date:20110601


On 11-06-01 7:26 AM, Jannis wrote:

Dear list members,


is it possible to retrieve the name of a function or the exact call to that function from 
the environment created inside this function? I wrote a function for 
debugging on a cluster that saves all the objects inside the whole stack of function 
environments and would also like to retrieve the names of the respective functions. 
Perhaps also a method to simply retrieve the name of a function inside this function 
could give me hints to the solution.


I'm not sure I understand what you're asking, but I think the answer is 
no.  You can retrieve the name of a function by calling sys.call() (or a 
related function for more information, see ?sys.call), but that has to 
be called in the context of the function.  The information is not stored 
in the active environment, you'll need to save more than that.  I think 
sys.calls() will give you the information you need.


Duncan Murdoch


Thanks
Jannis

__
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] Missing completely at random

2011-06-01 Thread Blaz Simcic

Hello!
I would like to sample 5 % of cases and from 1 to 3 variables within selected 
cases and set them as NA (MCAR-Missing completely at random). I managed to 
sample cases and variables, but I donât know how to set them as NA.
R code:
N - 1000 Â Â Â Â Â number of cases 
n - 12 Â Â Â Â Â Â Â Â Â Â number of variables
X - matrix(rnorm(N * n), N, n)Â Â Â  matrix
pMiss - 0.05Â Â Â Â  percent of missing values
idMiss - sample(1:N, N * pMiss)Â Â Â  sample cases
nMiss - length(idMiss)
m - 3Â Â Â  maximum number of missing variables within selected cases
howmanyMiss - sapply(idMiss, function(x) sample(1:m, 1))
howmanyMiss
lapply(howmanyMiss, function(x) sample(1:n, x))
Â 
Any suggestion?
Thanks,
BlaÅ¾
[[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.

Re: [R] Import/convert PMML to R model

2011-06-01 Thread Raji

Hi R-helpers,

 Can you please let me know if it is possible to import a PMML in R? If yes,
can you give me the command to do the same? If not, can you tell me the
reason why?

Many thanks,
Raji

--
View this message in context: 
http://r.789695.n4.nabble.com/Import-convert-PMML-to-R-model-tp3332772p3565260.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.

[R] treemaps

2011-06-01 Thread Richard Slater


Hello, 
I've created a treemap using the treemap package (tmPlot) showing distribution 
of my data across 3 countries. The boxes created within the country are 
proportionate to my data; however how do I make the 3 countries proportionate 
to the total of all 3 as well? Currently the 3 countries are the same size, 
this implies they all
have the same endowment so to speak, which is not the case.
Cheers
  
[[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.

[R] Problem with as.POSIXct()

2011-06-01 Thread circe

Hi, when I type in these words:

 a=c(2011-06-01 17:21:24.83, 2011-06-01 17:21:24.283)
 as.POSIXct(a)

the real output is:
   [1] 2011-06-01 17:21:24.830 CST 2011-06-01 17:21:24.283 CST
   
rather than the expected one:
   [1] 2011-06-01 17:21:24.083 CST 2011-06-01 17:21:24.283 CST

How can I deal with such milliseconds less than 100?

Thanks



--
View this message in context: 
http://r.789695.n4.nabble.com/Problem-with-as-POSIXct-tp3565424p3565424.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.

Re: [R] Identifying sequences



On Jun 1, 2011, at 6:27 AM, christiaan pauw wrote:


Hallo Everybody

Consider the following vector

a=1:10
b=20:30
c=40:50
x=c(a,b,c)

I need a function that can tell me that there are three set of  
continuos
sequences and that the first is from 1:10, the second from 20:30 and  
the

third from 40:50. In other words: a,b, and c.


You probably want something like

 which(diff(x) 1)
[1] 10 21

Or perhaps what the rle function provides.

?diff
?rle

--

David Winsemius, MD
West Hartford, CT

__
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] Identifying sequences

2011-06-01 Thread ONKELINX, Thierry

Something like this?

a=1:10
b=20:30
c=40:50
x=c(a,b,c)
borders - which(diff(x) != 1)
seqs - data.frame(start = c(1, borders + 1), end = c(borders, length(x)))

Best regards,

Thierry

 -Oorspronkelijk bericht-
 Van: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
 Namens christiaan pauw
 Verzonden: woensdag 1 juni 2011 12:27
 Aan: r-help@r-project.org
 Onderwerp: [R] Identifying sequences
 
 Hallo Everybody
 
 Consider the following vector
 
 a=1:10
 b=20:30
 c=40:50
 x=c(a,b,c)
 
 I need a function that can tell me that there are three set of continuos
 sequences and that the first is from 1:10, the second from 20:30 and the third
 from 40:50. In other words: a,b, and c.
 
 regards
 Christiaan
 
   [[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.

__
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] Identifying sequences

2011-06-01 Thread Jonathan Daily

I am assuming in this case that you are looking for continuity along
integers, so if you expect noninteger values this will not work.

You can get the index of where breaks can be found in your example using

which(diff(x)  1)

On Wed, Jun 1, 2011 at 6:27 AM, christiaan pauw cjp...@gmail.com wrote:
 Hallo Everybody

 Consider the following vector

 a=1:10
 b=20:30
 c=40:50
 x=c(a,b,c)

 I need a function that can tell me that there are three set of continuos
 sequences and that the first is from 1:10, the second from 20:30 and the
 third from 40:50. In other words: a,b, and c.

 regards
 Christiaan

        [[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.




-- 
===
Jon Daily
Technician
===
#!/usr/bin/env outside
# It's great, trust me.

__
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] Retaining MatchIt output as a dataset

2011-06-01 Thread Shane Phillips

Hi!

I am running a simulation testing group balance using different propensity 
score matching techniques.  I am using the MatchIt package to match and analyze 
the data.  Since I am running this analysis many, many times with different 
simulated data sets, I need to make the output from the MatchIt summary and put 
it into a dataset and then append each successive run to the end of that 
dataset.  How can I do that?

Any help would be appreciated.

Thanks!

Shane


[[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.

[R] graphical output - customization of x axis

2011-06-01 Thread Christine SINOQUET


Hello,

I do not understand why the following source does not success in 
customizing the x axis. The wished customization just consists in 
labelling the x ticks with 
0kb,1kb,5kb,10kb,20kb,50kb,100kb,inf.


pathFileName - test1.png


stats4H - c(0.8623, 0.7142, 0.7211, 0.6753, 0.6290, 0.6157, 0.5919, 0.5218)
stats4W - c(0.8724, 0.7889, 0.7253, 0.7235, 0.6860, 0.6630, 0.6424, 0.5228)

stats5H - c(0.7610, 0.6832, 0.6310, 0.5527, 0.5231, 0.5600, 0.6147, 0.5131)

stats5W - c(0.7740, 0.7641, 0.7099, 0.6918, 0.6483, 0.5769, 0.5539, 0.4934)


 plotType - b


 png(file=pathFileName, bg=transparent)

 xxf - c(0kb,1kb,5kb,10kb,20kb,50kb,100kb,inf)
 xf  - 1:8

 par(lab = c(length(xxf), 5, 7))
 #c(x, y, len)
 # ‘lab’ A numerical vector of the form ‘c(x, y, len)’ which modifies
 # the default way that axes are annotated. The values of ‘x’
 # and ‘y’ give the (approximate) number of tickmarks on the x
 # and y axes and ‘len’ specifies the label length.  The default
 # is ‘c(5, 5, 7)’.  Note that this only affects the way the
 # parameters ‘xaxp’ and ‘yaxp’ are set when the user coordinate
 # system is set up, and is not consulted when axes are drawn.
 # ‘len’ _is unimplemented_ in R.


 plot(stats4H, xlab = half-width of window, ylab=AUC, 
type=plotType, pch=7, col=blue, cex = 0.5, ylim=c(0.4,1),xlim=c(1,8), 
xaxt = n) # without x axis


 #If you are going to create a custom axis, you should suppress the 
axis automatically generated by your high level plotting function. The 
option axes=FALSE suppresses both x and y axes. xaxt=n and yaxt=n 
suppress the x and y axis respectively.


positions - 1:8
 axis(side = 1, at=positions, labels=xxf , pos=0, lty=1, 
col.axis=black, las=0)


 # side: bottom (1)
 # at: the points at which tick-marks are to be drawn
 # pos:coordinate of defined axis on the other axis
 # lty: linetype
 # col: line and tick marker color
 # las: labels are parallel (=0) or perpendicular(=2) to axis
 # tck: length of tick mark as fraction of plotting region (negative 
number is outside graph,
 #  positive number is inside, 0 suppresses ticks, 1 creates 
gridlines) default is -0.01


 lines(x = xf, y = stats4W, type=plotType, pch=15, col=green, cex = 0.5)
 lines(x = xf, y = stats5H, type=plotType, pch=16, col=red, cex = 0.5)
 lines(x = xf, y = stats5W, type=plotType, pch=17, col=magenta,cex = 0.5)


 legend(bottomright,
legend = c(H, statistic S1, W, statistic S1, H, statistic 
S2, W, statistic S2),

col = c(blue, green, red, magenta),
pch = c(7,15,16,17),
bty = n,
cex = 0.8,
inset = c(0.1, 0.1))

 dev.off()


I thank you in advance for any help.

Best regards,

Christine Sinoquet

__
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] Problem with as.POSIXct()

2011-06-01 Thread Stefan Peterson

circe tju_circe at hotmail.com writes:

 
 Hi, when I type in these words:
 
  a=c(2011-06-01 17:21:24.83, 2011-06-01 17:21:24.283)
  as.POSIXct(a)
 
 the real output is:
[1] 2011-06-01 17:21:24.830 CST 2011-06-01 17:21:24.283 CST
 
 rather than the expected one:
[1] 2011-06-01 17:21:24.083 CST 2011-06-01 17:21:24.283 CST
 
 How can I deal with such milliseconds less than 100?
 
 Thanks
 
 --
 View this message in context:
http://r.789695.n4.nabble.com/Problem-with-as-POSIXct-tp3565424p3565424.html
 Sent from the R help mailing list archive at Nabble.com.
 
 


Well, to me the R is behaving as expected. I.e. .83 = .830. If Your dates come
in a format like You describe, You need to manually add a leading zero to Your
milliseconds.

A suggestion to work from:

for(i in 1:length(a)){
  if(nchar(a[i]) == 23){
print(a[i])
  }else{
print( paste(substr(a[i],1,20),0,substr(a[i],21,22),sep='') ) 
  }
}

__
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] graphical output - customization of x axis


Take out the pos=0 argument in your call to axis.

--  
David.

On Jun 1, 2011, at 8:56 AM, Christine SINOQUET wrote:


Hello,

I do not understand why the following source does not success in  
customizing the x axis. The wished customization just consists in  
labelling the x ticks with  
0kb,1kb,5kb,10kb,20kb,50kb,100kb,inf.


pathFileName - test1.png


stats4H - c(0.8623, 0.7142, 0.7211, 0.6753, 0.6290, 0.6157, 0.5919,  
0.5218)
stats4W - c(0.8724, 0.7889, 0.7253, 0.7235, 0.6860, 0.6630, 0.6424,  
0.5228)


stats5H - c(0.7610, 0.6832, 0.6310, 0.5527, 0.5231, 0.5600, 0.6147,  
0.5131)


stats5W - c(0.7740, 0.7641, 0.7099, 0.6918, 0.6483, 0.5769, 0.5539,  
0.4934)



plotType - b


png(file=pathFileName, bg=transparent)

xxf - c(0kb,1kb,5kb,10kb,20kb,50kb,100kb,inf)
xf  - 1:8

par(lab = c(length(xxf), 5, 7))
#c(x, y, len)
# ‘lab’ A numerical vector of the form ‘c(x, y, len)’ which modifies
# the default way that axes are annotated. The values of ‘x’
# and ‘y’ give the (approximate) number of tickmarks on the x
# and y axes and ‘len’ specifies the label length.  The default
# is ‘c(5, 5, 7)’.  Note that this only affects the way the
# parameters ‘xaxp’ and ‘yaxp’ are set when the user coordinate
# system is set up, and is not consulted when axes are drawn.
# ‘len’ _is unimplemented_ in R.


plot(stats4H, xlab = half-width of window, ylab=AUC,  
type=plotType, pch=7, col=blue, cex = 0.5,  
ylim=c(0.4,1),xlim=c(1,8), xaxt = n) # without x axis


#If you are going to create a custom axis, you should suppress the  
axis automatically generated by your high level plotting function.  
The option axes=FALSE suppresses both x and y axes. xaxt=n and  
yaxt=n suppress the x and y axis respectively.


positions - 1:8
axis(side = 1, at=positions, labels=xxf , pos=0, lty=1,  
col.axis=black, las=0)


# side: bottom (1)
# at: the points at which tick-marks are to be drawn
# pos:coordinate of defined axis on the other axis
# lty: linetype
# col: line and tick marker color
# las: labels are parallel (=0) or perpendicular(=2) to axis
# tck: length of tick mark as fraction of plotting region (negative  
number is outside graph,
#  positive number is inside, 0 suppresses ticks, 1 creates  
gridlines) default is -0.01


lines(x = xf, y = stats4W, type=plotType, pch=15, col=green, cex =  
0.5)
lines(x = xf, y = stats5H, type=plotType, pch=16, col=red, cex =  
0.5)
lines(x = xf, y = stats5W, type=plotType, pch=17, col=magenta,cex  
= 0.5)



legend(bottomright,
   legend = c(H, statistic S1, W, statistic S1, H,  
statistic S2, W, statistic S2),

   col = c(blue, green, red, magenta),
   pch = c(7,15,16,17),
   bty = n,
   cex = 0.8,
   inset = c(0.1, 0.1))

dev.off()


I thank you in advance for any help.

Best regards,

Christine Sinoquet

__
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

__
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] Simulating SVAR Data

2011-06-01 Thread Downey, Patrick

Hello,

I'd like to simulate data according to an SVAR model in order to
demonstrate how other techniques (such as arima) yield biased estimates. I
am interested in a 2 variable SVAR with 2 lags (in the notation of the vars
vignette, K = 2, P = 2, where B = I_K). I'm using the {vars} package
outlined here:
http://cran.r-project.org/web/packages/vars/vignettes/vars.pdf

I thought that the following would generate the data and demonstrate the
accuracy of an SVAR compared to an arima, but the results are not what I
expected. I think the problem is the way that I'm generating the data, but
I don't understand what I could be doing wrong. Any guidance would be
greatly appreciated.

