On Tue, 2008-07-01 at 10:54 +1000, Jason Lee wrote:
Hi,
Currently I have a dataset of 2400*408. And I would like to apply PCR method
to study the any correlation between the tests.
My current data is in data.frame and I have formed horizontal(1-407) to be
the exact data, and (408) to be my
Hi;
It seems to me that has probably been asked in the past. But I cannot find
the track.
I usually need to extract elements from a list and contruct vector from
them; e.g., to create a table. Perhaps there is a way to directly extract
them without looping?
Simple example:
S.lst
$sublist.1
[EMAIL PROTECTED] wrote:
Hi;
It seems to me that has probably been asked in the past. But I cannot find
the track.
I usually need to extract elements from a list and contruct vector from
them; e.g., to create a table. Perhaps there is a way to directly extract
them without looping?
Simple
Perfect!!
Thanks a lot!
Javier
---
[EMAIL PROTECTED] wrote:
Hi;
It seems to me that has probably been asked in the past. But I cannot
find
the track.
I usually need to extract elements from a list and contruct vector from
them; e.g., to create a table. Perhaps there is a way to
Dear all,
I'm working with satellite images in R and plotting them via the code
below.
I was wondering whether coordinates (spatial[x], spatial[y]) are
used as centre coordinates of the pixels in the GRID? In this script;
spatial[x] spatial[y] are the centre coordinates of the satellite
image
Hi Jan,
In response to your ps. I think using rgdal is the best way to go. If
your data is for example in a GeoTIFF file, reading it is as simple as:
GRID = readGDAL(your_file.tif)
Saves you some hassel and a number of lines of code.
cheers and hth,
Paul
[EMAIL PROTECTED] wrote:
Dear all,
...and in addition:
The r-sig-geo mailing list is a better forum for your kind of questions,
you are much more likely to get good answers.
cheers,
Paul
[EMAIL PROTECTED] wrote:
Dear all,
I'm working with satellite images in R and plotting them via the code
below.
I was wondering whether
Gavin Simpson [EMAIL PROTECTED] writes:
You can do this another way though, that I feel is more natural. So lets
assume that your data frame contains columns that are named, and that
one of these is the response variable, the remaining columns are the
predictors. Further assume that this
I've been trying to install R into a user's home directory for them by
compiling from source code, on a machine for which neither of us has
administrative access. I've run configure using the --prefix option to specify
their home directory.
This seems to work OK up to a point, but when
Dear All,
Kindly tell me the steps to perform analysis of covariance (ANCOVA). Can I find
anything in the help(). My prime objective is to analyze the genotypic and
phenotypic correlation coefficients.
Thanks,
Partha
Partha Protim Banerjee, Ph. D.
Scientist - Corn Breeding
Hytech Seed
Hello everyone,
I need reshape an array. For example, if we have next array:
a - c(1,2,3,4,5,6,7,8,9,10,11,12)
dim(a) - c(2,2,3)
a
, , 1
[,1] [,2]
[1,]13
[2,]24
, , 2
[,1] [,2]
[1,]57
[2,]68
, , 3
[,1] [,2]
[1,]9 11
[2,] 10 12
I
Does this do what you want:
a - 1:12
dim(a) - c(2,2,3)
a
, , 1
[,1] [,2]
[1,]13
[2,]24
, , 2
[,1] [,2]
[1,]57
[2,]68
, , 3
[,1] [,2]
[1,]9 11
[2,] 10 12
dim(a) - c(4,3)
(b - t(a))
[,1] [,2] [,3] [,4]
[1,]1234
On 7/1/2008 4:57 AM, Partha wrote:
Dear All,
Kindly tell me the steps to perform analysis of covariance (ANCOVA). Can I find
anything in the help(). My prime objective is to analyze the genotypic and
phenotypic correlation coefficients.
Thanks,
Partha
RSiteSearch({analysis of covariance},
Hi R people
I am using a function to create a pdf device, then send a lot of plots
to it in a loop then a last lattice xyplot (itself within a function)
outside the loop and finally call dev.off() to write to the file.
This works well apart from the fact that the last plot does not get
Hello
Regressions with time series model is something more complicate than
usual, I recommend you to read more about it in any time series manual.
