Luke wrote:
Dear R Users,
When I installed e1071 use R CMD INSTALL, I got
configure: WARNING: g++ 2.96 cannot reliably be used with this package.
configure: error: Please use a different C++ compiler.
But how to let R CMD INSTALL use a different C++ compiler? and which
C++ compiler is
Is there any possibility to get high resolution plots in a windows xp
system?
I tried it with the device function png(filename =
c:/r/highresplot%d.png,pointsize=12, res=900)
but when I try to set: width = 480, height = 480 or pointsize = 12,
the text is not scaled in the same way as the
Knut Krueger wrote:
Is there any possibility to get high resolution plots in a windows xp
system?
I tried it with the device function png(filename =
c:/r/highresplot%d.png,pointsize=12, res=900)
but when I try to set: width = 480, height = 480 or pointsize = 12,
the text is not scaled
Hi, I would like to know whether it is possible to obtain a value of
significance for random effects when aplying the lme or related
functions. The default output in R is just a variance and standard
deviation measurement.
I feel it would be possible to obtain the significance of these random
On Mon, 11 Jul 2005, S.O. Nyangoma wrote:
Hi there,
Actually my aim was to compare anumber of extreme values (e.g. 39540)
with df1=1, df2=7025 via p-values.
If they have the same degrees of freedom, use the test statistic and not
the p value for comparing them.
Z
Spencer mentions that
Spencer == Spencer Graves [EMAIL PROTECTED]
on Mon, 11 Jul 2005 19:08:39 -0700 writes:
Spencer I'll bite: How does one detect bimodalidty from a boxplot?
One does not,
and that (whole area) was the main reason I mentioned violin
plots which do show features such as bimodality:
If they have the same degrees of freedom, use the test statistic
and not
the p value for comparing them.
Z
I appretiate your input to this discussion. Do you know of a reference
to your statement above?
I had actually used the test-statistic which in my case is r-squared
to compare them.
Rob Hyndman has written on a version of the boxplot that can show
bimodality. See, particularly, Figure 2 in:
Hyndman, R.J. (1996) Computing and graphing highest density regions,
Amer. Statist., 50, 120-126.
Andrew
On Tue, Jul 12, 2005 at 10:28:04AM +0200, Martin Maechler wrote:
Spencer ==
On Tue, 12 Jul 2005, S.O. Nyangoma wrote:
If they have the same degrees of freedom, use the test statistic
and not
the p value for comparing them.
Z
I appretiate your input to this discussion. Do you know of a reference
to your statement above?
?? Any basic statistics book?
I use a 32 bit machine. For those using 64 bit machines,
what is the .Machine$double.xmin for for machines?
regards. Stephen.
- Original Message -
From: Achim Zeileis [EMAIL PROTECTED]
Date: Tuesday, July 12, 2005 10:51 am
Subject: Re: [R] exact values for p-values - more
On 12-Jul-05 Spencer Graves wrote:
I'll bite: How does one detect bimodalidty from a boxplot?
spencer graves
Berton Gunter wrote:
[...]
Finally, a tidbit for boxplot afficianados: how does one detect
bimodality from a boxplot?
-- Bert Gunter
Not definitively
hi netters
i have a vector NAMES containing a series of variable names:
NAMES=c(x,r,z,m,st,qr,.nn).
i wanna fit a regression tree by using the code:
my.tree-tree(y~x+r+z+m+nn,my.dataframe)
but i don't want to type out x+r+z+m++nn one by one, as there are so
many
The default output in lmer also includes goodness of fit statistics which can
be used for this assessment. The anova() command invokes the likelihood ratio
test and can be used to compare models (under certain conditions).
The bVar slot in the lmer fitted model contains the posterior variance
I'm on R 2.1.0.
In the source function there is a bug preventing the proper use of the
chdir option (which simply doesn't work).
The problem is that in the function the following line occurs:
file - file(file, r, encoding = encoding)
This overwrites the variable file and later causes the check
Martin Maechler a écrit :
Spencer == Spencer Graves [EMAIL PROTECTED]
on Mon, 11 Jul 2005 19:08:39 -0700 writes:
Spencer I'll bite: How does one detect bimodalidty from a boxplot?
