Re: [R] Assign name to object for each iteration in a loop.

[Patrick Connolly]

On Thu, 01-Dec-2011 at 10:13AM -0800, lglew wrote:

|> Hi R-users,
|> 
|> I'm trying to produce decompositions of a multiple time-series, grouped by a
|> factor (called "area"). I'm modifying the code in the STLperArea function of
|> package ndvits, as this function only plots produces stl plots, it does not
|> return the underlying data. 
|> 
|> I want to extract the trend component of each decomposition
|> ("x$time.series[,trend]), assign a name based on the factor "area". 
|> 
|> My input data look like this:
|> Area is a factor, with three (but could be many more) levels.
|> area
|> 1
|> 2
|> 3
|> 
|> Ystart=2000
|> 
|> TS is a timeseries:
|> 
|>   X249   X265  X281  X297  X2000113 
|> 1 0.2080  0.2165  0.2149 0.2314  0.2028  
|> 2 0.1578  0.1671  0.1577 0.1593  0.1672   
|> 3 0.1897  0.1948  0.2290 0.2292  0.2067   
|> 
|> Here's the function:
|> 
|> STLpA<-function(TS, area, Ystart, period=23, nSG="5,5", DSG=0)
|> {
|> require (RTisean)
|> for(i in 1:unique(area)){
|> vi.metric=TS[area==i]
|> filt.vi<-sav_gol(vi.metric,n=nSG,D=DSG)
|> vi.sg<-ts(filt.vi[,1], start=Ystart,frequency=period)
|> stld.tmp<-stl(vi.sg, s.window="periodic", robust=TRUE, na.action=na.approx)
|> stld.trend<-stld.temp$time.series[,trend]
|> }
|> assign(paste("stld", i , sep= "."), stld.trend)
|> vi.trend<-ls(pattern= "^stld..$")
|> return(vi.trend)
|> }
|> 
|> When I call this function with signal=STLpA(TS,area,Ystart=2000,period=23,
|> nSG= "5,5", DSG="0"))
|> 
|> I get this error:
|> 
|> Error in cat(list(...), file, sep, fill, labels, append) : 
|>   argument 1 (type 'list') cannot be handled by 'cat'
|> In addition: Warning message:
|> In 1:unique(area) :
|>   numerical expression has 3 elements: only the first used
|> 
|> I'm guessing this is because I'm assigning names to each temporary stl.trend
|> file incorrectly. Can anyone 
|> improve on my rather poor efforts here?

I would suggest putting your individual trends into a list and
returning that at the end of your function rather than trying to
assing individual objects.  IMHO it's easier to do and more useful.
Something along these lines:


> trends <- list()
> for(i in 1:unique(area)){
...
+ trends[[i]] <- 

}

Then you'll have an element in the list trends that you do with what
you will.  It creates less clutter also.

HTH



-- 
~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.   
   ___Patrick Connolly   
 {~._.~}   Great minds discuss ideas
 _( Y )_ Average minds discuss events 
(:_~*~_:)  Small minds discuss people  
 (_)-(_)  . Eleanor Roosevelt
  
~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.

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Re: [R] rnorm command

[Jeff Newmiller]

No.

If you want a more informative answer, try following the recommendations in the 
posting guide mentioned below.
---
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Sent from my phone. Please excuse my brevity.

Cheryl Johnson  wrote:

>Hello,
>
>I use the command rnorm, and I feed these results into a lmer command.
>Since I am using the rnorm command I expect to get different results
>for
>each iteration, yet for each iteration I am getting the same answer. If
>someone understands why I am getting the same answer every time with a
>random number generator, I would appreciate help in understanding why
>this
>is happening. Is the lmer command causing the answer to be the same
>each
>time?
>
>Thanks
>
>   [[alternative HTML version deleted]]
>
>__
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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.

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Re: [R] upper bound in the integrate function depends on a parameter

[David Winsemius]

On Dec 3, 2011, at 10:23 PM, grttt nbbfg wrote:


Sorry for my English, is not my first language..

I have some trouble in terms of using integrate function in R.
fx is a function of m and x where m is supposed to be a unknown  
parameter.


R is not an algebraic solver (and R-help is not a homework list.).




fx=function(m,x){

+ x*2/(3*m)*(1-x/(3*m))
+ }

The problem is in upper bound, it depends on parameter m.


integrate(fx,lower=0,upper=3*m)$value


Is it possible to use the integrate function when bounds depend on a
parameter?


If it is known, yes. You could also do some investigation:

curve(x*2/(3*5)*(1-x/(3*5)), 0, 15)



How to define m to solve this integral? The result should be m.


As I said, no homework answers here.

--
David Winsemius, MD
West Hartford, CT

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Re: [R] density function always evaluating to zero

[David Winsemius]

On Dec 3, 2011, at 5:42 PM, napps22 wrote:


Dear R users,

I'm trying to carry out monte carlo integration of a posterior density
function which is the product of a normal and a gamma distribution.  
The
problem I have is that the density function always returns 0. How  
can I

solve this problem?


Your code throws errors due to several missing objects due in part to  
a missing "v", but moving that higher in the code does not cure  
everything.  A missing 's2_m1' is also listed as a source of error.  
You might want to try in a clean session and redo your debugging so  
that you know what variables are in your workspace and their values.


The usual debugging method is to sprinkle print() statements in your  
code to monitor where things might not be proceeding as planned.


--
David.


Here is my code

#generate data

x1 <- runif(100, min = -10, max = 10)
y <- 2 * x1^2 + rnorm(100)

# # # # # # # # Model 0 # # # # # # #

z <- x1^2
M <- sum(z^2)
MI <- 1/M
zy <- crossprod(z,y)
alpha.ols <- MI * zy
resid_m0 <- y - z * alpha.ols
s2_m0 <- sum(resid_m0^2)/v

# use gibbs sampler to sample from posterior densities

n <- length(y)
k <- 1
v <- n - k

# set up gibbs sampler

nrDraws <- 1
h_sample_m0 <- rgamma(nrDraws, v/2, v*s2_m0/2)
alpha_sample <- matrix(0, nrow = nrDraws, ncol = 1)

for(i in 1:nrDraws) {

alpha_sample[i] <- rnorm(1,alpha.ols,(1/h_sample_m0[i]) * MI)

}

# define posterior density for model 0

f <- function(alpha,h) { 

e <- y - alpha * x1^2
const <- (2*pi)^(-n/2) / ( gamma(v/2) * (v*s2_m1/2)^(-v/2) )
kernel <- h^((v-1)/2) * exp((-(h/2) * sum(e^2)) - (v*s2_m0)*h)
x <- const * kernel
return(x)
}

# calculate approximation of p(y|m_0)

m0 <-f(alpha_sample,h_sample_m0)
post_m0 <- sum(m0) / nrDraws


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David Winsemius, MD
West Hartford, CT

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[R] upper bound in the integrate function depends on a parameter

[grttt nbbfg]

Sorry for my English, is not my first language..

I have some trouble in terms of using integrate function in R.
fx is a function of m and x where m is supposed to be a unknown parameter.

>fx=function(m,x){
+ x*2/(3*m)*(1-x/(3*m))
+ }

The problem is in upper bound, it depends on parameter m.

>integrate(fx,lower=0,upper=3*m)$value

Is it possible to use the integrate function when bounds depend on a
parameter?

How to define m to solve this integral? The result should be m.

[[alternative HTML version deleted]]

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[R] Prediction from censReg?

[z2.0]

Hi - 

First post, so excuse any errors in protocol: 

Wanted to ask if there's an easy way to use 'predict' with objects of class
'censReg', 'maxLik', 'maxim' or 'list'.

Have a left-censored dataset, attempting to use a Tobit model and am working
with the censReg package. I like how easy it is to move from glm models to
predictions with 'predict' and wanted to ask if there was a way to do so
with objects output from 'censReg'.

Thanks,

Zack

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Re: [R] Shading the plot

[avinash barnwal]

Hi,

I apologies  for my naive doubts
I have been trying it from a while. I need the area between the curve based
on conditions i wrote (whenever red line is above the blue line then area
between curves to be grey color and whenever blue line is above red line
then area between curves to be red color )

x and y are vectors

Thank you for replying



On Sun, Dec 4, 2011 at 7:56 AM, David Winsemius wrote:

>
> On Dec 3, 2011, at 5:28 PM, avinash barnwal wrote:
>
>  Hi Weylandt,
>>
>>
>> I tried it but i was not successful.
>>
>> Here is the full code
>>
>> Any help would be great.
>> ##**##**
>> 
>> time<-c("02-Jan-05","09-Jan-**05","16-Jan-05","23-Jan-05","**
>> 30-Jan-05","06-Feb-05","13-**Feb-05","20-Feb-05","27-Feb-**
>> 05","06-Mar-05","13-Mar-05",
>> "20-Mar-05","27-Mar-05","03-**Apr-05","10-Apr-05","17-Apr-**
>> 05","24-Apr-05","01-May-05","**08-May-05","15-May-05")
>> time<-as.Date(time,"%d-%b-%y")
>> a2<-c("81.96392786","81.**96392786","82.16432866","82.**
>> 56513026","82.76553106","79.**96593186","78.16633267",
>> "75.56713427","74.76753507","**72.96793587","70.96793587","**
>> 45.96793587","83.96793587","**84.36873747",
>> "84.56913828","84.56913828","**84.56913828","84.56913828","**
>> 84.36873747","84.36873747")
>> a3<-c("81.96392786","81.**96392786","81.96392786","70.**
>> 96392786","68.16432866","66.**16432866",
>> "62.16432866","58.56513026","**56.56513026","54.56513026","**
>> 48.56513026","49.76553106","**52.76553106",
>> "54.76553106","57.96593186","**59.36673347","83.36673347","**
>> 83.36673347","83.36673347",
>> "83.56713427")
>> j<-0
>> for(i in 1:length(a3))
>> {
>> if(a2[i]>a3[i])
>> {
>> j<-j+1
>> x[j]<-a2[i]
>> y[j]<-a3[i]
>> temp_time[j]<-time[i]
>> }
>> }
>> plot(time,a2,type='l',col='**red',ylab='',xlab=" ",axes=FALSE)
>> lines(time,a3,col='blue',ylab=**'',xlab=" ")
>> polygon(c(temp_time,rev(temp_**time)),c(x,y),col="grey")
>> ##**##**
>> #
>>
>
> What are the rules that govern requesting outside help in homework at your
> institution?
>
> Your current code does not recognize  that there is line crossing. Nor do
> you describe what you mean by shading. If you wanted a rectangle within the
> lpot area it would drive one strategy and if you wanted only the area
> between the curves to be shaded it would drive a different strategy.
>
> AND you never define "x" or "y".
>
> --
> David.
>
>
>
>  On Sun, Dec 4, 2011 at 2:46 AM, R. Michael Weylandt <
>> michael.weyla...@gmail.com> wrote:
>>
>>  ? polygon
>>> example(polygon)
>>>
>>> Michael
>>>
>>> On Sat, Dec 3, 2011 at 4:08 PM, avinash barnwal
>>>  wrote:
>>>
 Hi all,

 I have been trying to shade the specific part of the plot

 Part to be shaded

 1. Any color whenever a2>a3

 2. Any other color( Not same as 1) whenever a2>>>
 Suggest me some code for this task

 PLOT CODE for reference


  ##**##**
>>> ##**
>>>

  a2<-c("81.96392786","81.**96392786","82.16432866","82.**
>>> 56513026","82.76553106","79.**96593186","78.16633267",
>>>

  "75.56713427","74.76753507","**72.96793587","70.96793587","**
>>> 45.96793587","83.96793587","**84.36873747",
>>>

  "84.56913828","84.56913828","**84.56913828","84.56913828","**
>>> 84.36873747","84.36873747")
>>>

  a3<-c("81.96392786","81.**96392786","81.96392786","70.**
>>> 96392786","68.16432866","66.**16432866",
>>>

  "62.16432866","58.56513026","**56.56513026","54.56513026","**
>>> 48.56513026","49.76553106","**52.76553106",
>>>

  "54.76553106","57.96593186","**59.36673347","83.36673347","**
>>> 83.36673347","83.36673347",
>>>
 "83.56713427")
 plot(a2,type='l',col='red',**ylab='',xlab=" ",axes=FALSE)
 lines(a3,col='blue',ylab='',**xlab=" ")

  ##**##**
>>> ##**
>>>

 Thank you in the advance
 --
 Avinash Barnwal
 Final year undergraduate student
 Statistics and informatics
 Department of Mathematics
 IIT Kharagpur

 __**
 R-help@r-project.org mailing list
 https://stat.ethz.ch/mailman/**listinfo/r-help
 PLEASE do read the posting guide

>>> http://www.R-project.org/**posting-guide.html
>>>
 and provide commented, minimal, self-contained, reproducible code.