The biggest problem comes from the so called potential spurious
regression, that is your regression can lead to errnoneous conclusions
(if you
[EMAIL PROTECTED]Hi, i'm tryng to build a function that take
some input from user and if the user doesn't provide that inputs the
function should set a deafult value. I have taken as example a function that
i found in the package dlm but with the same code i receive an error (the
component is
On Mon, 2008-06-30 at 04:43 -0700, mysimbaa wrote:
Dear R users,
Is it possible to add comments in a plot window?
I have 3 plots - this plots have to be commented on same window
Is possible to do the following :
http://www.nabble.com/file/p18193822/plot%2526comments.JPG
plot%26comments.JPG
Kathi [EMAIL PROTECTED] napsal dne 01.07.2008 13:39:18:
Thanks, Petr.
After fiddling around some more with my code I found out that this seems
to
have something to do
with the number of records the table I'm reading (here: column B112).
Somehow,
changing the
number of records for
Try this. mar is used to widen the right margin and
xpd allows display outside of the plot. See ?par and
?legend
op - par(mar = c(5, 4, 4, 10) + 0.1, xpd = TRUE)
plot(1:10)
legend(topright, inset = c(-.3, 0), abc\ndef, box.lty = 0)
par(op)
On Mon, Jun 30, 2008 at 7:43 AM, mysimbaa [EMAIL
Dear list,
I have some values like
Time2 4 8 24 48 72
UTR 82543 169105 207615 96633 31988 7005
UTRs82687 172934 205541 101842 31898 6950
of a twice repeated meassurement.
I know that the underlying function
Now that we have case cohort model , we have 1000 people and 50 cases
Let the first 10 cases occur at the same time
second 10
third 10
fourth 10
fifth10
How easy is it to randomly sample 50
use the function resid instead of lm.object$residuals to extract the
residuals:
resid(lm(a~b, na.action=na.exclude))
1 2 3 4
5 6
-2.533445e-17 4.222409e-17 -8.444818e-18 -8.444818e-18
NANA
lm(a~b,
Dear experts,
For the makeGenotype function I need a list as in the example. However,
since my list needs to be 184 long there must be an easy way to make it.
list(1:2,3:4,5:6,7:8)
[[1]]
[1] 1 2
[[2]]
[1] 3 4
[[3]]
[1] 5 6
[[4]]
[1] 7 8
I have tried
lis-1:184
dim(lis)=c(92,2,1)
Try to set your own starting values. Use the values that you would expect for
S0, mu and sigma.
HTH,
Thierry
ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek / Research Institute for Nature and
Forest
HiI need help about anderson-darlin test for a weibull distribution in R.Thanks.
René
_
[[elided Hotmail spam]]
e ready.
[[alternative HTML version deleted]]
__
I think there are many simple solutions, here is one:
lapply(1:92, function(x) c(2*x-1, 2*x))
Gabor
On Tue, Jul 01, 2008 at 02:46:07PM +0200, Boks, M.P.M. wrote:
Dear experts,
For the makeGenotype function I need a list as in the example. However,
since my list needs to be 184 long there
Dear UseRs,
I would like to know the way to find classes of each column of
data.frame().
Thank you in advance.
=
Dong-hyun Oh
Center of Excellence for Science and Innovation Studies
Royal Institute or Technology, Sweden
e-mail:
Use lapply on a costum function.
lapply(1:92, function(i){
(i - 1) * 2 + 1:2
})
HTH,
Thierry
ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek / Research Institute for Nature
and Forest
Cel biometrie,
A data frame is a special list:
d - data.frame( A=numeric(), B=logical(), C=character() )
lapply(d, class)
$A
[1] numeric
$B
[1] logical
$C
[1] factor
Gabor
On Tue, Jul 01, 2008 at 03:50:18PM +0200, Dong-hyun Oh wrote:
Dear UseRs,
I would like to know the way to find classes of each
another way:
split(1:184, rep(1:92, each=2))
On Tue, Jul 1, 2008 at 3:46 PM, Boks, M.P.M. [EMAIL PROTECTED]
wrote:
Dear experts,
For the makeGenotype function I need a list as in the example. However,
since my list needs to be 184 long there must be an easy way to make it.
