Re: [R] Need help using lattice

[Gerrit Draisma]

Ha David,
I do not know.
It must be that your AprtoDec object  is different from month.abb.
Anyhow it should contain the labels for all twelve months!
If I copy your command in an interactive session I just get the message
that AprtoDec is not defined.

I understand that you type the commands interactively in an R session?
I usually prefer using an editor to write the program, and execute with a
> source("dotplot.R",echo=TRUE)
command.
I like an editor with syntax highlighting and shows matching brackets.

I have attached my dotplot.R to this mail.

Gerrit

Op 09-02-2021 om 21:14 schreef Parkhurst, David F.:

When I enter your line:
df <- expand.grid(site=c("een","twee","drie","vier","vijf"),
   month=factor(3:12,levels=1:12,
    labels=month.abb, ordered=TRUE))
I get + signs in front of month and label, as expected, and it follows with <

When I convert that to use my own variable names to this:
df <- expand.grid(site=c("CrCr","NFSC","MFSC","SFSC","LMO","MCE","MUE","MLE"),
     monames=factor(3:12, levels=1:12,
        labels=AprToDec, ordered=TRUE))
  I get + signs in front of monames and labels, again as expected, but then it 
gives me another + sign.  And if I try to add another ), it keeps giving me + 
signs.  What is happening here?

From: Gerrit Draisma 
Date: Tuesday, February 9, 2021 at 6:14 AM
To: Parkhurst, David F. 
Cc: r-help@r-project.org 
Subject: Re: [R] Need help using lattice
Ha David,
Thanks for your reply.
For your last question, you have to change month into an ordered factor
variable:

library(lattice)
df <- expand.grid(site=c("een","twee","drie","vier","vijf"),
   month=factor(3:12,levels=1:12,
    labels=month.abb, ordered=TRUE))
df$conc <- rnorm(dim(df)[1])
dotplot(month~conc|site,data=df)

HTH, Gerrit



Op 08-02-2021 om 21:04 schreef Parkhurst, David F.:

That worked nicely;  thanks again.  Here’s what I used for our data:

library(lattice)

df <-
expand.grid(site=c("CrCr","NFSC","MFSC","SFSC","LMO","MCE","MUE","MLE","MoCe"),months=1:9)

dotplot(months~conc | site, data=df)

The result is attached.  Is there a way to replace the month numbers on
the y axes with the names Apr-Dec?

David

library(lattice)
# month is defined as a factor with 12 levels, of which only levels 4 to 12
# are present in the data. 
# Note that month.abb is a predefined constant in R.
# Here it provides the labels for all 12 levels. 
df <- expand.grid(site=c("CrCr","NFSC","MFSC","SFSC","LMO","MCE","MUE","MLE"),
  month=factor(4:12,levels=1:12,
   labels=month.abb, ordered=TRUE))
df$conc <- rnorm(dim(df)[1])
# introduce some missing values
i <- sample(1:dim(df)[1],6)
df$conc[i] <- NA
# so there some missing points in the plot
dotplot(month~conc|site,data=df)


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Re: [R] Need help using lattice

[Gerrit Draisma]

Ha David,
Thanks for your reply.
For your last question, you have to change month into an ordered factor 
variable:


library(lattice)
df <- expand.grid(site=c("een","twee","drie","vier","vijf"),
 month=factor(3:12,levels=1:12,
  labels=month.abb, ordered=TRUE))
df$conc <- rnorm(dim(df)[1])
dotplot(month~conc|site,data=df)

HTH, Gerrit



Op 08-02-2021 om 21:04 schreef Parkhurst, David F.:

That worked nicely;  thanks again.  Here’s what I used for our data:

library(lattice)

df <- 
expand.grid(site=c("CrCr","NFSC","MFSC","SFSC","LMO","MCE","MUE","MLE","MoCe"),months=1:9)


dotplot(months~conc | site, data=df)

The result is attached.  Is there a way to replace the month numbers on 
the y axes with the names Apr-Dec?


David



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[R] Need help using lattice

[Gerrit Draisma]

David,
Is this what you are looking for?
library(lattice)
df <- expand.grid(site=c("een","twee","drie","vier","vijf"),
 month=1:12)
df$conc <- rnorm(dim(df)[1])
dotplot(month~conc|site,data=df)

HTH Gerrit

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[R] turning a list of objects into columns

[Gerrit Draisma]

Ha Janet,
Does this reshape do what you want:
df = read.table(header=TRUE,text="
scen streamflowtrans evap  psn
1   0.019234 1.658967 0.002883 0.002391
1   0.019027 1.661192 0.002844 0.003142
2   0.018821 1.695623 0.003192 0.002167
2   0.018619 1.503481 0.002536 0.003059
3   0.018425 0.08 1.880355 0.002592
3   0.018369 0.100551 2.225793 0.006642
")
df$t <- 1:2
# reshape(df,idvar="scen",timevar="t",direction="wide")
reshape(df,idvar="t",timevar="scen",direction="wide")