One does not,
and that (whole area) was the main reason I mentioned violin
plots which do show
I have a situation where this is fine:
if (length(x)15) {
clever - rr.ATM(x, maxtrim=7)
} else {
clever - rr.ATM(x)
}
clever
$ATM
[1] 1848.929
$sigma
[1] 1.613415
$trim
[1] 0
$lo
[1] 1845.714
$hi
[1] 1852.143
But this variant, using ifelse(),
[EMAIL PROTECTED] wrote:
I'm on R 2.1.0.
In the source function there is a bug preventing the proper use of the
chdir option (which simply doesn't work).
The problem is that in the function the following line occurs:
file - file(file, r, encoding = encoding)
This overwrites the
Hello,
I use package tcltk to do some GUI programming, and want to find a
function which can do the operation click a button, just like using
a mouse to click. If tkevent.generate can do that? I tried it as
below, but failed. Please give me a hint!
tt - tktoplevel()
tkwm.title(tt,Simple Dialog)
Ajay Narottam Shah wrote:
I have a situation where this is fine:
if (length(x)15) {
clever - rr.ATM(x, maxtrim=7)
} else {
clever - rr.ATM(x)
}
clever
$ATM
[1] 1848.929
$sigma
[1] 1.613415
$trim
[1] 0
$lo
[1] 1845.714
$hi
[1]
S.O. Nyangoma [EMAIL PROTECTED] writes:
I use a 32 bit machine. For those using 64 bit machines,
what is the .Machine$double.xmin for for machines?
regards. Stephen.
Depends on the FPU which is basically independent of the address mode:
PIII, Fedora Core 4:
.Machine$double.xmin
[1]
Sorry i sent the answer not to the mailing list - here it is
Brian D Ripley schrieb:
On Tue, 12 Jul 2005, Knut Krueger wrote:
Prof Brian Ripley schrieb:
Please read carefully what `resolution' means for a png() device (and a
PNG file). It is a hint to the viewer in the file header
zhihua li wrote:
hi netters
i have a vector NAMES containing a series of variable names:
NAMES=c(x,r,z,m,st,qr,.nn).
i wanna fit a regression tree by using the code:
my.tree-tree(y~x+r+z+m+nn,my.dataframe)
but i don't want to type out x+r+z+m++nn one by one, as
Hi R users
Does anyone out there have a better/quicker way of adding a factor column
to a data frame based on levels of another factor?
I have a (large) data frame consisting of records for individual plants,
each represented by a unique ID number. The species of each plant is
indicated in the
You can do it by subsetting the dataframe
df - data.frame( y=rbinom(100, 1, prob=0.5),
x1=rnorm(100), x2=rnorm(100), x3=rnorm(100),
z1=rnorm(100), z2=rnorm(100) )
names - c(x1, x2, x3)
tree( y ~ . , data = df[ , c(y, names) ] )
This solution is
Please read the posting guide (as we do ask) and use the current version
as it asks to see if the `bug' has already been fixed.
This was fixed a while ago, definitely in 2.1.1.
On Tue, 12 Jul 2005 [EMAIL PROTECTED] wrote:
I'm on R 2.1.0.
In the source function there is a bug preventing the
It's a floating-point quantity: `64 bit' refers to the pointer size
(only).
Almost all current R platforms use IEC60559 arithmetic for real numbers
and 32-bit integers for integers, so differ only in the way the compiler
orders operations and stores to memory (thereby losing precision on some
R-2.1.1 on windows XP
I just noticed something unpleasant when using bwplot (from lattice).
In order to satisfy a wish from a client, I needed to produce sets of boxplots
conditioned by another factor. My client didn't like the look of the boxplots
(by default, they have a star to mark the
Hi
I want to write a little function that takes a vector of arbitrary
length n and returns a matrix of size n+1 by n+1.
I can't easily describe it, but the following function that works for
n=3 should convey what I'm trying to do:
f - function(x){
matrix(c(
1 , 0 , 0
Omar Lakkis wrote:
How can I define a static member of a class? not a static method,
rather a static field that would be accessed by all instances of the
class.
