I was surprised to see this unexpected behavior of subset in a for loop. I
looked in subset.data.frame and it seemed to me that both versions should
work, since the subset call should be evaluated in the global environment -
but perhaps I don't understand environments well enough. Can someone
Hi
r-help-boun...@r-project.org napsal dne 01.02.2010 18:14:04:
Dear Petr,
The intention is to get a ratio value for every observations (1400 obs)
for
every variable (4 variables).
Well, maybe that
mat-matrix(runif(12),4,3)
ratiomat-t(t(mat)/vec)
And from the ratios, I would like
Hi:
Try this for your second loop instead:
for(s in school.list){
print(s)
print(subset(input.data, sch == s))
}
[1] 1
sch pop
1 1 100
2 1 200
[1] 2
sch pop
3 2 300
4 2 400
Don't confound the 'sch' variable in your data frame with the
index in your loop :)
HTH,
Dennis
On Mon,
There is 'cov.wt'. (If you have univariate
data, then you need to use 'as.matrix'.)
On 01/02/2010 20:32, Thomas Lumley wrote:
On Mon, 1 Feb 2010, David Winsemius wrote:
On Feb 1, 2010, at 12:37 PM, Антон Морковин wrote:
Dear all,
what function can be used to calculate weighted SD or/and
Hi:
I played around with this earlier, but had only limited success. This is
what I
got; perhaps others can embellish this with more efficient (and correct)
code.
I couldn't get more than one expression in a line without overplotting.
Here's
an example:
plot(c(0, 1), c(0, 1))
text(0.5, 0.5,
Is there a better alternative to
x = c(1e7, 2e7)
x.lb = c(0,1e7,2e7)
s.lb = format(x.lb, scientific = FALSE, big.mark = ,)
barplot(x, yaxt = n, ylab = )
axis(side = 2, at = x.lb, labels = s.lb)
(I am sure there is a better alternative to line 2 :)).
Thank you.
--
View this message in
Dear all,
I just know how to solve an eaquation by using optim() or nlm(). But, now, I
have three nonlinear equations,
how could we use optim() or nlm() to solve a system of nonlinear equations in
R? Thank you so much.
Sincerely,
Joe
___
Hi David,
there is a solution using bquote instead of substitute
expr-bquote(italic(.(pname[1]))==.(params[1])~,
~.(as.name(pname[2]))==.(params[2]))
plot(1,1,main=expr)
hth.
dkStevens schrieb:
In trying to create a plotmath expression for plot labeling, such as
R = 6, beta = 15
where I
Dear R helpers,
Yesterday I had raised following query which was addressed by Mr Ellison. The
query and the wonderful solution as provided by Mr. Ellison are as given below.
## PROBLEM
I am calculating the 'Yield to Maturity' for the Bond with following
characteristics.
Its a $1000
That's the trick, Eik! Nice job.
To return to the problem I was struggling with, it works with the following
incantation:
names - c('R', 'P', 'k', 'alpha')
vals - c(13.34859, 2.53071, 4.06000, 0.00719)
expr - bquote(.(names[1])==.(vals[1])~,
~.(as.name(names[2]))[m]==.(vals[2])~,
Hi,
When I run the repeat loop in R for large dataset, I got Memory problem.
How can I solve these problem.
--
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The abstract submission deadline for the R Users Conference, useR! 2010,
is just one month away.
Submission deadline: 2010-03-01
Early registration will also close on 2010-03-01.
Now is the perfect time to prepare an abstract presenting innovations or
exciting applications of R. The call
Dear useRs,
we are happy to announce the release of mboost 2.0-0 on CRAN:
http://cran.r-project.org/package=mboost
This version contains major updates and changes to the implementation of
the main algorithm. Some slight changes to the user-interface where
necessary. Please consult the manual
Ruben Roa has kindly suggested using 'scipen' option - cf.
fixed notation will be preferred unless it is more than ‘scipen’ digits
wider.
However,
options(scipen = 50)
x = c(1e7, 2e7)
barplot(x)
still does not produce the desired result.
--
View this message in context:
Hello R-experts,
I am having difficulties with 3D plotting (i.e. the evolution of various
forward curves through time).
I have two comma seperated files both ordered by date (in the first column) one
containing contracts (meaning forward delivery months from YEAR_ Letter F
... January
Nai-Wei Chen wrote:
I just know how to solve an eaquation by using optim() or nlm(). But, now,
I have three nonlinear equations,
how could we use optim() or nlm() to solve a system of nonlinear
equations in R? Thank you so much.
