Try this:
x - c(2,5,9,4,5,6,7,8)
u - replicate(20, sample(x, replace = TRUE))
t(u)
The first argument of replicate() is the number of times to iterate the
process.
I believe you'll find that it does indeed do random sampling; each row
represents
a nonparametric bootstrap sample of x. There are
Dear Felipe,
Assuming you are not interested in the exact formulae that are used to
calculate each of these, from top to bottom it is:
minimum, lower quartile, median, upper quartile, maximum
If there are dots, these usually indicate outliers.
Please read ?boxplot for more details on what R
http://www.netmba.com/statistics/plot/box/
...found via Google search.
HTH,
Dennis
On Tue, Sep 28, 2010 at 10:06 PM, Luis Felipe Parra
felipe.pa...@quantil.com.co wrote:
Hello, does somebody know in a boxplot, what does each element in the
boxplot represent?
1. lines at the extremes of
My honor.
A short question: if there is something in the device that is sensitive to
the overlapping of the text, then is it possible to add a warning massage
output when the length of the text is longer then the device dimensions?
With much respect,
Tal
Contact
Hello,
The predict.glm is your friend.
Type ?predict.glm to see the help page.
Michael
On 29 September 2010 12:45, xinxin xx xxgr...@hotmail.com wrote:
Hi everyone:
I am new to R and I have a really basic question.
I have already got a generalized linear model from some dataset, say
x - rep(letters[1:3], 2)
Are there any ways to transform assign the above as the one shown below
to an object? (in exact format; i.e length of 1 class of character),
i.e
x
('a', 'b', 'c', 'a', 'b', 'c')
Highly appreciate for any advice.
On Wed, Sep 29, 2010 at 3:33 PM,
IC == Ivan Calandra ivan.calan...@uni-hamburg.de
on Mon, 27 Sep 2010 16:59:31 +0200 writes:
IC Hi,
IC I'm not sure it's even possible (and if it is I don't know how, but I'm
IC no expert).
yes it is possible (in some way), see below.
IC But I think it doesn't make much
Dear List,
I have developed two models i want to use to predict a response, one with a
binary response and one with a ordinal response.
My original plan was to divide the data into test (300 entries) and training
(1000 entries) and check the power of the model by looking at the % correct
Dear R-help list
The functions drop.terms and [.terms ignores if the intercept
has been explicitly removed. Is that a deliberate feature?
For instance,
drop.terms(terms(~a+b-1),1)
terms(~a+b-1)[2]
R version 2.12.0 Under development (unstable) (2010-09-13 r52905)
Best, Niels
--
Niels Richard
On Sep 29, 2010, at 10:29 , Niels Richard Hansen wrote:
The functions drop.terms and [.terms ignores if the intercept
has been explicitly removed. Is that a deliberate feature?
Perhaps rather an unimplemented one. The root cause is that both functions use
reformulate() on the term.labels
Leonardo Monasterio leonardo.monasterio at gmail.com writes:
Dear R users,
In the function bellow I want to find the maximum value of v,
subject to the constrain that the sum of x is equal to 1.
I want to maximize:
v-t(x)%*%distance%*%x
Subject to:
sum(x)=1
I do not see why you
Hi Soumen,
it depends what you mean by embedded
what i have done once is to run a R script in batch mode from an access
application,
the R code import an access table, with the RODBC package, produce some
table and graph
which are compiled on the fly in html output, using HWRITER or R2HTML
On 29/09/10 11.13, peter dalgaard wrote:
On Sep 29, 2010, at 10:29 , Niels Richard Hansen wrote:
The functions drop.terms and [.terms ignores if the intercept has been
explicitly removed. Is that a deliberate feature?
Perhaps rather an unimplemented one. The root cause is that both
http://r.789695.n4.nabble.com/file/n2718591/QQ%E6%88%AA%E5%9B%BE%E6%9C%AA%E5%91%BD%E5%90%8D--1.png
I am a beginner for R,and i could not solve this problem.who can help me to
solve it ?
how to rotate the x axis lable to an interested angle as the picture show?
