict three times with the same data and got three
differing predictions:
> library(e1071)
Lade nötiges Paket: class
> data(Glass, package = "mlbench")
> index <- 1:nrow(Glass)
> testindex <- sample(index, trunc(length(index)/5))
> testset <- Glass[testindex, ]
> trainset <- Gla
In addition to Simon Wood's reading of the methods I am adding a minor
note on terminology:
What was reported was the ratio in risk per a difference equal to the
interquartile range (of the predictor variable), and the 95% CI for
such an estimate. They were not reporting a 95% CI for the
Thank you everyone, that makes a lot more sense now. It's not at all what I
would have guessed! (I thought that it might have to do with scope)
It's one of those little things would add just enough confusion that I would
sort of "tune out" whenever I saw it. So, I really appreciate having this
li
On Wed, Feb 23, 2011 at 10:46 AM, mathijsdevaan wrote:
>
> Thanks Matthew that worked great. What a great forum this is: I am learning a
> lot!
>
> PS. I am now running both solutions on two similar computers. Let's see
> which is fastest.
If your data is small its pointless to be concerned with
Hi,
I have two questions:
1. How do I combine "DF$F =" and "DF$G =" into one function? (The original
dataset contains many more columns for which I want to execute the same
operation)
2. How do I improve the ave function so that the value DF(12,G) = 0 instead
of 1 (see bold font)? Both DF(12,B)=
I am helping someone calculate ICC using R. I know R has several packages
like irr, psy etc which provide options to calculate ICC (intraclass
correlation coefficient). When getting ICC, we need to use the model:
two-way mixed model with absolute agreement.
I only found that in irr package, it pr
On Wed, Feb 23, 2011 at 8:32 AM, mathijsdevaan wrote:
> Hi,
>
> I have two questions:
> 1. How do I combine "DF$F =" and "DF$G =" into one function? (The original
> dataset contains many more columns for which I want to execute the same
> operation)
Just define a function that can handle multiple
On Feb 23, 2011, at 5:47 AM, Michael Bach wrote:
Hello Everyone,
I am using R ver. 2.10.1 (2009-12-14). I installed lattice today via
install.packages.
I am doing this minimal example:
M = read.csv("/path/to/csv")
cloud(x ~ y * z, data = M)
this correctly gives me a scatterplot of the cs
Try this:
x1 <- transform(iris, g = iris[-ncol(iris)] * (as.numeric(iris$Species) + 1
))
On Wed, Feb 23, 2011 at 6:11 AM, Patrick Hausmann <
patrick.hausm...@uni-bremen.de> wrote:
> Dear list,
>
> this works fine:
>
> x <- split(iris, iris$Species)
> x1 <- lapply(x, function(L) transform(L, g =
Hello list,
Has anyone had any luck creating an M-step driver for negative
binomial regression for use with package flexmix? I've had a look
here: http://cran.r-project.org/web/packages/flexmix/vignettes/flexmix-intro.pdf
as well as poking around in the flexmix source, but I haven't had much
luck
People of R(th),
I have been ramming my head against this problem, and I wondered if
anyone could lend a hand. I want to parallelize a bootstrap of a linear
model on my 8-core mac. Below is the process that I want to parallelize
(namely, the m2.ph.rlm.boot<-boot(m2.ph,m2.ph.fun, R = nboot) com
Dear R users,
I have a question about optimization. Is that possible set a constraint that
only integer values are allowed for the optim in the R base and some other
optimizers such as psoptim in the package pso? For instance, my lower bound
is 0 and upper bound is 100 and I only want integer from
Thanks, learned something new again. About my second question:
You're right, the input rows are similar, but "cumsum(z) - z" generates
dissimilar output rows. The rule I want to define is: if input rows are
repeated, replace the output row with the minimum value of the output row.
So (the differe
Thanks guys,
I had to change two things to get it to work:
1. As Dennis pointed out, all of the variables in the model had to be
variable names in the preds data frame.
2. I had to use na.pass in the following statement to be able to merge the
predicted values with the preds data frame because of
> "RK" == Rumen Kostadinov
> on Sun, 13 Feb 2011 12:46:52 -0500 writes:
RK> Thanks Sarah,
RK> Yes, the function behaves Exactly as documented:
RK> check this out:
>> a = c(1,2,3,4,5)
>> a[which(a!=6)]
RK> [1] 1 2 3 4 5
>> a[!which(a==6)]
RK> numeric(0)
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
On Wed, Feb 23, 2011 at 08:02:16PM +0100, Chun Wang wrote:
> Dear R users,
>
> I have a question about optimization. Is that possible set a constraint that
> only integer values are allowed for the optim in the R base and some other
> optimizers such as psoptim in the package pso? For instance, my
Dear Céline,
First of all, this list is primarily for questions related to using R,
not statistical questions. Regardless, you would receive better
feedback if you provided more information and details. What type of
data are you modelling? You say your distribution is skewed---how is
it skewed?