Problems:

In the code below, the array means should show that SVAR parameters are
unbiased estiamtors, so the second column of means should be near 0, and
it's not any closer than the first column (the arima results). So my
results are no less biased than the arima results.

The first pair of plots should show the same: SVAR results are unbiased.
But they don't cluster around the red dots (true parameter values), so they
aren't.

Finally, the second pair of plots should show the parameters are
consistent: that MSE decreases as sample size increases. They don't really
show that. Perhaps they would if the smallest sample were smaller than 200,
but the SVAR tends not to converge with fewer observations. Further, the
MSE tends to be at least as high with the SVAR compared to arima, so it's
not any more accurate.

Program:

# Model #
# Y(t) = a0 + a1*Y(t-1) + a2*Y(t-2) + a3*X(t-1) + a4*X(t-2) + e(t)
# X(t) = b0 + b1*X(t-1) + b2*X(t-2) + b3*Y(t-1) + b4*Y(t-2) + b5*Y(t) +
d(t)
# e(t)  d(t) ~ N(0,s)
# So Y has a contemporaneous impact on X
# X only has an impact on future Ys
# So this is the setup of a SVAR

# Choosing parameters #

# Currently, all parameters are just random numbers less than 1 so that 
# it's a stationary series
# The standard deviations of the error terms are also random ~ U(0,2)

a0 - runif(1,-1,1)
a1 - runif(1,-0.9,0.9)
a2 - runif(1,-abs(a1),abs(a1))
a3 - runif(1,-0.9,0.9)
a4 - runif(1,-abs(a3),abs(a3))
s.e - runif(1,0,2)

b0 - runif(1,-1,1)
b1 - runif(1,-0.9,0.9)
b2 - runif(1,-abs(b1),abs(b1))
b3 - runif(1,-0.9,0.9)
b4 - runif(1,-abs(b3),abs(b3))
b5 - runif(1,-0.9,0.9)
s.d - runif(1,0,2)


### A Formal Test: Loop #

Z - 100
error - array(0,dim=c(Z,2,6))
N - runif(Z,200,1200)
# Let the sample size vary so that we can check for consistency


# Generating the data #

# Start with 2 initial values and then create a dataset with 1200
observations

X - rep(0,1202)
Y - rep(0,1202)

Y[1] - rnorm(1, a0 + a1*a0 + a2*a0 + a3*b0 + a4*b0, s.e)
X[1] - rnorm(1, b0 + b1*b0 + b2*b0 + b3*a0 + b4*a0 + b5*Y[1], s.d)

Y[2] - rnorm(1, a0 + a1*Y[1] + a2*a0 + a3*X[1] + a4*b0, s.e)
X[2] - rnorm(1, b0 + b1*X[1] + b2*b0 + b3*Y[1] + b4*a0 + b5*Y[2], s.d)

for(t in 3:1202){
 Y[t] - rnorm(1, a0 + a1*Y[t-1] + a2*Y[t-2] + 
 a3*X[t-1] + a4*X[t-2], s.e)
 X[t] - rnorm(1, b0 + b1*X[t-1] + b2*X[t-2] + 
 b3*Y[t-1] + b4*Y[t-2] + b5*Y[t], s.d)
}

L1.Y - rep(NA,1202)
L2.Y - rep(NA,1202)
for(t in 2:1202){
 L1.Y[t] - Y[t-1]
 L2.Y[t] - L1.Y[t-1]
}



for(z in 1:Z){

n - N[z]
x - X[3:(n+2)]
y - Y[3:(n+2)]
L1.y - c(NA,L1.Y[4:(n+2)])
L2.y - c(NA,NA,L2.Y[5:(n+2)])


# Modeling x with inclusion of y #

m2 - arima(x,order=c(2,0,0),xreg=cbind(L1.y,L2.y,y))
error[z,1,1] - m2$coef[3] - b0
error[z,1,2:3] - m2$coef[1:2] - c(b1,b2)
error[z,1,4:6] - m2$coef[4:6] - c(b3,b4,b5)

# SVAR of x and y #

m3 - VAR(cbind(y,x),p=2)
A - matrix(c(1,NA,0,1),ncol=2)
m4 - SVAR(m3,Amat=A)

error[z,2,1] - m4$var$varresult$x$coeff[5] - b0
error[z,2,2] - m4$var$varresult$x$coeff[2] - b1
error[z,2,3] - m4$var$varresult$x$coeff[4] - b2
error[z,2,4] - m4$var$varresult$x$coeff[1] - b3
error[z,2,5] - m4$var$varresult$x$coeff[3] - b4
error[z,2,6] - m4$A[2,1] - b5

}

mse - error^2

means - array(0,dim=c(6,2))
for(i in 1:2){
 means[,i] - apply(error[,i,],2,mean)
}
means



par(mfrow=c(2,1))
for(i in 1:2){
 plot(1+runif(Z,-0.4,0.4),error[,i,1],xlim=c(0.5,6.5),
xaxt='n',ylim=c(-1,1),ylab='',xlab='',main=paste(Model ,i))
  for(j in 2:6){
   points(j+runif(Z,-0.4,0.4),error[,i,j])
  }
 points(seq(1,6,1),c(b0,b1,b2,b3,b4,b5),col=2)
}
par(mfrow=c(1,1))



par(mfrow=c(2,1))
for(i in 1:2){
 plot(N,mse[,i,1],ylim=c(0,max(mse[,i,2:6])),main=paste(Model ,i))
  for(j in 2:6){
   points(N,mse[,i,j],col=j)
}}
par(mfrow=c(1,1))

__
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] error in model specification for cfa with lavaan-package

2011-06-01 Thread yrosseel


Dear Alain,

As for the first error (sample covariance can not be inverted): Mike 
is right: with only 10 observations and 16 variables, the ML estimation 
of the sample cov produces a covariance matrix that is not positive 
definite, and hence the inversion (deliberately) fails.


The lesson for me is that lavaan should simply check for this situation 
(p  N), and refuse to proceed (with a gentle and more informative error 
message).


As for the second error (Error in start.idx[i]:end.idx[i] : NA/NaN 
Argument): the first argument of cfa() must be a model syntax, not a 
lavaanified object.


The 'lavaanify' function is useful to learn about lavaan, and to export 
the model specification to other software packages, but you should not 
pass a 'lavaanified' object to the cfa/sem/growth functions directly 
(although this may work in the future).


Moreover, to get a 'complete' parameter list, you need to specify a 
number of additional arguments:


model - model - lavaanify(cfa.model, auto.fix.first = TRUE, auto.var = 
TRUE, auto.cov.lv.x = TRUE)


(see the 'Details' section in the man page of cfa() for a complete list).

Hope this helps,

Yves.


On 06/01/2011 12:31 PM, D. Alain wrote:

Dear R-List,

(I am not sure whether this list is the right place for my
question...)

I have a dataframe df.cfa

df.cfa-data.frame(x1=c(5,4,1,5,5,NA,4,NA,NA,5),x2=c(2,3,3,3,NA,1,2,1,2,1),x3=c(5,3,4,1,5,5,5,5,5,5),x4=c(5,3,4,5,5,5,5,5,5,5),x5=c(5,4,3,3,4,4,4,5,NA,5),x6=c(3,5,2,1,4,NA,NA,5,3,4),x7=c(5,4,3,4,4,3,4,NA,NA,4),x8=c(5,5,3,4,5,4,5,4,5,5),x9=c(5,1,1,1,NA,3,3,2,2,3),x10=c(5,5,2,3,4,3,1,3,2,5),x11=c(5,4,4,5,5,4,5,4,5,5),x12=c(5,4,5,5,5,4,5,5,5,5),x13=c(5,4,4,5,5,5,5,5,5,5),x14=c(5,3,3,NA,5,4,5,3,NA,5),x15=c(4,4,4,3,4,5,2,4,5,5),x16=c(5,4,3,3,4,4,3,3,4,5))





and I want to carry out a confirmatory factor analysis using the
cfa function (lavaan).

library(lavaan)

then I specify my model

cfa.model-'  f1=~x1+x2+x3+x4+x5 f2=~x6+x7+x8+x9+x10+x11
f3=~x12+x13+x14+x15+x16 '

and run the cfa routine

fit-cfa(cfa.model,data=df.cfa,missing=ml)

The output is an error message (here partly in german)

Error in chol.default(S) : der führende Minor der Ordnung 4 ist
nicht positiv definit Fehler in Sample(data = data, group = group,
sample.cov = sample.cov, sample.mean = sample.mean,  : sample
covariance can not be inverted

Then I tried to lavaanify my model specification first

cfa.model- lavaanify(cfa.model)

id lhs op rhs user group free ustart fixed.x   label eq.id
free.uncon 1   1  f1 =~  x11 11 NA   0  f1=~x1 0
1 2   2  f1 =~  x21 12 NA   0  f1=~x2 0
2 3   3  f1 =~  x31 13 NA   0  f1=~x3 0
3 4   4  f1 =~  x41 14 NA   0  f1=~x4 0
4 5   5  f1 =~  x51 15 NA   0  f1=~x5 0
5 6   6  f2 =~  x61 16 NA   0  f2=~x6 0
6 7   7  f2 =~  x71 17 NA   0  f2=~x7 0
7 8   8  f2 =~  x81 18 NA   0  f2=~x8 0
8 9   9  f2 =~  x91 19 NA   0  f2=~x9 0
9 10 10  f2 =~ x101 1   10 NA   0 f2=~x10 0
10 11 11  f2 =~ x111 1   11 NA   0 f2=~x11 0
11 12 12  f3 =~ x121 1   12 NA   0 f3=~x12 0
12 13 13  f3 =~ x131 1   13 NA   0 f3=~x13 0
13 14 14  f3 =~ x141 1   14 NA   0 f3=~x14 0
14 15 15  f3 =~ x151 1   15 NA   0 f3=~x15 0
15 16 16  f3 =~ x161 1   16 NA   0 f3=~x16 0
16

I run cfa again

fit-cfa(cfa.model,data=df.cfa,missing=ml)

And this time I get another error information

Error in start.idx[i]:end.idx[i] : NA/NaN Argument

I must admit that I am stuck, can anyone help?

Thanks, Alain [[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.


__
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] interpolation and extremum location of a surface‏

2011-06-01 Thread Clement LAUZIN



Hello,
 
I have a x,y,z file.Z is not corresponding to a simple analytical function of x 
and y (see below).
I am trying to find the minimum location of this surface and the z value 
corresponding to this location by a spline interpolation or from a polynomial 
fit.
I tried with the akima package but as the location of the point I am looking 
for (the minimum) is outside of the convex hull it's not giving any answer.
If someone has any idea or any suggestion?
 
Thank you in advance Pierre 
 
 
xyz
4.1 60 -152.1719593
4.1 75 -171.136801
4.1 90 -170.4604774
4.2 60 -168.7745552
4.2 75 -174.9667333
4.2 90 -172.1853334
4.3 60 -173.7736418
4.3 75 -171.6712745
4.3 90 -167.6662458
4.05 60 -137.8379387
4.15 60 -162.2264066
4.25 60 -172.4453286
4.35 60 -173.2123715
4.05 67.5 -158.8239625
4.1 67.5 -167.314534
4.15 67.5 -172.586182
4.2 67.5 -175.2217594
4.25 67.5 -175.7162683
4.3 67.5 -174.4890566
4.35 67.5 -171.8940061
4.05 75 -165.4388778
4.15 75 -174.1460392
4.25 75 -174.022344
4.35 75 -168.2149168
4.05 82.5 -166.4026077
4.1 82.5 -170.9199652
4.15 82.5 -172.9923449
4.2 82.5 -173.0803255
4.25 82.5 -171.5739101
4.3 82.5 -168.8024715
4.35 82.5 -165.0431276
4.05 90 -166.2592978
4.15 90 -172.2861302
4.25 90 -170.5383652
4.35 90 -163.8389615

__
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] Force Sweave to add figure filename extension

2011-06-01 Thread Renaud Gaujoux


Hi,

is there a way to force Sweave to add the extension (say .pdf) to the 
image names included by \includegraphics.
I understand that the objective of not putting the extension is to allow 
the compilation with both latex or pdflatex.
However, when option eps=false, the extension could optionally be forced 
to .pdf.
I am having troubles submitting a paper to a journal, whose latex system 
does not compile the file properly if the extensions are not explicitly 
given.

Thank you.

Renaud




###
UNIVERSITY OF CAPE TOWN 


This e-mail is subject to the UCT ICT policies and e-mai...{{dropped:5}}

__
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] Force Sweave to add figure filename extension


On 01/06/2011 10:13 AM, Renaud Gaujoux wrote:

Hi,

is there a way to force Sweave to add the extension (say .pdf) to the
image names included by \includegraphics.
I understand that the objective of not putting the extension is to allow
the compilation with both latex or pdflatex.
However, when option eps=false, the extension could optionally be forced
to .pdf.
I am having troubles submitting a paper to a journal, whose latex system
does not compile the file properly if the extensions are not explicitly
given.
Thank you.


Name the chunk and use include=false, then put in your own 
\includegrpahics call.  For example


figname, fig=true, include=false=
plot(1:10)
@

\includegraphics{figname.pdf}

Duncan Murdoch


Renaud




###
UNIVERSITY OF CAPE TOWN

This e-mail is subject to the UCT ICT policies and e-mai...{{dropped:5}}

__
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.

Re: [R] Force Sweave to add figure filename extension

2011-06-01 Thread Renaud Gaujoux


Hi,

thank you Duncan for your prompt response.
I was about to post another solution that worked fine for me (and the 
journal's system).

Add in the preamble the command:

\DeclareGraphicsExtensions{.pdf}

Bests,
Renaud

On 01/06/2011 16:18, Duncan Murdoch wrote:

On 01/06/2011 10:13 AM, Renaud Gaujoux wrote:

Hi,

is there a way to force Sweave to add the extension (say .pdf) to the
image names included by \includegraphics.
I understand that the objective of not putting the extension is to allow
the compilation with both latex or pdflatex.
However, when option eps=false, the extension could optionally be forced
to .pdf.
I am having troubles submitting a paper to a journal, whose latex system
does not compile the file properly if the extensions are not explicitly
given.
Thank you.