HTH, Gerrit

> Date: Sat, 29 Jun 2019 12:45:46 -0700
> From: Jeff Newmiller 
>
> Maybe look at reshape2? e.g. https://seananderson.ca/2013/10/19/reshape/
>
> There is a very capable base R reshape function, but I find it quite
> challenging to apply.
>
> Note that you normally want to retain one or more columns to identify
> each row in your result... but I am not sure I see anything suitable
> in your example data. You may have removed it, or it may have been
> implied and you may need to create something that will do the job.
>
> e.g. https://www.studytonight.com/dbms/database-key.php
>
> On June 29, 2019 11:27:40 AM PDT, Janet Choate  wrote:
> > Hi,
> > I have a data frame that looks something like this (but much longer):
> > df
> > scen streamflowtrans evap  psn
> > 1   0.019234 1.658967 0.002883 0.002391
> > 1   0.019027 1.661192 0.002844 0.003142
> > 2   0.018821 1.695623 0.003192 0.002167
> > 2   0.018619 1.503481 0.002536 0.003059
> > 3   0.018425 0.08 1.880355 0.002592
> > 3   0.018369 0.100551 2.225793 0.006642
> >
> > i want to end up with something like this (for each variable -
> > streamflow,
> > trans, evap, psn). using the variable trans here as an example:
> > trans1trans2  trans3
> > 1.658967  1.695623  0.08
> > 1.661192  1.503481  0.100551
> >
> > so that each variable (streamflow, trans, evap, psn) is in a separate
> > column based on the scen #.
>

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Re: [R] Problem with Example with Lattice

[Gerrit Draisma]

Hallo Lorenzo,
In addition to Berts advice
try
> trellis.par.get("superpose.symbol")
which shows the default for superposed graphics
and set them with the par.settings parameter
> dotplot(VADeaths, type = "o",
+ auto.key = list(lines = TRUE, space = "right"),
+ main = "Death Rates in Virginia - 1940",
+ par.settings=list(superpose.symbol=list(pch=1:4)),
+ xlab = "Rate (per 1000)")

Gerrit

On 05/28/2017 12:00 PM, r-help-requ...@r-project.org wrote:

Re: Problem with Example with Lattice


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Re: [R] Multiple Histograms in R

[Gerrit Draisma]

Prateek,
Does lattice do what you want?
HTH Gerrit

ppdat<-read.table(text="mou_mean,totalmrc_mean,rev_range,mou_range,Churn
23,24,25,27,1
45,46,47,49,1
43,44,45,47,1
45,46,47,49,0
56,57,58,60,0
67,68,69,71,1
67,68,69,71,0
44,45,46,48,1
33,34,35,37,0
90,91,92,94,1
87,88,89,91,1
76,77,78,80,1
33,34,35,37,1
44,45,46,48,1",
sep=",",header=TRUE)
library(lattice)
ppdat <- reshape(ppdat,times=names(ppdat)[1:4],
 varying=list(names(ppdat)[1:4]),
 direction="long",
 timevar="Characteristic",
 v.name="Value")
histogram(Churn~Value|Characteristic,nint=10,data=ppdat)



 Message: 1
Date: Thu, 20 Apr 2017 16:54:35 +0530
From: prateek pande 
To: Hasan Diwan 
Cc: r-help mailing list 
Subject: Re: [R] Multiple Histograms in R
Message-ID:

Content-Type: text/plain; charset="UTF-8"

HI Hasan,

Thanks for sharing the solution. Really appreciate it.

But i was reading somewhere that we cannot use par with ggplot 2 . we can
only use grid extra to have multiple plots in a single view.

Is it right?

Regards
Prateek
Message: 2
Date: Thu, 20 Apr 2017 12:26:47 +
From: Ulrik Stervbo 
To: prateek pande , Hasan Diwan

Cc: r-help mailing list 
Subject: Re: [R] Multiple Histograms in R
Message-ID:

Content-Type: text/plain; charset="UTF-8"

Hi Prateek,

maybe facet_* with ggplot is what you are looking for

HTH
Ulrik



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Re: [R] Scaling x axis

[Gerrit Draisma]

Ha Fabio,
With lattice' xyplot you can do
-
library(lattice)
x<-as.Date(rnorm(10)*10,origin="2016-1-1")
y<-5+rnorm(10)
xyplot(y~x,type="h",scales=list(x=list(at=x,rot=90)))
-

And yes, labels may overlap, even with rotation.

Gerrit.


Message: 5
Date: Thu, 25 Feb 2016 12:31:00 +
From: Fabio Monteiro 
To: r-help@r-project.org
Subject: [R] Scaling x axis
Message-ID:

[R] attribute and main value

[Gerrit Draisma]

Just a curiosity question:

In the documentation for the nlm procedure
a find this example of defining a function
with a gradient attribute:
---
 f - function(x, a)
 {
 res - sum((x-a)^2)
 attr(res, gradient) - 2*(x-a)
 res
 }
---
I get the gradient with
 attr(f(3,2),gradient)
but how do I get the function value it self?