--
View this message in
Dear All,
I have made a scatter plot and placed a plane within it using scatterplot3d.
However, I have been asked for the data points to be a surface plot or have
the plane more closely resemble the data rather than show trends.
I have since tried to use the rgl package. Why doesn't this
Ok, I don't think I was specific enough.
The data originally came in this form
1 a 12
2 b 4
3 a 3
4 c 54
5 a 12
6 b 11
7 c 9
8 c 2
. . .
. . .
. . .
Where I sorted by the second column (NB the second column is the categories
and they have long names). I would then like
I am looking to do the same but am a bit confused
and apply the inverse link function for your model.
i understand up to this point and i understand what this means, however i am
unsure why it needs to be done and how you do it - i.e i use family=binomial
is this wrong if i use this method
Bill, Michael,
good to see I'm not the only one who sees potential for improvements
in the regexpr domain. Adding a subpattern argument is certainly a
step in the right direction and would make my life much easier.
However, in my application I need to know not only the position of one
group but
Dear users,
When I boxplot(), the lines of the whiskers are dashed. However, when I
save in an svg file, the dashed lines of the whiskers are not dashed
anymore.
How can I have the dashed lines in the svg file?
I don't have this problem with a ps file, but I cannot edit such file as
easily
Hi,
Something like
bb =
data.frame(label=c(a,b,a,b,c,a,b,c),val=c(4,2,1,6,4,3,2,
1))
l = split(bb,bb$label)
par(mfrow=c(2,2))
lapply(l,function(a) {boxplot(a$val)})
might be what you are looking for
Martyn
-Original Message-
From: r-help-boun...@r-project.org
Hi,
I did a aov and used summary to obtain the p-value. I tried many ways to
extract the p-value from
the summary result but failed. Among others I tried the following:
test.summary -
summary(aov(data[,1]~time.points+Error(subject/time.points)))
test.summary
Error: subject
Df
Hi,
Fairly new to R - have done basic plots but now faced with plotting a
matrix/table of results -I know what I want but cannot find out how to
do it.
Basically have individual questions ( x) to which an organization can
rate themselves 1-10 (y) what I want to show is a matrix/density type
Hi
I have two data frames that contains the same sort of data and I want do add
them together to get one big data frame, anyone know how to do that?
ex:
data.frame1
A B C
1 2 3
4 5 6
7 8 9
and
data.frame2
A B C
9 8 7
6 5 4
3 2 1
Would then become one big set:
data.frame3
A B C
1 2 3
4 5 6
7
Hi,
Take a look at rbind()
HTH,
Ivan
Le 9/29/2010 13:24, Joel a écrit :
Hi
I have two data frames that contains the same sort of data and I want do add
them together to get one big data frame, anyone know how to do that?
ex:
data.frame1
A B C
1 2 3
4 5 6
7 8 9
and
data.frame2
A
I'd definitely be a customer for it Titus. And it does seem like an
obvious hole in regex processing in R that cries out to be filled.
Um, ggregexpr isn't the sexiest of function names :) Perhaps we can
think of something a little easier ?
How is your C coding ? Bill ? Anyone else ? I could
On Wed, Sep 29, 2010 at 1:58 PM, Michael Bedward
michael.bedw...@gmail.com wrote:
How is your C coding ? Bill ? Anyone else ? I could have a got at
writing some prototype code to test in the next few days, though if
someone else with decent C skills is itching to do it please speak up.
We
On 29/09/2010 6:00 AM, JoH wrote:
Dear All,
I have made a scatter plot and placed a plane within it using scatterplot3d.
However, I have been asked for the data points to be a surface plot or have
the plane more closely resemble the data rather than show trends.