>>>
>>
>>
>> --
>> Avinash Barnwal
>> Final year undergraduate student
>> Statistics and informatics
>> Department of Mathematics
>> IIT Kharagpur
>>
>>[[alternative HTML version deleted]]
>>
>>
>> __**
>> R-help@r-project.org mailing list

[R] Logistic Regression with genetic component

[Danielle Duncan]

Greetings, I have a question that I'd like to get input on. I have a
classic toxicology study where I artificially fertilized and exposed
embryos to a chemical and counted defects. In addition, I kept track of
male-female pairs that I used to artificially fertilize and generate
embryos with. I need to use logistic regression to model the response, but
also check that the genetics of the pairings did or did not have an effect
on the response. My data looks a bit like this:

response matrix chemical concentration  Genetic Matrix
Present AbsentMale Female
2 152   0.13 a 1
6 121  1 a 2
21 92  2 b 3
24 89  5 b 4
0141 10 c 5
5 95 15 c  6

R code:

DA<-cbind(Present, Absent)
glm<-(DA ~ chemical concentration)

If I do glm<-(DA ~ chemical concentration + Male + Female, I get every
possible combination, but I only want specific pairs. So, I am thinking
about doing:

MF<-cbind(Male, Female)
glm<-(DA ~ chemical concentration + MF)

Is this correct? Any help on how to model this would be greatly
appreciated! Thanks in advance!

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[R] density function always evaluating to zero

[napps22]

Dear R users,

I'm trying to carry out monte carlo integration of a posterior density
function which is the product of a normal and a gamma distribution. The
problem I have is that the density function always returns 0. How can I
solve this problem?

Here is my code

#generate data

x1 <- runif(100, min = -10, max = 10)
y <- 2 * x1^2 + rnorm(100)

# # # # # # # # Model 0 # # # # # # # 

z <- x1^2
M <- sum(z^2)
MI <- 1/M
zy <- crossprod(z,y)
alpha.ols <- MI * zy
resid_m0 <- y - z * alpha.ols
s2_m0 <- sum(resid_m0^2)/v

# use gibbs sampler to sample from posterior densities

n <- length(y)
k <- 1
v <- n - k

# set up gibbs sampler

nrDraws <- 1
h_sample_m0 <- rgamma(nrDraws, v/2, v*s2_m0/2) 
alpha_sample <- matrix(0, nrow = nrDraws, ncol = 1)

for(i in 1:nrDraws) {

alpha_sample[i] <- rnorm(1,alpha.ols,(1/h_sample_m0[i]) * MI)

}

# define posterior density for model 0

f <- function(alpha,h) {

e <- y - alpha * x1^2
const <- (2*pi)^(-n/2) / ( gamma(v/2) * (v*s2_m1/2)^(-v/2) )  
kernel <- h^((v-1)/2) * exp((-(h/2) * sum(e^2)) - (v*s2_m0)*h)
x <- const * kernel
return(x)
}

# calculate approximation of p(y|m_0)

m0 <-f(alpha_sample,h_sample_m0)
post_m0 <- sum(m0) / nrDraws
 

--
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[R] rnorm command

[Cheryl Johnson]

Hello,

I use the command rnorm, and I feed these results into a lmer command.
Since I am using the rnorm command I expect to get different results for
each iteration, yet for each iteration I am getting the same answer. If
someone understands why I am getting the same answer every time with a
random number generator, I would appreciate help in understanding why this
is happening. Is the lmer command causing the answer to be the same each
time?

Thanks

[[alternative HTML version deleted]]

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Re: [R] Shading the plot

[David Winsemius]

On Dec 3, 2011, at 5:28 PM, avinash barnwal wrote:


Hi Weylandt,


I tried it but i was not successful.

Here is the full code

Any help would be great.

time<-c("02-Jan-05","09-Jan-05","16-Jan-05","23-Jan-05","30- 
Jan-05","06-Feb-05","13-Feb-05","20-Feb-05","27-Feb-05","06- 
Mar-05","13-Mar-05",
"20-Mar-05","27-Mar-05","03-Apr-05","10-Apr-05","17-Apr-05","24- 
Apr-05","01-May-05","08-May-05","15-May-05")

time<-as.Date(time,"%d-%b-%y")
a2<- 
c 
("81.96392786 
","81.96392786 
","82.16432866 
","82.56513026","82.76553106","79.96593186","78.16633267",
"75.56713427 
","74.76753507 
","72.96793587 
","70.96793587","45.96793587","83.96793587","84.36873747",
"84.56913828 
","84.56913828 
","84.56913828","84.56913828","84.36873747","84.36873747")
a3<- 
c 
("81.96392786 
","81.96392786 
","81.96392786","70.96392786","68.16432866","66.16432866",
"62.16432866 
","58.56513026 
","56.56513026 
","54.56513026","48.56513026","49.76553106","52.76553106",
"54.76553106 
","57.96593186 
","59.36673347","83.36673347","83.36673347","83.36673347",

"83.56713427")
j<-0
for(i in 1:length(a3))
{
if(a2[i]>a3[i])
{
j<-j+1
x[j]<-a2[i]
y[j]<-a3[i]
temp_time[j]<-time[i]
}
}
plot(time,a2,type='l',col='red',ylab='',xlab=" ",axes=FALSE)
lines(time,a3,col='blue',ylab='',xlab=" ")
polygon(c(temp_time,rev(temp_time)),c(x,y),col="grey")
#


What are the rules that govern requesting outside help in homework at  
your institution?


Your current code does not recognize  that there is line crossing. Nor  
do you describe what you mean by shading. If you wanted a rectangle  
within the lpot area it would drive one strategy and if you wanted  
only the area between the curves to be shaded it would drive a  
different strategy.


AND you never define "x" or "y".

--
David.




On Sun, Dec 4, 2011 at 2:46 AM, R. Michael Weylandt <
michael.weyla...@gmail.com> wrote:


? polygon
example(polygon)

Michael

On Sat, Dec 3, 2011 at 4:08 PM, avinash barnwal
 wrote:

Hi all,

I have been trying to shade the specific part of the plot

Part to be shaded

1. Any color whenever a2>a3

2. Any other color( Not same as 1) whenever a2
##


a2<- 
c 
("81.96392786 
","81.96392786 
","82.16432866 
","82.56513026","82.76553106","79.96593186","78.16633267",


"75.56713427 
","74.76753507 
","72.96793587 
","70.96793587","45.96793587","83.96793587","84.36873747",


"84.56913828 
","84.56913828 
","84.56913828","84.56913828","84.36873747","84.36873747")


a3<- 
c 
("81.96392786 
","81.96392786 
","81.96392786","70.96392786","68.16432866","66.16432866",


"62.16432866 
","58.56513026 
","56.56513026 
","54.56513026","48.56513026","49.76553106","52.76553106",


"54.76553106 
","57.96593186 
","59.36673347","83.36673347","83.36673347","83.36673347",

"83.56713427")
plot(a2,type='l',col='red',ylab='',xlab=" ",axes=FALSE)
lines(a3,col='blue',ylab='',xlab=" ")


##


Thank you in the advance
--
Avinash Barnwal
Final year undergraduate student
Statistics and informatics
Department of Mathematics
IIT Kharagpur

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R-help@r-project.org mailing list
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http://www.R-project.org/posting-guide.html

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







--
Avinash Barnwal
Final year undergraduate student
Statistics and informatics
Department of Mathematics
IIT Kharagpur

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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.


David Winsemius, MD
West Hartford, CT

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Re: [R] help in removal of fixed pattern

[D. Schruth]

A great function for extracting pattern matches is 'm()'

library(caroline)
vect <- m('([xX][0-9])',df$Input)
toupper(vect)  #in case you really want all upper case x's

It does the hard work of using 'sub' to remove the non-matching parts 
(sub, grep, regexpr, etc aren't very good for this sort of thing)


It also can return a data.frame if you have multiple patterns you wish to 
match in each string vector element.


-Dave


On Fri, 2 Dec 2011, arunkumar wrote:


Hi

I have column name as given below

If the variable is in log(X1 + 1) pattern it should be removed and i need
only X1

Input
log(x1 + 1)
x2
log(X3 +1)

Expected Output X1 X2 X3

Please help me


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Re: [R] pivot table help

[jim holtman]

Forgot, you can also do this:

> dcast(x.m, Cluster ~ value, fun = length)
  Cluster ind1 ind2 ind3
1   1110
2   2101
3   3110
>


