On 2024-06-20 2:01 a.m., DynV Montrealer wrote:
I need to do a non-English report including the red line from plot(aov,
which=1), which I don't know how to reproduce. I'm thinking that if I
replace the labels that it would be good (so no English remains). What I
have so far is almost what I need,
Dear Rui,
I really thank you a lot for your precious R help. It is exactly what I was
trying to do! Once more, many thanks!
Best,
Sacha
Le vendredi 27 octobre 2023 à 09:36:18 UTC+2, Rui Barradas
a écrit :
Às 19:23 de 26/10/2023, varin sacha via R-help escreveu:
> Dear R-Experts,
>
Às 19:23 de 26/10/2023, varin sacha via R-help escreveu:
Dear R-Experts,
Here below my R code working but I don't know how to complete/finish my R code
to get the final plot with the extrapolation for the10 more years.
Indeed, I try to extrapolate my data with a linear fit over the next 10 yea
Incidentally, if all you wanted to do was plot fitted values, the
predict method is kinda overkill, as it's just the fitted line from
the model. But I assume you wanted to plot CI's/PI's also, as the
example illustrated.
-- Bert
On Thu, Oct 26, 2023 at 1:56 PM Bert Gunter wrote:
>
> from ?predic
from ?predict.lm:
"predict.lm produces a vector of predictions or a matrix of
predictions and bounds with column names fit, lwr, and upr if interval
is set. "
ergo:
predict(model, dfuture, interval = "prediction")[,"fit"] ## or [,1]
as it's the first column in the returned matrix
is your vector
Thanks for the suggestion. I believe it is exactly what I need. I will
try this function. Thanks!
Regards,
Shu Fai
On Mon, Oct 16, 2023 at 3:39 AM Paul Murrell wrote:
>
> Hi
>
> You could also look at dev.capture(), depending on which screen device
> you are using.
>
> Paul
>
> On 16/10/23 05:24
Thanks a lot for introducing these functions! I am not aware of them
but it seems that they can help me to do what I want to do.
Regards,
Shu Fai
Regards,
Shu Fai Cheung (張樹輝)
On Mon, Oct 16, 2023 at 12:24 AM Duncan Murdoch
wrote:
>
> On 15/10/2023 12:05 p.m., Shu Fai Cheung wrote:
> > Let me
Hi
You could also look at dev.capture(), depending on which screen device
you are using.
Paul
On 16/10/23 05:24, Duncan Murdoch wrote:
On 15/10/2023 12:05 p.m., Shu Fai Cheung wrote:
> Let me clarify my question:
>
> plot.new()
> polygon(c(.5, .5, .75, .8), c(.25, .3, .4, .5))
>
> If t
On 15/10/2023 12:05 p.m., Shu Fai Cheung wrote:
Let me clarify my question:
plot.new()
polygon(c(.5, .5, .75, .8), c(.25, .3, .4, .5))
If the device is an on-screen device, can I check whether a particular area
has anything drawn on it, or, to be precise, whether the color of a
particular area
Let me clarify my question:
plot.new()
polygon(c(.5, .5, .75, .8), c(.25, .3, .4, .5))
If the device is an on-screen device, can I check whether a particular area
has anything drawn on it, or, to be precise, whether the color of a
particular area has all pixels equal to the background color. That
Sorry that I did not make my question clear enough.
If the device is file based (e.g., a PNG file), then I believe I can do
that, by using functions that can inspect an image. This is the solution I
originally wanted to try. However, it requires creating a file in the
process.
I would like to see
This question is not clear to me. What is it you hope to retrieve from the
device?
Note that the type of device in your example is system-dependent. The content
in a png() would be different than the content in a win.graph() device.
On October 15, 2023 8:04:00 AM PDT, Shu Fai Cheung
wrote:
>H
box(col='grey')
> > axis(1)
> >
> > Best,
> > Wolfgang
>
>
> That's exactly what I needed.
>
>
> Thanks, Spencer
> >
>
> >> -Original Message-
> >> From: R-help [mailto:r-help-boun...@r-project.org] On Behal
X.[, 3], ylab='accel (%)', axes=FALSE)
axis(2)
box(col='grey')
axis(1)
Best,
Wolfgang
That's exactly what I needed.