I have since tried to use the
Hi:
Try this:
test.summary[[1]][, 5]
It should return a vector of p-values, the last being NA. In your case,
since there is only one non-NA p-value, it is enough to do
test.summary[[1]][, 5][1]
You mean that wasn't obvious? :)
Explanation:
summary(aovobj) actually returns a list, but
On 09/29/2010 07:19 PM, komais wrote:
http://r.789695.n4.nabble.com/file/n2718591/QQ%E6%88%AA%E5%9B%BE%E6%9C%AA%E5%91%BD%E5%90%8D--1.png
I am a beginner for R,and i could not solve this problem.who can help me to
solve it ?
how to rotate the x axis lable to an interested angle as the picture
Thx for the help works very well for me.
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__
R-help@r-project.org mailing list
Hi:
Are you thinking of a levelplot/heatmap? Some places to look:
?levelplot in the lattice package
?geom_tile in the ggplot2 package
?heatmap2 in the gplots package
You can find many more heatmap functions in R; try
library(sos)# install first if you don't have it - *very* handy
Split sample validation is highly unstable with your sample size.
The rms package can help with bootstrapping or cross-validation, assuming
you have all modeling steps repreated for each resample.
Frank
-
Frank Harrell
Department of Biostatistics, Vanderbilt University
--
View this
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1
## from ?glm
counts - c(18,17,15,20,10,20,25,13,12)
outcome - gl(3,1,9)
treatment - gl(3,3)
d.AD - data.frame(treatment, outcome, counts)
glm.D93 - glm(counts ~ outcome + treatment, family=poisson,
data=d.AD)
## predict on 'link' or
On Wed, Sep 29, 2010 at 7:52 AM, Struve, Juliane
j.str...@imperial.ac.uk wrote:
I will post the example again to see if its readable now. My question is why
does read.zoo(file=filenames,) work and lapply(filenames, read.zoo,...)
does not ? Since I am reading the same file in both
HH-
I'm not familiar with the plots you mention, but the following is a quick
attempt to create the plot you describe.
data-data.frame(
org=1:10,
q1=sample(1:10,replace=T),
q2=sample(1:10,replace=T),
q3=sample(1:10,replace=T))
# This generates a random data set like the one you
On Sep 29, 2010, at 14:25 , Dennis Murphy wrote:
test.summary[[1]][, 5][1]
You mean that wasn't obvious? :)
Worse, it doesn't actually work...
test.summary - summary(npk.aovE)
test.summary[[1]][, 5]
Error in `[.default`(test.summary[[1]], , 5) :
incorrect number of dimensions
Thanks for this,
I had used
validate(model0, method=boot,B=200)
To get a index.corrected Brier score,
However i am also wanting to bootstrap the predicted probabilities output from
predict(model1, type = response) to get a idea of confidence, or am i best
just using se.fit = TRUE and
Hi,
For my dissertation, I've made copious use of xtable. I've just gotten
stumped however. I'm a fan of extended captions explaining the table, but
now I have to assemble a a list of tables and the captions are unwieldy. I
presume xtable calls LaTeX's \caption function. Is there a way to
Hi,
The form of the data is not terribly important neither is whether it
was sorted as boxplots are not order dependent are category a is a
sorted or not. See below for individual plots with your new data.
# read in data
dat - read.table(textConnection(
1 a 12
2 b 4
3 a 3
4 c 54
5 a
Duncan Murdoch murdoch.duncan at gmail.com writes:
On 29/09/2010 6:00 AM, JoH wrote:
Are there anyother ways in which I can create a surface plot?
persp() (in base R) is less flexible than rgl::persp3d,
doesn't do hidden line removal, etc., but does appear in the
regular R graphics
Dear mailing list,
I have following array:
X2 Y2
[1,] 422.7900 6.0
[2,] 469.8007 10.5
[3,] 483.9428 11.0
[4,] 532.4917 25.5
[5,] 596.1942 33.5
[6,] 630.8496 40.5
[7,] 733.2996 45.0
[8,] 946.4779 32.0
[9,] 996.8068 35.5
[10,]
komais liutao1986 at yahoo.com.cn writes:
how to rotate the x axis lable to an interested angle as the picture show?
http://cran.r-project.org/doc/FAQ/R-FAQ.html#How-can-I-create-rotated-axis-labels_003f
__
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Thank you very much.