To define a static field in an Object class (the R.oo package), I
recommend you to use a private field .field and then create a
wu sz [EMAIL PROTECTED] writes:
Hello,
I use package tcltk to do some GUI programming, and want to find a
function which can do the operation click a button, just like using
a mouse to click. If tkevent.generate can do that? I tried it as
below, but failed. Please give me a hint!
tt -
I apologize for not having investigated enough.
However I want to bring up the point that upgrading can be a very
tedious thing. I administer a cluster for which upgrading brings it to a
halt for several hours stopping all calculations going on. At the moment
I have no way to go from 2.1.0 to
Maybe not elegant, but I guess this function does what you want:
f - function( x ) {
n - length( x ) + 1
m - matrix( 0, nrow = n, ncol = n )
diag(m) - 1
for( j in 1:( n-1 ) ) {
for( i in ( j+1 ):n ) {
m[ i, j ] - prod( x[ j:(i-1) ] )
}
}
return( m )
}
Best
First create a dataframe with the translation you want, i.e.
one column with the species and another with the number you want in the end.
Then merge these two dataframes using 'merge' and voila..
I would start with looking at ?merge
Cheers, Henrik Andersson
Karen Kotschy wrote:
Hi R users
On Tue, 12 Jul 2005 [EMAIL PROTECTED] wrote:
I apologize for not having investigated enough.
However I want to bring up the point that upgrading can be a very
tedious thing. I administer a cluster for which upgrading brings it to a
halt for several hours stopping all calculations going on.
This is working exactly as documented. Nothing `breaks'!
What does the help page say?
'ifelse' returns a value with the same shape as 'test' which is
filled with elements selected from either 'yes' or 'no' depending
on whether the element of 'test' is 'TRUE' or 'FALSE'.
[EMAIL PROTECTED] wrote:
I apologize for not having investigated enough.
However I want to bring up the point that upgrading can be a very
tedious thing. I administer a cluster for which upgrading brings it to a
halt for several hours stopping all calculations going on. At the moment
I have
[EMAIL PROTECTED] writes:
I apologize for not having investigated enough.
However I want to bring up the point that upgrading can be a very
tedious thing. I administer a cluster for which upgrading brings it to a
halt for several hours stopping all calculations going on. At the moment
I
Roy Sanderson wrote:
Hello
I'm using logistic regression from the Design library (lrm), then fastbw to
undertake a backward selection and create a reduced model, before trying to
make predictions against an independent set of data using predict.lrm with
the reduced model. I wouldn't
Robin Hankin wrote:
What about
foo - function(a){
n - length(a)
X - diag(n+1)
X[lower.tri(X)] - unlist(lapply(seq(n),
function(x) cumprod(c(1, a)[-seq(x)])))
X
}
foo(c(10,7,2))
Uwe Ligges
Hi
I want to write a little function that takes a vector of
On 7/12/05, Robin Hankin [EMAIL PROTECTED] wrote:
Hi
I want to write a little function that takes a vector of arbitrary
length n and returns a matrix of size n+1 by n+1.
I can't easily describe it, but the following function that works for
n=3 should convey what I'm trying to do:
f -
hi all
why does R do this:
(-8)^(1/3)=NaN
the answer should be : -2
a silly question but i kept on getting errors in some of my code due to this
problem.
i solve the problem as follows:
say we want : (-a)^(1/3)
then : sign(a)*(a^(1/3)) works
but there has to be a simpler way of soing such
Actually, glm() does not estimate the dispersion at all, so you will need
to be more specific.
For example, summary.glm() and predict.glm() use the Pearson statistic if
dispersion=NULL (the default) for most families. You can supply any other
value you choose, and the MASS package makes use of
Hi,
I hope I'm not totally Off Topic, but I'm actually working on binary
classifier with probabilistic output [0 - 1]. I tested my methods
with some sample datasets from UCI Database but I'm still in need of
some samples. Especially, I'm looking for binary datasets with a
probabilistic
(-8+0i)^(1/3)
[1] 1+1.732051i
ie complex...
On 12/07/05, [EMAIL PROTECTED]
[EMAIL PROTECTED] wrote:
hi all
why does R do this:
(-8)^(1/3)=NaN
the answer should be : -2
a silly question but i kept on getting errors in some of my code due to this
problem.
i solve the problem as
Hi
I find that one often needs to keep reals real and complexes complex.