You don't use general optimize routines to solve a
Erwin Kalvelagen-2 wrote:
Hans W Borchers hwborchers at googlemail.com writes:
# Prepare inputs for MILP solver
obj - c(rep(0, n), 0, 1, 1, 0)
typ - c(rep(B, n), B, C, C, B)
mat - matrix(c(s, -z, -1, 1, 0,# a = a_p + a_m
rep(0, n), 1, 0, 0,
Hi Madhavi,
the error message means, that your function returns NA evaluated at the
lower limit of the search interval.
try f.ytm(0) to check that.
I think,
for (i in 1 : (tenure * no_comp - 1))
E = NULL
F = NULL
{
E[i] = cash_flow[i]/(1+ytm)^i
F = (sum(E) + (face_value +
hello,
i'd be happy if someone could provide help with the following problem:
i have a dist.matrix that comes from vegdist() function of the vegan
package. the used method = horn is not accepted as argument in
hvcluster(...,dist.method=...).
is there a way to incorporate the method horn in
sorry, it should say pvclust(), of course...
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__
Dear Sir,
Thank you for valuable guidance. Though I have been using R occassionally, it
was limited to some basics and that way I am new to R. As suggested by you, I
have gone through the said chapter of Introduction to R manual, though I have
some urgent comittments to meet.
I have tried
On 02.02.2010 11:33, Meenakshi wrote:
Hi,
When I run the repeat loop in R for large dataset, I got Memory problem.
How can I solve these problem.
buy more memory, bigger machine, more efficient programming, import of
only relevant data, use of specific tools, .. or in other words:
Since your files did not come through the mail list, I'd suggest to put
them on some webside and provide a link.
Uwe Ligges
On 02.02.2010 13:01, walter.dju...@chello.at wrote:
Hello R-experts,
I am having difficulties with 3D plotting (i.e. the evolution of various
forward curves through
Hi,
I was wondering if there existed a package in R that would bicluster /
co-cluster in more than 2 dimensions.
thanks!
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
Hi all,
I've got a time series object (xts) called table. Table contains closing
price and volume for each market day of 2009 on a given equity.
I'd like to get another xts object, say table2, that contains for each
market day holds the close of that day minus the close of the day before and
It is a strange behaviour in that I did not expect it... but I am sure
there is a simple explanation for it and it'll have to do with the way
numbers are stored in R, but it's caught me by surprise and I don't
find it obvious.
Here's a simplified example reproducing the behaviour I
FAQ 7.31
j.delashe...@ed.ac.uk wrote:
It is a strange behaviour in that I did not expect it... but I am sure
there is a simple explanation for it and it'll have to do with the way
numbers are stored in R, but it's caught me by surprise and I don't find
it obvious.
Here's a simplified
Hi Tim,
maybe you are looking for something like this:
http://www.nature.com/nbt/journal/v26/n5/full/nbt1397.html
It is relatively easy to implement in R, I think.
Best,
Gabor
On Tue, Feb 2, 2010 at 3:44 PM, Tim Smith tim_smith_...@yahoo.com wrote:
Hi,
I was wondering if there existed a
On Tue, 2 Feb 2010, Nick Torenvliet wrote:
Hi all,
I've got a time series object (xts) called table. Table contains closing
price and volume for each market day of 2009 on a given equity.
I'd like to get another xts object, say table2, that contains for each
market day holds the close of
Unluckily I dela with miRNA files whose name may contain the character *.
Because of the special meaning of * I have to remove it.
I found out how to make list.files() extract only those file names which
contain a *
Namely:
# list.files(pattern=\\*)
Now I have to process all files whose name
Dear all,
I have a simple question for which I cannot find the answer.
I need to make an easy plot, but for the x axis I need to be able to specify
by myself the division of x axis from x[,1] either every single observation,
or every 5th, or 10th or 20th
x - matrix(data=NA, nrow=100, ncol=2)
mau...@alice.it wrote:
Unluckily I dela with miRNA files whose name may contain the character *.
Because of the special meaning of * I have to remove it.
I found out how to make list.files() extract only those file names which
contain a *
Namely:
# list.files(pattern=\\*)
Now I have to
mau...@alice.it wrote:
Unluckily I dela with miRNA files whose name may contain the character *.
Because of the special meaning of * I have to remove it.