Name the chunk and use include=false, then put in your own 
\includegrpahics call.  For example


figname, fig=true, include=false=
plot(1:10)
@

\includegraphics{figname.pdf}

Duncan Murdoch


Renaud




###
UNIVERSITY OF CAPE TOWN

This e-mail is subject to the UCT ICT policies and e-mai...{{dropped:5}}

__
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.






###
UNIVERSITY OF CAPE TOWN 


This e-mail is subject to the UCT ICT policies and e-mai...{{dropped:5}}

__
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] In a formula, what is the interaction of the intercept and a factor?

2011-06-01 Thread Kevin Wright

On Wed, Jun 1, 2011 at 1:19 AM, Prof Brian Ripley rip...@stats.ox.ac.ukwrote:

[text deleted]


 Note that you usually want to do '*' when you say 'interact with':

 model.matrix(y ~ (1+x)*group, data = dat)

   (Intercept)  x groupB x:groupB
 11  1  11
 ...


Thanks for that suggestion.  To follow up,

R model.matrix(y ~ (1+x)*group, data = dat)
   (Intercept)  x groupB x:groupB
11  1  00
21  2  12
...

but if I delete 'x',

R model.matrix(y ~ (1)*group, data = dat)
   (Intercept)
11
21
...

Why is group simply vanishing in the second model?  I tried to dig into
this until I hit .Internal(model.matrix


Kevin





 On Tue, 31 May 2011, Kevin Wright wrote:

  For a pedagogical purpose, I was trying to show how the formula for a
 simple
 regression line (~1+x) could be crossed with a factor (~1:group + x:group)
 to fit separate regressions by group.  For example:

 set.seed(201108)
 dat - data.frame(x=1:15, y=1:15+rnorm(15),
   group = sample(c('A','B'), size=15,
   replace=TRUE))

 m1 - lm(y~ 1 + x, data=dat)
 m2 - lm(y ~ group + x:group, data=dat)
 m3 - lm(y ~ 1:group + x:group, data=dat)
 m4 - lm(y ~ 1 + x:group, data=dat)

 The simple regression is model m1.

 The usual way to write the by-group regression is model m2.

 In model m3 was trying to be explicitly clear and interact 1+x with
 group.

 Looking only at the coefficients, it appears that model m3 is simplified
 to
 model m4.

 R coef(m3)
 (Intercept)groupA:xgroupB:x
  0.3775140   0.9213835   0.9879690

 R coef(m4)
 (Intercept)x:groupAx:groupB
  0.3775140   0.9213835   0.9879690

 I wonder if anyone can shed some light on what R is doing with the
 1:group
 term.

 Kevin

[[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.


 --
 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


[[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.

Re: [R] In a formula, what is the interaction of the intercept and a factor?

2011-06-01 Thread Kevin Wright

Perhaps of interest...in S-Plus 7.0

 model.matrix(y ~ (1+x)*group, data = dat)
   (Intercept)  x group x:group
 1   1  1 1   1
 2   1  2 1   2
...

 model.matrix(y ~ (1)*group, data = dat)
   (Intercept) group
 1   1 1
 2   1 1
 ...

Kevin



On Wed, Jun 1, 2011 at 9:44 AM, Kevin Wright kw.s...@gmail.com wrote:


 On Wed, Jun 1, 2011 at 1:19 AM, Prof Brian Ripley 
 rip...@stats.ox.ac.ukwrote:

 [text deleted]


 Note that you usually want to do '*' when you say 'interact with':

 model.matrix(y ~ (1+x)*group, data = dat)

   (Intercept)  x groupB x:groupB
 11  1  11
 ...


 Thanks for that suggestion.  To follow up,

 R model.matrix(y ~ (1+x)*group, data = dat)

(Intercept)  x groupB x:groupB
 11  1  00
 21  2  12
 ...

 but if I delete 'x',

 R model.matrix(y ~ (1)*group, data = dat)
(Intercept)
 11
 21
 ...

 Why is group simply vanishing in the second model?  I tried to dig into
 this until I hit .Internal(model.matrix


 Kevin





 On Tue, 31 May 2011, Kevin Wright wrote:

  For a pedagogical purpose, I was trying to show how the formula for a
 simple
 regression line (~1+x) could be crossed with a factor (~1:group +
 x:group)
 to fit separate regressions by group.  For example:

 set.seed(201108)
 dat - data.frame(x=1:15, y=1:15+rnorm(15),
   group = sample(c('A','B'), size=15,
   replace=TRUE))

 m1 - lm(y~ 1 + x, data=dat)
 m2 - lm(y ~ group + x:group, data=dat)
 m3 - lm(y ~ 1:group + x:group, data=dat)
 m4 - lm(y ~ 1 + x:group, data=dat)

 The simple regression is model m1.

 The usual way to write the by-group regression is model m2.

 In model m3 was trying to be explicitly clear and interact 1+x with
 group.

 Looking only at the coefficients, it appears that model m3 is simplified
 to
 model m4.

 R coef(m3)
 (Intercept)groupA:xgroupB:x
  0.3775140   0.9213835   0.9879690

 R coef(m4)
 (Intercept)x:groupAx:groupB
  0.3775140   0.9213835   0.9879690

 I wonder if anyone can shed some light on what R is doing with the
 1:group
 term.

 Kevin

[[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.


 --
 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




[[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.

Re: [R] Identifying sequences

2011-06-01 Thread christiaan pauw

Thanks to David, Thierry and Jonathan for your help.
I have been able to put this function together

a=1:10
b=20:30
c=40:50
x=c(a,b,c)

seq.matrix - function(x){
lower- x[which(diff(x) != 1)]
upper - x[which(diff(x) != 1)+1]
extremities - c(1,lower, upper,x[length(x)])
m - 
data.frame(matrix(extremities[order(extremities)],ncol=2,byrow=TRUE,dimnames=list(rows=paste(group,1:(length(lower)+1),sep=),cols=c(lower,upper
m$length=m$upper-m$lower+1
m
}

s.m=seq.matrix(x)
s.m
   lower upper length
group1 110 10
group22030 11
group34050 11

One can then make a test to see if a certain value (say 9) falls
within one of the groups and use that to find the group name or lower
or upper border

s.m.test=function(s.m,i){which(s.m[,1] i  i  s.m[,2])}
s.m.test(s.m,i=9)
[1] 1
e.g.
row.names(s.m)[s.m.test(s.m,i=9)]
[1] group1

Cheers
Christiaan

On 1 June 2011 14:31, Jonathan Daily biomathjda...@gmail.com wrote:

 I am assuming in this case that you are looking for continuity along
 integers, so if you expect noninteger values this will not work.

 You can get the index of where breaks can be found in your example using

 which(diff(x)  1)

 On Wed, Jun 1, 2011 at 6:27 AM, christiaan pauw cjp...@gmail.com wrote:
  Hallo Everybody
 
  Consider the following vector
 
  a=1:10
  b=20:30
  c=40:50
  x=c(a,b,c)
 
  I need a function that can tell me that there are three set of continuos
  sequences and that the first is from 1:10, the second from 20:30 and the
  third from 40:50. In other words: a,b, and c.
 
  regards
  Christiaan
 
         [[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.
 



 --
 ===
 Jon Daily
 Technician
 ===
 #!/usr/bin/env outside
 # It's great, trust me.

__
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] Create Matrix with Float32 values

2011-06-01 Thread Chris English


Dear R_Help:
The following gives me a matrix with integer values.
z= matrix(rep(10:1, each= 10), ncol= 10, byrow=TRUE) str(z) int [1:10, 1:10] 
10 9 8 7 6 5 4 3 2 1 ...
How do I specify that I want Float32 values instead.
Thanks,Chris  
__
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] RES: conversion of matrix into list

2011-06-01 Thread Filipe Leme Botelho

---BeginMessage---
Hi Martin, I believe this can do the trick. Let me know otherwise.
Cheers

testing - matrix(rnorm(10),50,2)
testing_list - lapply(1:nrow(testing), function(j) testing[j,])


-Mensagem original-
De: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
Em nome de Martin Spindler
Enviada em: quarta-feira, 1 de junho de 2011 04:46
Para: R-help@r-project.org
Assunto: [R] conversion of matrix into list

Dear all,



I have a matrix X which consists of 2 columns. I would like to convert
this
matrix into a list where every entry of the list consists of a single
row of
the matrix.

Does anyone have a suggestions how to manage this?



Thank you for your efforts in advance!



Best,



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.
---End Message---
This message and its attachments may contain confidential and/or privileged 
information. If you are not the addressee, please, advise the sender 
immediately by replying to the e-mail and delete this message.

Este mensaje y sus anexos pueden contener información confidencial o 
privilegiada. Si ha recibido este e-mail por error por favor bórrelo y envíe un 
mensaje al remitente.

Esta mensagem e seus anexos podem conter informação confidencial ou 
privilegiada. Caso não seja o destinatário, solicitamos a imediata notificação 
ao remetente e exclusão da mensagem.__
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] Overlaying two matrices to create a heatmap

2011-06-01 Thread Mark Aquino

Hi,

I'm trying to construct a heatmap using two matrices rather than one (Is this 
possible?)

For example, matrix1: (color range black--- blue)
acr DPA1-0103   DPA1-0104   DPA1-0201   DPA1-0202   
DPA1-0301
DPA1-0103   17  0   2   1   0
DPA1-0104   0   0   0   0
DPA1-0201   1   0   0
DPA1-0202   0   0
DPA1-0301   0

matrix 2: (color range black--- magenta)
noacr   DPA1-0103   DPA1-0104   DPA1-0201   DPA1-0202   
DPA1-0301
DPA1-0103   14  1   5   0   1
DPA1-0104   0   0   0   0
DPA1-0201   1   0   0
DPA1-0202   0   0
DPA1-0301   0

I'd like to create a separate color range for each and then overlay the two 
using some kind of additive color scheme so that if there are equally high 
numbers in a certain field the heatmap would display whatever color blue and 
magenta is (yellow I think).
__
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] re-write plot function for ggplot

2011-06-01 Thread rmje

Hi, I have the following function that i use to plot graphs.

plot_mi_time = function(mdata, miname) {
   mdata2 = mdata[row.names(mir_test3) == miname, ]
   # print(mdata2)
   xcoords - c(1,1,2,2,3,3)
   plot(xcoords, mdata2, xaxt=n, ylab=Expression, xlab=Time(h), ,
main=miname)
   axis(1, at=xcoords, labels=c(a,a,b,b,c,c))
}




I want to edit this function so that i can plot it using ggplot.
Any suggestions on how to re-write it?

--
View this message in context: 
http://r.789695.n4.nabble.com/re-write-plot-function-for-ggplot-tp3565868p3565868.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.

Re: [R] re-write plot function for ggplot

2011-06-01 Thread Lamke

You should post it on the ggplot2 mailing list.  For the long run I would
suggest getting the book as it's extremely helpful. 

--
View this message in context: 
http://r.789695.n4.nabble.com/re-write-plot-function-for-ggplot-tp3565868p3565950.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.

[R] Plot a network with edges of specified length

2011-06-01 Thread PhDGuy

Hello,

I would like to plot a network with given edge lengths by using the plot
function of the network package. Assume my network has the following
adjacency matrix:

  a b c
a 0 0 0
b 1 0 0
c 1 0 0

so that b and c are linked to a. I would like to specify that the length of
the edge from b to a is 10 times longer than the one from c to a. So I set
the following edge attribute:

 edgesLength=matrix(0,nrow=3,ncol=3)
 edgesLength[2,1]=10
 edgesLength[3,1]=1
 set.edge.attribute(network,edgesLength,1,e=1:length(network$mel))
 set.edge.value(network,edgesLength,edgesLength,e=1:length(network$mel))

and then use the plot command:

plot.network(network,attrname=edgesLength)

but this does not take into account the fact that one edge should be 10
times longer than the other one. In fact, both edges have approximately the
same length.
Is this possible with the network package?

Thanks.

--
View this message in context: 
http://r.789695.n4.nabble.com/Plot-a-network-with-edges-of-specified-length-tp3565986p3565986.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.

Re: [R] Identifying sequences

2011-06-01 Thread Mike Marchywka

 Date: Wed, 1 Jun 2011 17:12:29 +0200
 From: cjp...@gmail.com
 To: r-help@r-project.org
 Subject: Re: [R] Identifying sequences

 Thanks to David, Thierry and Jonathan for your help.
 I have been able to put this function together

 a=1:10
 b=20:30
 c=40:50
 x=c(a,b,c)

 seq.matrix - function(x){
 lower- x[which(diff(x) != 1)]
 upper - x[which(diff(x) != 1)+1]
 extremities - c(1,lower, upper,x[length(x)])
 m - 
 data.frame(matrix(extremities[order(extremities)],ncol=2,byrow=TRUE,dimnames=list(rows=paste(group,1:(length(lower)+1),sep=),cols=c(lower,upper
 m$length=m$upper-m$lower+1
 m
 }

 s.m=seq.matrix(x)
 s.m
 lower upper length
 group1 1 10 10
 group2 20 30 11
 group3 40 50 11

 One can then make a test to see if a certain value (say 9) falls
 within one of the groups and use that to find the group name or lower
 or upper border

As I understand, you are looking for large derivatives
or approx discontinuity against smooth signal.
This seems like a natural application for wavelets,
try the haar wavelet and use package 
wavelets,

 library(wavelets)
 f=wt.filter(c(-1,1),modwt=T)
 z-modwt(X=as.numeric(x),filter=f,n.levels=1)
 z@W
$W1
  [,1]
 [1,]   49
 [2,]   -1
 [3,]   -1
 [4,]   -1
 [5,]   -1
 [6,]   -1
 [7,]   -1
 [8,]   -1
 [9,]   -1
[10,]   -1
[11,]  -10
[12,]   -1
[13,]   -1
[14,]   -1
[15,]   -1
[16,]   -1
[17,]   -1
[18,]   -1
[19,]   -1
[20,]   -1
[21,]   -1
[22,]  -10
[23,]   -1
[24,]   -1
[25,]   -1
[26,]   -1
[27,]   -1
[28,]   -1
[29,]   -1
[30,]   -1
[31,]   -1
[32,]   -1

 s.m.test=function(s.m,i){which(s.m[,1]  s.m.test(s.m,i=9)
 [1] 1
 e.g.
 row.names(s.m)[s.m.test(s.m,i=9)]
 [1] group1

 Cheers
 Christiaan

 On 1 June 2011 14:31, Jonathan Daily wrote:

  I am assuming in this case that you are looking for continuity along
  integers, so if you expect noninteger values this will not work.