On Sat, Dec 3, 2011 at 7:02 PM, jim holtman  wrote:
> try this:
>
>> x <- read.table(text = "Cluster   Member1    Member2
> + 1            ind1           ind2
> + 2            ind3           ind1
> + 3            ind2           ind1", as.is = TRUE, header = TRUE)
>> require(reshape2)
>> x.m <- melt(x, id = "Cluster")
>> x.m
>  Cluster variable value
> 1       1  Member1  ind1
> 2       2  Member1  ind3
> 3       3  Member1  ind2
> 4       1  Member2  ind2
> 5       2  Member2  ind1
> 6       3  Member2  ind1
>> table(x.m$Cluster, x.m$value)
>
>    ind1 ind2 ind3
>  1    1    1    0
>  2    1    0    1
>  3    1    1    0
>>
>
>
> On Sat, Dec 3, 2011 at 5:53 PM, Richard M. Heiberger  wrote:
>> Pivot tables are an Excel concept, not an R concept.
>>
>> That means you must give an example of your starting pivot table as an R
>> object (use dump() so we can pick it up from the email and execute it
>> immediately).
>> and an example of the R object you want as the result.
>> Use a trivial but complete example.
>>
>> An example of dump
>>
>> tmp <- matrix(1:6,2,3)
>> tmp
>> dump("tmp","")
>> Be sure to read the posting guide before sending your revised request.
>>
>> PLEASE do read the posting guide
>> http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>>
>> Rich
>> On Sat, Dec 3, 2011 at 5:27 PM, set  wrote:
>>
>>> Hello R-users,
>>>
>>> I've got a huge table with about 20.00 rows and 50 columns. The table now
>>> has headers as Members1, Members2 etc. My data are 8 different individuals.
>>> And I've got a column with clusters. So each individual belongs to
>>> different
>>> clusters and can occurs multiple times within a cluster (that's the reason
>>> that there can be more than 8 members). I want a presence/ absence table
>>> for
>>> each individual within each cluster.
>>> So I want to go from:
>>> Cluster   Member1    Member2  etc.
>>> 1            ind1           ind2
>>> 2            ind3           ind1
>>> 3            ind2           ind1
>>>
>>> to
>>>
>>> cluster    ind1          ind2      ind3
>>> 1            1               1          0
>>> 2            1                0         1
>>> 3            1               1          0
>>>
>>> Has anybody any idea how I can do this? I already tried alot of things with
>>> pivottables (using cast()) But I think I'm missing out on something.
>>> thank you
>>>
>>> --
>>> View this message in context:
>>> http://r.789695.n4.nabble.com/pivot-table-help-tp4155144p4155144.html
>>> Sent from the R help mailing list archive at Nabble.com.
>>>
>>> __
>>> R-help@r-project.org mailing list
>>> https://stat.ethz.ch/mailman/listinfo/r-help
>>> PLEASE do read the posting guide
>>> http://www.R-project.org/posting-guide.html
>>> and provide commented, minimal, self-contained, reproducible code.
>>>
>>
>>        [[alternative HTML version deleted]]
>>
>> __
>> R-help@r-project.org mailing list
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>
>
>
> --
> Jim Holtman
> Data Munger Guru
>
> What is the problem that you are trying to solve?
> Tell me what you want to do, not how you want to do it.



-- 
Jim Holtman
Data Munger Guru

What is the problem that you are trying to solve?
Tell me what you want to do, not how you want to do it.

__
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Re: [R] pivot table help

[jim holtman]

try this:

> x <- read.table(text = "Cluster   Member1Member2
+ 1ind1   ind2
+ 2ind3   ind1
+ 3ind2   ind1", as.is = TRUE, header = TRUE)
> require(reshape2)
> x.m <- melt(x, id = "Cluster")
> x.m
  Cluster variable value
1   1  Member1  ind1
2   2  Member1  ind3
3   3  Member1  ind2
4   1  Member2  ind2
5   2  Member2  ind1
6   3  Member2  ind1
> table(x.m$Cluster, x.m$value)

ind1 ind2 ind3
  1110
  2101
  3110
>


On Sat, Dec 3, 2011 at 5:53 PM, Richard M. Heiberger  wrote:
> Pivot tables are an Excel concept, not an R concept.
>
> That means you must give an example of your starting pivot table as an R
> object (use dump() so we can pick it up from the email and execute it
> immediately).
> and an example of the R object you want as the result.
> Use a trivial but complete example.
>
> An example of dump
>
> tmp <- matrix(1:6,2,3)
> tmp
> dump("tmp","")
> Be sure to read the posting guide before sending your revised request.
>
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
> Rich
> On Sat, Dec 3, 2011 at 5:27 PM, set  wrote:
>
>> Hello R-users,
>>
>> I've got a huge table with about 20.00 rows and 50 columns. The table now
>> has headers as Members1, Members2 etc. My data are 8 different individuals.
>> And I've got a column with clusters. So each individual belongs to
>> different
>> clusters and can occurs multiple times within a cluster (that's the reason
>> that there can be more than 8 members). I want a presence/ absence table
>> for
>> each individual within each cluster.
>> So I want to go from:
>> Cluster   Member1    Member2  etc.
>> 1            ind1           ind2
>> 2            ind3           ind1
>> 3            ind2           ind1
>>
>> to
>>
>> cluster    ind1          ind2      ind3
>> 1            1               1          0
>> 2            1                0         1
>> 3            1               1          0
>>
>> Has anybody any idea how I can do this? I already tried alot of things with
>> pivottables (using cast()) But I think I'm missing out on something.
>> thank you
>>
>> --
>> View this message in context:
>> http://r.789695.n4.nabble.com/pivot-table-help-tp4155144p4155144.html
>> Sent from the R help mailing list archive at Nabble.com.
>>
>> __
>> R-help@r-project.org mailing list
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide
>> http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>>
>
>        [[alternative HTML version deleted]]
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.



-- 
Jim Holtman
Data Munger Guru

What is the problem that you are trying to solve?
Tell me what you want to do, not how you want to do it.

__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

Re: [R] pivot table help

[Richard M. Heiberger]

Pivot tables are an Excel concept, not an R concept.

That means you must give an example of your starting pivot table as an R
object (use dump() so we can pick it up from the email and execute it
immediately).
and an example of the R object you want as the result.
Use a trivial but complete example.

An example of dump

tmp <- matrix(1:6,2,3)
tmp
dump("tmp","")
Be sure to read the posting guide before sending your revised request.

PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

Rich
On Sat, Dec 3, 2011 at 5:27 PM, set  wrote:

> Hello R-users,
>
> I've got a huge table with about 20.00 rows and 50 columns. The table now
> has headers as Members1, Members2 etc. My data are 8 different individuals.
> And I've got a column with clusters. So each individual belongs to
> different
> clusters and can occurs multiple times within a cluster (that's the reason
> that there can be more than 8 members). I want a presence/ absence table
> for
> each individual within each cluster.
> So I want to go from:
> Cluster   Member1Member2  etc.
> 1ind1   ind2
> 2ind3   ind1
> 3ind2   ind1
>
> to
>
> clusterind1  ind2  ind3
> 11   1  0
> 210 1
> 31   1  0
>
> Has anybody any idea how I can do this? I already tried alot of things with
> pivottables (using cast()) But I think I'm missing out on something.
> thank you
>
> --
> View this message in context:
> http://r.789695.n4.nabble.com/pivot-table-help-tp4155144p4155144.html
> Sent from the R help mailing list archive at Nabble.com.
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>

[[alternative HTML version deleted]]

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Re: [R] Data Analysis for Gas Prices

[B77S]

use a < ? > to get help on a function; example:

?read.table

If you do this you will see an option called "header"...  
use header=T if your top row contains column names.  

Learn how to read these help pages.  Also, read thru a few beginner R
manuals and see this website:
http://www.statmethods.net/interface/io.html

As for the rest of your questions. 
Perhaps this is your sign to begin reading a intro-stats book, or inquire on
a different forum.
see:
http://stats.stackexchange.com/questions

good luck


inferno846 wrote
> 
> Hi there,
> 
> I'm looking to analyze a set of data on local gas prices for a single day.
> I'm wondering what kind of questions I should be looking to ask and how to
> find and answer to them with R. Examples would be: 
> 
> Do prices differ between brands?
> Does location affect (NE, NW, SE, SW) price?
> Does the number of nearby (within .25 miles) competitors affect price?
> Do gas stations near shopping centers or highways have different prices?
> 
> Also, could anyone help me figure out how to import a data table to R?
> When I try to create a .txt file from a word document and read it in R,
> the format of the first column always messes up. Any/all help is
> appreciated.
> 


--
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[R] pivot table help

[set]

Hello R-users,

I've got a huge table with about 20.00 rows and 50 columns. The table now
has headers as Members1, Members2 etc. My data are 8 different individuals.
And I've got a column with clusters. So each individual belongs to different
clusters and can occurs multiple times within a cluster (that's the reason
that there can be more than 8 members). I want a presence/ absence table for
each individual within each cluster.
So I want to go from:
Cluster   Member1Member2  etc.
1ind1   ind2
2ind3   ind1
3ind2   ind1

to

clusterind1  ind2  ind3
11   1  0
210 1
31   1  0

Has anybody any idea how I can do this? I already tried alot of things with
pivottables (using cast()) But I think I'm missing out on something.
thank you

--
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http://r.789695.n4.nabble.com/pivot-table-help-tp4155144p4155144.html
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__
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Re: [R] Shading the plot

[avinash barnwal]

Hi Weylandt,


I tried it but i was not successful.

Here is the full code

Any help would be great.

time<-c("02-Jan-05","09-Jan-05","16-Jan-05","23-Jan-05","30-Jan-05","06-Feb-05","13-Feb-05","20-Feb-05","27-Feb-05","06-Mar-05","13-Mar-05",
"20-Mar-05","27-Mar-05","03-Apr-05","10-Apr-05","17-Apr-05","24-Apr-05","01-May-05","08-May-05","15-May-05")
time<-as.Date(time,"%d-%b-%y")
a2<-c("81.96392786","81.96392786","82.16432866","82.56513026","82.76553106","79.96593186","78.16633267",
"75.56713427","74.76753507","72.96793587","70.96793587","45.96793587","83.96793587","84.36873747",
"84.56913828","84.56913828","84.56913828","84.56913828","84.36873747","84.36873747")
a3<-c("81.96392786","81.96392786","81.96392786","70.96392786","68.16432866","66.16432866",
"62.16432866","58.56513026","56.56513026","54.56513026","48.56513026","49.76553106","52.76553106",
"54.76553106","57.96593186","59.36673347","83.36673347","83.36673347","83.36673347",
"83.56713427")
j<-0
for(i in 1:length(a3))
{
if(a2[i]>a3[i])
{
j<-j+1
x[j]<-a2[i]
y[j]<-a3[i]
temp_time[j]<-time[i]
}
}
plot(time,a2,type='l',col='red',ylab='',xlab=" ",axes=FALSE)
lines(time,a3,col='blue',ylab='',xlab=" ")
polygon(c(temp_time,rev(temp_time)),c(x,y),col="grey")
#
On Sun, Dec 4, 2011 at 2:46 AM, R. Michael Weylandt <
michael.weyla...@gmail.com> wrote:

> ? polygon
> example(polygon)
>
> Michael
>
> On Sat, Dec 3, 2011 at 4:08 PM, avinash barnwal
>  wrote:
> > Hi all,
> >
> > I have been trying to shade the specific part of the plot
> >
> > Part to be shaded
> >
> > 1. Any color whenever a2>a3
> >
> > 2. Any other color( Not same as 1) whenever a2 >
> > Suggest me some code for this task
> >
> > PLOT CODE for reference
> >
> >
> ##
> >
> a2<-c("81.96392786","81.96392786","82.16432866","82.56513026","82.76553106","79.96593186","78.16633267",
> >
> "75.56713427","74.76753507","72.96793587","70.96793587","45.96793587","83.96793587","84.36873747",
> >
> "84.56913828","84.56913828","84.56913828","84.56913828","84.36873747","84.36873747")
> >
> a3<-c("81.96392786","81.96392786","81.96392786","70.96392786","68.16432866","66.16432866",
> >
> "62.16432866","58.56513026","56.56513026","54.56513026","48.56513026","49.76553106","52.76553106",
> >
> "54.76553106","57.96593186","59.36673347","83.36673347","83.36673347","83.36673347",
> > "83.56713427")
> > plot(a2,type='l',col='red',ylab='',xlab=" ",axes=FALSE)
> > lines(a3,col='blue',ylab='',xlab=" ")
> >
> ##
> >
> > Thank you in the advance
> > --
> > Avinash Barnwal
> > Final year undergraduate student
> > Statistics and informatics
> > Department of Mathematics
> > IIT Kharagpur
> >
> > __
> > R-help@r-project.org mailing list
> > https://stat.ethz.ch/mailman/listinfo/r-help
> > PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> > and provide commented, minimal, self-contained, reproducible code.
> >
>



-- 
Avinash Barnwal
Final year undergraduate student
Statistics and informatics
Department of Mathematics
IIT Kharagpur

[[alternative HTML version deleted]]

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[R] Data Analysis for Gas Prices

[inferno846]

Hi there,

I'm looking to analyze a set of data on local gas prices for a single day.
I'm wondering what kind of questions I should be looking to ask and how to
find and answer to them with R. Examples would be: 

Do prices differ between brands?
Does location affect (NE, NW, SE, SW) price?
Does the number of nearby (within .25 miles) competitors affect price?
Do gas stations near shopping centers or highways have different prices?