... or try this:
n = 184
split(1:n, rep(1:(n/2), each=2))
vQ
ONKELINX, Thierry wrote:
Use lapply on a costum function.
lapply(1:92, function(i){
(i - 1) * 2 + 1:2
})
HTH,
Thierry
ir. Thierry
sapply(your_data, class)
On Tue, Jul 1, 2008 at 10:50 AM, Dong-hyun Oh [EMAIL PROTECTED] wrote:
Dear UseRs,
I would like to know the way to find classes of each column of
data.frame().
Thank you in advance.
=
Dong-hyun Oh
Try this also:
unname(split(1:184, rep(1:(184/2), each = 2)))
On Tue, Jul 1, 2008 at 9:46 AM, Boks, M.P.M. [EMAIL PROTECTED]
wrote:
Dear experts,
For the makeGenotype function I need a list as in the example. However,
since my list needs to be 184 long there must be an easy way to make it.
Many thanks.
On Jul 1, 2008, at 3:57 PM, Romain Francois wrote:
sapply( iris, class )
Dong-hyun Oh wrote:
Dear UseRs,
I would like to know the way to find classes of each column of
data.frame().
Thank you in advance.
=
many thanks.
On Jul 1, 2008, at 4:00 PM, Henrique Dallazuanna wrote:
sapply(your_data, class)
On Tue, Jul 1, 2008 at 10:50 AM, Dong-hyun Oh [EMAIL PROTECTED]
wrote:
Dear UseRs,
I would like to know the way to find classes of each column of
data.frame().
Thank you in advance.
Nina,
In which way have you obtained your CSV file? Did you export data
from an Excel spreadsheet?
Regards,
Paulo Barata
---
Paulo Barata
Fundacao Oswaldo Cruz
Rua Leopoldo Bulhoes 1480 - 8A
21041-210 Rio de Janeiro - RJ
Dong-hyun Oh wrote:
Dear UseRs,
I would like to know the way to find classes of each column of
data.frame().
sapply(your-data-frame, class)
vQ
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the
Hi,
Currently I have a data set of 2 classes and I am interedted to use emerging
patterns algorithm. I wondered if there is any packages or recommended code
available for this?
Please advise.Thanks.
[[alternative HTML version deleted]]
__
Simply add a definition for this variable to the source for your package,
e.g.;
.Last.make.date - NULL
And the warning should go away.
-g
On 6/30/08 11:44PM , Rolf Turner [EMAIL PROTECTED] wrote:
I have written a function make.fun(), which I keep in my personal
``miscellaneous''
So, I used df - data.frame( x=I(coef(cancerv1(,2:407))),
y=cancerv1[,408])before feeding to PCR.
However, I get the below error.
Error in coef(cancerv1(, 2:407)) : could not find function cancerv1.
It's because you have the syntax error 'cancerv1(...)', which looks to
R, and me, a function
I solved the problem with
plot(0:1,0:1, type = n, axes=FALSE,xlab=,ylab=)
text(...)
Thanks,
Adel
mysimbaa wrote:
Dear R users,
Is it possible to add comments in a plot window?
I have 3 plots - this plots have to be commented on same window
Is possible to do the following :
Hi, i'm tryng to build a function that take some input from user and if the
user doesn't provide that inputs the function should set a deafult value. I
have taken as example a function that i found in the package dlm but with
the same code i receive an error (the component is missing...).
I'm trying to do realize the following:
I have 4 condtions.
If all conditions are satisfied I will paste(PASS)
If any of these is not satisfied I will paste(FAIL). But I have to paste
the corresponding failure.
ifelse is a good solution but for a 2 conditions. Maybe switch or something
like
Hello everyone,
I need reshape an array. For example, if we have next array:
a - c(1,2,3,4,5,6,7,8,9,10,11,12)
dim(a) - c(2,2,3)
a
, , 1
[,1] [,2]
[1,]13
[2,]24
, , 2
[,1] [,2]
[1,]57
[2,]68
, , 3
[,1] [,2]
[1,]9 11
[2,] 10 12
I
Hi. With the previous version of R, it was possible to execute the function
'select.list()' in BATCH mode. In this way, I could write my R code in a .R
file and execute that with a double click on a .bat file, that contain the
instruction tu run my R code in batch mode (R CMD BATCH myRcode.R).