  You can get the index of where breaks can be found in your example using

  which(diff(x)  1)

  On Wed, Jun 1, 2011 at 6:27 AM, christiaan pauw wrote:
   Hallo Everybody

   Consider the following vector

   a=1:10
   b=20:30
   c=40:50
   x=c(a,b,c)

   I need a function that can tell me that there are three set of continuos
   sequences and that the first is from 1:10, the second from 20:30 and the
   third from 40:50. In other words: a,b, and c.

   regards
   Christiaan

   [[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.

  --
  ===
  Jon Daily
  Technician
  ===
  #!/usr/bin/env outside
  # It's great, trust me.

 __
 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.

Re: [R] interpolation and extremum location of a surface‏

2011-06-01 Thread Hans W Borchers

Clement LAUZIN lauzin at hotmail.com writes:

 Hello,
 
 Hello,
 
 I have a x,y,z file.Z is not corresponding to a simple analytical function
 of x and y (see below). I am trying to find the minimum location of this
 surface and the z value corresponding to this location by a spline
 interpolation or from a polynomial fit. I tried with the akima package but
 as the location of the point I am looking for (the minimum) is outside of
 the convex hull it's not giving any answer.
 If someone has any idea or any suggestion?
 
 Thank you in advance Pierre 

According to a 2-dimensional barycentric Lagrange interpolation that
I am implementing right now in R, the minimum appears to be at

x0 = (4.230490, 68.52776); fval = -175.7959

I certainly would be interested to hear from other reasonable locations
of a minimum, as this may be a good test case.

--  Hans Werner

 xyz
 4.1 60 -152.1719593
 [ ... ]
 4.35 90 -163.8389615


__
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] Anyone have experience with kinship pedigree plot?

2011-06-01 Thread Ben Bimber

Hello,

I am using the Kinship package to draw a pedigree plot.  I am trying
to control the spacing between individuals in the pedigree.
Currently, it is drawing a pedigree with too little space, so the text
is overlapping.  I'd like to increase the distance between
individuals.  According to the documentation, I would expect the
following to be relevant:

http://rss.acs.unt.edu/Rdoc/library/kinship/html/plot.pedigree.html

symbolsize  controls symbolsize. Default=1.
width   default=8. For a packed pedigree, the minimum width allowed in
the realignment of pedigrees.
density defines density used in the symbols. Takes up to 4 different 
values.
mar no comment(s)

my pedigree command is similar to:

plot(ptemp, width=5, symbolsize=1);

however, 'width', doesnt seem to actually do anything as far as I can
see (ie. the plot looks the same no matter what value I use).  the
example command in the kinship doc uses:

plot(x, id=x$id, sex=x$sex, status=x$status,
affected=x$affected, cex=1, col=rep(1, length(x$id)),
symbolsize=1, branch=0.6, packed=T, align=packed, width=8,
density=c(-1, 50,70,90),
mar=c(4.1,1,4.1,1), angle=c(90, 70, 50, 0), keep.par=F, ...)

when I try density params similar to their example, it also does not
obviously do anything.  I could not find documentation on exactly what
those do either.  does anyone have any experience with pedigree plots,
other suggestions or tips on where else I should be looking?

thanks for the help.

-ben

__
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] interpolation and extremum location of a surface‏



On Jun 1, 2011, at 9:24 AM, Clement LAUZIN wrote:




Hello,

I have a x,y,z file.Z is not corresponding to a simple analytical  
function of x and y (see below).
I am trying to find the minimum location of this surface and the z  
value corresponding to this location by a spline interpolation or  
from a polynomial fit.
I tried with the akima package but as the location of the point I am  
looking for (the minimum) is outside of the convex hull it's not  
giving any answer.


I do not think that is the reason akima::interp is failing. (For one  
thing the minimum is not outside the convex hull of x and y.)  Note  
that the help page says that x and y cannot be collinear but yours  
appear to be so.


 table(ZZ$x)

4.05  4.1 4.15  4.2 4.25  4.3 4.35
   5555555
 table(ZZ$y)

  60 67.5   75 82.5   90
   77777

I took a shot at jittering the x and y and reapplying interpp ... and  
ended up with what appeared to be useless junk.


If you look at the data with contourplot you can see it would place   
the minimum at around z=-175 at x=4.24, y=67. However, the contourplot  
and associated help pages say there is no documentation for the  
algorithm used.


contourplot(z ~ x + y, data=ZZ, at=seq(-140, -180 , by= -0.5),  
interp=TRUE)





If someone has any idea or any suggestion?

Thank you in advance Pierre


xyz
4.1 60 -152.1719593
4.1 75 -171.136801
4.1 90 -170.4604774
4.2 60 -168.7745552
4.2 75 -174.9667333
4.2 90 -172.1853334
4.3 60 -173.7736418
4.3 75 -171.6712745
4.3 90 -167.6662458
4.05 60 -137.8379387
4.15 60 -162.2264066
4.25 60 -172.4453286
4.35 60 -173.2123715
4.05 67.5 -158.8239625
4.1 67.5 -167.314534
4.15 67.5 -172.586182
4.2 67.5 -175.2217594
4.25 67.5 -175.7162683
4.3 67.5 -174.4890566
4.35 67.5 -171.8940061
4.05 75 -165.4388778
4.15 75 -174.1460392
4.25 75 -174.022344
4.35 75 -168.2149168
4.05 82.5 -166.4026077
4.1 82.5 -170.9199652
4.15 82.5 -172.9923449
4.2 82.5 -173.0803255
4.25 82.5 -171.5739101
4.3 82.5 -168.8024715
4.35 82.5 -165.0431276
4.05 90 -166.2592978
4.15 90 -172.2861302
4.25 90 -170.5383652
4.35 90 -163.8389615

__
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

__
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] Memory management

2011-06-01 Thread Michael Conklin

I am trying to run a very large Bradley-Terry model using the BradleyTerry2 
package.  (There are 288 players in the BT model).

My problem is that I ran the model below successfully.
WLMat is a win-loss matrix that is 288 by 288
WLdf-countsToBinomial(WLMat)
  mod1-BTm(cbind(win1,win2),player1,player2,~player,id=player,data=WLdf)

Then I needed to run the same model with a subset of the observations that went 
into the win-loss matrix.  So I created my new win-loss matrix and tried to run 
a new model.

Now I get:  Error: cannot allocate vector of size 90.5 Mb

I found this particularly puzzling because the actual input data is the same 
size as the original model, just different values.

I tried increasing memory size, I tried running it in a clean workspace and the 
error message is always the same (sometimes the vector it is trying to allocate 
is 181.0MB (twice as large)) but it is always one of those two numbers no 
matter what I have done to the available memory.

To further complicate this...I cannot get the system to re-run my first model 
either . Same errors.

traceback indicates that the error occurs when the program is trying to do a qr 
decomposition.

R 2.13.0
Windows XP

Any suggestions?

W. Michael Conklin
Chief Methodologist
Google Voice: (612) 56STATS

MarketTools, Inc. | www.markettools.comhttp://www.markettools.com
6465 Wayzata Blvd | Suite 170 |  St. Louis Park, MN 55426.  PHONE: 952.417.4719 
| CELL: 612.201.8978
This email and attachment(s) may contain confidential and/or proprietary 
information and is intended only for the intended addressee(s) or its 
authorized agent(s). Any disclosure, printing, copying or use of such 
information is strictly prohibited. If this email and/or attachment(s) were 
received in error, please immediately notify the sender and delete all copies


[[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.

Re: [R] MAP datafile

2011-06-01 Thread Jim Silverton

Hi,
I have a MAP datafile with SNP data and would like to open it and run say
fisher's exact test and save the p-values. Anyone has any idea how this can
be done?


-- 
Thanks,
Jim.

[[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.

Re: [R] Recode numbers

2011-06-01 Thread Pete Brecknock


Lisa wrote:
 
 Dear all,
 
 I have two sets of numbers that look like
 
 a - c(1, 2, 3, 3, 4, 4, 5, 6, 1, 2, 2, 3, 1, 2, 1, 2, 3, 3, 4, 5, 1, 2,
 3, 4)
 
 b - c(1, 5, 8, 9, 14, 20, 3, 10, 12, 6, 16, 7, 11, 13, 17, 18, 2, 4, 15,
 19)
 
 I just want to use “b” to encode “a” so that “a” looks like
 
 a1- c(1, 5, 8, 8, 9, 9, 14, 20, 3, 10, 10, 12, 6, 16, 7, 11, 13, 13, 17,
 18, 2, 4, 15, 19)
 
 Does anyone have a suggestion how to deal with this? Thank you in advance.
 
 Lisa
 

is a1 = b[a] what you are looking for?

HTH

Pete


--
View this message in context: 
http://r.789695.n4.nabble.com/Recode-numbers-tp3566395p3566505.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.

[R] Recode numbers

Dear all,

I have two sets of numbers that look like

a - c(1, 2, 3, 3, 4, 4, 5, 6, 1, 2, 2, 3, 1, 2, 1, 2, 3, 3, 4, 5, 1, 2, 3,
4)

b - c(1, 5, 8, 9, 14, 20, 3, 10, 12, 6, 16, 7, 11, 13, 17, 18, 2, 4, 15,
19)

I just want to use “b” to encode “a” so that “a” looks like

a1- c(1, 5, 8, 8, 9, 9, 14, 20, 3, 10, 10, 12, 6, 16, 7, 11, 13, 13, 17,
18, 2, 4, 15, 19)

Does anyone have a suggestion how to deal with this? Thank you in advance.

Lisa


--
View this message in context: 
http://r.789695.n4.nabble.com/Recode-numbers-tp3566395p3566395.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.

[R] lattice panel fine control

2011-06-01 Thread maxbre

Hello R experts,

what follows is my reproducible example:

mydata-structure(list(ped.avg = c(335.9, 110.8, 645.7, 638.9, 1468.1, 
126.4, 4811.1, 88.5, 868.5, 656.6, 723.6, 654, 2.8, 15, 14.2, 
17.5, 15.4, 112.1, 424.7, 18.3, 19.9, 28.6, 25.6, 23.5, 15.4, 
27, 62.1, 15.6, 74.6), ped.erst = c(96, 53.2, 615.2, 616.5, 512.9, 
56.2, 1851.8, 57.1, 579.5, 613.2, 601.1, 613.6, 1.3, 6.3, 6.5, 
6.1, 6.3, 42, 166.4, 6.5, 6.5, 7.6, 8, 7, 6.3, 8.8, 24.6, 6.3, 
35.6), tv.avg = c(670.4, 320, 282.4, 266.6, 2077.3, 383.1, 7116, 
335.1, 642.9, 291, 405.1, 280.8, 5, 18.5, 16.5, 28.6, 24.8, 150.2, 
366, 40.3, 38.4, 51.9, 52.9, 48.1, 27.8, 73.8, 168.3, 32.8, 151.2
), tv.erst = c(233.8, 243.4, 245.9, 246.4, 389.9, 240.4, 1530.7, 
242.8, 232.4, 245.1, 239.7, 245.7, 2.2, 6.9, 6, 11.5, 9.5, 69.5, 
139.5, 17.9, 16.7, 22.9, 27.3, 24.5, 12.7, 46.3, 102.2, 15.7, 
71.2), family = structure(c(1L, 1L, 1L, 1L, 5L, 5L, 5L, 5L, 5L, 
5L, 5L, 5L, 13L, 13L, 13L, 13L, 13L, 13L, 13L, 20L, 20L, 20L, 
20L, 20L, 20L, 20L, 20L, 20L, 20L), .Label = c(non-orto PCB, 
non-orto PCB, non-orto PCB, non-orto PCB, mono-orto PCB, 
mono-orto PCB, mono-orto PCB, mono-orto PCB, mono-orto PCB, 
mono-orto PCB, mono-orto PCB, mono-orto PCB, PCDD, PCDD, 
PCDD, PCDD, PCDD, PCDD, PCDD, PCDF, PCDF, PCDF, 
PCDF, PCDF, PCDF, PCDF, PCDF, PCDF, PCDF), class = factor), 
name = structure(c(28L, 29L, 22L, 26L, 18L, 19L, 20L, 21L, 
23L, 24L, 25L, 27L, 1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L, 
11L, 12L, 13L, 14L, 15L, 16L, 17L), .Label = c(D4, D5, 
D6a, D6b, D6c, D7, D8, F4, F5a, F5b, F6a, 
F6b, F6c, F6d, F7a, F7b, F8, P105, P114, 
P118, P123, P126, P156, P157, P167, P169, P189, 
P77, P81), class = factor), ped.lower = c(239.9, 57.6, 
30.5, 22.400, 955.2, 70.2, 2959.3, 31.4, 289, 43.4, 122.5, 
40.4, 1.5, 8.7, 7.7, 11.4, 9.1, 70.1, 258.3, 11.8, 13.4, 
21, 17.6, 16.5, 9.1, 18.2, 37.5, 9.3, 39), ped.upper = c(431.9, 
164, 1260.9, 1255.4, 1981, 182.6, 6662.9, 145.6, 1448, 1269.8, 
1324.7, 1267.6, 4.1, 21.3, 20.7, 23.6, 21.7, 154.1, 591.1, 
24.8, 26.4, 36.2, 33.6, 30.5, 21.7, 35.8, 86.7, 21.9, 110.2
), tv.lower = c(436.6, 76.6, 36.5, 20.2, 1687.4, 142.7, 5585.3, 
92.3, 410.5, 45.9, 165.4, 35.1, 2.8, 11.6, 10.5, 17.1, 15.3, 
80.7, 226.5, 22.4, 21.7, 29, 25.6, 23.6, 15.1, 27.5, 66.1, 
17.1, 80), tv.upper = c(904.2, 563.4, 528.3, 513, 2467.2, 
623.5, 8646.7, 577.9, 875.3, 536.1, 644.8, 526.5, 7.2, 25.4, 
22.5, 40.1, 34.3, 219.7, 505.5, 58.2, 55.1, 74.8, 80.2, 72.6, 
40.5, 120.1, 270.5, 48.5, 222.4)), .Names = c(ped.avg, 
ped.erst, tv.avg, tv.erst, family, name, ped.lower, 
ped.upper, tv.lower, tv.upper), row.names = c(NA, -29L), class =
data.frame)


and this is the code I worked so far (in order to clear out what is the
final result I would like to get)

mydata$family - factor(mydata$family, levels=as.character(mydata$family))

library(lattice)

myplot-xyplot(ped.avg ~ tv.avg | family, 
data=mydata, 
strip=strip.custom(bg='white'), col.line=1, main=title,
xlab=tv [fg/m3], ylab=ped [fg/m3],
scales= list(x=list(relation=free, log=TRUE), 
y=list(relation=free,
log=TRUE)),
)

update(myplot, 
xlim=list(c(1.9,3),c(2,3.8),c(0.5,2.7),c(1.2,2.4)),
ylim=list(c(1.9,3),c(2,3.8),c(0.5,2.7),c(1.2,2.4)), 
cex=0.6, aspect=iso,
panel=function(...){
panel.abline(a=0, b=1, lty=1, col=1)
}
)

obviously there is something wrong in the resulting chart and therefore
these are my questions:

1-  how to draw a diagonal lines in each of the panels without “deleting”
points? (in my example the use of panel.abline is somehow getting rid of the
points)
2-  how to label points with the “name”? (no clue for this)
3-  how to draw error bar points referring of both x and y axes? (no clue 
for
this)
4-  how to format axes tickmark labels with a scientific notation? (no clue
for this)
5-  how to list all these commands inside xyplot() without the need to use
update()? (for some reasons I do not fully understand I can not ‘move’ all
commands inside the main plotting– xyplot() - without loosing some relevant
features of the chart)

any help much appreciated
sorry for such trivial question but I’m using R since a very short time:
please keep your replies as much simple and self explanatory as possible
thank you

maxbre


--
View this message in context: 
http://r.789695.n4.nabble.com/lattice-panel-fine-control-tp3566347p3566347.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.