Thanks, Spencer
-Original Message-
From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Spencer Graves
Sent: Wednesday, 31 Ma
box(col='grey')
axis(1)
Best,
Wolfgang
>-Original Message-
>From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Spencer Graves
>Sent: Wednesday, 31 May, 2023 17:45
>To: Eric Berger
>Cc: r-help
>Subject: Re: [R] plot level, velocity, acceleration wit
On 5/31/23 9:20 AM, Eric Berger wrote:
I sent you an updated response to deal with the redundant copies of the x-axis.
Re-sending.
par(mfrow=c(3,1))
plot(DAX.[, 1], log='y', ylab='DAX', xaxt="n")
plot(DAX.[, 2], ylab='vel (%)', xaxt="n")
plot(DAX.[, 3], ylab='accel (%)')
I got that. Th
I sent you an updated response to deal with the redundant copies of the x-axis.
Re-sending.
par(mfrow=c(3,1))
plot(DAX.[, 1], log='y', ylab='DAX', xaxt="n")
plot(DAX.[, 2], ylab='vel (%)', xaxt="n")
plot(DAX.[, 3], ylab='accel (%)')
On Wed, May 31, 2023 at 4:27 PM Spencer Graves
wrote:
>
>
>
>
On 5/30/23 10:23 AM, Eric Berger wrote:
What if you just precede these commands as follows:
par(mfrow=c(3,1))
plot(DAX.[, 1], log='y', ylab='DAX')
plot(DAX.[, 2], ylab='vel (%)')
plot(DAX.[, 3], ylab='accel (%)')
Most of the space is consumed with two extraneous copies of the axis.
We
Slight modification to have the xaxt ticks and labels only appear on
the bottom chart
par(mfrow=c(3,1))
plot(DAX.[, 1], log='y', ylab='DAX', xaxt="n")
plot(DAX.[, 2], ylab='vel (%)', xaxt="n")
plot(DAX.[, 3], ylab='accel (%)')
On Tue, May 30, 2023 at 6:23 PM Eric Berger wrote:
>
> What if you j
What if you just precede these commands as follows:
par(mfrow=c(3,1))
plot(DAX.[, 1], log='y', ylab='DAX')
plot(DAX.[, 2], ylab='vel (%)')
plot(DAX.[, 3], ylab='accel (%)')
On Tue, May 30, 2023 at 5:45 PM Spencer Graves
wrote:
>
>
>
> On 5/30/23 8:48 AM, Eric Berger wrote:
> > I am a bit confuse
On 5/30/23 8:48 AM, Eric Berger wrote:
I am a bit confused as to what you are trying to achieve - and even
if I could guess it is not clear what the interpretation would be.
head(DAX)
1628.75 1613.63 1606.51 1621.04 1618.16 1610.61
Including the leading NA's, what would be the 6 leading te
I am a bit confused as to what you are trying to achieve - and even
if I could guess it is not clear what the interpretation would be.
> head(DAX)
1628.75 1613.63 1606.51 1621.04 1618.16 1610.61
Including the leading NA's, what would be the 6 leading terms of the 3
series that you want to plot,
a
On 5/30/23 6:16 AM, Eric Berger wrote:
My code assumes that DAX is a ts object, as in your original post.
On Tue, May 30, 2023 at 2:06 PM Eric Berger wrote:
Untested but why not
a <- cbind(log(DAX), exp(diff(log(DAX))), exp(diff(diff(log(DAX)
colnames(a) <- c("logDAX", "vel", "accel")
My code assumes that DAX is a ts object, as in your original post.