Hi all,
I've tried to plot a vector which has two peaks in the density.
This link shows the figure.
http://docs.google.com/View?docid=dcvdrfrh_1dk9r2rc7
The red line is normal curve and green line is gamma curve.
Notice that red line can correctly fit the histogram that has two peaks
(i.e. red
on 07/01/2008 04:58 AM Francisco Javier Santos Alamillos wrote:
Hello everyone,
I need reshape an array. For example, if we have next array:
a - c(1,2,3,4,5,6,7,8,9,10,11,12)
dim(a) - c(2,2,3)
a
, , 1
[,1] [,2]
[1,]13
[2,]24
, , 2
[,1] [,2]
[1,]57
[2,]
Hello,
I'm trying to produce graphs automatically from data stored in database.
Before saving the graphs, I would like to maximize the size of the graphs.
The best would be to directly open maximized windows with x11() but up to
now I failed doing it.
I tried different widths and heighs but I
This is actually more like a Statistics problem:
I have a dataset with two dummy variables controlling three levels. The
problem is, one level does not have many observations compared with other
two levels (a couple of data points compared with 1000+ points on other
levels). When I run the
Ok, the mistake was in the pca(x)-princomp(SampleD[i,j]), should've used
pca(x)-princomp(SampleD) instead.
Now, is there anyway to keep track of the matrix index, so in the end of all
PCAs, I can tell which score/loading belongs to which sample?
Thanks everyone!
On Mon, Jun 30, 2008 at 9:08 PM,
Thanks!
I tried already. Still the same error message...
wishes
Gunther
- Original Message -
From: ONKELINX, Thierry [EMAIL PROTECTED]
To: Gunther Höning [EMAIL PROTECTED]; R-help@r-project.org
Sent: Tuesday, July 01, 2008 3:08 PM
Subject: RE: [R] Regression and fitting
Try to set
on 07/01/2008 07:40 AM mysimbaa wrote:
I'm trying to do realize the following:
I have 4 condtions.
If all conditions are satisfied I will paste(PASS)
If any of these is not satisfied I will paste(FAIL). But I have to paste
the corresponding failure.
ifelse is a good solution but for a 2
Hi All,
I'm working with R and want to ignore the warning messages given, is there a
way to stop R from giving out warning messages any more?
an example:
tt = test
as.numeric(tt)
would give me the following message:
[1] NA
Warning message:
NAs introduced by coercion
I decide to ignore the
Hi,
you can use the jpeg(), pdf(), bmp() functions. In this function you can
determinate the size of the graph, background color, quality of the image.
Leandro Marino
-Mensagem original-
De: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
nome de Ptit_Bleu
Enviada em: terca-feira, 1 de
Dear All,
I have found in the poly help this sentence:
The orthogonal polynomial is summarized by the coefficients, which can be
used to evaluate it via the three-term recursion given in Kennedy Gentle
(1980, pp. 3434), and used in the predict part of the code.
My question: which type of
Dear Hua,
Try
options(warn=-1)
tt = test
as.numeric(tt)
[1] NA
See also ?options for more info.
HTH,
Jorge
On Tue, Jul 1, 2008 at 12:11 PM, Hua Li [EMAIL PROTECTED] wrote:
Hi All,
I'm working with R and want to ignore the warning messages given, is there
a way to stop R from giving
Hi,
I'm trying to make a boxplot with the data at the end of the message, and when I
try to execute the command
boxplot(Diatoms) (or for any other field instead of Diatoms)
I get the following error message:
Error in oldClass(stats) - cl : adding class factor to an invalid object
Any advice
You should consult An Introduction to R, section 11, Writing your own
functions, along with any beginner books on R.
http://www.r-project.org/doc/bib/R-books.html
test - function(x = 2) {
x^2
}
test() ## will give 4
test(3) ## will give 9
Marco Salvi wrote:
Hi, i'm tryng to build a
On Wed, 2008-07-02 at 00:58 +1000, Jason Lee wrote:
Hi,
Thanks for the reply.
Basically I dont have any label for my data except column 1 which is labeled
with Sample1, Sample2...etc...(vertical).