What you want to do is create a new dataframe that includes all the years.
> newdata <- data.frame( Year=1981:1988 )
> merge(fire, newdata, all=TRUE)
If you don't include the all=TRUE, then you only get the rows that are
contained in both datasets.
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Hello R folks,
Reproducible code below - I'm trying to do a weighted mean by a factor and
can't figure it out. Thanks in advance for your assistance.
Mike
data<-data.frame(c(5,5,1,1,1),
c(10,8,9,5,3),
c("A","A","A","B","B"))
names
I withdraw this question, I was able to accomplish this by creating a new
function.
Now if only I could get the by output into a dataframe...
Mike
On Wed, Feb 23, 2011 at 2:25 PM, Mike Schumacher
wrote:
> Hello R folks,
>
> Reproducible code below - I'm trying to do a weighted mean by a factor
Hello everyone,
I have the following problem, I have a dataframes that looks like this:
fire$Year fire$Size
1 19811738.0
2 19842228.1
3 1985 38963.3
4 19862223.4
5 19873594.6
6 19881520.0
...
What I would like to do is copy the values from t
Here is the party line, perhaps
by(data, data$TYPE, function(dat)
with(dat, weighted.mean(MEASURE, COUNT)))
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Mike Schumacher
Sent: Thursday, 24 February 2011 9:40 AM
To: r-help@r-p
Hello,
I'm trying to use NLME for a non-linear model with one random effect
that is the same as the one grouping level (chamber). Using the following
statement:
test <- nlme(flux ~ b0 * exp((b1*soiltemp) - (b2*soiltemp^2))
* vpd^b3 * mstsoil2^b4 * airtemp^b5, data = ac5,
fixed = list(b0 + b1 + b
Hello. Does anybody know how to estimate nested logit models in R? I know
that the package mlogit does it, but It doesn´t report the logsum
parameters. I would like to have the logsum parameters, and the elasticites
if possible.
Thank you
Felipe Parra
[[alternative HTML version deleted]]
x<-data.frame(id=c(1,1,2,2,3,3,4,4), v1=c(1:8), V2=c(9:16))
> x
id v1 V2
1 1 1 9
2 1 2 10
3 2 3 11
4 2 4 12
5 3 5 13
6 3 6 14
7 4 7 15
8 4 8 16
1)
I want to split the data into 2 group with one unique ID. I should use the
split function but i don't know how to write it out.
2
Suspect that this is easier than I realize, but taking some figuring out
currently. Any help would be appreciated.
I have a data frame (testhm) with many rows such as:
ProbeSet.ID.F ProbeSet.ID Feature.ID G.S X0030V120810.14 X0143V120110.14
X0258V111710.14 X0283V111710.14 X0430V120710.14 X047
Hi Sandra,
Terribly ugly, but this way at least you don't have to know which years are
missing, etc.:
dat <- data.frame(year = c(1981,1984,1985,1986), size = c(1,2,3,4))
year_ <- data.frame(year = seq(1981,1986,1))
year_dat <- merge(year_, dat, by = "year", all=T)
na_s <- subset(year_dat, is.n
hi,R users,
Now I have some scatter figures, is there some method can plot equalprobable
error ellipse ?
--
TANG Jie
Email: totang...@gmail.com
Tel: 0086-2154896104
Shanghai Typhoon Institute,China
[[alternative HTML version deleted]]
__
R-
On Wed, Feb 23, 2011 at 1:02 PM, Martin Maechler
wrote:
>> "RK" == Rumen Kostadinov
>> on Sun, 13 Feb 2011 12:46:52 -0500 writes:
>
> RK> Thanks Sarah,
> RK> Yes, the function behaves Exactly as documented:
>
> RK> check this out:
> >> a = c(1,2,3,4,5)
> >> a[which(a!=6
On Feb 23, 2011, at 10:06 PM, Jie TANG wrote:
hi,R users,
Now I have some scatter figures, is there some method can plot
equalprobable
error ellipse ?
Consider searching?
http://search.r-project.org/cgi-bin/namazu.cgi?query=ellipse+density+2d&max=100&result=normal&sort=score&idxname=fun
Hi:
Is this what you want?
x$gpvar <- rep(c(1, 2), 4)
split(x, x$gpvar)
$`1`
id v1 V2 gpvar
1 1 1 9 1
3 2 3 11 1
5 3 5 13 1
7 4 7 15 1
$`2`
id v1 V2 gpvar
2 1 2 10 2
4 2 4 12 2
6 3 6 14 2
8 4 8 16 2
HTH,
Dennis
On Wed, Feb 23, 2011 at 5:16 P
Thanks a lot for David and Simon! Got it~
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