Dear all,
I found the last example of this link (
https://sites.ualberta.ca/~lkgray/uploads/7/3/6/2/7362679/6c_-_line_plots_with_error_bars.pdf)
very similar to the one I need to make for my paper, and I think I got what
I wanted by applying some of the suggestions of this mail list.
Here it is t
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Hi, Petr,
Indeed, your code is much better than the one I presented. I made small
editings.
par(mfrow = c(2, 1))
par(cex = 0.6)
par(mar = c(0, 0, 0, 0), oma = c(7, 7, 0.5, 5))
par(tcl = -0.25)
par(mgp = c(2, 0.6, 0))
plot(BUD~YEAR, type="o", ann=F, axes=F, pch=19, ylim=c(60,100),data=g1)
axis(2,
Dear expert friends,
I'm pretty young of this world and my question at your eyes can be petty
easy.
I'll need to change the name of the levels inside a column of my data-frame
levels(ind.davis$Ageclass) <- c("adult", "Juvanile", "sub-adult")
names(ind.davis$Ageclass) <- c("Adult", "Juvenile", "Su
I'm trying to use the following command.
arima (x, order = c(p,d,q), seasonal =list(order=c(P,D,Q), period=s)
How can I write an estimated seasonal ARIMA model from the outputs. To be
specifically, which sign to use? I know R uses a different signs from S plus.
Is it correct that the model is:
(
Hi all,
I want to use tabu search to solve my minimization problem. but tabu
search in R is for maximization, so I turn my function from f to -f， but the
eUtilityKeep always be 0 from the second position. I have go through a part of
source code found that it always give the default value
Dear All,
I am learning R so it's a very simple problem but I do not understand while
I am not able to generate a graph from two vectors.
when I type this code, it generates a very nice graph.
pdf("mygraph.pdf")
> attach(mtcars)
> plot(wt,mpg)
> abline(lm(mpg~wt))
> title("Regreesion of mpg")
>
Leonardo--
R-help can be a very useful resource. Some suggestions to use it well:
1. use an informative subject line, not "help"
2. include a "minimal working example:" a *little* data, the code that,
with those data, reproduces your problem, and the error message that
resulted.
As to your parti
One solution, among many, involving recoding. There is a function in
package QCA called recode()
(similar, but in my opinion more flexible than the same function recode()
in package car)
> library(QCA)
> ind.davis$Ageclass <- recode(ind.davis$Ageclass, "adult = Adult; Juvanile
= Juvenile; sub-adu
Zitat von Mars Xu :
Hi all,
I want to use tabu search to solve my minimization problem. but
tabu search in R is for maximization, so I turn my function from f
to -f， but the eUtilityKeep always be 0 from the second position. I
have go through a part of source code found that it always gi
This is an excellent exercise for you, the beginner. If you explicitly want a
line-by-line translation and don't want to use the strengths of R
(vectorization/functions) then there isn't much point in asking us to read the
Introduction to R document that comes with the software for you.
Please
Hi
Did you close pdf device by
dev.off()
?
Cheers
Petr
> -Original Message-
> From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Yogesh Gupta
> Sent: Tuesday, June 20, 2017 7:18 AM
> To: r-help@r-project.org
> Subject: [R] error while creating a simple graph
>
> Dear All,
Hello,
You don't need dev.new, but you need to tell R that the plot is finished
by calling dev.off after the plot command.
Hope this helps,
Rui Barradas
Em 20-06-2017 06:18, Yogesh Gupta escreveu:
Dear All,
I am learning R so it's a very simple problem but I do not understand while
I am no
Cc'd back to the list... always use reply-all.
You say res should have been LSS but LSS is a scalar so the for loop will only
run once. What does a successful output look like for a sample input? How do
you (we) know when success has been achieved?
In fact, what is the formula you want to imple
As Chris points out, we do not know what the factor labels were before you
tried to change them. The levels() function should have worked as long as it
included all of the factor levels in the same order. The names() function lists
the names of an object. For a data frame that is the column head
I am trying to transpose a dataframe by its first column using the by statement
using the t function. When I use the by function, I get a message,
Error in FUN(X[[i]], ...) : could not find function "FUN"
I don't think I have a syntax error in my by statement because the by statment
works using
> On 20 Jun 2017, at 07:18, Yogesh Gupta wrote:
>
> Dear All,
>
> I am learning R so it's a very simple problem but I do not understand while
> I am not able to generate a graph from two vectors.
>
> when I type this code, it generates a very nice graph.