Gerrit

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Re: [R] attribute and main value

[Gerrit Draisma]

Thanks Duncan.

But my question was how to extract
simply the function value from value,
without the gradient attribute?

I see that things like value2 give the right answer.
I was curiosity. I found now that value[1]
gives strips the attributes from value:
--
 value
[1] 1
attr(,gradient)
[1] 2
 value[1]
[1] 1
--
Is that the way?

Gerrit


On 12/29/2014 05:05 PM, Duncan Murdoch wrote:

On 29/12/2014 10:32 AM, Gerrit Draisma wrote:

Just a curiosity question:

In the documentation for the nlm procedure
a find this example of defining a function
with a gradient attribute:
---
   f - function(x, a)
   {
   res - sum((x-a)^2)
   attr(res, gradient) - 2*(x-a)
   res
   }
---
I get the gradient with
   attr(f(3,2),gradient)
but how do I get the function value it self?


value - f(3,2)
gradient - attr(value, gradient)

Duncan Murdoch



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Re: [R] attribute and main value

[Gerrit Draisma]

Thanks Bill,
That is what I was looking for.
Gerrit

On 12/29/2014 05:53 PM, William Dunlap wrote:

as.vector(x) will return x without any attributes and
structure(x, attrA=NULL, attrB=NULL) will return x
without the named attributes.
 z - f(1:3, 4)
 z
[1] 14
attr(,gradient)
[1] -6 -4 -2
 as.vector(z)
[1] 14
 structure(z, gradient=NULL)
[1] 14

as.vector is a generic function, so for certain classes
(e.g., factor) it may do more than just strip attributes.
Matrix dimensions are attributes and will be removed by
as.vector.



Bill Dunlap
TIBCO Software
wdunlap tibco.com http://tibco.com

On Mon, Dec 29, 2014 at 8:17 AM, Gerrit Draisma gdrai...@xs4all.nl
mailto:gdrai...@xs4all.nl wrote:

Thanks Duncan.

But my question was how to extract
simply the function value from value,
without the gradient attribute?

I see that things like value2 give the right answer.
I was curiosity. I found now that value[1]
gives strips the attributes from value:
--
  value
[1] 1
attr(,gradient)
[1] 2
  value[1]
[1] 1
--
Is that the way?

Gerrit



On 12/29/2014 05:05 PM, Duncan Murdoch wrote:

On 29/12/2014 10:32 AM, Gerrit Draisma wrote:

Just a curiosity question:

In the documentation for the nlm procedure
a find this example of defining a function
with a gradient attribute:
---
f - function(x, a)
{
res - sum((x-a)^2)
attr(res, gradient) - 2*(x-a)
res
}
---
I get the gradient with
attr(f(3,2),gradient)
but how do I get the function value it self?


value - f(3,2)
gradient - attr(value, gradient)

Duncan Murdoch



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http://www.R-project.org/posting-guide.html
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Re: [R] lattice question: removing strips

[Gerrit Draisma]

Dear Martin,
Would
-
library(latticeExtra)
useOuterStrips(p)
-
be useful for you?
Gerrit.


--

Message: 1
Date: Mon, 10 Feb 2014 14:12:45 +0200 (EET)
From: Martin Ivanov tra...@abv.bg
To: Pascal Oettli kri...@ymail.com
Cc: r-help@r-project.org
Subject: Re: [R] lattice question: removing strips
Message-ID:
999135689.229399.1392034366300.javamail.apa...@mail21.abv.bg
Content-Type: text/plain; charset=UTF-8


Dear Pascal,
Thank You very much for Your reply. Here is a minimal working example:

library(lattice);
# this is with both strips plotted:
data - data.frame(x=1:5, y=6:10, f1=c(a1, a1, a1, a2, a3), f2=c(b1, b2, b2, 
b1, b2));
p - xyplot(y ~ x | f1 + f2, data=data)
print(p);

# and this is my attempt to only plot the strips for the second conditioning 
variable, i.e. for f2:
my.strip - function(which.given, strip.levels, ...) {
  if(which.given == 1) {
return(FALSE);
  }
  else {
strip.default(which.given=which.given, strip.levels=c(TRUE, FALSE), ...);
  }
}

p1 - xyplot(y ~ x | f1 + f2, data=data, strip=my.strip)
print(p1);

#As You can see, here I do not get the strips for f1 potted, but there is an 
empty space below
the strips for f1, which I cannot get rid of. I only want to show the strips 
for f2. Directly
below them I want to see the panels themselves, instead of an empty space for 
the missing f1 strips.

Any suggestions?