Well that is going to cause you problems, not the one you report below,
but it will bite you
options(warn = -1)
or
suppressWarnings(as.numeric(test))
On Tue, Jul 1, 2008 at 1:11 PM, Hua Li [EMAIL PROTECTED] wrote:
Hi All,
I'm working with R and want to ignore the warning messages given, is there
a way to stop R from giving out warning messages any more?
an example:
tt = test
Does anybody have any suggestions on how I might simulate from fitted GAM
model? I am using the gam function in the mgcv package to fit a variable
coefficient model like the following from the examples. I would like simulate
based on the fitted model like the simulate function in the stats
The advice is to store numeric data as numeric, and not factors. Try
str(yourdata) and you'll probably see that Diatoms is stored as a
'factor'. Are those commas used as decimal points? read.table can deal
with that if so. If not, you do not need them, and should convert those
fields to
HI Miriam,
the problem is that you have the Date and some other factors in your
database.
try to use the
lapply(your_base,class)
and see what appears. If you have any kind of factor or date you cannot use
boxplot.
If you verify that only V1, V2 and V3 are numeric you can do the box plot
with
Hi,
I am extracting data from a table where the rows have different column lengths,
and empty columns have NA in them. Whenever I extract a row with some empty
columns, the resulting vector carries all the NAs. Is there a way to ignore the
empty columns?
Thanks,
-Nina
It depends what you mean by 'ignore'. Some functions have an na.rm
argument which throws out NAs before computing the statistic.
if the vector 'x' has NAs, then
x - x[!is.na(x)]
may be what you're looking for. This removes NAs from x and reassigns
the value to x.
[EMAIL PROTECTED] wrote:
Can someone please enlighten me as to why the following happens?
-2.7^8.6
[1] -5125.407
p- -2.7
q- 8.6
p^q
[1] NaN
R seems perfectly able to calculate -2.7^8.6, but fails when the exact same
values are assigned to variables and then the computation is repeated.
Thanks in advance for any
On Tue, Jul 01, 2008 at 11:15:58AM -0700, [EMAIL PROTECTED] wrote:
Can someone please enlighten me as to why the following happens?
-2.7^8.6
[1] -5125.407
p- -2.7
q- 8.6
p^q
[1] NaN
R seems perfectly able to calculate -2.7^8.6, but fails when the exact same
values are assigned to
Dear Nina,
If you have a matrix X like this
# Data set
set.seed(123)
X=matrix(rnorm(10*5),ncol=5)
X[1,1]-NA
X[2,1]-NA
X[1,5]-NA
X[5,2]-NA
X
and you'd like to remove the NA values for a particular row (for example row
1), you can try something like:
X[1,!is.na(X[1,])]
Now, if you have a
Hi,
I am interested in using a bayesian network as a predictor (machine
learning); however, I can't get any of the implementations (deal, nblearn)
to learn predict stuff.
Shouldn't there also be probabilites for each node after the learning phase,
how can I access these?
Cheers,
Stephan
--
Thank you very much for your precise help.
I will try to implement this tomorrow.
It looks like the best solution which could be done.
Greetings,
Adel Tekari
Marc Schwartz wrote:
on 07/01/2008 07:40 AM mysimbaa wrote:
I'm trying to do realize the following:
I have 4 condtions.
If all
on 07/01/2008 01:15 PM [EMAIL PROTECTED] wrote:
Can someone please enlighten me as to why the following happens?
-2.7^8.6
[1] -5125.407
p- -2.7 q- 8.6 p^q
[1] NaN
R seems perfectly able to calculate -2.7^8.6, but fails when
the exact same values are assigned to variables and then the
Nina,
read.csv() will default fill = TRUE,
or add to your read.table() argument list.
If that does not help, you will need to use col.names argument. see
?read.table
HTH,
Jim Porzak
Responsys, Inc.
San Francisco, CA
http://www.linkedin.com/in/jimporzak
On Mon, Jun 30, 2008 at 11:16 AM, [EMAIL
poolloopus at yahoo.com poolloopus at yahoo.com writes:
Can someone please enlighten me as to why the following happens?
-2.7^8.6
[1] -5125.407
p- -2.7
q- 8.6
p^q
[1] NaN
R seems perfectly able to calculate -2.7^8.6, but fails when the exact same
values are assigned to
variables
Hi,
the problem is what you want!