Also, could anyone help me figure out how to import a data table to R? When
I try to create a .txt file from a word document and read it in R, the
format of the first column always messes up. Any/all help is appreciated.

--
View this message in context: 
http://r.789695.n4.nabble.com/Data-Analysis-for-Gas-Prices-tp4155078p4155078.html
Sent from the R help mailing list archive at Nabble.com.

__
R-help@r-project.org mailing list
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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.

Re: [R] Shading the plot

[R. Michael Weylandt]

? polygon
example(polygon)

Michael

On Sat, Dec 3, 2011 at 4:08 PM, avinash barnwal
 wrote:
> Hi all,
>
> I have been trying to shade the specific part of the plot
>
> Part to be shaded
>
> 1. Any color whenever a2>a3
>
> 2. Any other color( Not same as 1) whenever a2
> Suggest me some code for this task
>
> PLOT CODE for reference
>
> ##
> a2<-c("81.96392786","81.96392786","82.16432866","82.56513026","82.76553106","79.96593186","78.16633267",
> "75.56713427","74.76753507","72.96793587","70.96793587","45.96793587","83.96793587","84.36873747",
> "84.56913828","84.56913828","84.56913828","84.56913828","84.36873747","84.36873747")
> a3<-c("81.96392786","81.96392786","81.96392786","70.96392786","68.16432866","66.16432866",
> "62.16432866","58.56513026","56.56513026","54.56513026","48.56513026","49.76553106","52.76553106",
> "54.76553106","57.96593186","59.36673347","83.36673347","83.36673347","83.36673347",
> "83.56713427")
> plot(a2,type='l',col='red',ylab='',xlab=" ",axes=FALSE)
> lines(a3,col='blue',ylab='',xlab=" ")
> ##
>
> Thank you in the advance
> --
> Avinash Barnwal
> Final year undergraduate student
> Statistics and informatics
> Department of Mathematics
> IIT Kharagpur
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>

__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

[R] Shading the plot

[avinash barnwal]

Hi all,

I have been trying to shade the specific part of the plot

Part to be shaded

1. Any color whenever a2>a3

2. Any other color( Not same as 1) whenever a2<>__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

Re: [R] Moving column averaging

[B77S]

I don't know the answer, but would suppose not.
You could test this for yourself using:
system.time()
example: 
system.time(rnorm(10,0,1)) 


Chega wrote
> 
> This solved my problem - Thanks a lot for your help! Please allow me one
> more question: Works zoo's rollapply on a plain matrix faster than on a
> zoo object? I am asking since I wanted to apply the provided averaging
> code to a larger matrix (2500 rows x 200 cols), which is quite time
> consuming...
> 
> Thanks again!
> 
> Chega
> 
> 
> From: B77S [via R] 
> To: Chega  
> Sent: Friday, December 2, 2011 6:16 AM
> Subject: Re: Moving column averaging
> 
> 
> Sorry for that, and thanks Gabor, 
> I could have sworn that it wouldn't. 
> 
> 
> 
> Gabor Grothendieck wrote
>>On Thu, Dec 1, 2011 at 7:13 PM, B77S <[hidden email]> wrote: 
>>> # need zoo to use rollapply() 
>>> 
>>> # your data (I called df) 
>>> df <- structure(list(a = 1:2, b = 2:3, c = c(5L, 9L), d = c(9L, 6L), 
>>>    e = c(1L, 5L), f = c(4, 7)), .Names = c("a", "b", "c", "d", 
>>> "e", "f"), class = "data.frame", row.names = c(NA, -2L)) 
>>> 
>>> # transpose and make a zoo object 
>>> df2 <- zoo(t(df)) 
>>> 
>>> #rollapply to get means and transpose back 
>>> means <- t(rollapply(df2, width=2, by=2, FUN=mean)) 
>>> 
>>> # adding the combined column names you requested 
>>> colnames(means) <- apply(matrix(names(df), nrow=2), 2, paste,
>>> collapse=", ") 
>>> 
>>
>>Note that zoo's rollapply also works on plain matrices and vectors. 
>>
>>-- 
>>Statistics & Software Consulting 
>>GKX Group, GKX Associates Inc. 
>>tel: 1-877-GKX-GROUP 
>>email: ggrothendieck at gmail.com 
>>
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Re: [R] Iteration in R

[R. Michael Weylandt]

Because lapply() tries to pass an argument to FUN and there's none
that it can receive.

This would work however:

lapply(rep(100, 6), rnorm, mean = 1, sd = 2)

Michael

On Sat, Dec 3, 2011 at 2:42 PM, B77S  wrote:
> Interesting and thank you; I'm confused as to why this doesn't work with:
>
> lapply(rep(1,6), FUN=rnorm, n=10, mean=1.0, sd=1)
>
>
>
> andrija djurovic wrote
>>
>> Hi Brad. Maybe something like this:
>>
>> lapply(rep(1,6), function(x) rnorm(10,0,1))
>>
>> Andrija
>>
>> On Sat, Dec 3, 2011 at 8:21 PM, B77S  wrote:
>>
>>> Hi Michael,
>>> How would you do this with lapply to return a list?
>>> I can't seem to get that to work (I haven't used these much and am trying
>>> to
>>> learn).
>>> Thanks
>>> Brad
>>>
>>>
>>> Michael Weylandt wrote
>>> >
>>> > ? replicate
>>> >
>>> > or a for loop
>>> >
>>> > or do all one hundred simulations at once
>>> >
>>> > x <- matrix(rnorm(100^2, 1, 2), 100)
>>> >
>>> > It's going to depend on what you want to do with the numbers.
>>> >
>>> > Michael
>>> >
>>> > On Sat, Dec 3, 2011 at 1:10 PM, Martin Zonyrah 
>>> wrote:
>>> >> Hi,
>>> >> I need help. I am trying to iterate this command  x <- rnorm(100, 1.0,
>>> >> 2.0) one hundred times in R but I don't seem to have a clue.
>>> >> Can anyone help?
>>> >> Your help is very much appreciated.
>>> >>
>>> >> Martin
>>> >>
>>> >>        [[alternative HTML version deleted]]
>>> >>
>>> >>
>>> >> __
>>> >> R-help@ mailing list
>>> >> https://stat.ethz.ch/mailman/listinfo/r-help
>>> >> PLEASE do read the posting guide
>>> >> http://www.R-project.org/posting-guide.html
>>> >> and provide commented, minimal, self-contained, reproducible code.
>>> >>
>>> >
>>> > __
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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.
>>> >
>>>
>>>
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>>>
>>
>>       [[alternative HTML version deleted]]
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>>
>
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Re: [R] Iteration in R

[R. Michael Weylandt]

And with replicate:

replicate(100, rnorm(100, 1,2), simplify = FALSE)


Michael

On Sat, Dec 3, 2011 at 2:38 PM, andrija djurovic  wrote:
> Hi Brad. Maybe something like this:
>
> lapply(rep(1,6), function(x) rnorm(10,0,1))
>
> Andrija
>
> On Sat, Dec 3, 2011 at 8:21 PM, B77S  wrote:
>
>> Hi Michael,
>> How would you do this with lapply to return a list?
>> I can't seem to get that to work (I haven't used these much and am trying
>> to
>> learn).
>> Thanks
>> Brad
>>
>>
>> Michael Weylandt wrote
>> >
>> > ? replicate
>> >
>> > or a for loop
>> >
>> > or do all one hundred simulations at once
>> >
>> > x <- matrix(rnorm(100^2, 1, 2), 100)
>> >
>> > It's going to depend on what you want to do with the numbers.
>> >
>> > Michael
>> >
>> > On Sat, Dec 3, 2011 at 1:10 PM, Martin Zonyrah 
>> wrote:
>> >> Hi,
>> >> I need help. I am trying to iterate this command  x <- rnorm(100, 1.0,
>> >> 2.0) one hundred times in R but I don't seem to have a clue.
>> >> Can anyone help?
>> >> Your help is very much appreciated.
>> >>
>> >> Martin
>> >>
>> >>        [[alternative HTML version deleted]]
>> >>
>> >>
>> >> __
>> >> R-help@ mailing list
>> >> https://stat.ethz.ch/mailman/listinfo/r-help
>> >> PLEASE do read the posting guide
>> >> http://www.R-project.org/posting-guide.html
>> >> and provide commented, minimal, self-contained, reproducible code.
>> >>
>> >
>> > __
>> > R-help@ mailing list
>> > https://stat.ethz.ch/mailman/listinfo/r-help
>> > PLEASE do read the posting guide
>> > http://www.R-project.org/posting-guide.html
>> > and provide commented, minimal, self-contained, reproducible code.
>> >
>>
>>
>> --
>> View this message in context:
>> http://r.789695.n4.nabble.com/Iteration-in-R-tp4154433p4154479.html
>> Sent from the R help mailing list archive at Nabble.com.
>>
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>> https://stat.ethz.ch/mailman/listinfo/r-help
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>> http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>>
>
>        [[alternative HTML version deleted]]
>
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Re: [R] Iteration in R

[B77S]

Interesting and thank you; I'm confused as to why this doesn't work with: 

lapply(rep(1,6), FUN=rnorm, n=10, mean=1.0, sd=1) 



andrija djurovic wrote
> 
> Hi Brad. Maybe something like this:
> 
> lapply(rep(1,6), function(x) rnorm(10,0,1))
> 
> Andrija
> 
> On Sat, Dec 3, 2011 at 8:21 PM, B77S  wrote:
> 
>> Hi Michael,
>> How would you do this with lapply to return a list?
>> I can't seem to get that to work (I haven't used these much and am trying
>> to
>> learn).
>> Thanks
>> Brad
>>
>>
>> Michael Weylandt wrote
>> >
>> > ? replicate
>> >
>> > or a for loop
>> >
>> > or do all one hundred simulations at once
>> >
>> > x <- matrix(rnorm(100^2, 1, 2), 100)
>> >
>> > It's going to depend on what you want to do with the numbers.
>> >
>> > Michael
>> >
>> > On Sat, Dec 3, 2011 at 1:10 PM, Martin Zonyrah 
>> wrote:
>> >> Hi,
>> >> I need help. I am trying to iterate this command  x <- rnorm(100, 1.0,
>> >> 2.0) one hundred times in R but I don't seem to have a clue.
>> >> Can anyone help?
>> >> Your help is very much appreciated.
>> >>
>> >> Martin
>> >>
>> >>[[alternative HTML version deleted]]
>> >>
>> >>
>> >> __
>> >> R-help@ mailing list
>> >> https://stat.ethz.ch/mailman/listinfo/r-help
>> >> PLEASE do read the posting guide
>> >> http://www.R-project.org/posting-guide.html
>> >> and provide commented, minimal, self-contained, reproducible code.
>> >>
>> >
>> > __
>> > R-help@ mailing list
>> > https://stat.ethz.ch/mailman/listinfo/r-help
>> > PLEASE do read the posting guide
>> > http://www.R-project.org/posting-guide.html
>> > and provide commented, minimal, self-contained, reproducible code.
>> >
>>
>>
>> --
>> View this message in context:
>> http://r.789695.n4.nabble.com/Iteration-in-R-tp4154433p4154479.html
>> Sent from the R help mailing list archive at Nabble.com.
>>
>> __
>> R-help@ mailing list
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide
>> http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>>
> 
>   [[alternative HTML version deleted]]
> 
> __
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> 