Best regards,

Martin


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Re: [R] eliminating white space in lattice plot

[Gerrit Draisma]

Thanks Duncan,
I will have a look at the print command.
Gerrit.

op 1/18/2014 2:23 PM Duncan Mackay schreef:

I forgot to mention for multiple plots

? print.trellis

Duncan

-Original Message-
From: Gerrit Draisma [mailto:gdrai...@xs4all.nl]
Sent: Saturday, 18 January 2014 21:49
To: arun; Duncan Mackay
Subject: Re: [R] eliminating white space in lattice plot

Ha Arun and Duncan,
Thanks for your answers.
They helped to solve my problem.
I found that the width and height parameters can be specified directly in
the pdf() command
source file, as =
pdf(wideplot.pdf,width=10,height=3)
xyplot(y~x,data=X,type=l,aspect=0.25)
dev.off()
=
or more useful for me in the R-code chunk in a Sweave
=
fig=TRUE,echo=FALSE,height=10,width=7=
xyplot(y~x,data=X,type=l,aspect=0.25)
@
=
Have a good day. Mine is already.
Gerrit.


op 1/17/2014 5:37 PM arun schreef:
   Hi,
   May be this helps:
 x11(width=10,height=3)
 xyplot(y~x,data=X,type=l,aspect=0.25)
 dev.copy2pdf(file=wideplot.pdf)
   dev.off()
   A.K.
  
  
  
  
   On Friday, January 17, 2014 11:26 AM, Gerrit Draisma gdrai...@xs4all.nl
wrote:
   Dear R-users,
   How do I get rid of the white space above and under   a plot made with
lattice xyplot?
  
   I searched the documentation but could not find how to   do it.
  
   Standard lattice plots are square,
   but sometimes I want a wide but low graph   and use aspect parameter
to obtain such a graph.
   But then the plot has a lot of white space above and below   the
graph.
   That makes it more difficult to include in a LaTeX document.
  
   Thanks,
   Gerrit Draisma
  
   =
   library(lattice)
   X-data.frame(x=1:100,y=runif(100))
   # pdf(wideplot.pdf)
   xyplot(y~x,data=X,type=l,aspect=0.25)
   # dev.off()
  
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[R] eliminating white space in lattice plot

[Gerrit Draisma]

Dear R-users,
How do I get rid of the white space above and under
a plot made with lattice xyplot?

I searched the documentation but could not find how to
do it.

Standard lattice plots are square,
but sometimes I want a wide but low graph
and use aspect parameter to obtain such a graph.
But then the plot has a lot of white space above and below
the graph.
That makes it more difficult to include in a LaTeX document.

Thanks,
Gerrit Draisma

=
library(lattice)
X-data.frame(x=1:100,y=runif(100))
# pdf(wideplot.pdf)
xyplot(y~x,data=X,type=l,aspect=0.25)
# dev.off()

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[R] Lattice: how to change the canvas size with Sweave

[Gerrit Draisma]

Dear all,
I would like to include a wide graph with narrow height in my LaTeX 
output. Up to now I only get it done using Stangle and including

  pdf(pietje-001.pdf,width=7,height=4)
  xyplot(...)
  dev.off()
What is the correct way of setting the canvas size?
Thanks.
Gerrit Draisma

==pietje.tex
\documentclass{article}
\title{How to define the size of the plotting canvas?}
\author{Gerrit Draisma}
\begin{document}
\maketitle
I would like to make this a wide graph with a small height.
How do I do that?
=
library(lattice)
@
\raggedbottom
\begin{figure}[!h]
\centering
echo=FALSE,fig=TRUE=
xyplot(runif(11)~(0:10),type=l, aspect=0.4,
scales=list(x=list(tick.number=11),y=list(tick.number=3)))
@
\caption{The canvas is too big!}
\end{figure}
\end{document}
==

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Re: [R] Lattice: how to change the canvas size with Sweave

[Gerrit Draisma]

Oops,

I forget to write the I included the lines pdf()..(dev.off()
in the file pietje.R generated by Stangle.
And now I see that included pietje.tex in reality was named pietje.rnw
with pietje.tex the result from Sweave.

Apologies for this confusion
Gerrit.


op 11/17/2013 12:43 PM Gerrit Draisma schreef:

Dear all,
I would like to include a wide graph with narrow height in my LaTeX
output. Up to now I only get it done using Stangle and including
   pdf(pietje-001.pdf,width=7,height=4)
   xyplot(...)
   dev.off()
What is the correct way of setting the canvas size?
Thanks.
Gerrit Draisma

==pietje.tex
\documentclass{article}
\title{How to define the size of the plotting canvas?}
\author{Gerrit Draisma}
\begin{document}
\maketitle
I would like to make this a wide graph with a small height.
How do I do that?
=
library(lattice)
@
\raggedbottom
\begin{figure}[!h]
\centering
echo=FALSE,fig=TRUE=
xyplot(runif(11)~(0:10),type=l, aspect=0.4,
scales=list(x=list(tick.number=11),y=list(tick.number=3)))
@
\caption{The canvas is too big!}
\end{figure}
\end{document}
==


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Re: [R] how to combine apply and which or alternative ways to, do so

[Gerrit Draisma]

Ha Guillaume,
Is this what you are looking for?