[R] Qs on vector of lists, etc.

2011-06-01 Thread Roy Shimizu

I'm pretty new to R, so please forgive the cluelessness of these questions.

Is it true that it is not possible to have a vector or lists in R?

Is a list of lists the closest one can get to a vector of lists?

Is it possible to have a matrix of lists in R?  Or an array of lists?

Thanks!

Roy

[[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.

[R] xtable with conditional formatting using \textcolor

2011-06-01 Thread Walmes Zeviani

Hello list,

I'm doing a table with scores and I want include colors to represent status
of an individual. I'm using sweave results=tex= and xtable but I can't
get a result I want. My attemps are

#-
# code R

da - data.frame(id=letters[1:5], score=1:5*2)

col - function(x){
  ifelse(x7,
 paste(\textcolor{blue}{, formatC(x, dig=2, format=f), }),
 paste(\textcolor{red}{, formatC(x, dig=2, format=f), }))
}

da$score.string - col(da$score)

require(xtable)
xtable(da[,c(id,score.string)])

#-

actual result
#-
\begin{tabular}{rll}
  \hline
  id  score.string \\
  \hline
1  a  extcolor\{red\}\{ 2.00 \} \\
  2  b  extcolor\{red\}\{ 4.00 \} \\
  3  c  extcolor\{red\}\{ 6.00 \} \\
  4  d  extcolor\{blue\}\{ 8.00 \} \\
  5  e  extcolor\{blue\}\{ 10.00 \} \\
   \hline
\end{tabular}
#-

desired result (lines omited to save space)
#-
1  a  \textcolor{red}{ 2.00 } \\
2  b  \textcolor{red}{ 4.00} \\
#-

Any contribution will be useful. Thanks.
Walmes.

==
Walmes Marques Zeviani
LEG (Laboratório de Estatística e Geoinformação, 25.450418 S, 49.231759 W)
Departamento de Estatística - Universidade Federal do Paraná
fone: (+55) 41 3361 3573
VoIP: (3361 3600) 1053 1173
e-mail: wal...@ufpr.br
twitter: @walmeszeviani
homepage: http://www.leg.ufpr.br/~walmes
linux user number: 531218
==

[[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.

Re: [R] Recode numbers

Thank you for your help, Pete. I tried b[a], but it is not a1. 

--
View this message in context: 
http://r.789695.n4.nabble.com/Recode-numbers-tp3566395p3566534.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.

[R] Replacing variables in one dataset with those from another

2011-06-01 Thread mcginnis21

Hoping someone out there can help me...this seems like an easy task but I
can't figure it out...

I want to replace variables in one dataset (Dataset1) with a variable from
another dataset (Dataset2).
All the values for variables x1 and x2 in Dataset1 have a unique match to
the variable uniquenum in Dataset2. For example, in Dataset1 grpnum A has a
value of 343 for variable x1. In Dataset2, 343 will only be found once for
variable uniquenum. I would like to create a new dataset where the variable
x1 from Dataset1 will be replaced by uniqueid from Dataset2. So variable x1
for grpnum A now would equal GHU82RK02HD7D6 rather than 343. Below is an
abbreviated example from my datasets to show what I’m trying to do rather
than explain in words because I’m not sure my question is coming across very
clearly. I have about 100 columns in Dataset1 (x1, x2, etc.) and the
variable uniquenum in Dataset2 has about 4,000 values.

Dataset1
grpnum x1 x2
A 343 334
B 313 898
C 40764077
D 275 276

Dataset2
unique num uniqueid
343 GHU82RK02HD7D6
4076GHU82RK02IXPC7
4077GHU82RK02HICZ2
261 GHU82RK02FLOFU
4100GHU82RK02IDO7K
4101GHU82RK02HQYQL
4102GHU82RK02GADV1
4103GHU82RK02G7RQH
4104GHU82RK02HQD2E

NewDataset
grpnum x1 x2
a GHU82RK02HD7D6Na
b Na Na
c GHU82RK02IXPC7 GHU82RK02HICZ2
d Na Na

Thanks for your time and thoughts. I hope someone can help!

--
View this message in context:
http://r.789695.n4.nabble.com/Replacing-variables-in-one-dataset-with-those-from-another-tp3566552p3566552.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.

Re: [R] lattice panel fine control

2011-06-01 Thread Bert Gunter

It is perhaps not entirely clear from ?update, but update replaces
your original lattice call with your new call and re-drawsthe plot.
This means that the panel function used for the update() plot is the
one in update(), which only draws the line. That's why you see no
points.

Obviously, then, you need to include both points and abline in your
panel call to get both. However, if you would like to follow the
strategy that you have indicated, you can do so by using the layer()
functionality in the lattceExtra package, in which you would do
something similar to what you have shown.

library(latticeExtra)  ## after you have installed it from CRAN, possibly
?layer

for further info.

Your numerous other questions can be answered by close reading of the
?xyplot Help file and/or adding additional functionality to your panel
call. For example, text can be added via the panel.text() function.

?lattice

will get you info on this and other low level graphics functionality,
as well as references to consult. I would also suggest Deepayan
Sarkar's book, if you intend to use lattice seriously.

Cheers,
Bert

On Wed, Jun 1, 2011 at 10:00 AM, maxbre mbres...@arpa.veneto.it wrote:
 Hello R experts,

 what follows is my reproducible example:

 mydata-structure(list(ped.avg = c(335.9, 110.8, 645.7, 638.9, 1468.1,
 126.4, 4811.1, 88.5, 868.5, 656.6, 723.6, 654, 2.8, 15, 14.2,
 17.5, 15.4, 112.1, 424.7, 18.3, 19.9, 28.6, 25.6, 23.5, 15.4,
 27, 62.1, 15.6, 74.6), ped.erst = c(96, 53.2, 615.2, 616.5, 512.9,
 56.2, 1851.8, 57.1, 579.5, 613.2, 601.1, 613.6, 1.3, 6.3, 6.5,
 6.1, 6.3, 42, 166.4, 6.5, 6.5, 7.6, 8, 7, 6.3, 8.8, 24.6, 6.3,
 35.6), tv.avg = c(670.4, 320, 282.4, 266.6, 2077.3, 383.1, 7116,
 335.1, 642.9, 291, 405.1, 280.8, 5, 18.5, 16.5, 28.6, 24.8, 150.2,
 366, 40.3, 38.4, 51.9, 52.9, 48.1, 27.8, 73.8, 168.3, 32.8, 151.2
 ), tv.erst = c(233.8, 243.4, 245.9, 246.4, 389.9, 240.4, 1530.7,
 242.8, 232.4, 245.1, 239.7, 245.7, 2.2, 6.9, 6, 11.5, 9.5, 69.5,
 139.5, 17.9, 16.7, 22.9, 27.3, 24.5, 12.7, 46.3, 102.2, 15.7,
 71.2), family = structure(c(1L, 1L, 1L, 1L, 5L, 5L, 5L, 5L, 5L,
 5L, 5L, 5L, 13L, 13L, 13L, 13L, 13L, 13L, 13L, 20L, 20L, 20L,
 20L, 20L, 20L, 20L, 20L, 20L, 20L), .Label = c(non-orto PCB,
 non-orto PCB, non-orto PCB, non-orto PCB, mono-orto PCB,
 mono-orto PCB, mono-orto PCB, mono-orto PCB, mono-orto PCB,
 mono-orto PCB, mono-orto PCB, mono-orto PCB, PCDD, PCDD,
 PCDD, PCDD, PCDD, PCDD, PCDD, PCDF, PCDF, PCDF,
 PCDF, PCDF, PCDF, PCDF, PCDF, PCDF, PCDF), class = factor),
    name = structure(c(28L, 29L, 22L, 26L, 18L, 19L, 20L, 21L,
    23L, 24L, 25L, 27L, 1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L,
    11L, 12L, 13L, 14L, 15L, 16L, 17L), .Label = c(D4, D5,
    D6a, D6b, D6c, D7, D8, F4, F5a, F5b, F6a,
    F6b, F6c, F6d, F7a, F7b, F8, P105, P114,
    P118, P123, P126, P156, P157, P167, P169, P189,
    P77, P81), class = factor), ped.lower = c(239.9, 57.6,
    30.5, 22.400, 955.2, 70.2, 2959.3, 31.4, 289, 43.4, 122.5,
    40.4, 1.5, 8.7, 7.7, 11.4, 9.1, 70.1, 258.3, 11.8, 13.4,
    21, 17.6, 16.5, 9.1, 18.2, 37.5, 9.3, 39), ped.upper = c(431.9,
    164, 1260.9, 1255.4, 1981, 182.6, 6662.9, 145.6, 1448, 1269.8,
    1324.7, 1267.6, 4.1, 21.3, 20.7, 23.6, 21.7, 154.1, 591.1,
    24.8, 26.4, 36.2, 33.6, 30.5, 21.7, 35.8, 86.7, 21.9, 110.2
    ), tv.lower = c(436.6, 76.6, 36.5, 20.2, 1687.4, 142.7, 5585.3,
    92.3, 410.5, 45.9, 165.4, 35.1, 2.8, 11.6, 10.5, 17.1, 15.3,
    80.7, 226.5, 22.4, 21.7, 29, 25.6, 23.6, 15.1, 27.5, 66.1,
    17.1, 80), tv.upper = c(904.2, 563.4, 528.3, 513, 2467.2,
    623.5, 8646.7, 577.9, 875.3, 536.1, 644.8, 526.5, 7.2, 25.4,
    22.5, 40.1, 34.3, 219.7, 505.5, 58.2, 55.1, 74.8, 80.2, 72.6,
    40.5, 120.1, 270.5, 48.5, 222.4)), .Names = c(ped.avg,
 ped.erst, tv.avg, tv.erst, family, name, ped.lower,
 ped.upper, tv.lower, tv.upper), row.names = c(NA, -29L), class =
 data.frame)


 and this is the code I worked so far (in order to clear out what is the
 final result I would like to get)

 mydata$family - factor(mydata$family, levels=as.character(mydata$family))

 library(lattice)

 myplot-xyplot(ped.avg ~ tv.avg | family,
                data=mydata,
                strip=strip.custom(bg='white'), col.line=1, main=title,
                xlab=tv [fg/m3], ylab=ped [fg/m3],
                scales= list(x=list(relation=free, log=TRUE), 
 y=list(relation=free,
 log=TRUE)),
                                )

 update(myplot,
                xlim=list(c(1.9,3),c(2,3.8),c(0.5,2.7),c(1.2,2.4)),
                ylim=list(c(1.9,3),c(2,3.8),c(0.5,2.7),c(1.2,2.4)),
                cex=0.6, aspect=iso,
                panel=function(...){
                panel.abline(a=0, b=1, lty=1, col=1)
                                }
                )

 obviously there is something wrong in the resulting chart and therefore
 these are my questions:

 1-      how to draw a diagonal lines in each of the panels without “deleting”
 points? (in my example the use of panel.abline is somehow getting rid of the
 points)

Re: [R] xtable with conditional formatting using \textcolor

2011-06-01 Thread Marc Schwartz

On Jun 1, 2011, at 1:33 PM, Walmes Zeviani wrote:

 Hello list,
 
 I'm doing a table with scores and I want include colors to represent status
 of an individual. I'm using sweave results=tex= and xtable but I can't
 get a result I want. My attemps are
 
 #-
 # code R
 
 da - data.frame(id=letters[1:5], score=1:5*2)
 
 col - function(x){
  ifelse(x7,
 paste(\textcolor{blue}{, formatC(x, dig=2, format=f), }),
 paste(\textcolor{red}{, formatC(x, dig=2, format=f), }))
 }
 
 da$score.string - col(da$score)
 
 require(xtable)
 xtable(da[,c(id,score.string)])
 
 #-
 
 actual result
 #-
 \begin{tabular}{rll}
  \hline
  id  score.string \\
  \hline
 1  a  extcolor\{red\}\{ 2.00 \} \\
  2  b  extcolor\{red\}\{ 4.00 \} \\
  3  c  extcolor\{red\}\{ 6.00 \} \\
  4  d  extcolor\{blue\}\{ 8.00 \} \\
  5  e  extcolor\{blue\}\{ 10.00 \} \\
   \hline
 \end{tabular}
 #-
 
 desired result (lines omited to save space)
 #-
 1  a  \textcolor{red}{ 2.00 } \\
 2  b  \textcolor{red}{ 4.00} \\
 #-
 
 Any contribution will be useful. Thanks.
 Walmes.