On Tue, May 30, 2023 at 2:06 PM Eric Berger wrote:
>
> Untested but why not
>
> a <- cbind(log(DAX), exp(diff(log(DAX))), exp(diff(diff(log(DAX)
> colnames(a) <- c("logDAX", "vel", "accel")
> plot(a)
>
>
> On Tue, May 30, 2023
Untested but why not
a <- cbind(log(DAX), exp(diff(log(DAX))), exp(diff(diff(log(DAX)
colnames(a) <- c("logDAX", "vel", "accel")
plot(a)
On Tue, May 30, 2023 at 1:46 PM Spencer Graves
wrote:
>
>
>
> On 5/29/23 2:37 AM, Eric Berger wrote:
> > How about this:
> >
> > a <- cbind(AirPassengers,
On 5/29/23 2:37 AM, Eric Berger wrote:
How about this:
a <- cbind(AirPassengers, diff(log(AirPassengers)),
diff(diff(log(AirPassengers
colnames(a)[2:3] <- c("percent increase", "acceleration")
plot(a, xlab="year", main="AirPassengers")
My real problem is more difficult: I'm analyzi
On 5/29/23 2:37 AM, Eric Berger wrote:
How about this:
a <- cbind(AirPassengers, diff(log(AirPassengers)),
diff(diff(log(AirPassengers
colnames(a)[2:3] <- c("percent increase", "acceleration")
plot(a, xlab="year", main="AirPassengers")
That's it. Thanks. sg
HTH,
Eric
On Mon, May
How about this:
a <- cbind(AirPassengers, diff(log(AirPassengers)),
diff(diff(log(AirPassengers
colnames(a)[2:3] <- c("percent increase", "acceleration")
plot(a, xlab="year", main="AirPassengers")
HTH,
Eric
On Mon, May 29, 2023 at 7:57 AM Spencer Graves
wrote:
>
> Hello, All:
>
>
>
Alternatively, you can put R-studio server on AWS by going to
https://www.louisaslett.com/RStudio_AMI/. -- H
On Thu, 20 Apr 2023 at 04:58, Duncan Murdoch
wrote:
> On 20/04/2023 7:43 a.m., Naresh Gurbuxani wrote:
> > In my Amazon Web Services (AWS) account, I use R via emacs launched from
> termi
On 20/04/2023 7:43 a.m., Naresh Gurbuxani wrote:
In my Amazon Web Services (AWS) account, I use R via emacs launched from
terminal. While R computations work well, viewing graphs is inconvenient. I am
not able to use screen device. I can send graphs to a png or pdf file, then
open the file.
A number of problems.
The variable names are not helpful. PointEstx is not a point it is a value. I
need an x and a y coordinate for a point.
row1 is not defined
While there is a function data.frame(), there is no data_frame(). Making this
change gives an error that row1 is not defined. I solved
Please run your code before you send it. Your example is not reproducible.
library(ggplot2)
linedata<- data.frame(
PointEstx = c( 1, 2 )
, PointEsty = c( 1, 1.5 )
)
# make sure we have a data frame
str(linedata)
#plot the data
ggplot(
linedata
, aes( x = PointEstx
, y = PointEsty
)
Hi ektaraK,
Here is a step by step way to create an example and get your plot:
# make up a lot of dates
dates<-as.Date("2021-1-1")+sample(0:364,100)
# make up the temperatures
temps<-runif(100,0,40)
# create a data frame
mydf<-data.frame(date=dates,temp=temps)
# create a month variable
mydf$month<
I suggest you post this in the r-sig-geo list rather than here. The
expertise you seek is more likely to be there.
Bert Gunter
"The trouble with having an open mind is that people keep coming along and
sticking things into it."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
Hello,
I don't know if the following is what you want but it gives confidence
bars. The error in your code is to have Esp as id.
dat$Id <- as.integer(factor(dat$Esp))
m5 <- geeglm(
formula = tim ~ Pa*Pt,
family = Gamma(link = log),
data = dat,
id = Id,
corstr = "exchangeable"
)
plo
Hi
For your example
sel <- c(1,4, 9, 11)
text(B, C, A, col= c("black", "red")[(A %in% A[sel])+1])
gives you required colouring.Not sure if it works with basicPlotteR.
Cheers
Petr
-Original Message-
From: R-help On Behalf Of varin sacha via R-help
Sent: Wednesday, May 19, 2021 3:14 PM
I did not know about matplot, but it did the work. using lines is my
default procedure but it can be time-consuming. Thank you all.