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Re: [R] Iteration in R

[andrija djurovic]

Hi Brad. Maybe something like this:

lapply(rep(1,6), function(x) rnorm(10,0,1))

Andrija

On Sat, Dec 3, 2011 at 8:21 PM, B77S  wrote:

> Hi Michael,
> How would you do this with lapply to return a list?
> I can't seem to get that to work (I haven't used these much and am trying
> to
> learn).
> Thanks
> Brad
>
>
> Michael Weylandt wrote
> >
> > ? replicate
> >
> > or a for loop
> >
> > or do all one hundred simulations at once
> >
> > x <- matrix(rnorm(100^2, 1, 2), 100)
> >
> > It's going to depend on what you want to do with the numbers.
> >
> > Michael
> >
> > On Sat, Dec 3, 2011 at 1:10 PM, Martin Zonyrah 
> wrote:
> >> Hi,
> >> I need help. I am trying to iterate this command  x <- rnorm(100, 1.0,
> >> 2.0) one hundred times in R but I don't seem to have a clue.
> >> Can anyone help?
> >> Your help is very much appreciated.
> >>
> >> Martin
> >>
> >>[[alternative HTML version deleted]]
> >>
> >>
> >> __
> >> R-help@ mailing list
> >> https://stat.ethz.ch/mailman/listinfo/r-help
> >> PLEASE do read the posting guide
> >> http://www.R-project.org/posting-guide.html
> >> and provide commented, minimal, self-contained, reproducible code.
> >>
> >
> > __
> > R-help@ mailing list
> > https://stat.ethz.ch/mailman/listinfo/r-help
> > PLEASE do read the posting guide
> > http://www.R-project.org/posting-guide.html
> > and provide commented, minimal, self-contained, reproducible code.
> >
>
>
> --
> View this message in context:
> http://r.789695.n4.nabble.com/Iteration-in-R-tp4154433p4154479.html
> Sent from the R help mailing list archive at Nabble.com.
>
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> R-help@r-project.org mailing list
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>

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Re: [R] Iteration in R

[B77S]

Hi Michael, 
How would you do this with lapply to return a list?
I can't seem to get that to work (I haven't used these much and am trying to
learn).
Thanks
Brad


Michael Weylandt wrote
> 
> ? replicate
> 
> or a for loop
> 
> or do all one hundred simulations at once
> 
> x <- matrix(rnorm(100^2, 1, 2), 100)
> 
> It's going to depend on what you want to do with the numbers.
> 
> Michael
> 
> On Sat, Dec 3, 2011 at 1:10 PM, Martin Zonyrah  wrote:
>> Hi,
>> I need help. I am trying to iterate this command  x <- rnorm(100, 1.0,
>> 2.0) one hundred times in R but I don't seem to have a clue.
>> Can anyone help?
>> Your help is very much appreciated.
>>
>> Martin
>>
>>        [[alternative HTML version deleted]]
>>
>>
>> __
>> R-help@ mailing list
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide
>> http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>>
> 
> __
> R-help@ mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
> 


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Re: [R] Iteration in R

[andrija djurovic]

Hi. One approach is using replicate. See ?replicate:

replicate(3,rnorm(100,1,2))

Andrija

On Sat, Dec 3, 2011 at 7:10 PM, Martin Zonyrah wrote:

> Hi,
> I need help. I am trying to iterate this command  x <- rnorm(100, 1.0,
> 2.0) one hundred times in R but I don't seem to have a clue.
> Can anyone help?
> Your help is very much appreciated.
>
> Martin
>
>[[alternative HTML version deleted]]
>
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
>

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Re: [R] Iteration in R

[David Winsemius]

On Dec 3, 2011, at 1:10 PM, Martin Zonyrah wrote:


Hi,
I need help. I am trying to iterate this command  x <- rnorm(100,  
1.0, 2.0) one hundred times in R but I don't seem to have a clue.


?replicate




==

David Winsemius, MD
West Hartford, CT

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Re: [R] Counting the occurences of a charater within a string

[Hadley Wickham]

On Thu, Dec 1, 2011 at 10:32 AM, Douglas Esneault
 wrote:
> I am new to R but am experienced SAS user and I was hoping to get some help 
> on counting the occurrences of a character within a string at a row level.
>
> My dataframe, x,  is structured as below:
>
> Col1
> abc/def
> ghi/jkl/mno
>
> I found this code on the board but it counts all occurrences of "/" in the 
> dataframe.
>
> chr.pos <- which(unlist(strsplit(x,NULL))=='/')
> chr.count <- length(chr.pos)
> chr.count
> [1] 3
>
> I'd like to append a column, say cnt, that has the count of "/" for each row.

Here's an easy way from stringr:

library(stringr)
str_count( c("abc/def", "ghi/jkl/mno"), "/")
# [1] 1 2

Hadley

-- 
Assistant Professor / Dobelman Family Junior Chair
Department of Statistics / Rice University
http://had.co.nz/

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Re: [R] Iteration in R

[R. Michael Weylandt]

? replicate

or a for loop

or do all one hundred simulations at once

x <- matrix(rnorm(100^2, 1, 2), 100)

It's going to depend on what you want to do with the numbers.

Michael

On Sat, Dec 3, 2011 at 1:10 PM, Martin Zonyrah  wrote:
> Hi,
> I need help. I am trying to iterate this command  x <- rnorm(100, 1.0, 2.0) 
> one hundred times in R but I don't seem to have a clue.
> Can anyone help?
> Your help is very much appreciated.
>
> Martin
>
>        [[alternative HTML version deleted]]
>
>
> __
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> and provide commented, minimal, self-contained, reproducible code.
>

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Re: [R] Problem with loop

[Komine]

Hi, 
Thank Michael for your help.
Komine

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[R] Iteration in R

[Martin Zonyrah]

Hi,
I need help. I am trying to iterate this command  x <- rnorm(100, 1.0, 2.0) one 
hundred times in R but I don't seem to have a clue.
Can anyone help?
Your help is very much appreciated.

Martin

[[alternative HTML version deleted]]

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Re: [R] problems using the thin plate spline method

[Jeff Newmiller]

I can tell that you are puzzled and confused. Unfortunately, I am not psychic, 
so I cannot tell what you did, and therefore cannot tell where you went astray.

The solution is for you to read the posting guide mentioned at the bottom of 
every R-help message. Spend a little time to create a small bit of data like 
yours if your actual data is large (subset and dput are useful for this). 
Remember to include the output of sessionInfo, and so on. Many times you are 
likely to find the answer yourself by going through these steps, but they are 
essential for communication.

Good luck.
---
Jeff NewmillerThe .   .  Go Live...
DCN:Basics: ##.#.   ##.#.  Live Go...
  Live:   OO#.. Dead: OO#..  Playing
Research Engineer (Solar/BatteriesO.O#.   #.O#.  with
/Software/Embedded Controllers)   .OO#.   .OO#.  rocks...1k
--- 
Sent from my phone. Please excuse my brevity.

Mintewab Bezabih  wrote:

>Dear R users, 
>
>I am a beginner in R trying to apply the thin plate spline method to my
>climate data. I used the example in R to do so, and the lines seem to
>run fine ( I am not getting errors) but I am not getting any output in
>the form of graph or anything. I got a warning message saying that
>'surface extends beyond box'. 
>
>Any help is much appreciated. 
>thanks 
>minti 
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Re: [R] Problem installing Rtools

[stillerfan]

Great, thanks for answering this basic question.

On Sat, Dec 3, 2011 at 5:12 AM, Duncan Murdoch-2 [via R] <
ml-node+s789695n4152880...@n4.nabble.com> wrote:

> On 11-12-02 10:25 PM, stillerfan wrote:
> > I'm a new user to R and recently installed R 2.14  for Windows 7 (64
> bit).  I
> > have downloaded and ran Rtools 2.14 but do not see any new packages in
> the
> > install packages section of R.  I installed Rtools in a subdirectory to
> the
> > /library/ directory.  Does anyone have any idea what I could be doing
> wrong?
> >
>
> Rtools doesn't contain R packages, it contains various compilers and
> tools that you could use to build R and R packages.  So you should
> install it in its own directory.  I usually use c:/Rtools.
>
> Duncan Murdoch
>
> __
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Re: [R] replace values

[David Winsemius]

On Dec 3, 2011, at 8:41 AM, syrvn wrote:


Hello,


imagine the following data.frame:

ID name
1 *_A
2 *_A
3 *_B
4 *_B

* = can be any pattern

I want to replace every row which ends with _A by 1 and every row  
which ends

by _B with a 0


You can use grep(patt, x, value=TRUE) to return vlaues that can be  
used in the prior answers to your earlier question today.


I would have worked it out and tested the strategy, but you continue  
to fail to produce workable examples and then apologize that  
people are "confused" by them. We are NOT confused... only annoyed ...  
that you fail to use dput() or code for constructing workable code.


--
David.


so that the data.frame looks like the following:

ID name
1 1
2 1
3 0
4 0


Which function do I use best to achieve this?

Cheers,
Syrvn



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David Winsemius, MD
West Hartford, CT

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[R] replace values

[syrvn]

Hello,


imagine the following data.frame:

ID name
1 *_A
2 *_A
3 *_B
4 *_B

* = can be any pattern

I want to replace every row which ends with _A by 1 and every row which ends
by _B with a 0

so that the data.frame looks like the following:

ID name
1 1
2 1
3 0
4 0


Which function do I use best to achieve this?

Cheers,
Syrvn



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Re: [R] Downloading tab separated data from internet

[HC]

Thanks for your reply.

I tried to use the  postForm function of RCurl as below but do not have much
clue as to how to go further with what it does. 

library(RCurl)
stn<-"03015795"
myurl<-paste("http://ida.water.usgs.gov/ida/available_records.cfm?sn=",stn,sep="";)
mypage1 = readLines(myurl)

# Getting the start and end dates
mypattern = '([^<]*)'
datalines = grep(mypattern, mypage1[124], value=TRUE)

getexpr = function(s,g)substring(s,g,g+attr(g,'match.length')-1)
gg = gregexpr(mypattern, datalines)
matches = mapply(getexpr,datalines,gg)
result = gsub(mypattern,'\\1',matches)
names(result)=NULL
mydates<-result[1:2]

result = postForm(myurl,fromdate=mydates[1], todate=mydates[2],rtype="Save
to File", submit1="Retrieve Data")


I tried to read the RCurl's documentation but do not know what functions I
should be using. Are there any  examples available that could be helpful.
Could you point me to those please.