#---
x1-sample(1:10,5,replace=FALSE)
x2-sample(1:10,100,replace=TRUE)
#This
tapply(x1,1:5,FUN=function(i){which(x2==i)})
#or this
tapply(x1,x1,FUN=function(i){which(x2==i)})
#---

Gerrit

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Re: [R] integrate / Vectorize question

[Gerrit Draisma]

Dag Rui,
Yes, it helped a lot!
Thanks and greetings,
Gerrit.

Op 10/4/2012 12:06 AM, Rui Barradas schreef:

Hello,

Try


Fnum - function(x, beta){
sapply(x, function(x) integrate(f,lower=-x,upper=0,beta=beta)$value)
}


And all results were the same.

Hope this helps,

Rui Barradas
Em 03-10-2012 21:59, Gerrit Draisma escreveu:

Dear R-users,

I want to use the function Fnum below
in another integrate procedure.

How do I write Fnum so that it returns a vector of values?
And will the last integrate return the right answer?

(I tried Vectorize() in several places,
but got all sorts of errors.)

Thanks for helping out.

Gerrit.


This my code
=
beta=c(3,0.3)*1E-2
mu-5

f-function(x,beta){beta[1]+beta[2]*x}

Fnum-function(x,beta){
   integrate(f,lower=-x,upper=0,beta=beta)$value
   }

Fsym-function(x,beta){x*(beta[1]-beta[2]/2*x)}

y-1:4
Fsym(y,beta)
Fnum(y,beta)

g-function(y,mu){exp(-y/mu)/mu}

integrate((function(y,beta,mu){Fsym(y,beta)*g(y,mu)}),lower=0,upper=2*mu,
  beta=beta,mu=mu)

#This fails
integrate((function(y,beta,mu){Fnum(y,beta)*g(y,mu)}),lower=0,upper=2*mu,
  beta=beta,mu=mu)

t-seq(0,2*mu, by=0.2)
0.2*sum(tapply(t,t,function(y,beta,mu){Fnum(y,beta)*g(y,mu)},
   beta=beta,mu=mu))

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[R] integrate / Vectorize question

[Gerrit Draisma]

Dear R-users,

I want to use the function Fnum below
in another integrate procedure.

How do I write Fnum so that it returns a vector of values?
And will the last integrate return the right answer?

(I tried Vectorize() in several places,
but got all sorts of errors.)

Thanks for helping out.

Gerrit.


This my code
=
beta=c(3,0.3)*1E-2
mu-5

f-function(x,beta){beta[1]+beta[2]*x}

Fnum-function(x,beta){
   integrate(f,lower=-x,upper=0,beta=beta)$value
   }

Fsym-function(x,beta){x*(beta[1]-beta[2]/2*x)}

y-1:4
Fsym(y,beta)
Fnum(y,beta)

g-function(y,mu){exp(-y/mu)/mu}

integrate((function(y,beta,mu){Fsym(y,beta)*g(y,mu)}),lower=0,upper=2*mu,
  beta=beta,mu=mu)

#This fails
integrate((function(y,beta,mu){Fnum(y,beta)*g(y,mu)}),lower=0,upper=2*mu,
  beta=beta,mu=mu)

t-seq(0,2*mu, by=0.2)
0.2*sum(tapply(t,t,function(y,beta,mu){Fnum(y,beta)*g(y,mu)},
   beta=beta,mu=mu))

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Re: [R] how to include integrate in a function that can be solved with uniroot?

[Gerrit Draisma]

Thanks Michael,
Op 11/14/2011 3:30 PM, R. Michael Weylandt schreef:

You need to explicitly pass th to your function with the ... argument
of integrate.

That was a point I was missing!
Thanks again,
This solved my problems for this time.
Gerrit.




E- function(th){
 integrate(function(x,th) x*g(x, th), 0, Inf, th)$value
}

Also, it's value, not Value, which might be producing errors of another sort.

Michael

On Mon, Nov 14, 2011 at 9:16 AM, Gerrit Draismagdrai...@xs4all.nl  wrote:

Thanks Michael,

I see now how to include integrate function in the EV function.
And apologies:
I now realize that my code was sloppy.
I intended to write

E- function(th) {
+  integrate( f = function(x,th){x*g(x,th)},
+  0,Inf)$Value}
E(1/10)

But that does not work either,

Gerrit.

Op 11/14/2011 2:50 PM, R. Michael Weylandt schreef:


Try this:

EV- function(lamb){
  fnc- function(x) x * dexp(x, lamb)
  integrate(fnc, 0, Inf)$value
}

Your problem is that there's nothing to translate th to lambda in your
code for E.

Michael

On Mon, Nov 14, 2011 at 5:32 AM, Gerrit Draismagdrai...@xs4all.nl
  wrote:


Hallo,
I am trying to define expectation as an integral
and use uniroot to find the distribution parameter
for a given expectation.

However I fail to understand how to define properly
the functions involved and pass the parameters correctly.

Can anyone help me out?

Thanks,
Gerrit Draisma.