On Sun, Jan 24, 2021 at 6:45 PM Rui Barradas wrote:
>
> Hello,
>
> Base R:
>
> colrs <- rainbow(length(unique(ROI$Z)))
> roi_wide <- reshape(ROI, v.names = "Y", time
Hello,
Base R:
colrs <- rainbow(length(unique(ROI$Z)))
roi_wide <- reshape(ROI, v.names = "Y", timevar = "Z", idvar = "X",
direction = "wide")
matplot(roi_wide, type = "l", lwd = 3, lty = "solid", col = colrs)
Hope this helps,
Rui Barradas
Às 14:48 de 24/01/21, Luigi Marongiu escreveu:
He
No. I was wrong. ?plot.default says only one line color (the first) will be
used. So it appears that you need to use lines().
Bert Gunter
"The trouble with having an open mind is that people keep coming along and
sticking things into it."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic
1. lines() *is* base R.
2. See the col argument of ?plot.default.
Bert Gunter
"The trouble with having an open mind is that people keep coming along and
sticking things into it."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
On Sun, Jan 24, 2021 at 6:48 AM Luigi Marongiu
OK, but that's not the point of my comment.
Bert Gunter
"The trouble with having an open mind is that people keep coming along and
sticking things into it."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
On Fri, Jan 22, 2021 at 5:03 PM Abby Spurdle wrote:
> Sorry, Bert.
>
Sorry, Bert.
The fitdistr function estimates parameters via maximum likelihood.
(i.e. The "lognormal" part of this, is not a kernel).
On Fri, Jan 22, 2021 at 5:14 AM Bert Gunter wrote:
>
> In future, you should try to search before posting. I realize that getting
> good search terms can sometime
Hello,
A solution based on Marc's first one, maybe easier? (It doesn't rely on
multiplying the dlnorm values by 400 since it plots the histogram with
freq = FALSE.)
set.seed(2020)
data <- rlnorm(100, meanlog = 1, sdlog = 1)
library(MASS)
f <- fitdistr(data, "lognormal")
f$estimate
p <- pr
In future, you should try to search before posting. I realize that getting
good search terms can sometimes be tricky, but not in this case: 'plot
density with histogram in R' or similar on rseek.org or just on your usual
search platform brought up several ways to do it.
As a (slightly offtopic) si
Two solutions not exactly equivalent ;
data <- rlnorm(100, meanlog = 1, sdlog = 1)
histdata <- hist(data, ylim=c(0, 100))
library(MASS)
f <- fitdistr(data, "lognormal")
f$estimate
lines(x=seq(from=0, to=50, by=0.1),
� y=dlnorm(x=seq(from=0, to=50, by=0.1), meanlog =
f$estimate["meanlog"], sdl
Thank you, all solutions work!
On Fri, Aug 28, 2020 at 9:02 AM Deepayan Sarkar
wrote:
>
> On Thu, Aug 27, 2020 at 5:46 PM Luigi Marongiu
> wrote:
> >
> > Hello,
> > I have a dataframe as follows:
> > ```
> > x = c("0 pmol", "10 pmol", "100 pmol", "1000 pmol")
> > y = c(0.9306, 1.8906, 2.2396, 2
On Thu, Aug 27, 2020 at 5:46 PM Luigi Marongiu wrote:
>
> Hello,
> I have a dataframe as follows:
> ```
> x = c("0 pmol", "10 pmol", "100 pmol", "1000 pmol")
> y = c(0.9306, 1.8906, 2.2396, 2.7917)
> df = data.frame(x, y)
>
> > str(df)
> 'data.frame': 4 obs. of 2 variables:
> $ x: chr "0 pmol"
Hi Luigi,
Maybe just:
plot(as.numeric(factor(x,levels=x)),y,xaxt="n",
main="Concentration by effect",
xlab="Concentration",ylab="Effect")
axis(1,at=1:4,labels=x)
Jim
On Thu, Aug 27, 2020 at 10:16 PM Luigi Marongiu
wrote:
>
> Hello,
> I have a dataframe as follows:
> ```
> x = c("0 pmol", "10
Hello,
The plots that you say give bars (or my equivalent version below) don't
give bars, what they give are boxplots with just one value and the
median Q1 and Q3 are all equal.
plot(y ~ factor(x), df, pch = 16) # boxplot
Is the following what you are looking for?
plot(y ~ as.integer(fac
Thank you, better than before...