Try this:
cuberooti -
function (x)
{
if (is.complex(x)) {
return(sqrt(x + (0+0i)))
}
sign(x)* abs(x)^(1/3)
}
best wishes
[see that (0+0i) sitting there!]
Robin
On 12 Jul 2005, at 14:11, [EMAIL
Gabor
I cannot begin to tell you how much value you have added to my research
with your observation.
A real eureka moment for me.
[oh, and it answered my question as well]
kia ora
Robin
On 12 Jul 2005, at 13:35, Gabor Grothendieck wrote:
On 7/12/05, Robin Hankin [EMAIL PROTECTED]
On 7/12/2005 9:29 AM, Robin Hankin wrote:
Hi
I find that one often needs to keep reals real and complexes complex.
Try this:
cuberooti -
function (x)
{
if (is.complex(x)) {
return(sqrt(x + (0+0i)))
}
sign(x)* abs(x)^(1/3)
}
best wishes
[see that (0+0i)
hi all
i simply wanted to work with real numbers and thought that (-8)^(1/3) should
work.
sorry for not making the question clearer.
/
allan
Quoting Duncan Murdoch [EMAIL PROTECTED]:
On 7/12/2005 9:29 AM, Robin Hankin wrote:
Hi
I find that one often needs to keep reals real and
On 12 Jul 2005, at 14:51, Duncan Murdoch wrote:
On 7/12/2005 9:29 AM, Robin Hankin wrote:
[bogus function snipped]
I don't understand this.
1. I don't think you meant to use sqrt() there, did you??
2. What effect does the 0+0i have? x has already been determined
to be
complex.
On 7/12/2005 9:53 AM, [EMAIL PROTECTED] wrote:
hi all
i simply wanted to work with real numbers and thought that (-8)^(1/3) should
work.
It might work in an ideal world, but not in the R floating point world.
There's no way to express (1/3) exactly. Since (-8)^(1/3 + epsilon)
In general, x^y is evaluated as exp(y*log(x)). In your case, x is
negative, so log(x) is NaN. Note also that 1/3 is not represented
exactly in your computer anyway, so you would not get an exact cube root
this way; e.g.:
R format((1234567891112^3)^(1/3),digits=16)
[1] 1234567891112.001
(Probably
Hi R users,
Maybe the question is too simple.
In a IF ... ELSE ... statement if(cond) cons.expr else alt.expr, IF and
ELSE should be at the same line?
For example,
if (x1==12)
{
y1 - 5
}else
{
y1 - 3
}
is right, while
if (x1==12)
{
y1 - 5
}
else # Error: syntax error
{
y1 - 3
}
is wrong?
Hi,
I got 1000 NxN matrices grouped in one array. I want one matrix in which p_ij
is the average of all the 1000 matrices in the array. Here's what I'm trying to
do:
# P is the NxNx1000 array
for(i in 1:N)
for(j in 1:N)
for(k in 1: 1000)
mymat[ i, j ] - mean( P [i , j , k ] )
Otherwise, I
Try:
mymat - rowMeans(P, dims=2)
Andy
From: Dimitri Joe
Hi,
I got 1000 NxN matrices grouped in one array. I want one
matrix in which p_ij is the average of all the 1000 matrices
in the array. Here's what I'm trying to do:
# P is the NxNx1000 array
for(i in 1:N)
for(j in 1:N)
I have a formula from which I want to deduce the name of the response
variable. One way of doing so is as follows:
my.form - as.formula(y ~ x + z)
all.vars(my.form)[1]
[1] y
Is there a better way and/or preferrred method of determining y from
my.form than this one? In messing around with
This is a theorem for maximum likelihood tests. See:
Theorem 12.2 (presented without proof), page 391, in John E. Freund's
Mathematical Statistics with Applications, Seventh Edition, by Irwin
Miller and Marylees Miller. Upper Saddle River, N.J.: Pearson Prentice
Hall, 2004.
Theorem 12.2:
On Tue, 2005-07-12 at 16:22 +0200, ecoinfo wrote:
Hi R users,
Maybe the question is too simple.
In a IF ... ELSE ... statement if(cond) cons.expr else alt.expr, IF and
ELSE should be at the same line?