I found out how to make list.files() extract only those file names which contain a
*
Namely:
# list.files(pattern=\\*)
Now I have to
Hello everyone,
I have two vectors having only one element different:
vector1 vector2
vector1
TWC TWC
TWC
VFC TWX
David,
Now the code is:
for (j in seq_along(rwy)) { # subset the data and merge them
ar4rw = ar4rw - subset(arrgnd, arrgnd$Runway==rwy[j])
if(j == 1) {
arrw = ar4rw
}
else {
arrw = merge(arrw, ar4rw)
}
}
I attach the data. I needed 500 rows to get both runways in rwy.
The suggestions did not
On 02.02.2010 16:01, walter.dju...@chello.at wrote:
Here they are for now for you.
And I will put them somewhere in a few minutes and poost the link.
No need to do so. If you really want to use 3D plots (which I doubt you
really want since you could do better in 2D using dotcharts or so):
On 02-Feb-10 15:18:28, Peter Dalgaard wrote:
mau...@alice.it wrote:
Unluckily I dela with miRNA files whose name may contain the
character *.
Because of the special meaning of * I have to remove it.
I found out how to make list.files() extract only those file
names which contain a *
Namely:
options(scipen = 50, digits = 5)
x = c(1e7, 2e7)
barplot(x)
Still scientific...
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Dear Madam/Sir,
I read carefully The R Installation and Administration paper. I downloaded
Rtools11.exe and I run it. As a result there is a directory R on my PC.
Could you please tell me what should I do next in order to run R?
Thank you in advance.
BR
Dragomir Nedeltchev
On Mon, Feb 1, 2010 at 13:19, Michael Friendly frien...@yorku.ca wrote:
[Env: WinXp, R 2.9.2]
In my .Rprofile, I define a number of utility functions I'd like to have
available in my R session, but don't want them
to be *normally* listed by ls(), or more importantly, saved if I save my
a link to the 3D plot data files
http://members.chello.at/gwd
the files are NG_contracts and NG_values
--
mfg
WD
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
Download the R installer from CRAN, and run that. A desktop icon
should pop up, and you should be set.
Stephen
On Tue, Feb 2, 2010 at 8:16 AM, dragomir nedeltchev
dnedelche...@yahoo.com wrote:
Dear Madam/Sir,
I read carefully The R Installation and Administration paper. I downloaded
Strange, the following works reproducibly on my machine (Windows 2000
Pro):
options(scipen = 50, digits = 5)
x = c(1e7, 2e7)
?barplot
barplot(x)
while I also get scientific with your code. After I called ?barplot
once, I'm incapable of getting the scientific notation again, though...
But I
Hi
Option one
vec1-sample(letters[1:10])
vec2-vec1[-5]
vec2-c(vec2,NA)
# missing position
mis-which(!(vec1 %in% vec2))
c(vec2[1:(mis-1)], NA,vec2[mis:((length(vec2)-1))])
[1] f e g d NA j c i b h
gives you desired vector. For two or more values missing it shall be
rephrased probably by rle
When you create your plot us the argument xaxt='n'. That will suppress the
default axis labels and tick marks. Then use the axis function which lets you
specify exactly where you want the tick marks and exactly what you want the
labels to be. You can also specify tck=1 as an argument if you
Hi, I have a data frame datas with half of the columns with the same name
A. I want to delete all those columns from the data frame so here is what
I did:
datas$A - NULL
The problem is that it deleted only one column, I would have to do it as
many times as there are A columns. Is there a way to
Hi David,
Thanks for your answer.
But I don't really see how I can extend it with my real data.
The thing is that I have more than 3 names and 1 value for each name.
Moreover, each is different from one run to another. That is why I was
trying with a modification of names(). Also to be noted
Hi,
On Mon, Feb 1, 2010 at 11:16 PM, Eunjung Kim eunjung.haw...@gmail.com wrote:
*Dear R users,
*I'm facing a trivial problem but I cannot solve it.