On Thu, Aug 27, 2020 at 2:37 PM PIKAL Petr wrote:
>
> Hi.
> It is probably somewhere in docs, but factors are actually numerics vith
> labels.
>
> So with your original data frame
>
> df$x <- factor(df$x)
> plot(as.numeric(df$x), df$y)
>
> gives you points. You ne
Hi.
It is probably somewhere in docs, but factors are actually numerics vith
labels.
So with your original data frame
df$x <- factor(df$x)
plot(as.numeric(df$x), df$y)
gives you points. You need to set labels to x axis though.
Cheers
Petr
> -Original Message-
> From: R-help On Behalf
On 2020-08-18 11:02 +1200, Paul Murrell wrote:
| On 18/08/20 9:54 am, Bert Gunter wrote:
| | On Mon, Aug 17, 2020 at 2:14 PM wrote:
| | |
| | | Plotmath seems to be the right way
| | | to do it. But without reading
| | | plotmath I'd have gone with this:
| | |
| | | plot(y, main=paste("data",
I think that comment is fair *on graphics devices that can handle unicode*.
So that is true for Cairo-based graphics devices, but not for the pdf()
or postscript() devices, for example.
Paul
On 18/08/20 9:54 am, Bert Gunter wrote:
"Plotmath seems to be the right way to do it."
Not sure I
"Plotmath seems to be the right way to do it. "
Not sure I agree with that. Paul Murrell put together plotmath around 2000
prior to the widescale development and adoption of the unicode standard
(corrections/modifications welcome!). So at the time, there really was no
other way to handle this for
On 2020-08-17 22:14 +0100, cpolw...@chemo.org.uk wrote:
| On 2020-08-17 03:13, Rasmus Liland wrote:
| | On Sun, Aug 16, 2020 at 3:18 PM Bert wrote:
| | |
| | | ?plotmath
| |
| | Dear John, read ?plotmath, it is
| | good, I was not aware of its
| | existence; then backquote s like
| | so:
|
| P
On 2020-08-17 03:13, Rasmus Liland wrote:
On Sun, Aug 16, 2020 at 3:18 PM Bert wrote:
| On Sun, Aug 16, 2020, 14:53 John wrote:
| |
| | I would like to make plots with
| | titles for different data sets and
| | different parameters. The first
| | title doesn't show sigma as a math
| | symbol, whi
Thanks to Dunkan, Rasmus and Bert. Will keep the very useful tips. Best!
On Sun, Aug 16, 2020 at 9:13 PM Rasmus Liland wrote:
> On Sun, Aug 16, 2020 at 3:18 PM Bert wrote:
> | On Sun, Aug 16, 2020, 14:53 John wrote:
> | |
> | | I would like to make plots with
> | | titles for different data sets
On Sun, Aug 16, 2020 at 3:18 PM Bert wrote:
| On Sun, Aug 16, 2020, 14:53 John wrote:
| |
| | I would like to make plots with
| | titles for different data sets and
| | different parameters. The first
| | title doesn't show sigma as a math
| | symbol, while the second one
| | doesn't contain
Specifically, see the "how to combine "math" and numeric variables" in the
Examples therein.
Bert Gunter
"The trouble with having an open mind is that people keep coming along and
sticking things into it."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
On Sun, Aug 16, 2020 a
?plotmath
On Sun, Aug 16, 2020, 14:53 John Smith wrote:
> Dear Helpers,
>
> I would like to make plots with titles for different data sets and
> different parameters. So a useful title should combine data name
> and parameter for clarity. The following is a simplified code example with
> two plo
You can do this with PerformanceAnalytics.
library(PerformanceAnalytics)
data(edhec)
chart.RelativePerformance(edhec, 0, legend = "topleft")
Also note that there's a finance-specific mailing list: R-SIG-Finance.
Best,
Josh
On Sun, Jul 19, 2020 at 1:46 PM Pedro páramo wrote:
>
> Hi all,
>
> I a
I understand. I Will send my Code tomorrow, I am outside home now.
Many thanks David, sorry about all.