For example,
if (x1==12)
{
y1 - 5
}else
{
y1 - 3
}
is right, while
if
Hi list,
I'm looking for a function or a combination of functions to do panel plotting
of mixed graph types with the same x axis.
I would like to construct a panel with 3 stacked windows with on top a
histogram, below that 2 cdf plots. They all have the same x axis value but
different y axis
Marc,
I see. Thanks.
Xiaohua
On 7/12/05, Marc Schwartz (via MN) [EMAIL PROTECTED] wrote:
On Tue, 2005-07-12 at 16:22 +0200, ecoinfo wrote:
Hi R users,
Maybe the question is too simple.
In a IF ... ELSE ... statement if(cond) cons.expr else alt.expr, IF
and
ELSE should be at the
David Kane wrote:
I have a formula from which I want to deduce the name of the response
variable. One way of doing so is as follows:
my.form - as.formula(y ~ x + z)
all.vars(my.form)[1]
[1] y
Is there a better way and/or preferrred method of determining y from
my.form than this
Hi,
I am trying to implement the Adaboost.M1. algorithm as described in
The Elements of Statistical Learning p.301
I don't use Dtettling 's library boost because :
- I don't understande the difference beetween Logitboost and L2boost
- I 'd like to use larger trees than stumps.
By using
is it what you want?
dat-data.frame(x=rnorm(10),y=rnorm(10),z=rnorm(10))
my.form - as.formula(y ~ x + z)
my.form
y ~ x + z
m-model.frame(my.form,data=dat)
model.extract(m,response)
1 2 3 4 5 6 7
-0.3434826 1.0145622 -0.4749584
Hello,
I want to fit a tree parameter distribution to given data. I tried it with
sample data using the fitdistr function.
Here my workflow that didn't had any result:
I started with the generalized gamma distr, which is:
r*dgamma(x^r,shape,rate)
The R-function is:
ggamma = function
Hi Karen
I am not sure if I understand correctly your question. If no,
please ignore this answer.
Do you want a new factor group which contains the same
information like species, just with other names, e.g
1,2,... or A,B,... ?
If yes you can do it like this
## Your data.frame (without ht
Hi,
I am trying to implement the Adaboost.M1. algorithm as described in
The Elements of Statistical Learning p.301
I don't use Dtettling 's library boost because :
- I don't understande the difference beetween Logitboost and L2boost
- I 'd like to use larger trees than stumps.
It also
Carsten == Carsten Steinhoff [EMAIL PROTECTED]
on Tue, 12 Jul 2005 17:49:34 +0200 writes:
Carsten Hello,
Carsten I want to fit a tree parameter distribution to
Carsten given data. I tried it with sample data using the
Carsten fitdistr function.
Carsten Here my
Hi list,
For my research, I started using R as a statistics engine, driven by
perl scripts that manage my data and computational jobs. I recently
upgraded my cluster to Suse 9.2, and obviously wanted to upgrade R to
it's latest version, 2.1.1. Installation of R-2.1.1 went flawlessly
thanks to
On 7/12/05, Luc Vereecken [EMAIL PROTECTED] wrote:
Hi list,
For my research, I started using R as a statistics engine, driven by
perl scripts that manage my data and computational jobs. I recently
upgraded my cluster to Suse 9.2, and obviously wanted to upgrade R to
it's latest version,
XSolutions Corp (www.xlsolutions-corp.com) is proud to announce
our Advanced R/Splus programming course taught by R Development
Core Team Guru!
www.xlsolutions-corp.com/Radv.htm
*Atlanta July 28th-29th, 2005
Ask for group discount and reserve your seat Now (payment due
Dear list,
I will like to learn how to read a lower triangular matrix in R. The
input file *.txt have the following format:
A B C D E
A 0
B 10
C 250
D 3680
E 479 100
How this can be done?
Thanks in advance for your help
I have the following data:
gene_name microarray expression
A 2323
B 1983
.
I have about 10,000 observations.
I would like to know if there is a way to
I know R has a steep learning curve, but from where I stand the slope
looks like a sheer cliff. I'm pawing through the available docs and
have come across examples which come close to what I want but are
proving difficult for me to modify for my use.