When the '\t' is being cat()'d to the TeX file (or console) by print.xtable(), 
it is being interpreted as a tab character. You need to escape it with 
additional backslashes and then adjust the sanitize.text.function in 
print.xtable() so that it does not touch the backslashes:


da - data.frame(id=letters[1:5], score=1:5*2)

col - function(x){
 ifelse(x7,
paste(\\textcolor{blue}{, formatC(x, dig=2, format=f), }),
paste(\\textcolor{red}{, formatC(x, dig=2, format=f), }))
}

da$score.string - col(da$score)


 da
  id score   score.string
1  a 2   \\textcolor{red}{ 2.00 }
2  b 4   \\textcolor{red}{ 4.00 }
3  c 6   \\textcolor{red}{ 6.00 }
4  d 8  \\textcolor{blue}{ 8.00 }
5  e10 \\textcolor{blue}{ 10.00 }


require(xtable)

print(xtable(da[,c(id,score.string)]), sanitize.text.function = 
function(x){x})


That will give you:

% latex table generated in R 2.13.0 by xtable 1.5-6 package
% Wed Jun  1 13:44:54 2011
\begin{table}[ht]
\begin{center}
\begin{tabular}{rll}
  \hline
  id  score.string \\ 
  \hline
1  a  \textcolor{red}{ 2.00 } \\ 
  2  b  \textcolor{red}{ 4.00 } \\ 
  3  c  \textcolor{red}{ 6.00 } \\ 
  4  d  \textcolor{blue}{ 8.00 } \\ 
  5  e  \textcolor{blue}{ 10.00 } \\ 
   \hline
\end{tabular}
\end{center}
\end{table}


HTH,

Marc Schwartz

__
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] Recode numbers


(The attributions are a little messed up here:)


I have two sets of numbers that look like

a- c(1, 2, 3, 3, 4, 4, 5, 6, 1, 2, 2, 3, 1, 2, 1, 2, 3, 3, 4, 5, 1, 2, 3,
4)

b- c(1, 5, 8, 9, 14, 20, 3, 10, 12, 6, 16, 7, 11, 13, 17, 18, 2, 4, 15,
19)

I just want to use “b” to encode “a” so that “a” looks like

a1- c(1, 5, 8, 8, 9, 9, 14, 20, 3, 10, 10, 12, 6, 16, 7, 11, 13, 13, 17,
18, 2, 4, 15, 19)

Does anyone have a suggestion how to deal with this? Thank you in advance.



is a1 = b[a] what you are looking for?

HTH

Pete


Thank you for your help, Pete. I tried b[a], but it is not a1.



I think you'll need to explain the rule used for a1[9], a1[10], etc.

Duncan Murdoch

__
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] xtable with conditional formatting using \textcolor

2011-06-01 Thread Walmes Zeviani

Marc,

Thank you very much. You gave exactly what I wanted.

Bests.
Walmes.

==
Walmes Marques Zeviani
LEG (Laboratório de Estatística e Geoinformação, 25.450418 S, 49.231759 W)
Departamento de Estatística - Universidade Federal do Paraná
fone: (+55) 41 3361 3573
VoIP: (3361 3600) 1053 1173
e-mail: wal...@ufpr.br
twitter: @walmeszeviani
homepage: http://www.leg.ufpr.br/~walmes
linux user number: 531218
==

[[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.

[R] BiodiversityR GUI on macosx

2011-06-01 Thread János Korponai

Dear List!

I installed R and quite a few packages I use. When I try to start
BiodiversityR the library loads without any problems but GUI do not start.
Rcmdr loads without any problems. I am using R 2.13.0 64 bit.

Downgrade to R 2.12.2 works.

Thanks,
Janos

__
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] Qs on vector of lists, etc.

2011-06-01 Thread Uwe Ligges




On 01.06.2011 18:59, Roy Shimizu wrote:

I'm pretty new to R, so please forgive the cluelessness of these questions.

Is it true that it is not possible to have a vector or lists in R?

Is a list of lists the closest one can get to a vector of lists?


Actually a list is a vector of mode list.

Hence if you have a list of lists, it *is* a vector of lists and behaves 
consistantly (e.g. when subsetiing with [] rather than [[]]).


L - list(list(a=1, b=2), list(c=3, d=4))

L[1] # L can be considered a vector of mode list.




Is it possible to have a matrix of lists in R?  Or an array of lists?


Yes:

 X - matrix(L, nrow=2)
 X[1,1]
 X[2,1]

or even:

 dim(X) - c(1,2,1)
 mode(X)  # list
 class(X) # array

Uwe Ligges





Thanks!

Roy

[[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.


__
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] weird error from MASS::eqcsplot with postscript driver

2011-06-01 Thread Michael Friendly


[Env: R 2.12.2, Win XP]

I'm creating figures using MASS::eqcsplot to provide equal scaling of 
the axes.  My figures work OK
when I plot to the screen, but when I try to do the same plot as a 
postscript file, I get an

unexplicable error,

 figframe()
Error in if (yuin  xuin * ratio) yuin - xuin * ratio else xuin - 
yuin/ratio :

  missing value where TRUE/FALSE needed
 traceback()
2: eqscplot(x = 0, y = 0, xlim = xlim, ylim = ylim, xlab = xlab,
   ylab = ylab, type = n)
1: figframe()


Here is a minimal source script exhibiting the problem:

library(MASS)

eps - function(file=Rplot.eps, horizontal=FALSE, paper=special, ...) {
postscript(file=file, onefile=FALSE, horizontal=horizontal, 
paper=paper, ...)

  }


# Figure frame
figframe - function(  xlim = c(-3,3), ylim = c(-3,3), xlab = '', ylab = 
'') {
eqscplot( x=0,y=0, xlim = xlim, ylim = ylim, xlab = xlab, ylab = 
ylab, type = 'n')

abline( v=0, col=gray)
abline( h=0, col=gray)
}

# screen: OK
op - par(mar=c(3, 3, 1, 1) + 0.4)
figframe()
par(op)

# eps: fails
eps(file=test1.eps)
op - par(mar=c(3, 3, 1, 1) + 0.4)
figframe()
par(op)
dev.off()

I know this worked under an earlier version, probably ~ R 2.9.x

--
Michael Friendly Email: friendly AT yorku DOT ca
Professor, Psychology Dept.
York University  Voice: 416 736-5115 x66249 Fax: 416 736-5814
4700 Keele StreetWeb:   http://www.datavis.ca
Toronto, ONT  M3J 1P3 CANADA

__
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] as.character limits length of result for formula

2011-06-01 Thread Terrence Ireland

If you want a character representation of a long formula (or

a formula with long names), you can use:

 

as.character(my.formula)

 

However restriction on length of an as.character result

returns only the beginning of a long formula, and without comment.

 

In most cases, the following expression provides the complete

result:

 

paste(my.formula[[2]],   ~ , paste(attr(terms(my.formula),
term.labels),collapse=  + ),sep=)

 

It would be better to make as.character handle any size formula.


[[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.

[R] Force the for loop to stop

Hi,
I am looking for a command in R that would force the for loop to stop after
it finds what it is looking for.
As an example
for(i in 1:5){
   for(j in 3:6){
  if(i==j)
# do something...
   break;
}
}
And i don't want the loop to execute once i = 3 and stop.

Is there a way to do this?

best,
salih

[[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.

Re: [R] Force the for loop to stop

2011-06-01 Thread Stephan Kolassa


Hi,

you could set a dummy variable to FALSE outside the outermost loop. If 
the break condition is met in the inner loop, set the dummy variable to 
TRUE before breaking and test its truth status in the outer loop.


HTH
Stephan


Am 01.06.2011 21:25, schrieb Salih Tuna:

Hi,
I am looking for a command in R that would force the for loop to stop after
it finds what it is looking for.
As an example
for(i in 1:5){
for(j in 3:6){
   if(i==j)
 # do something...
break;
}
}
And i don't want the loop to execute once i = 3 and stop.

Is there a way to do this?

best,
salih

[[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.



__
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] as.character limits length of result for formula

2011-06-01 Thread William Dunlap

as.character() doesn't give a faithful character
representation of its input for lots of language-
related inputs.  E.g.,
   f - reformulate(paste(sep=,X,1:500), quote(log(Y)))
   cat(strwrap(as.character(f), 60), sep=\n)
  ~
  log(Y)
  X1 + X2 + X3 + X4 + X5 + X6 + X7 + X8 + X9 + X10 + X11 +
  X12 + X13 + X14 + X15 + X16 + X17 + X18 + X19 + X20 + X21 +
  X22 + X23 + X24 + X25 + X26 + X27 + X28 + X29 + X30 + X31 +
  X32 + X33 + X34 + X35 + X36 + X37 + X38 + X39 + X40 + X41 +
  X42 + X43 + X44 + X45 + X46 + X47 + X48 + X49 + X50 + X51 +
  X52 + X53 + X54 + X55 + X56 + X57 + X58 + X59 + X60 + X61 +
  X62 + X63 + X64 + X65 + X66 + X67 + X68 + X69 + X70 + X71 +
  X72 + X73 + X74 + X75 + X76 + X77 + X78 + X79 + X80 + X81 +
  X82 + X83 + X84 + X85 +
(Yes, there is nothing after the X85 +.)

deparse() works better for such objects.  You may want to use
paste(collapse=\n) on its output to make a single string.
   deparse(f)
   [1] log(Y) ~ X1 + X2 + X3 + X4 + X5 + X6 + X7 + X8 + X9 + X10 + X11
+  
   [2] X12 + X13 + X14 + X15 + X16 + X17 + X18 + X19 + X20 + X21 +
   
   ... lots of text omitted ...
  [53] X480 + X481 + X482 + X483 + X484 + X485 + X486 + X487 + X488
+ 
  [54] X489 + X490 + X491 + X492 + X493 + X494 + X495 + X496 + X497
+ 
  [55] X498 + X499 + X500 


Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com  

 -Original Message-
 From: r-help-boun...@r-project.org 
 [mailto:r-help-boun...@r-project.org] On Behalf Of Terrence Ireland
 Sent: Wednesday, June 01, 2011 12:19 PM
 To: r-help@r-project.org
 Subject: [R] as.character limits length of result for formula
 
 If you want a character representation of a long formula (or
 
 a formula with long names), you can use:
 
  
 
 as.character(my.formula)
 
  
 
 However restriction on length of an as.character result
 
 returns only the beginning of a long formula, and without comment.
 
  
 
 In most cases, the following expression provides the complete
 
 result:
 
  
 
 paste(my.formula[[2]],   ~ , paste(attr(terms(my.formula),
 term.labels),collapse=  + ),sep=)
 
  
 
 It would be better to make as.character handle any size formula.
 
 
   [[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.
 

__
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] Contributed Packages - Hmisc survey

2011-06-01 Thread Muhuri, Pradip (SAMHSA/CBHSQ)

Hello List,

Could someone tell  why I can't install the Himsc and survey packages for R 
version 2.13.0 (2011-04-13)? What am I doing wrong here?

Thanks,

Pradip


 install.packages (Hmisc, dependencies=TRUE)
--- Please select a CRAN mirror for use in this session ---
Warning: unable to access index for repository 
http://watson.nci.nih.gov/cran_mirror/bin/windows/contrib/2.13
Warning message:
In getDependencies(pkgs, dependencies, available, lib) :
  package 'Hmisc' is not available (for R version 2.13.0)



 install.packages (survey, dependencies=TRUE)
Warning: unable to access index for repository 
http://watson.nci.nih.gov/cran_mirror/bin/windows/contrib/2.13
Warning message:
In getDependencies(pkgs, dependencies, available, lib) :
  package 'survey' is not available (for R version 2.13.0)

Pradip K. Muhuri, PhD
Statistician
Substance Abuse  Mental Health Services Administration
The Center for Behavioral Health Statistics and Quality
Division of Population Surveys
1 Choke Cherry Road, Room 7-1023
Rockville, MD 20857

Tel: 240-276-1070
Fax: 240-276-1260
e-mail: pradip.muh...@samhsa.hhs.govmailto:pradip.muh...@samhsa.hhs.gov

The Center for Behavioral Health Statistics and Quality your feedback.  Please 
click on the following link to complete a brief customer survey:   
http://cbhsqsurvey.samhsa.gov

[[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.

Re: [R] Recode numbers

Thank you, Duncan,

Here “a” has the length of 24, and “b” has the length of 20 with numbers
from 1 to 20 uniquely. I just want encode “a” from 1 to 20 based on “a”
current order using “b”. So, a1[1] = b[1] = 1
a1[2] = b[2] = 5
a1[3] = a1[4] = b[3] = 8 (since third and fourth numbers are the same in
“a”)
a1[5] = a1[6] = b[4] = 9
a1[7] = b[5] = 14
a1[8] = b[6] = 20
a1[9] = b[7] = 3
a1[10] = a1[11] = b[8] = 10
...
a1[23] = b[19] = 15
a1[24] = b[20] = 19


--
View this message in context: 
http://r.789695.n4.nabble.com/Recode-numbers-tp3566395p3566681.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.

[R] problems with copula

2011-06-01 Thread ANGELA86

Hi, I'd like to know why using the program R I can't add a number of
margins 3,  I have a problem with the graphics. 
Post here my script:
 myCop.norm - ellipCopula(family = normal, dim = 3, param = 0.4)
 myMvd - mvdc(copula = myCop.norm, margins = c(norm, norm,norm),
 paramMargins = list(list(mean = 0, sd = 2), list(mean = 0, sd =
 1),list(mean=0,sd=2)))
persp(myMvd, dmvdc, xlim = c(-4, 4), ylim=c(-4, 4),main=Normal copula)

I  obtain the following  message of error :
Error in eval  index out of bound 


Help me thanks 

http://r.789695.n4.nabble.com/file/n3566629/Cattura.jpg 

--
View this message in context: 
http://r.789695.n4.nabble.com/problems-with-copula-tp3566629p3566629.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.

[R] RES: Recode numbers

2011-06-01 Thread Filipe Leme Botelho

---BeginMessage---
I think this is proper.

a - c(1, 2, 3, 3, 4, 4, 5, 6, 1, 2, 2, 3, 1, 2, 1, 2, 3, 3, 4, 5, 1, 2,
3, 4)
b - c(1, 5, 8, 9, 14, 20, 3, 10, 12, 6, 16, 7, 11, 13, 17, 18, 2, 4,
15, 19)

ref - 1
for (i in 2:length(a)) {
   if (a[i]!=a[i-1]) ref - c(ref, ref[length(ref)]+1)
   if (a[i]==a[i-1]) ref - c(ref, ref[length(ref)])
   }

a2 -b[ref]
a1 - c(1, 5, 8, 8, 9, 9, 14, 20, 3, 10, 10, 12, 6, 16, 7, 11, 13, 13,
17, 18, 2, 4, 15, 19)

a1=¢


-Mensagem original-
De: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
Em nome de Duncan Murdoch
Enviada em: quarta-feira, 1 de junho de 2011 15:51
Para: Lisa
Cc: r-help@r-project.org
Assunto: Re: [R] Recode numbers

(The attributions are a little messed up here:)

 I have two sets of numbers that look like
 a- c(1, 2, 3, 3, 4, 4, 5, 6, 1, 2, 2, 3, 1, 2, 1, 2, 3, 3, 4, 5, 1,
2, 3,
 4)

 b- c(1, 5, 8, 9, 14, 20, 3, 10, 12, 6, 16, 7, 11, 13, 17, 18, 2, 4,
15,
 19)

 I just want to use b to encode a so that a looks like

 a1- c(1, 5, 8, 8, 9, 9, 14, 20, 3, 10, 10, 12, 6, 16, 7, 11, 13,
13, 17,
 18, 2, 4, 15, 19)

 Does anyone have a suggestion how to deal with this? Thank you in
advance.


 is a1 = b[a] what you are looking for?