My question is:
I want to make data set like this.
i_lon1 i_lat1
i_lon2 i_lat2
i_lon3 i_lat3
i_lon4 i_lat4
i+1_lon1 i+1_lat1
i+1_lon2
This is what I just found now but I guess there is a simpler way:
datas[which(names(datas)==A)]-list(rep(NULL,length(which(names(datas)==A
but it worked
-
Anna Lippel
--
View this message in context:
This is an interesting approach but it doesn't quite get what I want. The
sample below is what I'm looking for where Eik's Title is compared with the
text line within the plot. My text line was produced by the explicit
text(0.6,1.2,expression(R *'= 13.35, ' *P[m] *'= 2.531, ' *alpha[r] *'=
Thanks, that helps! Subset creates a new context where a name clash can
occur. So if I don't want to check for that possibility, I should use a
special kind of index like .sch, or avoid subset:
for(sch in school.list){
print(sch)
print(input.data[input.data[,school.var] == sch,])}
which
Have a look at ?par and check out the 'xaxp' parameter
par(mfrow=c(1,2))
plot(x[,2] ~ x[,1], xaxp=c(0,100,5))
plot(x[,2] ~ x[,1], xaxp=c(0,100,10), cex.axis=0.5)
--
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Sent from the R
Try:
newdf - datas[names(datas) != A]
On Tue, Feb 2, 2010 at 11:47 AM, anna lippelann...@hotmail.com wrote:
This is what I just found now but I guess there is a simpler way:
datas[which(names(datas)==A)]-list(rep(NULL,length(which(names(datas)==A
but it worked
-
Anna Lippel
--
Try this:
leg - function(pNames, vparams) parse(text = paste(pNames, vparams,
sep = ==))
plot(0, main = leg(Sigma, 3.2))
legend(topleft, legend = leg(c(alpha[beta], delta), 1:2))
On Tue, Feb 2, 2010 at 11:44 AM, dkStevens david.stev...@usu.edu wrote:
This is an interesting approach but it
Romain, to keep it simple, I reproduced the example that you can find in the
toString function help and I still got the same error:
x - c(a, b, aaa)
toString(x)
Error in toString(x) : could not find function .jcall
-
Anna Lippel
--
View this message in context:
Thanks for the replies.
Apparently you cannot adjust the extension factor used by yaxs (which is set
at 4%), so what I did is
wrote a function (barplotCovered) to get y-axis limits based on the range of
the data and a user-defined axis expansion factor (axExFact).
I included an example below
On Mon, Oct 19, 2009 at 12:53 PM, Kingsford Jones
kingsfordjo...@gmail.com wrote:
sum((x-mean(x))^2)/(n)
[1] 0.4894708
((n-1)/n) * var(x)
[1] 0.4894708
But this is not a built-in function in R to do so, right?
hth,
Kingsford
On Mon, Oct 19, 2009 at 9:30 AM, Peng Yu pengyu...@gmail.com
Anna -
You could also look at the problem from the other direction:
data[,names(datas) != 'A']
- Phil Spector
Statistical Computing Facility
Department of Statistics
What packages are loaded? On the basic system, it works fine:
toString(123)
[1] 123
toString(c('a', 'b', 'zz'))
[1] a, b, zz
On Tue, Feb 2, 2010 at 12:17 PM, anna lippelann...@hotmail.com wrote:
Romain, to keep it simple, I reproduced the example that you can find in the
toString
Here is one way with an example:
datas=data.frame(x=1:3,A=1:3,A=1:3)
names(datas)=c(x,A,A)
datas
datas=datas[,names(datas)!=A,drop=FALSE]
datas
On 2/2/2010 8:35 AM, anna wrote:
Hi, I have a data frame datas with half of the columns with the same name
A. I want to delete all those columns from
datas[ , A != colnames(datas)]
Uwe Ligges
On 02.02.2010 17:35, anna wrote:
Hi, I have a data frame datas with half of the columns with the same name
A. I want to delete all those columns from the data frame so here is what
I did:
datas$A- NULL
The problem is that it deleted only one column, I
On Tue, 2 Feb 2010, David Katz wrote:
Thanks, that helps! Subset creates a new context where a name clash can
occur. So if I don't want to check for that possibility, I should use a
special kind of index like .sch, or avoid subset:
for(sch in school.list){
print(sch)
Dear R community,
I'm a beginner with Cluster Analysis. I would like to know if there is a
criterion to select the best set of clusters to do this analysis.
Thanks in advance,
[[alternative HTML version deleted]]
__
R-help@r-project.org
Hi,
On Tue, Feb 2, 2010 at 11:47 AM, anna lippelann...@hotmail.com wrote:
This is what I just found now but I guess there is a simpler way:
datas[which(names(datas)==A)]-list(rep(NULL,length(which(names(datas)==A
but it worked
For what it's worth, you could also have done:
clean -
anna wrote:
Romain, to keep it simple, I reproduced the example that you can find in the
toString function help and I still got the same error:
x - c(a, b, aaa)
toString(x)
Error in toString(x) : could not find function .jcall
.jcall() is an rJava function. toString() is a base R
This might give you a hint as to how to do it since I have no idea of
what your data looks like:
lon - lat - matrix(1:100, nrow=5)
output - file('/tempxx.txt', open='w')
for (i in 1:5){
x - cbind(lon[,i], lat[,i])
write.table(x, row.names=FALSE, col.names=FALSE, file=output)
cat('\n',
Hola, ¿qué tal?