Calculating simple group means is fairly
I'm updating the loess routines to allow for, among other things,
arbitrary local polynomial degree and number of predictors. For now,
I've given the updated package its own namespace. The trouble is,
panel.loess still calls the original code in package:stats instead of
the new loess package,
First call par(mfrow = c(2,2)) to get four plots in one panel
then plot your top figures using
plot(x, xaxt=n)#this will generate the plot withoput displaying the x axis
and then plot your bottom figures keeping the axis (without using xaxt=n)
You may want to fix the x axis limits to match
MASS (MODERN APPLIED STATISTICS WITH S) by Venables and Ripley.
-- Bert Gunter
Genentech Non-Clinical Statistics
South San Francisco, CA
The business of the statistician is to catalyze the scientific learning
process. - George E. P. Box
-Original Message-
From: [EMAIL
Take a look at ?aggregate ?ave and ?tapply
Cheers
Francisco
From: [EMAIL PROTECTED]
To: R-Help r-help@stat.math.ethz.ch
Subject: [R] Calculation of group summaries
Date: Tue, 12 Jul 2005 10:51:03 -0700
I know R has a steep learning curve, but from where I stand the slope
looks like a sheer
Benjamin Tyner wrote:
I'm updating the loess routines to allow for, among other things,
arbitrary local polynomial degree and number of predictors. For now,
I've given the updated package its own namespace. The trouble is,
panel.loess still calls the original code in package:stats instead
A quick workaround, kudos to Deepayan Sarkar, is to use grid:: for both
grid.lines AND gpar in panel.loess:
grid::grid.lines(x = smooth$x, y = smooth$y, default.units =
native,
gp = grid::gpar(col = col.line, lty = lty, lwd = lwd))
Then write a new panel function as you
On Tue, 2005-07-12 at 14:37 -0300, Rogério Rosa da Silva wrote:
Dear list,
I will like to learn how to read a lower triangular matrix in R. The
input file *.txt have the following format:
A B C D E
A 0
B 10
C 250
D 3680
E 47
mark salsburg [EMAIL PROTECTED] wrote:
[...]
Mainly one that shows all the points, but when a point is clicked on
it shows what gene_name it is
I think ?identify will help you.
--
Sebastian P. Luque
__
R-help@stat.math.ethz.ch mailing list
Hi,
The following may sound stupid so please forgive my stupidness. I have a
question which I don't know how to name it so I have to start from the
beginning. In an attempt to gain better understand how a photochemical
air qaulity model works, I plotted hourly ozone concentration contour
from
See http://biostat.mc.vanderbilt.edu/twiki/bin/view/Main/SasByMeansExample
for one example.
Frank
[EMAIL PROTECTED] wrote:
I know R has a steep learning curve, but from where I stand the slope
looks like a sheer cliff. I'm pawing through the available docs and
have come across examples
I had a similar problem when trying to modify a function from lattice.
I second the opinion that it
would be nice if lattice exported more things. My soulution was to
give my function the same
environment as the one I had copied, i.e. try:
environment(panel.mybwplot) -
This has been answered at the following URL
http://tolstoy.newcastle.edu.au/~rking/R/help/04/11/6695.html
On Tue, 2005-07-12 at 14:37 -0300, Rogério Rosa da Silva wrote:
Dear list,
I will like to learn how to read a lower triangular matrix in R. The
input file *.txt have the following
Dear Adaikalavan and Marc,
Thanks for advices. The answer in R-help archives is exactly what I'm
needing.
Best regards,
Rogério
Adaikalavan Ramasamy wrote:
This has been answered at the following URL
http://tolstoy.newcastle.edu.au/~rking/R/help/04/11/6695.html
On Tue, 2005-07-12 at 14:37
Good day:
I am trying to use
readcsvIts(nwr_data_qc.txt,informat=its.format(%Y%m%d%h%M
%Y),header=TRUE,sep=,skip=0,row.names=NULL,as.is=TRUE,dec=.)
to read in a file (nwr_data_qc.txt) that looks like this:
Time Y M D H MinCO2
2000.18790 2000. 3. 9. 18. 30.373.60
You would probably do better to read in your data and reformat the
dates.
What date format is 2000.18790? The Its package uses POSIXct dates and
readcsvIts expects the formats to be %Y-%m-%d unless you've changed
the default format.