 HTH

 Pete

 Thank you for your help, Pete. I tried b[a], but it is not a1.


I think you'll need to explain the rule used for a1[9], a1[10], etc.

Duncan Murdoch

__
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.
---End Message---
This message and its attachments may contain confidential and/or privileged 
information. If you are not the addressee, please, advise the sender 
immediately by replying to the e-mail and delete this message.

Este mensaje y sus anexos pueden contener información confidencial o 
privilegiada. Si ha recibido este e-mail por error por favor bórrelo y envíe un 
mensaje al remitente.

Esta mensagem e seus anexos podem conter informação confidencial ou 
privilegiada. Caso não seja o destinatário, solicitamos a imediata notificação 
ao remetente e exclusão da mensagem.__
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] Function to save plots

2011-06-01 Thread Merik Nanish

Hello,

I'm using ROCR to plot ROC Curves and I want to automate the saving of plots
into PNG files using a custom function.

My data frames are named like test1, test2, test3. Each data frame has three
variables: method1, method2, goldstandard.

Right now, for each plot I have to run:

png('test1_method1.png')
plot(performance(prediction(test1$method1, test1$goldstandard), tpr,
fpr))
dev.off()

Here is the function I tried to create but I failed:

roc - function(arg1, arg2){
  png(paste(arg1, arg2, png, sep=_))
  plot(performance(prediction(arg1$arg2, arg1$goldstandard), tpr, fpr))
  dev.off()
}

I wanted to pass the test name as arg1, and the method name as arg2. Here
are my problems:

1) arg1$arg2 causes an error message. How can I use a function argument to
address a variable in a dataframe?

2) paste(arg1, arg2, png, sep=.) doesn't output something like
'test1.method1.png'. How should I deal with that?

Thanks in advance.

[[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.

Re: [R] Force the for loop to stop

Hi Stephan,
Thanks a lot. But i am not very good at R yet so i dont know how to set the
dummy variable to FALSE. Can you please help me with that as well?

On Wed, Jun 1, 2011 at 8:34 PM, Stephan Kolassa stephan.kola...@gmx.dewrote:

 Hi,

 you could set a dummy variable to FALSE outside the outermost loop. If the
 break condition is met in the inner loop, set the dummy variable to TRUE
 before breaking and test its truth status in the outer loop.

 HTH
 Stephan


 Am 01.06.2011 21:25, schrieb Salih Tuna:

 Hi,
 I am looking for a command in R that would force the for loop to stop
 after
 it finds what it is looking for.
 As an example
 for(i in 1:5){
for(j in 3:6){
   if(i==j)
 # do something...
break;
 }
 }
 And i don't want the loop to execute once i = 3 and stop.

 Is there a way to do this?

 best,
 salih

[[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.




[[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.

Re: [R] Force the for loop to stop

2011-06-01 Thread Stephan Kolassa


Hi Salih,

here you go:


dummy - FALSE
for ( ii in 1:5 ) {
  for ( jj in 3:6 ) {
cat(ii=,ii,; jj=,jj,\n,sep=)
if ( ii == jj ) {
  dummy - TRUE
  break
}
  }
  if ( dummy ) break
}
###

Note that I am using ii and jj as loop indices, not i and j. 
This makes it a lot easier to search for the loop counter in more 
complex scripts - if you just search for i, most of your hits will be 
something else than the loop counter.


HTH,
Stephan


Am 01.06.2011 22:06, schrieb Salih Tuna:

Hi Stephan,
Thanks a lot. But i am not very good at R yet so i dont know how to set the
dummy variable to FALSE. Can you please help me with that as well?

On Wed, Jun 1, 2011 at 8:34 PM, Stephan Kolassastephan.kola...@gmx.dewrote:


Hi,

you could set a dummy variable to FALSE outside the outermost loop. If the
break condition is met in the inner loop, set the dummy variable to TRUE
before breaking and test its truth status in the outer loop.

HTH
Stephan


Am 01.06.2011 21:25, schrieb Salih Tuna:


Hi,
I am looking for a command in R that would force the for loop to stop
after
it finds what it is looking for.
As an example
for(i in 1:5){
for(j in 3:6){
   if(i==j)
 # do something...
break;
}
}
And i don't want the loop to execute once i = 3 and stop.

Is there a way to do this?

best,
salih

[[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.








__
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] Recode numbers

2011-06-01 Thread William Dunlap



Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com  

 -Original Message-
 From: r-help-boun...@r-project.org 
 [mailto:r-help-boun...@r-project.org] On Behalf Of Lisa
 Sent: Wednesday, June 01, 2011 12:28 PM
 To: r-help@r-project.org
 Subject: Re: [R] Recode numbers
 
 Thank you, Duncan,
 
 Here a has the length of 24, and b has the length of 20 
 with numbers
 from 1 to 20 uniquely. I just want encode a from 1 to 20 
 based on a
 current order using b. So, a1[1] = b[1] = 1
 a1[2] = b[2] = 5
 a1[3] = a1[4] = b[3] = 8 (since third and fourth numbers are 
 the same in
 a)
 a1[5] = a1[6] = b[4] = 9
 a1[7] = b[5] = 14
 a1[8] = b[6] = 20
 a1[9] = b[7] = 3
 a1[10] = a1[11] = b[8] = 10
 ...
 a1[23] = b[19] = 15
 a1[24] = b[20] = 19

I don't understand the verbal description, but the the
following makes your example work:
   b[cumsum(c(TRUE,a[-1]!=a[-length(a)]))]
   [1]  1  5  8  8  9  9 14 20  3 10 10 12  6 16  7 11 13 13
  [19] 17 18  2  4 15 19
   identical(.Last.value, a1)
  [1] TRUE
Is that what you want?  The cumsum(...) increments each
at each changepoint in a.

Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com 

 
 
 --
 View this message in context: 
 http://r.789695.n4.nabble.com/Recode-numbers-tp3566395p3566681.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.
 

__
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] Function to save plots

2011-06-01 Thread Joshua Wiley

Hi Merik,

I believe this does what you want (and maybe a little more):

## define the function, a1 and a2 correspond to your arguments
## d is the directory to save the file in (by default the working directory)
## ... are additional arguments that can be passed to png() to control
things like quality, etc.
roc - function(a1, a2, d = getwd(), ...) {

## someone wiser may know a better way, but I put dev.off() into
on.exit() because
## otherwise, if the plot fails (say due to an error in prediction()
or performance())
## the device does not get shutdown otherwise, and the image is empty
but still used by R
  on.exit(dev.off())

## I use get() to deal with the fact that the dataset name is passed
as a character
  dat - get(a1)

  png(filename = paste(d, /, a1, _, a2, .png, sep = ''), ...)
  plot(performance(prediction(dat[, a2], dat[, goldstandard]),
tpr, fpr))
## I like titles (if you don't need it, just delte this line)
  title(main = paste(a2, in dataset:, a1))
}


## Test dataset
test1 - data.frame(method1 = rbinom(40, 1, .4),
  method2 = rbinom(40, 1, .4), goldstandard = rbinom(40, 1, .8))


## Example Usage
roc(a1 = test1, a2 = method1)
roc(a1 = test1, a2 = method1, d = DRIVE:/AlternatePath)
roc(a1 = test1, a2 = method1, width = 960, height = 960)

Hope this helps,

Josh

On Wed, Jun 1, 2011 at 12:39 PM, Merik Nanish merik.nan...@gmail.com wrote:
 Hello,

 I'm using ROCR to plot ROC Curves and I want to automate the saving of plots
 into PNG files using a custom function.

 My data frames are named like test1, test2, test3. Each data frame has three
 variables: method1, method2, goldstandard.

 Right now, for each plot I have to run:

 png('test1_method1.png')
 plot(performance(prediction(test1$method1, test1$goldstandard), tpr,
 fpr))
 dev.off()

 Here is the function I tried to create but I failed:

 roc - function(arg1, arg2){
  png(paste(arg1, arg2, png, sep=_))
  plot(performance(prediction(arg1$arg2, arg1$goldstandard), tpr, fpr))
  dev.off()
 }

 I wanted to pass the test name as arg1, and the method name as arg2. Here
 are my problems:

 1) arg1$arg2 causes an error message. How can I use a function argument to
 address a variable in a dataframe?

 2) paste(arg1, arg2, png, sep=.) doesn't output something like
 'test1.method1.png'. How should I deal with that?

 Thanks in advance.

        [[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.




-- 
Joshua Wiley
Ph.D. Student, Health Psychology
University of California, Los Angeles
http://www.joshuawiley.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.

Re: [R] Force the for loop to stop

That is great Stephan,
thanks a lot for your help.
best,
salih

On Wed, Jun 1, 2011 at 9:18 PM, Stephan Kolassa stephan.kola...@gmx.dewrote:

 Hi Salih,

 here you go:

 
 dummy - FALSE
 for ( ii in 1:5 ) {
  for ( jj in 3:6 ) {
cat(ii=,ii,; jj=,jj,\n,sep=)
if ( ii == jj ) {
  dummy - TRUE
  break
}
  }
  if ( dummy ) break
 }
 ###

 Note that I am using ii and jj as loop indices, not i and j. This
 makes it a lot easier to search for the loop counter in more complex scripts
 - if you just search for i, most of your hits will be something else than
 the loop counter.


 HTH,
 Stephan


 Am 01.06.2011 22:06, schrieb Salih Tuna:

 Hi Stephan,
 Thanks a lot. But i am not very good at R yet so i dont know how to set
 the
 dummy variable to FALSE. Can you please help me with that as well?

 On Wed, Jun 1, 2011 at 8:34 PM, Stephan Kolassastephan.kola...@gmx.de
 wrote:

  Hi,

 you could set a dummy variable to FALSE outside the outermost loop. If
 the
 break condition is met in the inner loop, set the dummy variable to TRUE
 before breaking and test its truth status in the outer loop.

 HTH
 Stephan


 Am 01.06.2011 21:25, schrieb Salih Tuna:

  Hi,
 I am looking for a command in R that would force the for loop to stop
 after
 it finds what it is looking for.
 As an example
 for(i in 1:5){
for(j in 3:6){
   if(i==j)
 # do something...
break;
 }
 }
 And i don't want the loop to execute once i = 3 and stop.

 Is there a way to do this?

 best,
 salih

[[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.







[[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.

[R] Problems Dating....

2011-06-01 Thread Struckmeier, Nathanael

I'm trying to convert a column in a data frame with dates from a
Factor type to a Date Object but I am encountering and error. (I am
having trouble plotting an x,y scatter and I suspect it's something with
my data format). I have a table with two columns and 8,000 rows. 

 dsort=read.delim(C:\\Documents and Settings\\E066582\\My
Documents\\R\\R-2.13.0\\bin\\dsort.txt)

 

dsort#name of
data.frame

 colnames(dsort)[1] #name of column 1

[1] Date

 colnames(dsort)[2] #name of column 2

[1] Qty

 class(dsort$Date) #checked data type of column
Date and it came back as a factor

[1] factor

 Date2=as.Date(dsort$Date)#attempt at changing the data type from a
factor to a date object (see error below).

Error in charToDate(x) : 

  character string is not in a standard unambiguous format

 

Dates in my table are listed in 3/4/2007 format. 

StatBat2

 


[[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.

Re: [R] re-write plot function for ggplot

2011-06-01 Thread John Ramey

A useful way to get a feel for ggplot2 is to check out
http://yeroon.net/ggplot2/

This site allows for a point-and-click ggplot2 graphs. You will not
have all of the options from ggplot2 (e.g. themes), but it's a good
place to get started.

Also, there are some wonderful video tutorials in the help section.

Cheers!

-- 
John A. Ramey, M.S.
Ph.D. Candidate
Department of Statistics
Baylor University
http://www.johnramey.net



On Wed, Jun 1, 2011 at 6:43 AM, rmje robinmje...@gmail.com wrote:
 Hi, I have the following function that i use to plot graphs.

 plot_mi_time = function(mdata, miname) {
   mdata2 = mdata[row.names(mir_test3) == miname, ]
   # print(mdata2)
   xcoords - c(1,1,2,2,3,3)
   plot(xcoords, mdata2, xaxt=n, ylab=Expression, xlab=Time(h), ,
 main=miname)
           axis(1, at=xcoords, labels=c(a,a,b,b,c,c))
 }




 I want to edit this function so that i can plot it using ggplot.
 Any suggestions on how to re-write it?

 --
 View this message in context: 
 http://r.789695.n4.nabble.com/re-write-plot-function-for-ggplot-tp3565868p3565868.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.




-- 
John A. Ramey, M.S.
Ph.D. Candidate
Department of Statistics
Baylor University
http://www.johnramey.net

__
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] finding numbers in a range

Hi,
Is there a quick way of finding if numbers in a vector falls in between a
range defined in another matrix.
As an example let
x   y
0  1   3
2  6   9
2515  18

I want to check one by one whether 0,2 and 25 falls in any of the ranges in
y.
I am using 2 for loops but it is taking a huge amount of time.
is there any other alternative way to do this?

best,
salih

[[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.

Re: [R] weird error from MASS::eqcsplot with postscript driver

2011-06-01 Thread Rolf Turner



The problem is the ``paper=special '' specification in your eps() 
function.


This makes par(pin), which eqcsplot() uses in its calculations, undefined.
The ``value'' thus produced is:

[1] NaN NaN

Which makes perfect sense if you think about it.