No sé si la conoces, pero te invito a participar en la lista oficial de
ayuda de R en español:
https://stat.ethz.ch/mailman/listinfo/r-help-es
A través de ella, además de atender dudas de usuarios, tratamos de crear
una comunidad. De hecho, hace no mucho organizamos las
Here is the list of the loaded packages:
package:lattice package:fSeries
[5] package:fCalendarpackage:fEcofin
package:fUtilities package:MASS
[9] package:robustbase package:caTools
No I just did it on the R console, I built the vector just before.
-
Anna Lippel
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__
Try this:
subset(DF, select = setdiff(names(DF), A))
On Tue, Feb 2, 2010 at 2:47 PM, anna lippelann...@hotmail.com wrote:
This is what I just found now but I guess there is a simpler way:
datas[which(names(datas)==A)]-list(rep(NULL,length(which(names(datas)==A
but it worked
-
Dear all,
I have a simple question: How can I retrieve a list with all
properties of an object?
Example:
If I fit a regression with
model - coxph(Surv()~...)
I know that I can access the coefficients with
model$coefficients[...] afterwards, but how do I know which other
variables are
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of Steve Lianoglou
Sent: Tuesday, February 02, 2010 9:05 AM
To: anna
Cc: r-help@r-project.org
Subject: Re: [R] Deleting many columns of a data frame with
the same name ina row
I was trying to save a data frame to an excel file using the following
command:
write.table(myData, file=myData.csv,sep=,, row.names=F)
The command works for some data frames, but for other data frames, I get the
following error:
Error in if (inherits(X[[j]], data.frame) ncol(xj) 1L) X[[j]]
OK, I need help plotting. I have column headings of Day, Wgt, Foodin, Rep,
Grp and Tanks. Rep=c(1,2,3) and Tanks=c(a1,a2,a3,a4,a5,a6,
c1,c2,c3,c4,c5,c6, h1,h2,h3,h4,h5,h6).
I created a subset where I only would like Rep=2, and Tanks=c(a4,c4,h4) and
would like to graph (points) of Wgt and Day.
Hello everyone,
I am trying to retrieve the list of distinct values within a whole data
frame. I tried to use unique() function but it retrieves the distinct values
within each column or row, I want it for the entire data frame, any idea?
-
Anna Lippel
--
View this message in context:
thanks this is actually shorter :)
-
Anna Lippel
--
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Sent from the R help mailing list archive at Nabble.com.
yes it looks really simpler, thank you!
-
Anna Lippel
--
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Sent from the R help mailing list archive at Nabble.com.
Hi,
On Tue, Feb 2, 2010 at 1:19 PM, anna lippelann...@hotmail.com wrote:
Hello everyone,
I am trying to retrieve the list of distinct values within a whole data
frame. I tried to use unique() function but it retrieves the distinct values
within each column or row, I want it for the entire
How about a MINIMAL example of a data.frame that produces the error
message you see? That will help track down what's going on...
jlwoodard wrote:
I was trying to save a data frame to an excel file using the following
command:
write.table(myData, file=myData.csv,sep=,, row.names=F)
The
ok it worked so the key is unlist() from what I see :working:
-
Anna Lippel
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On Tue, Feb 2, 2010 at 11:44, Ivan Calandra
ivan.calan...@uni-hamburg.de wrote:
Hi David,
Thanks for your answer.
But I don't really see how I can extend it with my real data.
The thing is that I have more than 3 names and 1 value for each name.
Moreover, each is different from one run to
Anna,
Try omitting pkg:RBloomberg.
If you really need to use that package, you will have to
install the non-CRAN package RDCOMClient from omegahat.
I still don't see why toString() wouldn't do its job, even
with RBloomberg loaded.