This what I tried:
===


# exponential density
g- function(x,lambda){ lambda *exp(-lambda*x) }

# expectation with lambda=1/10
integrate(f = function(x,lambda=1/10) {x*g(x,lambda)}, 0,Inf)


10 with absolute error6.7e-05


# *how to write this as a function?*
E- function(lambda) {


+  integrate( f = function(x,th){x*g(x,lambda)},
+  0,Inf)$Value}


E(1/10)


NULL


# *how to include this function in uniroot to find lambda*
# *for a given expectation?*
mu- 10
uniroot(f-function(th){E(th)-mu},lower=1,upper=100)


Error in if (is.na(f.lower)) stop(f.lower = f(lower) is NA) :
  argument is of length zero





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and provide commented, minimal, self-contained, reproducible code.

[R] how to include integrate in a function that can be solved with uniroot?

[Gerrit Draisma]

Hallo,
I am trying to define expectation as an integral
and use uniroot to find the distribution parameter
for a given expectation.

However I fail to understand how to define properly
the functions involved and pass the parameters correctly.

Can anyone help me out?

Thanks,
Gerrit Draisma.


This what I tried:
===
 # exponential density
 g - function(x,lambda){ lambda *exp(-lambda*x) }

 # expectation with lambda=1/10
 integrate(f = function(x,lambda=1/10) {x*g(x,lambda)}, 0,Inf)
10 with absolute error  6.7e-05

 # *how to write this as a function?*
 E - function(lambda) {
+  integrate( f = function(x,th){x*g(x,lambda)},
+  0,Inf)$Value}
 E(1/10)
NULL

 # *how to include this function in uniroot to find lambda*
 # *for a given expectation?*
 mu - 10
 uniroot(f-function(th){E(th)-mu},lower=1,upper=100)
Error in if (is.na(f.lower)) stop(f.lower = f(lower) is NA) :
  argument is of length zero



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Re: [R] understanding output of tapply/by cumsum

[Gerrit Draisma]

Thanks Jim,
Ave does what I wanted.
It is simpler and probably more efficient
than unlisting Sn as I tried.

Still I remain puzzled with the structure
of the by() or tapply() output and how
to access the individual cumsums.

Yes the split command is useful for checking
the result.
Gerrit.


Op 12/7/2010 1:43 PM, jim holtman schreef:

Maybe 'ave' is what you were looking for:


d$cum- ave(d$n, d$a, d$c, FUN = cumsum)
d

a b cn   cum
1  1 1 1 11.1  11.1
2  2 1 1 21.1  21.1
3  3 1 1 31.1  31.1
4  4 1 1 41.1  41.1
5  5 1 1 51.1  51.1
6  1 2 1 12.1  23.2
7  2 2 1 22.1  43.2
8  3 2 1 32.1  63.2
9  4 2 1 42.1  83.2
10 5 2 1 52.1 103.2
11 1 3 1 13.1  36.3
12 2 3 1 23.1  66.3
13 3 3 1 33.1  96.3
14 4 3 1 43.1 126.3
15 5 3 1 53.1 156.3
16 1 1 2 11.2  11.2
17 2 1 2 21.2  21.2
18 3 1 2 31.2  31.2
19 4 1 2 41.2  41.2
20 5 1 2 51.2  51.2
21 1 2 2 12.2  23.4
22 2 2 2 22.2  43.4
23 3 2 2 32.2  63.4
24 4 2 2 42.2  83.4
25 5 2 2 52.2 103.4
26 1 3 2 13.2  36.6
27 2 3 2 23.2  66.6
28 3 3 2 33.2  96.6
29 4 3 2 43.2 126.6
30 5 3 2 53.2 156.6





On Tue, Dec 7, 2010 at 6:39 AM, Gerrit Draismagdrai...@xs4all.nl  wrote:

Dear R-users,

I have a dataset with categories and numbers.
I would like to compute and add cumulative numbers
to the dataset.
I do not understand the structure of by(...) or
tapply(...) output enough to handle it.

Here a small example
--
d-expand.grid(a=1:5,b=1:3,c=1:2)
d$n = 10 * d$a + d$b +0.1* d$c
Sn-by(d$n,list(d$a,d$c),cumsum)
str(Sn)
-
List of 10
  $ : num [1:3] 11.1 23.2 36.3
  $ : num [1:3] 21.1 43.2 66.3
  $ : num [1:3] 31.1 63.2 96.3
  $ : num [1:3]  41.1  83.2 126.3
  $ : num [1:3]  51.1 103.2 156.3
  $ : num [1:3] 11.2 23.4 36.6
  $ : num [1:3] 21.2 43.4 66.6
  $ : num [1:3] 31.2 63.4 96.6
  $ : num [1:3]  41.2  83.4 126.6
  $ : num [1:3]  51.2 103.4 156.6
  - attr(*, dim)= int [1:2] 5 2
  - attr(*, dimnames)=List of 2
  ..$ : chr [1:5] 1 2 3 4 ...
  ..$ : chr [1:2] 1 2
  - attr(*, call)= language by.default(data = d$n, INDICES = list(d$a,
d$c), FUN = cumsum)
  - attr(*, class)= chr by
-
# these give (a) lists of one numerical vector(a)
Sn[5,2]
Sn[cbind(d$a,d$c)]
# how to access the individual cumsum values?
# and assign them to d$Sn?
--

Thanks,
Gerrit.