El dom., 19 jul. 2020 21:13, David Winsemius
escribió:
>
> On 7/19/20 11:18 AM, Pedro páramo wrote:
> > Hi all,
> >
> > I am trying to make a plot based on stock market prices and the library
Hi Pedro,
Assuming that you get a vector of daily percentage changes:
dpc<-c(+3.1, -2.8, +1.7, +2.1)
get the cumulative change and add 100:
cumsum(dpc)+100
[1] 103.1 100.3 102.0 104.1
You can then plot that. As Rui noted, your mail client is inserting
invisible characters that prevent cutting a
On 7/19/20 11:18 AM, Pedro páramo wrote:
Hi all,
I am trying to make a plot based on stock market prices and the library
quantmod, imagine
BatchGetSymbols(‘^IBEX’, first.date = ‘1999-12-31’,
last.date = ‘2020-12-07’)
The thing is I want to plot a plot that for each year on 31/12/year tthe
Dear Jim,
As I pointed out yesterday, setting ylim as you suggest still results in
"0e+00" as the smallest tick mark, as it should for evenly spaced ticks.
Best,
John
> On Jul 10, 2020, at 12:13 AM, Jim Lemon wrote:
>
> Hi Luigi,
> This is a result of the "pretty" function that calculates h
Thank you!
I reckon the main problem is the large data range, anyway. I should
stick with logarithmic scales...
Best regards
Luigi
On Fri, Jul 10, 2020 at 6:14 AM Jim Lemon wrote:
>
> Hi Luigi,
> This is a result of the "pretty" function that calculates hopefully
> good looking axis ticks automat
Hi Luigi,
This is a result of the "pretty" function that calculates hopefully
good looking axis ticks automatically. You can always specify
ylim=c(1.0E09,max(Y)) if you want.
Jim
On Thu, Jul 9, 2020 at 10:59 PM Luigi Marongiu wrote:
>
> Hello,
> I have these vectors:
> ```
> X <- 1:7
> Y <- c(14
Dear Bernard,
> On Jul 9, 2020, at 10:25 AM, Bernard Comcast
> wrote:
>
> Use the xlim option in the plot function?
I think you mean ylim, but as you'll find out when you try it, you still
(reasonably) get an evenly spaced tick mark at 0:
plot(Y ~ X, ylim=c(1e9, 6e11))
The "right" thing to
Hello,
Like this?
plot(Y~X, log="y")
Hope this helps,
Rui Barradas
Às 14:59 de 09/07/20, Luigi Marongiu escreveu:
Thank you,
but why it does not work in linear? With the log scale, I know it
works but I am not looking for it; is there a way to force a linear
scale?
Regards
Luigi
On Thu, J
Use the xlim option in the plot function?
Bernard
Sent from my iPhone so please excuse the spelling!"
> On Jul 9, 2020, at 10:06 AM, Luigi Marongiu wrote:
>
> Thank you,
> but why it does not work in linear? With the log scale, I know it
> works but I am not looking for it; is there a way to f
Please consult ?axis and follow its links (e.g. "axTicks" and "pretty") for
the details of the algorithm used to construct axis annotation.
Bert Gunter
"The trouble with having an open mind is that people keep coming along and
sticking things into it."
-- Opus (aka Berkeley Breathed in his "Bloom
Dear Luigi,
> On Jul 9, 2020, at 9:59 AM, Luigi Marongiu wrote:
>
> Thank you,
> but why it does not work in linear? With the log scale, I know it
> works but I am not looking for it; is there a way to force a linear
> scale?
The scale *is* linear and the choice of tick marks, which are evenly
Thank you,
but why it does not work in linear? With the log scale, I know it
works but I am not looking for it; is there a way to force a linear
scale?