-Peter Ehlers
anna wrote:
Here is the list of the loaded
Hi:
Taking David's and Gabor's contributions, I came up with a function that I
hope
satisfies David's needs:
parText - function(names, pars) {
p1 - paste(* '= , pars[-length(pars)], , '*, sep = )
p2 - paste(* '= , pars[length(pars)], ')
p - c(p1, p2)
txt - paste(names, p,
On 2/1/2010 5:51 PM, David Winsemius wrote:
I figured this out finally. I really believe that the R help write-ups
are sorely lacking. As soon as I looked at
http://www.statmethods.net/management/merging.html, it was obvious:
Adding Columns
To merge two dataframes (datasets) horizontally,
On Feb 2, 2010, at 1:08 PM, Philipp Rappold wrote:
Dear all,
I have a simple question: How can I retrieve a list with all
properties of an object?
Example:
If I fit a regression with
model - coxph(Surv()~...)
I know that I can access the coefficients with model
$coefficients[...]
On Feb 2, 2010, at 12:04 PM, Steve Lianoglou wrote:
Hi,
On Tue, Feb 2, 2010 at 11:47 AM, anna lippelann...@hotmail.com
wrote:
This is what I just found now but I guess there is a simpler way:
datas[which(names(datas)==A)]-
list(rep(NULL,length(which(names(datas)==A
but it worked
I tried your code and a funny thing happened: I got an
error
object 'Foodin' not found
Hmmm; could you try to post something _reproducible_
(and minimal, please)?
But here are some suggestions:
1. investigate '%in%'
2. put a comma after Wgt in your plot call
3. see what you get from
James Rome wrote:
On 2/1/2010 5:51 PM, David Winsemius wrote:
I figured this out finally. I really believe that the R help write-ups
are sorely lacking.
The help docs are probably not the best way to learn R, but they are
great for users of the functions. I have found that after going
Philipp,
Check ?str which displays the structure of R objects.
And do use extractor functions when available:
coef(yourModel) instead of yourModel$coef
-Peter Ehlers
Philipp Rappold wrote:
Dear all,
I have a simple question: How can I retrieve a list with all properties
of an object?
Probably a simple typo, but just to keep things straight: you want to
divide by n when describing the standard deviation of a sample, and
divide by n-1 when estimating a population standard deviation (your
initial description had it backwards I think).
On Tue, Feb 2, 2010 at 5:25 PM, Peng Yu
hello! i'm dealing with the following: i've collected a factor
covariable at irregularly placed sampling points along a line with
spatial informations, i.e.: dataset-c(x-coordinates, y-coordinates,
level-of-factor)
the factor describes the density of vegetation between 0 (no ground
cover) and
I agree. I have a foot of books on R now, for example the R Book by
Michael Crowly. But so far, Googling the archives of this list has been
the most help. Nonetheless, if I cannot understand the documentation of
a function, then the documentation needs to be updated. For example,
there needs to be
Yeah, sometimes the vocabulary we bring to a task does not match up
(or merge properly) with the vocabulary that the developers use. In
this case the merge operation is one that has a precise meaning in
database lingo, which apparently you do not have background in. My
experience in
Hi,
i'm trying to put a legend on some figures and they're coming out a
bit wonky. here's an example:
a - c(1:10)
par(mfrow=c(2,1))
plot(a,type=s,lwd=3)
leg - c(expression(paste(data1 (,rho,=1))),
expression(paste(data2 (,rho,=0.0
legend(bottomright,legend=leg,col=c(1,2),lwd=3)
I just did a search on StructTS on the same topic and ended up here. Did
this issue ever get resolved here or elsewhere?
Regards,
Myron
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Thanks Jim,
As I know nothing about Perl (is it really necessary?), I think that I will try
with David's approach.
What I want to do is to for a pseudo-panel with my survey data (that is
collected periodically), and process it with the survey package. The resulting
data set will be big (around
Thanks again, David
I think that this could work.
Final questions:
1. I have read that read.fwt could be slow for big tables (my tables have
aprox. 16 records, with 176 characters of recordlenght, almost 28MBytes).
Could that be a problem?
2. If using read.fwt is not a problem, wouldn't be
isn't there a way to specify from which library I am taking the method
from...like rjava.toString or base.toString..
-
Anna Lippel
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try putting sep= in your paste method
-
Anna Lippel
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__
On Feb 2, 2010, at 2:13 PM, George Locke wrote:
Hi,
i'm trying to put a legend on some figures and they're coming out a
bit wonky. here's an example:
a - c(1:10)
par(mfrow=c(2,1))
plot(a,type=s,lwd=3)
leg - c(expression(paste(data1 (,rho,=1))),
expression(paste(data2 (,rho,=0.0
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