---
Gerrit Draisma
Department of Public Health
Erasmus MC, University Medical Center Rotterdam
Room AE-235
P.O. Box 2040 3000 CA  Rotterdam The Netherlands
Phone: +31 10 7043787 Fax: +31 10 7038474
http://mgzlx4.erasmusmc.nl/pwp/?gdraisma

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and provide commented, minimal, self-contained, reproducible code.

[R] understanding output of tapply/by cumsum

[Gerrit Draisma]

Dear R-users,

I have a dataset with categories and numbers.
I would like to compute and add cumulative numbers
to the dataset.
I do not understand the structure of by(...) or
tapply(...) output enough to handle it.

Here a small example
--
d-expand.grid(a=1:5,b=1:3,c=1:2)
d$n = 10 * d$a + d$b +0.1* d$c
Sn-by(d$n,list(d$a,d$c),cumsum)
str(Sn)
-
List of 10
 $ : num [1:3] 11.1 23.2 36.3
 $ : num [1:3] 21.1 43.2 66.3
 $ : num [1:3] 31.1 63.2 96.3
 $ : num [1:3]  41.1  83.2 126.3
 $ : num [1:3]  51.1 103.2 156.3
 $ : num [1:3] 11.2 23.4 36.6
 $ : num [1:3] 21.2 43.4 66.6
 $ : num [1:3] 31.2 63.4 96.6
 $ : num [1:3]  41.2  83.4 126.6
 $ : num [1:3]  51.2 103.4 156.6
 - attr(*, dim)= int [1:2] 5 2
 - attr(*, dimnames)=List of 2
  ..$ : chr [1:5] 1 2 3 4 ...
  ..$ : chr [1:2] 1 2
 - attr(*, call)= language by.default(data = d$n, INDICES = list(d$a, 
d$c), FUN = cumsum)

 - attr(*, class)= chr by
-
# these give (a) lists of one numerical vector(a)
Sn[5,2]
Sn[cbind(d$a,d$c)]
# how to access the individual cumsum values?
# and assign them to d$Sn?
--

Thanks,
Gerrit.

---
Gerrit Draisma
Department of Public Health
Erasmus MC, University Medical Center Rotterdam
Room AE-235
P.O. Box 2040 3000 CA  Rotterdam The Netherlands
Phone: +31 10 7043787 Fax: +31 10 7038474
http://mgzlx4.erasmusmc.nl/pwp/?gdraisma

__
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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.

[R] combined plot of observed and expected fractions

[Gerrit Draisma]

Dear R-users,
I have a dataset of numbers of cases with a certain age
and stage.

I want to plot the observed stage distribution by age
and compare it to an expected or predicted one.
As I want to plot these for several populations
I want to use lattice plots.

Below is example code I made that produces a plot
that satisfies me. I prefer this to using stacked bars.
But I would like to improve on the code.

Specifically:
Question 1: How to get cumulative fractions from
observed numbers?

Question 2: How to divide each number by the
total number in each age group?

Question 3:  How to construct an indicator
for combined categories (here:
observed/expected x Stage)

Question 4: Lattice: could I use data values
to locate the text?

I would appreciate any comments.
Thank you,
Gerrit.

=
# stagexage.r
# compare observed and predicted stage by age.

library(lattice)
# simple data set
df-data.frame(Age=rep(1:5,each=3),Stage=1:3)
# expectation
df$mu-5+(df$Age-1)*df$Stage

# simulated data
df$N-rpois(15,df$mu)

# for lattice xyplot
df-reshape(df,direction=long,varying=c(N,mu),timevar=OE,v.name=N)

# cumulate numbers over Age group and OE (Observed, Expected)
# Q1: how to aply cumsum by age and stage?
i-df$Age+5*(df$OE-1)
P-unlist(tapply(df$N,i,cumsum))

# compute probabilities
# Q2: How to divide by the total number by age and stage?
j-3*(0:29%/%3) + 3
P-P/P[j]

# combining OE and Stage in a single index
# for superposing graphs in one plot
# Q3: How to create an index from Stage and number (obs or exp)?
ix-(df$Stage-1)*2+df$OE

# plot observed and predicted fractions in a area plot
j-df$Stage!=3
xyplot(P[j]~Age,groups=ix[j],data=df[j,],type=o,
panel=function(x,y,...){
   panel.xyplot(x,y,...)
# Q4: would it be possible to position labels based
# data values?
   panel.text(x=3,y=c(0.15,0.5,0.85),labels=paste(Stage, 1:3))
   },
scales=list(y=list(at=0:4/4)),
axs=i,ylim=c(0,1),
xlab=Age, ylab=Fraction,
par.settings=list(superpose.line=list(col=c(blue,red),lty=1:2),
superpose.symbol=list(col=c(blue,red),type=1)),
auto.key=list(text=c(Obs,Pred),
   points=F,lines=T,type=o,divide=1,columns=2)
)