Thanks for the help.
HC

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[R] problems using the thin plate spline method

[Mintewab Bezabih]

Dear R users, 

I am a beginner in R trying to apply the thin plate spline method to my climate 
data. I used the example in R to do so, and the lines seem to run fine ( I am 
not getting errors) but I am not getting any output in the form of graph or 
anything. I got a warning message saying that 'surface extends beyond box'. 

Any help is much appreciated. 
thanks 
minti 
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Re: [R] Data alignment

[syrvn]

Thanks for your suggestions. I will try them.

The "-" in my original post was actually only there to serve as a separator
so that it is easier for you to see the data structure but apparently it
rather confused you... sorry :)

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Re: [R] Sweave problem on Mac OS when using umlauts and summary()

[Mark Heckmann]

I just noticed that the question has already been answered on the R-help list 
before:

http://r.789695.n4.nabble.com/Sweave-font-problems-with-Signif-codes-lines-td977346.html

Adding the following R line solves the problem:

options(useFancyQuotes = FALSE)

Mark

> 
> 
> On Fri, Dec 2, 2011 at 4:08 PM, Mark Heckmann  wrote:
>> I have the following Sweave file which gets sweaved correctly.
>> 
>> <<>>=
>> m <- lm(y1 ~x1, anscombe)
>> summary(m)
>> @
>> 
>> I include the sweaved .tex file into another .tex file via include.
>> When I use a single umlaut in the .snw file a warning occurs.
>> As a result part of the summary output is not contained in the .tex file.
>> 
>> ä
>> <<>>=
>> m <- lm(y1 ~x1, anscombe)
>> summary(m)
>> @
>> 
>> You can now run (pdf)latex on 'ch1.tex'
>> Warnmeldungen:
>> 1: ‘ch1.Snw’ has unknown encoding: assuming Latin-1
>> 2: ungültige Zeichenkette in Konvertierung der Ausgabe (wrong character in 
>> conversion of output)
>> 
>> Interestingly, this error does NOT occur, when I omit the summary(m) 
>> statement.
>> 
>> ä
>> <<>>=
>> m <- lm(y1 ~x1, anscombe)
>> #summary(m)
>> @
>> 
>> You can now run (pdf)latex on 'ch1.tex'
>> Warnmeldung:
>> ‘ch1.Snw’ has unknown encoding: assuming Latin-1
>> 
>> I know that I can prevent this by adding a line at the beginning of the .snw 
>> file:
>> 
>> \usepackage[utf8]{inputenc}
>> 
>> ä
>> <<>>=
>> m <- lm(y1 ~x1, anscombe)
>> summary(m)
>> @
>> 
>> This gets sweaved correctly without warnings:
>> 
>> But this solution is not good as it is not the preamble of the .tex document 
>> where I add the usepackage line.
>> This will cause an error when processing the entire document with tex.
>> 
>> How can I achieve the last result in another way?
>> I tried:
>> 
>> Sweave('/Users/markheckmann/Desktop/test_sweave/ch1.Snw', encoding="UFT-8")
>> 
>> But this does not work either when the usepackage line is omitted.
>> 
>> I am stuck here. Can anyone help?
>> 
>> TIA
>> Mark
>> 
>> 
>> 
>> Mark Heckmann
>> Blog: www.markheckmann.de
>> R-Blog: http://ryouready.wordpress.com
>> 
>> 
>> 
>> 
>> 
>> 
>> 
>> 
>> 
>> 
>> 
>>[[alternative HTML version deleted]]
>> 
>> 
>> __
>> R-help@r-project.org mailing list
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>> 


Mark Heckmann
Blog: www.markheckmann.de
R-Blog: http://ryouready.wordpress.com











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Re: [R] Reading multiple text files and then combining into a dataframe

[jim holtman]

Does this do it for you:

> #Creating example data in similar format to data I have
> sub <- rep(1,10)
> trial <- seq(1,10,1)
> size <- rep(3,10)
> shortest <- rep(444,10)
> startlab <- rep(444,10)
> endlab <- rep(444,10)
> startconf <- rep(444,10)
> endconf <- rep(444,10)
> steps <- sample(1:30,10)
> time <- sample(1:300,10)
>
> #creating example of text file and saving into a shared director with other 
> txt files
>
> subject1 <-
+ cbind(sub,trial,size,shortest,startlab,endlab,startconf,endconf,steps,time)
> write.table(subject1, "/temp/subject1.txt", sep=" ")
>
> #2nd example of same text file
> sub <- rep(2,10)
> trial <- seq(1,10,1)
> size <- rep(3,10)
> shortest <- rep(444,10)
> startlab <- rep(444,10)
> endlab <- rep(444,10)
> startconf <- rep(444,10)
> endconf <- rep(444,10)
> steps <- sample(1:30,10)
> time <- sample(1:300,10)
>
> subject1 <-
+ cbind(sub,trial,size,shortest,startlab,endlab,startconf,endconf,steps,time)
> write.table(subject1, "/temp/subject2.txt", sep=" ")
>
>
> setwd("/temp")
>
> #Getting list of file names
> file_name <- list.files(pattern="subject*")
>
> mydata <- lapply (file_name, read.table, sep=" ", header=T, row.names=NULL)
>
> # rbind to dataframe
> mydf <- do.call(rbind, mydata)
>
> # or use the plyr package
> require(plyr)
> mydf.1 <- ldply(file_name, read.table, sep = ' ', header = TRUE, row.names = 
> NULL)
>
> str(mydf)
'data.frame':   20 obs. of  11 variables:
 $ row.names: chr  "1" "2" "3" "4" ...
 $ sub  : int  1 1 1 1 1 1 1 1 1 1 ...
 $ trial: int  1 2 3 4 5 6 7 8 9 10 ...
 $ size : int  3 3 3 3 3 3 3 3 3 3 ...
 $ shortest : int  444 444 444 444 444 444 444 444 444 444 ...
 $ startlab : int  444 444 444 444 444 444 444 444 444 444 ...
 $ endlab   : int  444 444 444 444 444 444 444 444 444 444 ...
 $ startconf: int  444 444 444 444 444 444 444 444 444 444 ...
 $ endconf  : int  444 444 444 444 444 444 444 444 444 444 ...
 $ steps: int  10 28 20 18 4 21 5 3 8 7 ...
 $ time : int  225 188 205 189 103 227 148 221 275 14 ...
> str(mydf.1)
'data.frame':   20 obs. of  11 variables:
 $ row.names: chr  "1" "2" "3" "4" ...
 $ sub  : int  1 1 1 1 1 1 1 1 1 1 ...
 $ trial: int  1 2 3 4 5 6 7 8 9 10 ...
 $ size : int  3 3 3 3 3 3 3 3 3 3 ...
 $ shortest : int  444 444 444 444 444 444 444 444 444 444 ...
 $ startlab : int  444 444 444 444 444 444 444 444 444 444 ...
 $ endlab   : int  444 444 444 444 444 444 444 444 444 444 ...
 $ startconf: int  444 444 444 444 444 444 444 444 444 444 ...
 $ endconf  : int  444 444 444 444 444 444 444 444 444 444 ...
 $ steps: int  10 28 20 18 4 21 5 3 8 7 ...
 $ time : int  225 188 205 189 103 227 148 221 275 14 ...
>


On Fri, Dec 2, 2011 at 11:51 PM, James Holland  wrote:
> I have a multiple text files, separated by a single space, that I need to
> combine into a single data.frame.  Below is the track I'm on using
> list.files to capture the names of the files and then lapply with
> read.table.  But I run into problems making a usable dataframe out of the
> data.
>
>
> #Creating example data in similar format to data I have
> sub <- rep(1,10)
> trial <- seq(1,10,1)
> size <- rep(3,10)
> shortest <- rep(444,10)
> startlab <- rep(444,10)
> endlab <- rep(444,10)
> startconf <- rep(444,10)
> endconf <- rep(444,10)
> steps <- sample(1:30,10)
> time <- sample(1:300,10)
>
> #creating example of text file and saving into a shared director with other
> txt files
>
> subject1 <-
> cbind(sub1,trial,size,shortest,startlab,endlab,startconf,endconf,steps,time)
> write.table(subject1, "C:Folder/R/subject1.txt", sep=" ")
>
> #2nd example of same text file
> sub <- rep(2,10)
> trial <- seq(1,10,1)
> size <- rep(3,10)
> shortest <- rep(444,10)
> startlab <- rep(444,10)
> endlab <- rep(444,10)
> startconf <- rep(444,10)
> endconf <- rep(444,10)
> steps <- sample(1:30,10)
> time <- sample(1:300,10)
>
> subject1 <-
> cbind(sub1,trial,size,shortest,startlab,endlab,startconf,endconf,steps,time)
> write.table(subject1, "C:Folder/R/subject2.txt", sep=" ")
>
>
> setwd("C:Folder/R/")
>
> #Getting list of file names
> file_name <- list.files(pattern="subject*")
>
> mydata <- lapply (file_name, read.table, sep=" ", header=T, row.names=NULL)
>
>
>
> Thank you,
> James
>
>        [[alternative HTML version deleted]]
>
> __
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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.
>



-- 
Jim Holtman
Data Munger Guru

What is the problem that you are trying to solve?
Tell me what you want to do, not how you want to do it.

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Re: [R] Data alignment

[jim holtman]

try this:

> match(x$Name> x <- read.table(text = "Name - Value
+ A - 400
+ A - 300
+ B - 200
+ B - 350
+ C - 500
+ C - 350
+ D - 450
+ D - 600
+ E - 700
+ E - 750
+ F - 630
+ F - 650", header = TRUE, as.is = TRUE)
> map <- data.frame(key =   c("A", "B", "C", "D", "E", "F")
+ , value = c( 1,   1,   2,   2,   3,   3)
+ , stringsAsFactors = FALSE
+ )
> x$ID <- map$value[match(x$Name, map$key)]
> x
   Name X. Value ID
1 A  -   400  1
2 A  -   300  1
3 B  -   200  1
4 B  -   350  1
5 C  -   500  2
6 C  -   350  2
7 D  -   450  2
8 D  -   600  2
9 E  -   700  3
10E  -   750  3
11F  -   630  3
12F  -   650  3


On Sat, Dec 3, 2011 at 7:06 AM, Sarah Goslee  wrote:
> I doubt your data frame looks like that, with all the -, but regardless you
> can use ifelse() to construct your column.
>
> Sarah
>
> On Saturday, December 3, 2011, syrvn  wrote:
>> Hello!
>>
>> I have a data.frame which looks like:
>>
>> Name - Value
>> A - 400
>> A - 300
>> B - 200
>> B - 350
>> C - 500
>> C - 350
>> D - 450
>> D - 600
>> E - 700
>> E - 750
>> F - 630
>> F - 650
>>
>> I want to add another column where all A,B should get an index 1, all C,D
> an
>> index of 2 and all E,F an index of 3 so that the data.frame looks like:
>>
>> ID - Name - Value
>> 1 - A - 400
>> 1 - A - 300
>> 1 - B - 200
>> 1 - B - 350
>> 2 - C - 500
>> 2 - C - 350
>> 2 - D - 450
>> 2 - D - 600
>> 3 - E - 700
>> 3 - E - 750
>> 3 - F - 630
>> 3 - F - 650
>>
>> My data.frame is quite big so I cannot add all values by hand.
>>
>>
>> Cheers
>>
>>
>>
>>
>> --
>> View this message in context:
> http://r.789695.n4.nabble.com/Data-alignment-tp4153024p4153024.html
>> Sent from the R help mailing list archive at Nabble.com.
>>
>> __
>> R-help@r-project.org mailing list
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>>
>
> --
> Sarah Goslee
> http://www.stringpage.com
> http://www.sarahgoslee.com
> http://www.functionaldiversity.org
>
>        [[alternative HTML version deleted]]
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>



-- 
Jim Holtman
Data Munger Guru

What is the problem that you are trying to solve?
Tell me what you want to do, not how you want to do it.