Regards
Luigi
On Thu, Jul 9, 2020 at 3:44 PM Fox, John wrote:
>
> Dear Luigi,
>
> > On Jul 9, 2020, at 8:59 AM, Luigi Marongiu wrote:
> >
> > He
Dear Luigi,
> On Jul 9, 2020, at 8:59 AM, Luigi Marongiu wrote:
>
> Hello,
> I have these vectors:
> ```
> X <- 1:7
> Y <- c(1438443863, 3910100650, 10628760108, 28891979048, 78536576706,
> 213484643920, 580311678200)
> plot(Y~X)
> ```
> The y-axis starts at 0e0, but the first value is 1.4 billi
On Fri, 27 Sep 2019, Rolf Turner wrote:
You may better off using base R graphics, unless there are considerations
which demand the use of lattice. Something like this, maybe:
plot(maxtemp ~ sampdate, data=watertemp, type="h", col="red",
)
points(maxtemp ~ sampdate, data=
On Fri, 27 Sep 2019, David Winsemius wrote:
Instead of trying to mix lattice and base functions, you might try using
the formula:
maxtemp+mintemp ~ sampdate
And then: col= c(“red”, “blue”)
David,
I certainly will. I've not needed R for projects for many months and when I
looked through prev
On 27/09/19 10:27 AM, Rich Shepard wrote:
I want to plot maximum and minimum water temperatures on the same axes and
thought I had the correct syntax:
watertemp <- read.table("../../data/hydro/water-temp.dat", header =
TRUE, sep =",")
watertemp$sampdate <- as.Date(as.character(watertemp$sam
Instead of trying to mix lattice and base functions, you might try using the
formula:
maxtemp+mintemp ~ sampdate
And then: col= c(“red”, “blue”)
Sent from my iPhone, so make sure those quotes are ordinary double quotes.
—
David
> On Sep 27, 2019, at 6:27 AM, Rich Shepard wrote:
>
> I wan
I'm not sure I understand exactly what you mean.
> I want to EXTRACT FOR ONE HAND A TABLE WITH MEAN BY CENTER, STD DESV.
Do you want a table giving the mean and sd of Cost, grouped by Center?
> I want a plot or smooth plot replicating a gauss line distribution by
> center.
Do you want one plot
Dear Sarah,
everything worked out! Thank You!!!
--
Sarah Goslee :
> Well, you don't provide a reproducible example, so there's only so
> much we can do. The help for par is a lot, but I told you which option
> to use. Did you try reading the examples for ?axis at all
Well, you don't provide a reproducible example, so there's only so
much we can do. The help for par is a lot, but I told you which option
to use.
Did you try reading the examples for ?axis at all?
plot (cox, col=1:2, xscale=1, xlab="OS", ylab="Probability", xaxt="n")
axis(1, at=seq(0, 48, by=12)
Thanks, but too hard for me
Sarah Goslee :
> You can presumably use xaxt="n" in your plot() statement (see ?par for
> details), and then use axis() to make anything you'd like (see ?axis
> for details).
--
>> Medic wrote:
>> In this code:
>> plot (cox, col=1:2, xsc
You can presumably use xaxt="n" in your plot() statement (see ?par for
details), and then use axis() to make anything you'd like (see ?axis
for details).
Sarah
On Mon, Apr 15, 2019 at 12:51 PM Medic wrote:
>
> In this code:
>
> plot (cox, col=1:2, xscale=1, xlab="OS", ylab="Probability")
>
> th
This is the wrong place for this question.
https://stat.ethz.ch/mailman/listinfo/r-sig-geo
On February 18, 2019 12:57:38 AM PST, Miluji Sb wrote:
>Dear all,
>
>I am trying to plot coordinates on a world map with Robinson CRS. While
>the
>world map is generated without any issues, when I try to p
On 12/14/18 1:44 AM, Luigi Marongiu wrote:
Thank you, that worked good. I tried to read the help for
layout/split.screen but I found it confusing.
Me too. I conjecture that *everybody* finds it confusing, except maybe
Paul M., and I'm not so sure about him! :-)
However it *is possible*
pages to understand what we are doing:
?read.table
?rownames
?as.dist
?dput
?as.numeric
?attributes
?dist
?plot
?Extract (to understand what the "$" is all about.
--------
David L Carlson
Department of Anthropology
Texas A&M University
College Station,
Below a similar example, using sf and leaflet; plotting the trajectory
on a background map.
library(leaflet)
library(sf)
library(dplyr)
# Generate example data
gen_data <- function(id, n) {
data.frame(
id = id,
date = 1:n,
lat = runif(10, min = -90, max = 90),
lon = runif(10,
Hello,
The following uses ggplot2.