If you're going to use paper=special, you need to specify height and 
width.


cheers,

Rolf Turner

On 02/06/11 07:17, Michael Friendly wrote:

[Env: R 2.12.2, Win XP]

I'm creating figures using MASS::eqcsplot to provide equal scaling of 
the axes.  My figures work OK
when I plot to the screen, but when I try to do the same plot as a 
postscript file, I get an

unexplicable error,

 figframe()
Error in if (yuin  xuin * ratio) yuin - xuin * ratio else xuin - 
yuin/ratio :

  missing value where TRUE/FALSE needed
 traceback()
2: eqscplot(x = 0, y = 0, xlim = xlim, ylim = ylim, xlab = xlab,
   ylab = ylab, type = n)
1: figframe()


Here is a minimal source script exhibiting the problem:

library(MASS)

eps - function(file=Rplot.eps, horizontal=FALSE, paper=special, 
...) {
postscript(file=file, onefile=FALSE, horizontal=horizontal, 
paper=paper, ...)

  }


# Figure frame
figframe - function(  xlim = c(-3,3), ylim = c(-3,3), xlab = '', ylab 
= '') {
eqscplot( x=0,y=0, xlim = xlim, ylim = ylim, xlab = xlab, ylab = 
ylab, type = 'n')

abline( v=0, col=gray)
abline( h=0, col=gray)
}

# screen: OK
op - par(mar=c(3, 3, 1, 1) + 0.4)
figframe()
par(op)

# eps: fails
eps(file=test1.eps)
op - par(mar=c(3, 3, 1, 1) + 0.4)
figframe()
par(op)
dev.off()

I know this worked under an earlier version, probably ~ R 2.9.x



__
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] Problems Dating....

2011-06-01 Thread Stephan Kolassa


Hi Nat,

I guess something like
as.Date(as.character(3/4/2007),format=%d/%m/%Y)
should work - as.character() coerces the factors to characters, which 
the as.Date() function can work with, given the right format argument.


HTH
Stephan


Am 01.06.2011 22:59, schrieb Struckmeier, Nathanael:

I'm trying to convert a column in a data frame with dates from a
Factor type to a Date Object but I am encountering and error. (I am
having trouble plotting an x,y scatter and I suspect it's something with
my data format). I have a table with two columns and 8,000 rows.


dsort=read.delim(C:\\Documents and Settings\\E066582\\My

Documents\\R\\R-2.13.0\\bin\\dsort.txt)



dsort#name of
data.frame


colnames(dsort)[1] #name of column 1


[1] Date


colnames(dsort)[2] #name of column 2


[1] Qty


class(dsort$Date) #checked data type of column

Date and it came back as a factor

[1] factor


Date2=as.Date(dsort$Date)#attempt at changing the data type from a

factor to a date object (see error below).

Error in charToDate(x) :

   character string is not in a standard unambiguous format



Dates in my table are listed in 3/4/2007 format.

StatBat2




[[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.



__
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] Problems Dating....

2011-06-01 Thread Marc Schwartz

On Jun 1, 2011, at 3:59 PM, Struckmeier, Nathanael wrote:

 I'm trying to convert a column in a data frame with dates from a
 Factor type to a Date Object but I am encountering and error. (I am
 having trouble plotting an x,y scatter and I suspect it's something with
 my data format). I have a table with two columns and 8,000 rows. 
 
 dsort=read.delim(C:\\Documents and Settings\\E066582\\My
 Documents\\R\\R-2.13.0\\bin\\dsort.txt)
 
 
 
 dsort#name of
 data.frame
 
 colnames(dsort)[1] #name of column 1
 
 [1] Date
 
 colnames(dsort)[2] #name of column 2
 
 [1] Qty
 
 class(dsort$Date) #checked data type of column
 Date and it came back as a factor
 
 [1] factor
 
 Date2=as.Date(dsort$Date)#attempt at changing the data type from a
 factor to a date object (see error below).
 
 Error in charToDate(x) : 
 
  character string is not in a standard unambiguous format
 
 
 
 Dates in my table are listed in 3/4/2007 format. 
 
 StatBat2


After seeing your subject line and your e-mail address, I was going to suggest 
a dozen roses and some Moose Munch...  ;-)

When you convert a character/factor to a Date, you need to specify the format 
of the object to be converted:

 as.Date(3/4/2007, format = %m/%d/%Y)
[1] 2007-03-04

See ?as.Date and ?strptime for more information and format specifications.

HTH,

Marc Schwartz

__
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] Contributed Packages - Hmisc survey

2011-06-01 Thread Nordlund, Dan (DSHS/RDA)

 -Original Message-
 From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
 project.org] On Behalf Of Muhuri, Pradip (SAMHSA/CBHSQ)
 Sent: Wednesday, June 01, 2011 1:01 PM
 To: 'r-help-requ...@r-project.org'; 'R-help@r-project.org'
 Subject: [R] Contributed Packages - Hmisc  survey

 Hello List,

 Could someone tell  why I can't install the Himsc and survey packages
 for R version 2.13.0 (2011-04-13)? What am I doing wrong here?

 Thanks,

 Pradip

  install.packages (Hmisc, dependencies=TRUE)
 --- Please select a CRAN mirror for use in this session ---
 Warning: unable to access index for repository
 http://watson.nci.nih.gov/cran_mirror/bin/windows/contrib/2.13
 Warning message:
 In getDependencies(pkgs, dependencies, available, lib) :
   package 'Hmisc' is not available (for R version 2.13.0)

  install.packages (survey, dependencies=TRUE)
 Warning: unable to access index for repository
 http://watson.nci.nih.gov/cran_mirror/bin/windows/contrib/2.13
 Warning message:
 In getDependencies(pkgs, dependencies, available, lib) :
   package 'survey' is not available (for R version 2.13.0)

I get a 502 Bad Gateway error when I try to go to that mirror.  Maybe it is 
down.  Have you tried another mirror?

Dan

Daniel J. Nordlund
Washington State Department of Social and Health Services
Planning, Performance, and Accountability
Research and Data Analysis Division
Olympia, WA 98504-5204

__
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] Problems Dating....

2011-06-01 Thread Struckmeier, Nathanael

I'll give this a try and mess with what format to convert it to. Thanks
again!

-Original Message-
From: Stephan Kolassa [mailto:stephan.kola...@gmx.de] 
Sent: Wednesday, June 01, 2011 2:10 PM
To: Struckmeier, Nathanael
Cc: R-help@r-project.org
Subject: Re: [R] Problems Dating

Hi Nat,

I guess something like
as.Date(as.character(3/4/2007),format=%d/%m/%Y)
should work - as.character() coerces the factors to characters, which 
the as.Date() function can work with, given the right format argument.

HTH
Stephan

Am 01.06.2011 22:59, schrieb Struckmeier, Nathanael:
 I'm trying to convert a column in a data frame with dates from a
 Factor type to a Date Object but I am encountering and error. (I
am
 having trouble plotting an x,y scatter and I suspect it's something
with
 my data format). I have a table with two columns and 8,000 rows.

 dsort=read.delim(C:\\Documents and Settings\\E066582\\My
 Documents\\R\\R-2.13.0\\bin\\dsort.txt)

 dsort#name of
 data.frame

 colnames(dsort)[1] #name of column 1

 [1] Date

 colnames(dsort)[2] #name of column 2

 [1] Qty

 class(dsort$Date) #checked data type of
column
 Date and it came back as a factor

 [1] factor

 Date2=as.Date(dsort$Date)#attempt at changing the data type from
a
 factor to a date object (see error below).

 Error in charToDate(x) :

character string is not in a standard unambiguous format

 Dates in my table are listed in 3/4/2007 format.

 StatBat2

   [[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.

__
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] Plotting from functions

2011-06-01 Thread Yang Zhang

I can plot to png's fine when i run this directly from the top-level
script/console:

png('diamonds.png');qplot(carat,price,data=diamonds);dev.off()

But for some reason it's not working when that's in a function:

(function(){png('diamonds.png');qplot(carat,price,data=diamonds);dev.off()})()

I suspect this is because the qplot's return value must be returned
to the top-level, because this works:

(function(){png('diamonds.png');qplot(carat,price,data=diamonds)})();dev.off()

Any hints on how I can plot from within a function?  Thanks in advance.

__
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] RES: Recode numbers

Thank you so much, Filipe. Your R script is what I am looking for. 

--
View this message in context: 
http://r.789695.n4.nabble.com/Recode-numbers-tp3566395p3566818.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.

Re: [R] Plotting from functions

2011-06-01 Thread Yang Zhang

On Wed, Jun 1, 2011 at 1:04 PM, Yang Zhang yanghates...@gmail.com wrote:
 I can plot to png's fine when i run this directly from the top-level
 script/console:

 png('diamonds.png');qplot(carat,price,data=diamonds);dev.off()

 But for some reason it's not working when that's in a function:

 (function(){png('diamonds.png');qplot(carat,price,data=diamonds);dev.off()})()

 I suspect this is because the qplot's return value must be returned
 to the top-level, because this works:

 (function(){png('diamonds.png');qplot(carat,price,data=diamonds)})();dev.off()

 Any hints on how I can plot from within a function?  Thanks in advance.


Actually, just found that things work if I do:

print(qplot(...))

Is this just what you need to do to plot from within functions?

__
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] RES: Recode numbers

Thank you very much, Bill. Your script works very well.

--
View this message in context: 
http://r.789695.n4.nabble.com/Recode-numbers-tp3566395p3566847.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.

Re: [R] Plotting from functions

2011-06-01 Thread Joshua Wiley

Hi,

Within functions, you often do need to wrap the plotting calls in
print() (you would see this if you read the FAQ).

Cheers,

Josh

On Wed, Jun 1, 2011 at 1:06 PM, Yang Zhang yanghates...@gmail.com wrote:
 On Wed, Jun 1, 2011 at 1:04 PM, Yang Zhang yanghates...@gmail.com wrote:
 I can plot to png's fine when i run this directly from the top-level
 script/console:

 png('diamonds.png');qplot(carat,price,data=diamonds);dev.off()

 But for some reason it's not working when that's in a function:

 (function(){png('diamonds.png');qplot(carat,price,data=diamonds);dev.off()})()

 I suspect this is because the qplot's return value must be returned
 to the top-level, because this works:

 (function(){png('diamonds.png');qplot(carat,price,data=diamonds)})();dev.off()

 Any hints on how I can plot from within a function?  Thanks in advance.


 Actually, just found that things work if I do:

 print(qplot(...))

 Is this just what you need to do to plot from within functions?

 __
 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.




-- 
Joshua Wiley
Ph.D. Student, Health Psychology
University of California, Los Angeles
http://www.joshuawiley.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.

Re: [R] Missing completely at random

2011-06-01 Thread Steven Kennedy

This works, but there might be a better way:

misscols-lapply(howmanyMiss, function(x) sample(1:n, x))

for (i in 1:nMiss){
for (j in misscols[[i]]){
X[idMiss[i],j]-NA
}
}

__
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] finding numbers in a range



On Jun 1, 2011, at 5:07 PM, Salih Tuna wrote:


Hi,
Is there a quick way of finding if numbers in a vector falls in  
between a

range defined in another matrix.
As an example let
x   y
0  1   3
2  6   9
2515  18

I want to check one by one whether 0,2 and 25 falls in any of the  
ranges in

y.
I am using 2 for loops but it is taking a huge amount of time.
is there any other alternative way to do this?


If this represents the full complexity of the problem then  
findInterval would work:


 findInterval(c(0,2,25), c(1,3,6,9,15,18) ) %in% c(1,3,5)
[1] FALSE  TRUE FALSE

On the other hand if the y intervals overlap then a different method  
may be needed.


 sapply( c(0, 2, 25) , function(x) {any( y[,1]  x  x  y[,2]) })
[1] FALSE  TRUE FALSE

--
David Winsemius, MD
West Hartford, CT

__
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] Create Matrix with Float32 values

2011-06-01 Thread Peter Ehlers


On 2011-06-01 09:16, Chris English wrote:


Dear R_Help:
The following gives me a matrix with integer values.
z= matrix(rep(10:1, each= 10), ncol= 10, byrow=TRUE)  str(z) int [1:10, 1:10] 
10 9 8 7 6 5 4 3 2 1 ...
How do I specify that I want Float32 values instead.
Thanks,Chris


Have you tried adding 0? Or you could use the seq() function.
I assume that you have a reason for wanting this.

Peter Ehlers


__
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.

Re: [R] Problems Dating....



On Jun 1, 2011, at 5:14 PM, Struckmeier, Nathanael wrote:

I'll give this a try and mess with what format to convert it to.  
Thanks

again!

You need to decide whether to use %d/%m/%Y or %m/%d/% , since the  
information you provided so far leaves that undetermined.


--
David.



Hi Nat,

I guess something like
as.Date(as.character(3/4/2007),format=%d/%m/%Y)
should work - as.character() coerces the factors to characters, which
the as.Date() function can work with, given the right format argument.

HTH
Stephan


Am 01.06.2011 22:59, schrieb Struckmeier, Nathanael:

I'm trying to convert a column in a data frame with dates from a
Factor type to a Date Object but I am encountering and error. (I

am

having trouble plotting an x,y scatter and I suspect it's something

with

my data format). I have a table with two columns and 8,000 rows.


dsort=read.delim(C:\\Documents and Settings\\E066582\\My

Documents\\R\\R-2.13.0\\bin\\dsort.txt)



dsort#name of
data.frame


colnames(dsort)[1] #name of column 1


[1] Date


colnames(dsort)[2] #name of column 2


[1] Qty


class(dsort$Date) #checked data type of

column

Date and it came back as a factor

[1] factor


Date2=as.Date(dsort$Date)#attempt at changing the data type from

a

factor to a date object (see error below).

Error in charToDate(x) :

  character string is not in a standard unambiguous format



Dates in my table are listed in 3/4/2007 format.

StatBat2




[[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.



__
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

__
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] Create Matrix with Float32 values


On 01/06/2011 12:16 PM, Chris English wrote:


Dear R_Help:
The following gives me a matrix with integer values.
z= matrix(rep(10:1, each= 10), ncol= 10, byrow=TRUE)  str(z) int [1:10, 1:10] 
10 9 8 7 6 5 4 3 2 1 ...
How do I specify that I want Float32 values instead.


You can't.  R doesn't support that type.

If you want to pass it to an external function that is expecting C 
single type, you can use as.single(x).  That won't convert it to single 
precision, but it will cause it to be copied to a single precision array 
whenever you call .C or .Fortran.


Duncan Murdoch

__
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] Plotting from functions