-Original Message-
From: [EMAIL PROTECTED]
On 7/12/05, Ritter, Christian C GSMCIL-GSTMS/2
[EMAIL PROTECTED] wrote:
R-2.1.1 on windows XP
I just noticed something unpleasant when using bwplot (from lattice).
In order to satisfy a wish from a client, I needed to produce sets of
boxplots conditioned by another factor. My client
On 7/12/05, [EMAIL PROTECTED] [EMAIL PROTECTED] wrote:
Hi list,
I'm looking for a function or a combination of functions to do panel plotting
of mixed graph types with the same x axis.
I would like to construct a panel with 3 stacked windows with on top a
histogram, below that 2 cdf
Hi user R,
I am try to calculate the spectrum function in two time series. But when plot a
single serie, the labels in axes x is in the range 0.1 to 0.6 (frequency), but
when calculate de spectrum with ts.union function, the labels x is in the range
1 to 6. I not understand why change the
On 7/12/05, Sheri Conner Gausepohl [EMAIL PROTECTED] wrote:
Good day:
I am trying to use
readcsvIts(nwr_data_qc.txt,informat=its.format(%Y%m%d%h%M
%Y),header=TRUE,sep=,skip=0,row.names=NULL,as.is=TRUE,dec=.)
to read in a file (nwr_data_qc.txt) that looks like this:
Time Y
On 7/12/05, Gabor Grothendieck [EMAIL PROTECTED] wrote:
On 7/12/05, Sheri Conner Gausepohl [EMAIL PROTECTED] wrote:
Good day:
I am trying to use
readcsvIts(nwr_data_qc.txt,informat=its.format(%Y%m%d%h%M
%Y),header=TRUE,sep=,skip=0,row.names=NULL,as.is=TRUE,dec=.)
to read in a file
Hi all,
I have jut installed the foreign package (v 0.8-8) on my OS X
machine, and have a bit of a problem writing out a data frame in SPSS
format. Specifically, the code file (the .sps format file) seems to
write 3 unprintable hex values instead of double quotes. For example, in
the
Simon,
An hour's worth of contour plots (12) should be able to be saved in far
less than 1 Gb, depending on the detail desired.
Here is one way to do your task. There are other, and probably better, ones.
At each time step (5 mins), you could have the Fortran program write a
data file and then
Michael Prager wrote:
I don't know if R can easily read Fortran unformatted (sometimes called
binary) data, or how.
Yes, it's fairly easy. You would use a file() connection, and
readBin(). The only tricky part might be figuring out exactly what is
the R equivalent to what your fortran is
Dear all,
My machine is SUN Java Workstation 2100 with 2 AMD Opteron CPUs and 16GB RAM.
R is compiled as 64bit by using SUN compilers.
I trying to fit quantile smoothing on my data and I got an message as below.
fit1-rqss(z1~qss(cbind(x,y),lambda=la1),tau=t1)
Error in
Dear all,
I really appreciate your help. I think I have a little advancement. ^_^
Now I use the package.skeleton() function to create a template. I type:
f - function(x,y) x+y
g - function(x,y) x-y
d - data.frame(a=1, b=2)
e - rnorm(1000)
hhc.exe is the Microsoft help compiler. You have to download it and put
it somewhere in your path.
On 7/12/05, Ivy_Li [EMAIL PROTECTED] wrote:
Dear all,
I really appreciate your help. I think I have a little advancement. ^_^
Now I use the package.skeleton() function to create a template. I
Hello,
I have a data set with 15 variables, and use pairs to plot the
scatterplot of this data set. Then I want to plot some circles on the
small pictures with high correlation(e.g. 0.9).
First, I use cor to obtain the corresponding correlation matrix (x)
for this scatterplot.
Second, use
Hello,
How to use the function plot to produce graphs as Matlab?
example in Matlab:
a = [1,2,5,3,6,8,1,7];
b = [1,7,2,9,2,3,4,5];
plot(a,'b')
hold
plot(b,'r')
How to make the same in R-package ?
I am trying something thus:
a - c(1,2,5,3,6,8,1,7)
c(1,7,2,9,2,3,4,5) - b
a;b
1 - 100 of 105 matches
Mail list logo