=

--
Gerrit Draisma
Department of Public Health
Erasmus MC, University Medical Center Rotterdam
Room AE-235
P.O. Box 2040 3000 CA  Rotterdam The Netherlands
Phone: +31 10 7043787 Fax: +31 10 7038474
http://mgzlx4.erasmusmc.nl/pwp/?gdraisma

__
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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.

[R] printing a dataframe by categories

[Gerrit Draisma]

I am looking for the following simple question.
I have a data frame with names and numbers, divided in categories.
I would like to produce a text file with page breaks,
listing the names and numbers by category,
and totalling the numbers.

Example:

  Name-LETTERS[1:6]
  Score-rep(5:8,length.out=6)
  Form-rep(1:2,each=3)
  x-data.frame(Name,Score,Form)

This gives approximately what I need
  by(x,x$Form,function(x){x[,1:2]})
but I would like a total added to the list.

Gerrit.

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and provide commented, minimal, self-contained, reproducible code.

Re: [R] latex.table for table with character and numeric columns

[Gerrit Draisma]

Ha Charlie,
It helped.
It is just what i was looking for.
Gerrit.


Gerrit Draisma wrote:


Hallo,
I have a dataset with one or two columns with character data
and the rest with numeric data.
Using latex.table from the quantreg package produced a table,
but I cannot set the decimals.
For instance:
---
  x-data.frame(Name=c(Jan,Piet,Jan), V=c(1,2.991,3))
  latex.table(as.matrix(x),file=x,caption=x)
  latex.table(as.matrix(x),file=x,caption=x,dec=2)
Error in round(x, dec) : Non-numeric argument to mathematical function
 
---
Am I not using the right command, or is there a way around?
Thanks,
Gerrit.




Hi Gerrit,

I haven't used latex.table from quantreg so I can't comment on this
approach. However, I would recommend using the xtable package to format your
data.frame into LaTeX:

  require( xtable )

  x-data.frame(Name=c(Jan,Piet,Jan), V=c(1,2.991,3))

  # Create a table object from the data.frame
  latexTable - xtable( x )

  # Set the number of digits in each column. Your example has two columns,
  # but we must specify three digits since the rownames count as a column.
  digits( latexTable ) - c( 0, 0, 2 )

  # Print the table object in order to produce the LaTeX code.
  print( latexTable )

You should also read the help page for print.xtable() as it discusses many
options that can affect the final LaTeX output.

Hope this helps!

-Charlie


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and provide commented, minimal, self-contained, reproducible code.

[R] latex.table for table with character and numeric columns

[Gerrit Draisma]

Hallo,
I have a dataset with one or two columns with character data
and the rest with numeric data.
Using latex.table from the quantreg package produced a table,
but I cannot set the decimals.
For instance:
---
 x-data.frame(Name=c(Jan,Piet,Jan), V=c(1,2.991,3))
 latex.table(as.matrix(x),file=x,caption=x)
 latex.table(as.matrix(x),file=x,caption=x,dec=2)
Error in round(x, dec) : Non-numeric argument to mathematical function

---
Am I not using the right command, or is there a way around?
Thanks,
Gerrit.

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and provide commented, minimal, self-contained, reproducible code.

[R] correct line types in lattice legends

[Gerrit Draisma]

Hallo R-users,
I do not understand how to specify the correct
line and symbol types in the legends of a lattice xyplot.
This is what I tried, but the line types and symbol  in the
graph are not seen in the legend.
Any help is appreciated.

Thanks,
Gerrit.

library(lattice)
s-rep(1:3,len=10)
x- 1:10
y- x+s+rnorm(10)
d-data.frame(s,x,y)
xyplot(y~x, groups=s,data=d, type=b, pch=1:3,lty=1:3, 
auto.key=list(text=c(A,B,C),points=FALSE,lines=TRUE,

 type=b,lty=1:3,divide=1,pch=1:3,columns=3))

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[R] Vectorize tapply(...,cumsum) output

[Gerrit Draisma]

I would like to graph cumulative counts
and use tapply(...,cumsum) to get
the cumulative counts.
For instance:
library(lattice)
t-rep(1:4,each=3)#time
i-rep(1:3,times=4)   #categories
n-rpois(length(t),t) #count
xyplot(n ~ t,groups=i,
type=o,key=simpleKey(as.character(1:3),x=0,y=1,lines=T))
N-tapply(n,t,cumsum)

Now, what is the best way to convert N
to a vector that can be plotted?
E.g.
xyplot(N ~ t,groups=i,
type=o,key=simpleKey(as.character(1:3),x=0,y=1,lines=T))

Or is there a better way to get cumulative graphs?

Thanks,
Gerrit.

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and provide commented, minimal, self-contained, reproducible code.