>
> pdf("mygraph.pdf")
>> attach(mtcar
> On Jun 20, 2017, at 7:19 AM, Sorkin, John wrote:
>
> I am trying to transpose a dataframe by its first column using the by
> statement using the t function. When I use the by function, I get a message,
>
> Error in FUN(X[[i]], ...) : could not find function "FUN"
>
> I don't think I have a
Hello,
Works fine also with me:
> by(phonydata2[,2:3],phonydata2[,1],t)
phonydata2[, 1]: CONTROL
1 2 3 5 6 7 8 11 12 13 17 20 22 23 25 26 27 29
HGDOM1_MW1 NA NA NA NA NA 17 30 16 16 14 12 NA NA 13.5 1 NA 12 NA
HGDOM1_W1 8 NA NA NA NA 18 NA 14 8 NA 15 9 NA 14.0 2 NA 18 NA
The attached png file shows the expected output. The loop is complicated
because the values multiplied shift in value and sign so your effort to sum the
values along the way fails. It we compute the first two values directly and
create arrays to handle the changing multipliers, the loop is prett
HelloI have three x,y,z vectors (lets say each is set as rnorm(360)). So each
one is having 360 elements each one correpsonding to angular coordinates (1
degree, 2 degrees, 3 degrees, 360 degrees) and I want to plot those on the
xyz axes that have degress.
Is there a function or library to
package rgl.
Best,
Uwe Ligges
On 20.06.2017 21:29, Alaios via R-help wrote:
HelloI have three x,y,z vectors (lets say each is set as rnorm(360)). So each
one is having 360 elements each one correpsonding to angular coordinates (1
degree, 2 degrees, 3 degrees, 360 degrees) and I want to
I took another look at the problem.
Essentially there isn't enough information in the data to estimate the
parameters. What there is
doesn't have the right shape for the model.
You can see this by plotting the data -- essentially a straight line.
Get the slope as follows:
rr <- mean(mydata
Hi R users,
I have a question about fitting a cosine curve. I don't know how to set the
approximate starting values. Besides, does the method work for sine curve
as well? Thanks.
Part of the dataset is in the following:
y=c(16.82, 16.72, 16.63, 16.47, 16.84, 16.25, 16.15, 16.83, 17.41, 17.67,
17.
Hi lily,
You can get fairly good starting values just by eyeballing the curves:
plot(y)
lines(supsmu(1:20,y))
lines(0.6*cos((1:20)/3+0.6*pi)+17.2)
Jim
On Wed, Jun 21, 2017 at 9:17 AM, lily li wrote:
> Hi R users,
>
> I have a question about fitting a cosine curve. I don't know how to set the
>
Thanks, that is cool. But would there be a way that can approximate the
curve by trying more starting values automatically?
On Tue, Jun 20, 2017 at 5:45 PM, Jim Lemon wrote:
> Hi lily,
> You can get fairly good starting values just by eyeballing the curves:
>
> plot(y)
> lines(supsmu(1:20,y))
>
On 20/06/17 17:21, Y S wrote:
I'm trying to use the following command.
arima (x, order = c(p,d,q), seasonal =list(order=c(P,D,Q), period=s)
How can I write an estimated seasonal ARIMA model from the outputs. To be
specifically, which sign to use? I know R uses a different signs from S plus.
Is
hello,
I have a graph and i use qgraph package to calculate centrality parameters.
Now I want to know the maximum value of shortest path for each vertex with
discarding the Inf value in short pathes. For this I use the
ShortestPathLengths of centrality function in qgraph. but when I want to
get th
What I did was to plot your initial values, then plot the smoothed
values and guess the constants. That is, I got an "eyeball" fit to the
smoothed values. As I have described this as "gross cheating" in the
past, you should either split your data, estimate on one subset and
then test on another, or
Hi, I am doing composite plots with the package plot3D. One of my variables is
qualitative and indexed to integers, and I would like the legend for it to have
labels located at the integer values (midpoints), and not at the breaks between
classes. In the example below, the Elev Classes legend
Thanks. I will do a trial first. Also, is it okay to have the datasets that
have only part of the cycle, or better to have equal or more than one
cycle? That is to say, I cannot have the complete datasets sometimes.
On Tue, Jun 20, 2017 at 7:37 PM, Don Cohen
wrote:
>
> If you know the period and
I'm trying the different parameters, but don't know what the error is:
Error in nlsModel(formula, mf, start, wts) :
singular gradient matrix at initial parameter estimates
Thanks for any suggestions.
On Tue, Jun 20, 2017 at 7:37 PM, Don Cohen
wrote:
>
> If you know the period and want to fit
On Tue, 20 Jun 2017, lily li wrote:
Hi R users,
I have a question about fitting a cosine curve. I don't know how to set the
approximate starting values.
See
Y.L. Tong (1976) Biometrics 32:85-94
The method is known as `cosinor' analysis. It takes advantage of the
*intrinsic* linear
If you know the period and want to fit phase and amplitude, this is
equivalent to fitting a * sin + b * cos
> >>> > I don't know how to set the approximate starting values.
I'm not sure what you meant by that, but I suspect it's related to
phase and amplitude.
> >>> > Besides, does the method
Hard to follow data analysis without data. Try making your example reproducible
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