First, make up a dataset, since you have not posted one.
lat0 <- 38.736946
lon0 <- -9.142685
n <- 10
set.seed(1)
Date <- seq(Sys.Date() - n + 1, Sys.Date(), by = "days")
Lat <- lat0 + cumsum(c(0, runif(n - 1)))
Lon <- lon0 + cumsum(c(0, runif(n - 1)))
Place
Hi Myriam,
This may not be the ideal way to do this, but I think it works:
mcdf<-read.table(text="41540 41540 41442 41599 41709 41823 41806 41837
41898 41848
41442 0.001
41599 0.002 0.001
41709 0.004 0.003 0.003
41823 0.002 0.001 0.002 0.001
41806 0.004 0.004 0.005 0.006 0.005
41837 0.004 0.004 0.
Probably
sort the data frame by date
Then
plot( mydf$geogr.longitude, mydf$geogr.latitude, type='l')
Search the web for some tutorials
See the help pages for
plot
plot.default
--
Don MacQueen
Lawrence Livermore National Laboratory
7000 East Ave., L-627
Livermore, CA 94550
925-423-1062
Dar Jim,
late, but it works now perfectly. I made a little function for
autoscaling elevation data.
https://stackoverflow.com/questions/50606797/plot-in-real-units-mm/52696705#52696705The
(paper-)printer typically also autoscales, which needs to be
deactivated.
Example for a plot with scale i
Pre-process your data into per-day or per-month records, then plot it. There
are many ways to do this... for example, base R has the aggregate function, and
the dplyr package has the group_by/summarise functions, and the data.table
package can do this as well (read the vignettes).
All of these
Normally, one turns off the x-axis tick marks and labels by supplying
xaxt='n' in the plot() call, and then adds a customized x-axis using the
axis() function.
But without more information, little help can be provided (a vague question
receives a vague answer).
I'd suggest reviewing the pos
om my Samsung Galaxy smartphone.
Original message From: Jim Lemon Date:
7/31/18 02:53 (GMT+03:00) To: ouedraogo_agathe
Cc: r-help mailing list Subject: Re: [R] Plot Rclimdex
or Climpact map with R
Hi Agathe,
You can start with the "maps" package:
# in an R session
i
Hi Agathe,
You can start with the "maps" package:
# in an R session
install.packages("maps")
# assume you want a simple map containing France
map("world",xlim=c(-6.0,9.6),ylim=c(42,51.5))
then plot your data by the coordinates of the stations. You will
probably want to plot graphical elements to
> Duncan Murdoch
> on Thu, 28 Jun 2018 20:57:19 -0400 writes:
> On 28/06/2018 5:29 PM, Jeff Reichman wrote:
>> R-Help
>>
>>
>>
>> Is there a way to make a rectangle transparent (alpha=0.1??)
>>
>>
>>
>> plot(c(100, 200), c(300, 450), type=
28, 2018 7:57 PM
To: reichm...@sbcglobal.net; R-help@r-project.org
Subject: Re: [R] Plot Rect Transparency
On 28/06/2018 5:29 PM, Jeff Reichman wrote:
> R-Help
>
>
>
> Is there a way to make a rectangle transparent (alpha=0.1??)
>
>
>
>plot(c(100, 200), c(3
On 28/06/2018 5:29 PM, Jeff Reichman wrote:
R-Help
Is there a way to make a rectangle transparent (alpha=0.1??)
plot(c(100, 200), c(300, 450), type= "n", xlab = "", ylab = "")
rect(110, 300, 175, 350, density = 5, border = "red")
Can't figure out how to take the changepo
How about this:
in2mm<-25.4 # scale factor to convert inches to mm
pdf("test.pdf",width=8.3,height=11.7)
pin<-par("pin")
plot(c(0,pin[1]*in2mm),c(0,pin[2]*in2mm), type="n", xaxs="i", yaxs="i")
lines(c(10,10),c(0,10))
text(11,5,"1 cm", adj=0)
lines(c(0,40),c(20,20))
text(20,24,"4 cm")
polygon(c(
Thanks a lot!
Jim Lemon schrieb am Do., 7. Juni 2018, 06:13:
> Hi Christian,
> Well, it almost worked. I suspect that the postscript device adds some
> padding to account for the printable area, so with a bit of
> experimentation, The following example seems to do what you want. When
> I printed
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