__
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Re: [R] help in removal of fixed pattern

[Sarah Goslee]

You can use gsub() to replace parts of strings.

Sarah

On Saturday, December 3, 2011, arunkumar  wrote:
> Hi
>
> I have column name as given below
>
> If the variable is in log(X1 + 1) pattern it should be removed and i need
> only X1
>
> Input
> log(x1 + 1)
> x2
> log(X3 +1)
>
> Expected Output X1 X2 X3
>
> Please help me
>
>
> --
> View this message in context:
http://r.789695.n4.nabble.com/help-in-removal-of-fixed-pattern-tp4152524p4152524.html
> Sent from the R help mailing list archive at Nabble.com.
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>

-- 
Sarah Goslee
http://www.stringpage.com
http://www.sarahgoslee.com
http://www.functionaldiversity.org

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Re: [R] Data alignment

[Sarah Goslee]

I doubt your data frame looks like that, with all the -, but regardless you
can use ifelse() to construct your column.

Sarah

On Saturday, December 3, 2011, syrvn  wrote:
> Hello!
>
> I have a data.frame which looks like:
>
> Name - Value
> A - 400
> A - 300
> B - 200
> B - 350
> C - 500
> C - 350
> D - 450
> D - 600
> E - 700
> E - 750
> F - 630
> F - 650
>
> I want to add another column where all A,B should get an index 1, all C,D
an
> index of 2 and all E,F an index of 3 so that the data.frame looks like:
>
> ID - Name - Value
> 1 - A - 400
> 1 - A - 300
> 1 - B - 200
> 1 - B - 350
> 2 - C - 500
> 2 - C - 350
> 2 - D - 450
> 2 - D - 600
> 3 - E - 700
> 3 - E - 750
> 3 - F - 630
> 3 - F - 650
>
> My data.frame is quite big so I cannot add all values by hand.
>
>
> Cheers
>
>
>
>
> --
> View this message in context:
http://r.789695.n4.nabble.com/Data-alignment-tp4153024p4153024.html
> Sent from the R help mailing list archive at Nabble.com.
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>

-- 
Sarah Goslee
http://www.stringpage.com
http://www.sarahgoslee.com
http://www.functionaldiversity.org

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Re: [R] partial mantel tests in ecodist with intential NA values.

[Sarah Goslee]

It is not possible in ecodist, and I'd be wary of doing it in other
packages. Instead, I would consider the model matrix approach described in
Legendre and Fortin 1989.

Sarah

On Saturday, December 3, 2011, Nevil Amos  wrote:
> I would like to perform partial mantel tests  on only within group
values, with "between group" values assigned to NA.
> This is possible in package ncf partial.mantel.test, however this sues a
different permutation to that used in ecodist.ecodist will not accept data
with NA values, returning a "matrix is not square error.
>
> is it possible to perform this test in ecodist?
>
> many thanks
>
> Nevil Amos
>
>> library(ecodist)
>> library(ncf)
>> x<-sample(1:1000,20)
>> y<-sample(1:1000,20)
>> z<-sample(1:1000,20)
>> M1<-as.matrix( distance(x))
>> M2 <-as.matrix( distance(y ))
>> M3<-as.matrix( distance(z ))
>> D1<-(lower(M1))
>> D2<-(lower(M2))
>> D3<-(lower(M3))
>> mantel(D1 ~ D2+D3, nperm=1000)
>   mantelr  pval1  pval2  pval3  llim.2.5% ulim.97.5%
> 0.09014696 0.1030 0.8980 0.1840 0.01857311 0.18468621
>> partial.mantel.test(M1,M2,M3,quiet=T)
> $MantelR
>r12 r13 r23   r12.3   r13.2
>  0.08977575  0.02170997 -0.01561346  0.09014696  0.02320821
>
> $p
> [1] 0.09590410 0.30769231 0.47552448 0.09490509 0.30169830
>
> $call
> [1] "partial.mantel.test(M1 = M1, M2 = M2, M3 = M3, quiet = T)"
>
> attr(,"class")
> [1] "partial.Mantel"
>> M1[1:10,11:20]<-NA
>> M1[11:20,1:10]<-NA
>> D1<-(lower(M1))
>> mantel(D1 ~ D2+D3, nperm=1000)
> Error in mantel(D1 ~ D2 + D3, nperm = 1000) : Matrix not square.
>> partial.mantel.test(M1,M2,M3,quiet=T)
> $MantelR
> r12  r13  r23r12.3r13.2
>  0.054906562  0.003446670 -0.015613460  0.054967403  0.004310979
>
> $p
> [1] 0.2837163 0.4275724 0.4555445 0.2857143 0.4235764
>
> $call
> [1] "partial.mantel.test(M1 = M1, M2 = M2, M3 = M3, quiet = T)"
>
> attr(,"class")
> [1] "partial.Mantel"
> Warning message:
> In partial.mantel.test(M1, M2, M3, quiet = T) :
>  Missing values exist; Pairwise deletion will be used
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>

-- 
Sarah Goslee
http://www.stringpage.com
http://www.sarahgoslee.com
http://www.functionaldiversity.org

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Re: [R] side-by-side map with different geographies using spplot

[Roger Bivand]

David Epstein  umich.edu> writes:

> 
> Hello,
> 
> I want to create side-by-side maps of similar attribute data in two 
> different cities using a single legend.
> 
> To simply display side-by-side census block group boundary 
> (non-thematic) maps for Minneapolis & Cleveland I do the following:
> 

> 
> I can display a single thematic map for a city using spplot as follows:
> 
> spplot(Minneapolis,"Thematic_Data_Column")
> 

The spplot function uses lattice graphics, so you may use the standard methods
for the "trellis" objects that it returns. If you did something like:

sp_out_Minneapolis <- spplot(Minneapolis, "your_var")
print(sp_out_Minneapolis, split=c(1, 1, 2, 1), more=TRUE)
sp_out_Cleveland <- spplot(Cleveland, "your_var")
print(sp_out_,Cleveland split=c(2, 1, 2, 1), more=FALSE)

you will get some of the way there. If in addition you would like only one
legend using the same breaks and colours, you'll need to use appropriate
arguments to spplot() in both calls to handle this.

It may be worth considering posting queries of this kind on R-sig-geo, as
response may be forthcoming more quickly there.

Hope this helps,

Roger

> 
> thank you,
> -david
> 
>

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[R] Data alignment

[syrvn]

Hello!

I have a data.frame which looks like:

Name - Value
A - 400
A - 300
B - 200
B - 350
C - 500
C - 350
D - 450 
D - 600
E - 700 
E - 750
F - 630
F - 650

I want to add another column where all A,B should get an index 1, all C,D an
index of 2 and all E,F an index of 3 so that the data.frame looks like:

ID - Name - Value
1 - A - 400
1 - A - 300
1 - B - 200
1 - B - 350
2 - C - 500
2 - C - 350
2 - D - 450 
2 - D - 600
3 - E - 700 
3 - E - 750
3 - F - 630
3 - F - 650

My data.frame is quite big so I cannot add all values by hand.


Cheers




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Re: [R] Project local libraries (reproducible research)

[Jim Lemon]

On 12/03/2011 06:04 AM, Hadley Wickham wrote:

Hi all,

I was wondering if any one had scripts that they could share for
capturing the current version of R packages used for a project. I'm
interested in creating a project local library so that you're safe if
someone (e.g. the ggplot2 author) updates a package you're relying on
and breaks your code.   I could fairly easily hack together, but I was
wondering if any one had any neat scripts they'd care to share.


Hi Hadley,
This makes quite a few assumptions, but I would build a list of packages 
called in R scripts:


# get all calls to library or require in a subdirectory of scripts
packagestrings<-c(system("grep library *.R",intern=TRUE),
 system("grep require *.R",intern=TRUE))
# get the unique package names
packagenames<-unique(sapply(strsplit(packagestrings,"[()]"),"[",2))
# get the information on all installed packages in the system
installedpackages<-installed.packages()
# get the indices of the packages used in the subdirectory
whichpackages<-which(match(rownames(installedpackages),packagenames,0)>0)
# get the version information for each used package
installedpackages[whichpackages,"Built"]

Jim

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Re: [R] Problem installing Rtools

[Duncan Murdoch]

On 11-12-02 10:25 PM, stillerfan wrote:

I'm a new user to R and recently installed R 2.14  for Windows 7 (64 bit).  I
have downloaded and ran Rtools 2.14 but do not see any new packages in the
install packages section of R.  I installed Rtools in a subdirectory to the
/library/ directory.  Does anyone have any idea what I could be doing wrong?



Rtools doesn't contain R packages, it contains various compilers and 
tools that you could use to build R and R packages.  So you should 
install it in its own directory.  I usually use c:/Rtools.


Duncan Murdoch

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Re: [R] Help! Big problem when using "browser()" to do R debugging?

[Liviu Andronic]

On Sat, Dec 3, 2011 at 2:38 AM, Michael  wrote:
> Hi all,
>
> Could you please help me?
>
> I am having the following weird problem when debugging R programs
> using "browser()":
>
> In my function, I've inserted a "browser()" in front of Step 1. My
> function has 3 steps and at the end of each step, it will print out
> the message "Step i is done"...
>
> However, after I hit  when the program stopped before Step 1
> and entered into the debugging mode, it not only executed the next
> line(i.e. the Step 1), but also all the (many) remaining lines in that
> function, as shown below:
>
>
> Browse[1]>
> [1] "Step 1 is done.."
> [1] "Step 2 is done.."
> [1] "Step 3 is done.."
>
> Then it automatically quited the debugging mode and when I tried to
> check the value of "myobj", I've got the following error message:
>
>> names(myobj)
> Error: object 'myobj' not found
> No suitable frames for recover()
>
> 
>
> So my question is: why did one key stroke  lead it to execute
> all the remaining lines in that function and then "returned" from the
> function and quited the debugging mode?
>
As soon as you get into 'debug' mode, type 'n', hit , then you
can simply hit  to evaluate the fun step by step. To exit the
step-by-step, type 'c' and hit enter: this will evaluate the rest of
the function.

Regards
Liviu


> Thanks a lot!
>
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> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.



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Re: [R] Downloading tab separated data from internet

[Prof Brian Ripley]

AFAICS what you mean is 'how can I fill in an HTML form using R'.
Answer: use package RCurl.

Do study the posting guide: none of the 'at a minimum' information was 
given here.



On 03/12/2011 04:47, HC wrote:

Hi all,

I am trying to download some tab separated data from the internet. The data
is not available directly at the URL that could be known apriori. There is
an intermediate form where start and end dates have to be given to get to
the required page.