With the -2.7^8.6 you are doing -(2.7^8.6) it exists...
But, if you try (-2.7)^8.6 the R gives you NaN. When you define p=-2.7 and
q=8.6 and do p^q you are doing that (-2.7)^8.6.
If you write p=2.7 and q=8.6 and use the -p^q it will work!
Leandro Marino
Dear R People:
I'm having some trouble with mpiexec and Rmpi.
I would like to be able to pass in the number of children via the
mpiexec command (from the command line).
this is in SUSE10.1, with R-2.7.1
Here are my files:
cat eb.R
library(Rmpi)
mpi.remote.exec(paste(i
Hi,
I am trying to load the library(ks), but I am getting the following error:
Loading required package: KernSmooth
KernSmooth 2.22 installed
Copyright M. P. Wand 1997
Loading required package: mvtnorm
Loading required package: rgl
Loading required package: misc3d
Error in lazyLoadDBfetch(key,
Hi list,
I want to know how can i creat a plot window with this configuration:
___
| |
| PLOT 1|
| |
|-|
| | |
| |
Try:
nf - layout(matrix(c(1,2, 4, 6, 1, 3, 5, 6), nc = 2))
layout.show(nf)
On Tue, Jul 1, 2008 at 4:15 PM, Leandro Marino [EMAIL PROTECTED]
wrote:
Hi list,
I want to know how can i creat a plot window with this configuration:
___
| |
Hi,
Sorry for the simple question. Is there a way to change the name of only one
column of an existing data frame?
I know colnames allows you to set the name of all the columns, but only one
column in the middle of my data frame needs a new name.
Thanks,
-Nina
x - data.frame(a=1,b=2,c=3)
x
a b c
1 1 2 3
colnames(x)[2] - done
x
a done c
1 12 3
On Tue, Jul 1, 2008 at 3:17 PM, [EMAIL PROTECTED] wrote:
Hi,
Sorry for the simple question. Is there a way to change the name of only one
column of an existing data frame?
I know colnames
Try
require(gregmisc)
rename(data, from=old_name, to=new_name)
A Smile costs Nothing
But Rewards Everything
Happiness is not perfected until it is shared
-Jane Porter
--- On Tue, 7/1/08, [EMAIL PROTECTED] [EMAIL
I will see the code! It is similar that i want!
Thanks a lot!
-Mensagem original-
De: Gabor Grothendieck [mailto:[EMAIL PROTECTED]
Enviada em: terça-feira, 1 de julho de 2008 16:22
Para: Leandro Marino
Cc: [EMAIL PROTECTED] Org
Assunto: Re: [R] plot window
Look at the code for the
You can use names(database)[position]
where position is the number of the column that you want to rename!
Leandro Marino
-Mensagem original-
De: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
nome de [EMAIL PROTECTED]
Enviada em: terca-feira, 1 de julho de 2008 16:18
Para:
Hi
I have a set of data like this:
*Time of Day* *Pct of Daily Volume* 9:45 7.50% 10 6.25% 10:15 4.45%
10:30 4.80% 10:45 4.45% 11:00 4.20% 11:15 2.50% 11:30 2.30% 11:45 2.25%
12:00 2.45% 12:15 2.60% 12:30 2.00% 12:45 2.05% 13:00 2.40% 13:15 1.90%
13:30 3.10% 13:45 2.90% 14:00 2.80% 14:15
?lm
lm(x[,1]~x[,2])
On Tue, Jul 1, 2008 at 2:28 PM, Shirin Safa [EMAIL PROTECTED] wrote:
Hi
I have a set of data like this:
*Time of Day* *Pct of Daily Volume* 9:45 7.50% 10 6.25% 10:15 4.45%
10:30 4.80% 10:45 4.45% 11:00 4.20% 11:15 2.50% 11:30 2.30% 11:45 2.25%
12:00 2.45% 12:15
does loess return the values that it produces- take that and then integrate
under the curve.
On Tue, Jul 1, 2008 at 4:28 PM, Shirin Safa [EMAIL PROTECTED] wrote:
no this is not what I want.
I was using loess function or smooth.spline.
but For loess I don't know how would I be able to get the
I think the previous answer (to use lm() ) is not necessarily the best
option.