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Hi:
On Wed, Sep 29, 2010 at 5:56 AM, peter dalgaard pda...@gmail.com wrote:
On Sep 29, 2010, at 14:25 , Dennis Murphy wrote:
test.summary[[1]][, 5][1]
You mean that wasn't obvious? :)
Worse, it doesn't actually work...
test.summary - summary(npk.aovE)
test.summary[[1]][, 5]
I will post the example again to see if its readable now. My question is why
does read.zoo(file=filenames,) work and lapply(filenames, read.zoo,...)
does not ? Since I am reading the same file in both statements I just do not
know how to interpret Error in strptime(x, format, tz = tz) :
Hi Ben and list,
Sorry to be a pain! I have followed your code, and modified it -
pp - predict(model1,se.fit=TRUE, type = response)
etaframe -
+ with(pp,cbind(fit,lower=fit-1.96*se.fit,upper=fit+1.96*se.fit))
pframe - plogis(etaframe)
pframe
My response variable is 0 or 1, the
Hi Ben and list,
Sorry to be a pain! I have followed your code, and modified it -
pp - predict(model1,se.fit=TRUE, type = response)
etaframe -
+ with(pp,cbind(fit,lower=fit-1.96*se.fit,upper=fit+1.96*se.fit))
pframe - plogis(etaframe)
pframe
My response variable is 0 or 1, the
Hi.
I want to create a plot with Pantone654 as the background. The RGB for this
color is (0,61,121), which corresponds to a hex of #003D79. How do I specify
the bg parameter for this?
All Best,
Ethan
[[alternative HTML version deleted]]
Try this:
par(bg = '#003D79')
On Wed, Sep 29, 2010 at 11:14 AM, Arenson, Ethan earen...@nbome.org wrote:
Hi.
I want to create a plot with Pantone654 as the background. The RGB for this
color is (0,61,121), which corresponds to a hex of #003D79. How do I specify
the bg parameter for this?
On 10-09-29 10:04 AM, Sam wrote:
Hi Ben and list,
Sorry to be a pain! I have followed your code, and modified it -
**You should not use type=response here.**
The point is that the (symmetric) confidence intervals are computed on
the link/linear predictor
scale, and then
Dear R-users,
Is there any R-function for fitting generalized additive mixed
models for ordinal data? Do they actually make some sense? I can fit a
generalized linear mixed model for ordinal data using the function
clmm(ordinal) and I'm able to cope with generalized additive model for
On Sep 29, 2010, at 8:31 AM, cuz wrote:
Hi,
For my dissertation, I've made copious use of xtable. I've just gotten
stumped however. I'm a fan of extended captions explaining the table, but
now I have to assemble a a list of tables and the captions are unwieldy. I
presume xtable calls
I am using both nlminb and optim to get MLEs from a likelihood function I have
developed. AFAIK, the model I has not been previously used in this way and so I
am struggling a bit to unit test my code since I don't have another data set to
compare this kind of estimation to.
The likelihood I
Right, that makes sense, thanks
The reason i used type= response was i wanted to convert the predicted
probabilities to the response scale, as surely this is the scale at which a
95CI value is most useful for?
I.e
pp - predict(model1,se.fit=TRUE, type = response)
1 0.68
Probability of
Hi,
I'm just learning to write R extensions in C and to embed R in C.
I was trying to get through the example in the help page on calling the .dll
directly (
http://cran.r-project.org/doc/manuals/R-exts.html#Calling-R_002edll-directly
).
When I compile I consistently get the error that
Ethan-
You need to be more explicit about what you mean by 'background'. Do you
mean:
(a) the entire plot including margins?, or
(b) only the plotting area?, or
(c) a different color for both margins and plotting area?
If you want (a), the solution is par(bg = '#003D79').
If you want (b), the
I am refering to a function call like this:
data(iris)
x - svmlight(Species ~ ., data = iris)
I tried to see the content of it by typing:
Species ~ .
but it gives nothing. How can I see it's content ?