For example, I want to download data for a station 03015795. The form for
this station is at:

http://ida.water.usgs.gov/ida/available_records.cfm?sn=03015795

I could get the start date and end date from this form using:

#
# Specifying station and reading from the opening form
stn<-"03015795"
myurl<-paste("http://ida.water.usgs.gov/ida/available_records.cfm?sn=",stn,sep="";)
mypage1 = readLines(myurl)

# Getting the start and end dates
mypattern = '([^<]*)'
datalines = grep(mypattern, mypage1[124], value=TRUE)
getexpr = function(s,g)substring(s,g,g+attr(g,'match.length')-1)
gg = gregexpr(mypattern, datalines)
matches = mapply(getexpr,datalines,gg)
result = gsub(mypattern,'\\1',matches)
names(result)=NULL
mydates<-result[1:2]

I want to know how I can feed these start and end dates to the form and
execute the button to go to the data page and then to download the data,
either as displayed in the browser or by saving as a file.

Any help on this is most appreciated.

Thanks.
HC



--
Brian D. Ripley,  rip...@stats.ox.ac.uk
Professor of Applied Statistics,  http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel:  +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UKFax:  +44 1865 272595

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[R] partial mantel tests in ecodist with intential NA values.

[Nevil Amos]

I would like to perform partial mantel tests  on only within group values, with 
"between group" values assigned to NA.
This is possible in package ncf partial.mantel.test, however this sues a 
different permutation to that used in ecodist.ecodist will not accept data with 
NA values, returning a "matrix is not square error.

is it possible to perform this test in ecodist?

many thanks

Nevil Amos

> library(ecodist)
> library(ncf)
> x<-sample(1:1000,20)
> y<-sample(1:1000,20)
> z<-sample(1:1000,20)
> M1<-as.matrix( distance(x))
> M2 <-as.matrix( distance(y ))
> M3<-as.matrix( distance(z ))
> D1<-(lower(M1))
> D2<-(lower(M2))
> D3<-(lower(M3))
> mantel(D1 ~ D2+D3, nperm=1000)
   mantelr  pval1  pval2  pval3  llim.2.5% ulim.97.5% 
0.09014696 0.1030 0.8980 0.1840 0.01857311 0.18468621 
> partial.mantel.test(M1,M2,M3,quiet=T)
$MantelR
r12 r13 r23   r12.3   r13.2 
 0.08977575  0.02170997 -0.01561346  0.09014696  0.02320821 

$p
[1] 0.09590410 0.30769231 0.47552448 0.09490509 0.30169830

$call
[1] "partial.mantel.test(M1 = M1, M2 = M2, M3 = M3, quiet = T)"

attr(,"class")
[1] "partial.Mantel"
> M1[1:10,11:20]<-NA
> M1[11:20,1:10]<-NA
> D1<-(lower(M1))
> mantel(D1 ~ D2+D3, nperm=1000)
Error in mantel(D1 ~ D2 + D3, nperm = 1000) : Matrix not square.
> partial.mantel.test(M1,M2,M3,quiet=T)
$MantelR
 r12  r13  r23r12.3r13.2 
 0.054906562  0.003446670 -0.015613460  0.054967403  0.004310979 

$p
[1] 0.2837163 0.4275724 0.4555445 0.2857143 0.4235764

$call
[1] "partial.mantel.test(M1 = M1, M2 = M2, M3 = M3, quiet = T)"

attr(,"class")
[1] "partial.Mantel"
Warning message:
In partial.mantel.test(M1, M2, M3, quiet = T) :
  Missing values exist; Pairwise deletion will be used
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[R] Reading multiple text files and then combining into a dataframe

[James Holland]

I have a multiple text files, separated by a single space, that I need to
combine into a single data.frame.  Below is the track I'm on using
list.files to capture the names of the files and then lapply with
read.table.  But I run into problems making a usable dataframe out of the
data.


#Creating example data in similar format to data I have
sub <- rep(1,10)
trial <- seq(1,10,1)
size <- rep(3,10)
shortest <- rep(444,10)
startlab <- rep(444,10)
endlab <- rep(444,10)
startconf <- rep(444,10)
endconf <- rep(444,10)
steps <- sample(1:30,10)
time <- sample(1:300,10)

#creating example of text file and saving into a shared director with other
txt files

subject1 <-
cbind(sub1,trial,size,shortest,startlab,endlab,startconf,endconf,steps,time)
write.table(subject1, "C:Folder/R/subject1.txt", sep=" ")

#2nd example of same text file
sub <- rep(2,10)
trial <- seq(1,10,1)
size <- rep(3,10)
shortest <- rep(444,10)
startlab <- rep(444,10)
endlab <- rep(444,10)
startconf <- rep(444,10)
endconf <- rep(444,10)
steps <- sample(1:30,10)
time <- sample(1:300,10)

subject1 <-
cbind(sub1,trial,size,shortest,startlab,endlab,startconf,endconf,steps,time)
write.table(subject1, "C:Folder/R/subject2.txt", sep=" ")


setwd("C:Folder/R/")

#Getting list of file names
file_name <- list.files(pattern="subject*")

mydata <- lapply (file_name, read.table, sep=" ", header=T, row.names=NULL)



Thank you,
James

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[R] help in removal of fixed pattern

[arunkumar1111]

Hi

I have column name as given below

If the variable is in log(X1 + 1) pattern it should be removed and i need
only X1

Input 
log(x1 + 1) 
x2
log(X3 +1)

Expected Output X1 X2 X3

Please help me


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[R] Problem installing Rtools

[stillerfan]

I'm a new user to R and recently installed R 2.14  for Windows 7 (64 bit).  I
have downloaded and ran Rtools 2.14 but do not see any new packages in the
install packages section of R.  I installed Rtools in a subdirectory to the
/library/ directory.  Does anyone have any idea what I could be doing wrong?

Thanks!

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[R] Downloading tab separated data from internet

[HC]

Hi all,

I am trying to download some tab separated data from the internet. The data
is not available directly at the URL that could be known apriori. There is
an intermediate form where start and end dates have to be given to get to
the required page.

For example, I want to download data for a station 03015795. The form for
this station is at:

http://ida.water.usgs.gov/ida/available_records.cfm?sn=03015795

I could get the start date and end date from this form using:

# 
# Specifying station and reading from the opening form
stn<-"03015795"
myurl<-paste("http://ida.water.usgs.gov/ida/available_records.cfm?sn=",stn,sep="";)
mypage1 = readLines(myurl)

# Getting the start and end dates
mypattern = '([^<]*)'
datalines = grep(mypattern, mypage1[124], value=TRUE)
getexpr = function(s,g)substring(s,g,g+attr(g,'match.length')-1)
gg = gregexpr(mypattern, datalines)
matches = mapply(getexpr,datalines,gg)
result = gsub(mypattern,'\\1',matches)
names(result)=NULL
mydates<-result[1:2]

I want to know how I can feed these start and end dates to the form and
execute the button to go to the data page and then to download the data,
either as displayed in the browser or by saving as a file.

Any help on this is most appreciated.

Thanks.
HC







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Re: [R] simple lm question

[Worik R]

>
>   Please note that your "df" and "M" are undoubtedly different
> objects by now:
>
> Right.  Not my most coherent day.

thanks
W



> > M <- matrix(runif(5*20), nrow=20)
> > colnames(M) <- c('a', 'b', 'c', 'd', 'e')
> > l1 <- lm(e~., data=as.data.frame(M))
> > l1
>
> Call:
> lm(formula = e ~ ., data = as.data.frame(M))
>
> Coefficients:
> (Intercept)abcd
>0.40139 -0.15032 -0.06242  0.13139  0.23905
>
>
> > l3 <- lm(M[,5]~M[,1]+M[,2]+M[,3]+M[,**4])
> > l3
>
>
> Call:
> lm(formula = M[, 5] ~ M[, 1] + M[, 2] + M[, 3] + M[, 4])
>
> Coefficients:
> (Intercept)   M[, 1]   M[, 2]   M[, 3]   M[, 4]
>0.40139 -0.15032 -0.06242  0.13139  0.23905
>
> As expected.
>
> --
> David.
>
>
>> On Sat, Dec 3, 2011 at 5:10 PM, R. Michael Weylandt <
>> michael.weyla...@gmail.com> wrote:
>>
>>  In your code by supplying a vector M[,"e"] you are regressing "e"
>>> against all the variables provided in the data argument, including "e"
>>> itself -- this gives the very strange regression coefficients you
>>> observe. R has no way to know that that's somehow related to the "e"
>>> it sees in the data argument.
>>>
>>>
>>  In the suggested way,
>>>
>>> lm(formula = e ~ ., data = as.data.frame(M))
>>>
>>> e is regressed against everything that is not e and sensible results are
>>> given.
>>>
>>>
>> But still 'l1 <- lm(e~., data=df)' is not the same as 'l3 <-
>> lm(M[,5]~M[,1]+M[,2]+M[,3]+M[,**4])'
>>
>>  M <- matrix(runif(5*20), nrow=20)
>>> colnames(M) <- c('a', 'b', 'c', 'd', 'e')
>>> l1 <- lm(e~., data=df)
>>> summary(l1)
>>>
>>
>> Call:
>> lm(formula = e ~ ., data = df)
>>
>> Residuals:
>>Min   1Q   Median   3Q  Max
>> -0.38343 -0.21367  0.03067  0.13757  0.49080
>>
>> Coefficients:
>>   Estimate Std. Error t value Pr(>|t|)
>> (Intercept)  0.285210.29477   0.9680.349
>> a0.092830.30112   0.3080.762
>> b0.239210.22425   1.0670.303
>> c   -0.160270.24154  -0.6640.517
>> d0.240250.20054   1.1980.250
>>
>> Residual standard error: 0.2871 on 15 degrees of freedom
>> Multiple R-squared: 0.1602,Adjusted R-squared: -0.06375
>> F-statistic: 0.7153 on 4 and 15 DF,  p-value: 0.5943
>>
>>  l3 <- lm(M[,5]~M[,1]+M[,2]+M[,3]+M[,**4])
>>> summary(l3)
>>>
>>
>> Call:
>> lm(formula = M[, 5] ~ M[, 1] + M[, 2] + M[, 3] + M[, 4])
>>
>> Residuals:
>>Min   1Q   Median   3Q  Max
>> -0.36355 -0.22679 -0.01202  0.18462  0.37377
>>
>> Coefficients:
>>   Estimate Std. Error t value Pr(>|t|)
>> (Intercept)  0.769720.24501   3.142  0.00672 **
>> M[, 1]  -0.238300.24123  -0.988  0.33890
>> M[, 2]  -0.020460.21958  -0.093  0.92699
>> M[, 3]  -0.295180.22559  -1.308  0.21040
>> M[, 4]  -0.315450.24570  -1.284  0.21866
>> ---
>> Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
>>
>> Residual standard error: 0.2668 on 15 degrees of freedom
>> Multiple R-squared: 0.2762,Adjusted R-squared: 0.08317
>> F-statistic: 1.431 on 4 and 15 DF,  p-value: 0.272
>>
>>
>>>
>>[[alternative HTML version deleted]]
>>
>> __**
>> R-help@r-project.org mailing list
>> https://stat.ethz.ch/mailman/**listinfo/r-help
>> PLEASE do read the posting guide http://www.R-project.org/**
>> posting-guide.html 
>> and provide commented, minimal, self-contained, reproducible code.
>>
>
> David Winsemius, MD
> West Hartford, CT
>
>

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