Since what you want is the definite integral (area under the curve), you
can just use one of the existing definite integration tools (sorry, I
don't recall the names because I don't use them).
If you want to get
This is FAQ 7.33
-thomas
On Tue, 1 Jul 2008, [EMAIL PROTECTED] wrote:
Can someone please enlighten me as to why the following happens?
-2.7^8.6
[1] -5125.407
p- -2.7
q- 8.6
p^q
[1] NaN
R seems perfectly able to calculate -2.7^8.6, but fails when the exact same
values are
On Tue, 2008-07-01 at 17:12 +, Mark Lyman wrote:
Does anybody have any suggestions on how I might simulate from fitted GAM
model? I am using the gam function in the mgcv package to fit a variable
coefficient model like the following from the examples. I would like simulate
based on the
Ow ow ow. I can't stand it any longer! Weird is one of those weird
exceptions in English to the i before e except after c rule and is spelled
W-E-I-R-D .
In case you're interested, the other exceptions are seize, inveigle ,
either, leisure, neither.
:-( -- Bert Gunter
-Original
I have constructed a Trellis style xyplot.
lengthf - factor(length)
xyplot(SLI$velocity ~ SLI$width | SLI$lengthf, layout = c(2,7), xlab =
Width (cm), ylab = Velocity (m/s^2), col = black)
This produces a lovely little plot. However, the grouping factor(lengthf)
isn't in the right order. My
On 2/07/2008, at 10:15 AM, Bert Gunter wrote:
Ow ow ow. I can't stand it any longer! Weird is one of those weird
exceptions in English to the i before e except after c rule and
is spelled
W-E-I-R-D .
In case you're interested, the other exceptions are seize, inveigle ,
either, leisure,
I would like to know the answer to this question now that I know what we are
getting at. integrate() looks like it is the right thing, but it has to use
a function- I would like to know how to just integrate the area under a
curve with just an input of x and y coordinates.
Stephen
On Tue, Jul
Here are some ways of rearranging panels:
library(lattice)
p - xyplot(Sepal.Length ~ Sepal.Width | Species, iris)
p
p[c(2, 1, 3)]
xyplot(Sepal.Length ~ Sepal.Width | Species, iris, as.table = TRUE)
On Tue, Jul 1, 2008 at 6:20 PM, Sam Albers [EMAIL PROTECTED] wrote:
I have constructed a
I would like to integrate the area under a curve without any smoothing or
the like- just on the raw numbers. I looked at integrate() but it requires
a function which I assume means something like x+x^2+x^3
is there a built in function in R for this?
#let's say
x - seq(1:50)
y - seq(1:50)
Hoisted by own petard! How appropriate.
1. I hereby invoke the codicil.
2. Check the dictionary. An accepted pronunciation of inveigle is
inveegle.
Does this allow me to wriggle through the escape clause?
-- Bert
-Original Message-
From: Rolf Turner [mailto:[EMAIL PROTECTED]
Sent:
On 7/1/08, Sam Albers [EMAIL PROTECTED] wrote:
I have constructed a Trellis style xyplot.
lengthf - factor(length)
xyplot(SLI$velocity ~ SLI$width | SLI$lengthf, layout = c(2,7), xlab =
Width (cm), ylab = Velocity (m/s^2), col = black)
As an aside, the recommended incantation is
There is an integrate.xy in sfsmic. Limitations discussed there.
On Tue, Jul 1, 2008 at 6:27 PM, stephen sefick [EMAIL PROTECTED] wrote:
I would like to know the answer to this question now that I know what we are
getting at. integrate() looks like it is the right thing, but it has to use
a
as.table =TRUE was exactly what I needed. Also thanks for the little aside.
Gabor Grothendieck wrote:
Here are some ways of rearranging panels:
library(lattice)
p - xyplot(Sepal.Length ~ Sepal.Width | Species, iris)
p
p[c(2, 1, 3)]
xyplot(Sepal.Length ~ Sepal.Width | Species, iris,
On 2/07/2008, at 10:38 AM, stephen sefick wrote:
I would like to integrate the area under a curve without any
smoothing or
the like- just on the raw numbers. I looked at integrate() but it
requires
a function which I assume means something like x+x^2+x^3
is there a built in function in R
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