- P.Dubois
__
R-help@r-project.org mailing
On 29/09/2010 10:51 AM, Gundala Viswanath wrote:
I am refering to a function call like this:
data(iris)
x- svmlight(Species ~ ., data = iris)
I tried to see the content of it by typing:
Species ~ .
but it gives nothing. How can I see it's content ?
str(Species ~ .) will tell you that
On Wed, 29 Sep 2010, Duncan Murdoch wrote:
On 29/09/2010 10:51 AM, Gundala Viswanath wrote:
I am refering to a function call like this:
data(iris)
x- svmlight(Species ~ ., data = iris)
I tried to see the content of it by typing:
Species ~ .
but it gives nothing. How can I see it's
Hej,
Calling newuoa (from the minqa package) makes R crash when the package rgl
is loaded first. This however only on certain selected data.
The data used for testing (saved to 'bugs.R'):
xvals =
c(1,2,4,5,7,8,9,10,11,12,14,15,16,18,19,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36)
Hi:
You could look into the gamm4 package. Its description is:
Fit generalized additive mixed models via a version of mgcv's gamm function,
using lme4 for estimation via Fabian Scheipl's trick.
HTH,
Dennis
On Wed, Sep 29, 2010 at 7:28 AM, Camarda, Carlo Giovanni
cama...@demogr.mpg.de wrote:
Apologies for bothering anyone who may not be interested in
this, but some of you will, for instance, have my current
email address in your address books, etc.
As result of a new policy by Manchester University, retired
former staff who no longer contribute actively to research
in their former
Thanks. I didn't see any obvious way to do it either. It looks like the
caption option has room for additional input that are as yet not designated.
I'll take a look at the latex() possibility.
Best,
cuz
--
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sehr geehrte Damen und Herren
Für unseren Statistik Unterricht brauchen wir das R-Programm.
Ich habe das Programm heruntergeladen, deutsch gewählt und installiert. Das
Problem ist nach 3mal neu installieren, dass das Programm immer auf englisch
kommt.
Können Sie mir helfen?
Vg
Tobias Keller
Hello all,
I have been meaning to learn R for a while and have just subscribed to this
list. I am planning to give R a shot at one of my live projects. I am looking
to explore graphical features of R on my data below.
Sample Data:
Cat1 - Cat2 - Cat3 - Cat4 - NumPeople - Salary
H - L - H
Hi:
Someone off-list (Josh Wiley - thank you) mentioned the Error() term in the
OP's ANOVA, which I missed in responding to the post - sorry for the
misinformation. Using the npk example with code that Josh showed me, we
have, for the following model,
npk.aov2 - aov(yield ~ N*P*K +
Version 4.63 of the caret package is now on CRAN.
caret can be used to tune the parameters of predictive models using
resampling, estimate variable importance and visualize the results.
There are also various modeling and helper functions that can be
useful for training models.
caret has
Hi:
This 'works' on the lapply end:
# Function to read one file from the list:
g - function(x) read.zoo(file = x, header = TRUE, FUN = as.chron, sep =
,,
colClasses = c(NULL, NULL, character,
numeric))
# Apply to all files in list:
lapply(filenames, g)
[[1]]
(01/01/04
Hello,
go to https://stat.ethz.ch/mailman/options/r-help/y...@email and if You do not
remember Your password, use Password reminder (down) to mail it to Your
address. Then login and do all needed changes. You can replace r-hep in link
above with r-sig-phylo, r-announce and so on. And of course
G'day Tobias,
On Wed, 29 Sep 2010 14:01:10 +
Keller Tobias (kelleto0) kelle...@students.zhaw.ch wrote:
Für unseren Statistik Unterricht brauchen wir das R-Programm.
Ich habe das Programm heruntergeladen, deutsch gewählt und
installiert. Das Problem ist nach 3mal neu installieren, dass das
Dear List and Ben
( I apologise if this has been sent twice, but it is not showing in my sent
folder and i have been having trouble with my email of late)
Right, that makes sense, thanks
The reason i used type= response was i wanted to convert the predicted
probabilities to the response
[I'm a little confused: are you Sam Smith or Chris Mcowen ... ?]
This is admittedly a bit confusing, but the best scale on which to
compute standard errors is the link scale.
It turns out (I hadn't realized this) that predict.glm does give
you not-crazy answers when you ask for
se.fit=TRUE
Thats great thanks very much for your help
On 29 Sep 2010, at 17:30, Ben Bolker wrote:
[I'm a little confused: are you Sam Smith or Chris Mcowen ... ?]
This is admittedly a bit confusing, but the best scale on which to
compute standard errors is the link scale.
It turns out (I hadn't
Hello.
How can I write this all in one line?
mydata is a zoo series, limit is a numeric vector of the same size
tmp - ave(coredata(mydata),as.Date(index(mydata)),FUN = function(x) (
(cummax(x)-x )) )
tmp - (tmp limit)
final - ave(coredata(tmp),as.Date(index(mydata)),FUN = function(x) cumprod(
On Sep 29, 2010, at 18:21 , Berwin A Turlach wrote:
G'day Tobias,
On Wed, 29 Sep 2010 14:01:10 +
Keller Tobias (kelleto0) kelle...@students.zhaw.ch wrote:
Für unseren Statistik Unterricht brauchen wir das R-Programm.
Ich habe das Programm heruntergeladen, deutsch gewählt und
It all depends on the ultimate use of the results.
Frank
-
Frank Harrell
Department of Biostatistics, Vanderbilt University
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Dear All;
I have searched to see if any code in R was written to calculate mean and
variance of a Ranked Set Sample, but did not find any. Is there any package
for RSS, or kindly can somebody share a code he/she wrote, I am very
grateful and willing to acknowledge that in my work. Thanks much
On 2010-09-28 18:39, Marlin Keith Cox wrote:
When using a wireframe, I need to move the colorkey from the right
position (default0 towards the plot. I have also needed to adjust the
height and used the code
colorkey=list(T,space='right',height=.5)
I have looked at documents (within levelplot)
Ted is well aware of how to change his list email. He was advising
people on the list who who have his old email address in their address
books to remove it.
On 09/29/2010 12:16 PM, Vojtěch Zeisek wrote:
Hello,
go to https://stat.ethz.ch/mailman/options/r-help/y...@email and if You do not
On Tue, 28 Sep 2010, L Brown wrote:
Hi, everyone. I have what I hope will be a simple coding question. It seems
this is a common job, but so far I've had trouble finding the answer in
searches.
I have two matrices (x and y) with a different number of observations in
each. I need to draw a
Dear useRs,
I am currently fitting an advanced failure time model using Göran
Broström's excellent eha library with the aftreg command.
My question: How do I interpret the Overall p-value, that is
reported at the very bottom of the output? I already figured out it
must be a chi-square test, but
Thanks for the help.
Sharad
On Mon, Sep 27, 2010 at 9:12 PM, Remko Duursma [via R]
ml-node+2716469-935075351-6...@n4.nabble.comml-node%2b2716469-935075351-6...@n4.nabble.com
wrote:
Try something like this:
dfr - read.table(textConnection(plate.id well.id Group HYB
rlt1
1
Thank you very much for your solution but it works only in a dataframe
object. If I am using an ftable object, it doesn't run.
I use, as a workaround, to fill with blank spaces the left of each number,
so when I print the table, it appears aligned to right.
But, obviously, this doesn't work for
Dear All,
I am trying to define a loop for a m*n matrix, where i=1:n and j=1:m. Unlike
the usual for loop, i should go in the following way:
For j=1,
i=1,2,3,n
For j=2,
i=n,n-1,n-2,..,1
For j=3,
i=1,2,3,.n etc.
which means i should go in either increasing or decreasing order
I am trying to turn several lines of information into a variable. I used
the filx function to input my file then the readlines to qualify what I
want. Essentially I have data in a file every 10 minutes through a day for
several years down a column:
date time value
9/28/10 02:00 13
9/28/10
for (j in 1:n)
{
if (j%%2==0)
{
iRange = c(n:1)
} else
iRange = c(1:n)
for (i in iRange)
{
your code
}
}
Peter
On Wed, Sep 29, 2010 at 10:40 AM, cassie jones cassiejone...@gmail.com wrote:
Dear All,
I am trying to define a loop for a m*n matrix, where
Hello
On Tue, Sep 28, 2010 at 1:39 PM, Soumen Pal soumen.4...@gmail.com wrote:
I need your kind help regarding the following:
I wish to know is there any way to use R in Visual Basic environment. I want
to develop a VB application where R can be embedded (R will work as a back
end
Is there an R function for evaluating the exponential integral
Ei(x) = INT(-inf,x) exp(t)/t dt
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PLEASE do read the posting
I am trying to get R to resample my dataset of two columns of age and length
data for fish. I got it to work, but it is not resampling every replicate.
Instead, it resamples my data once and then repeated it 5 times.
Here is my dataset of 9 fish samples with an age and length for each one:
Mishra, Srikanta MISHRAS at battelle.org writes:
Is there an R function for evaluating the exponential integral
Ei(x) = INT(-inf,x) exp(t)/t dt
Try the gsl package.
Also library(sos); findFn({exponential integral})
although admittedly it doesn't find the Expint page
in the gsl package
I think that you want:
replicate(5, growth[sample(9,12,replace=T),], simplify = FALSE)
On Wed, Sep 29, 2010 at 3:19 PM, Michael Larkin mlar...@rsmas.miami.eduwrote:
I am trying to get R to resample my dataset of two columns of age and
length
data for fish. I got it to work, but it is not
Hmmm. Maybe a documentation typo in ?spplot.
If you follow the documentation through to ?levelplot, you find
that
cuts: number of levels the range of ‘z’ would be divided into
(no mention of actual breakpoints) but:
at: numeric vector giving breakpoints along the range of ‘z’.
George Coyle wrote:
I am trying to turn several lines of information into a variable. I used
the filx function to input my file then the readlines to qualify what I
want. Essentially I have data in a file every 10 minutes through a day
for
several years down a column:
date time value
paste( '(', paste( ', rep(letters[1:3],2), ', sep=, collapse=','), ')',
sep= )
[1] ('a','b','c','a','b','c')
If you need the space after the comma then just change ',' to ', '.
The outer paste can be replaced with sprintf (and that may be more readable).
--
Gregory (Greg) L. Snow Ph.D.
Hi, everyone.
GAD package analyses complex ANOVA models with any combination of
orthogonal/nested and fixed/random factors, as described by Underwood
(1997). There are two restrictions: (i) data must be balanced; (ii)
fixed nested factors are not allowed. Homogeneity of variances is
checked using
Thanks,
Thats great just what I was trying to do.
HH
Thomas Stewart wrote:
HH-
I'm not familiar with the plots you mention, but the following is a
quick attempt to create the plot you describe.
data-data.frame(
org=1:10,
q1=sample(1:10,replace=T),
q2=sample(1:10,replace=T),
Dear R-helpers,
I'm trying to associate linear coefficients (intercept and slope) to tens of
thousands of observations based on a table with benchmark values.
#Example - Value table and their corresponding coefficients (intercept and
slope)
coef = data.frame(cbind(st=c(1:5),b =
Mathieu -
First of all, you can combine as many conditions as
you want in an if statement, using (and) and || (or).
So to say
coef$st[i-1] obs[t] coef$st[i]
use
coef$st[i-1] obs[t] obs[t] coef$st[i]
So following your logic, you could use
x = numeric(length(obs))
for(t in
Hi
On 29/09/2010 8:17 p.m., Tal Galili wrote:
My honor.
A short question: if there is something in the device that is sensitive
to the overlapping of the text, then is it possible to add a warning
massage output when the length of the text is longer then the device
dimensions?
The graphics
#I am trying to understand how R fits models for contrasts in a
#simple one-way anova. This is an example, I am not stupid enough to want
#to simultaneously apply all of these contrasts to real data. With a few
#exceptions, the tests that I would compute by hand (or by other software)
#will give
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