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,
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, the only thing that needs to change is
what's rig
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
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 years.
So I create a date sequence for the next 10 years and
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
Hi All,
I want to inspect the content of a plot generated by another function.
For example:
plot.new()
polygon(c(.5, .5, .75, .8), c(.25, .3, .4, .5))
A polygon will be drawn. If I do not know what has been done to generate
the plot, is it possible to query the content in the active device?
Re
Dear Ani,
A solution using my CRAN package HelpersMG; note that yellow is a
difficult color to manage correct overlay. It is easier to use red.
mean1 <-
c(122.194495954369, 118.955256282505, 115.540991140893, 113.116216840647,
111.24053267553, 109.827890459103, 108.523652505026, 107.03318302
Thank you very much for the solution. Absolutely will try this to see which
one produces better plot.
On Tue, Sep 19, 2023, 18:17 Jan van der Laan wrote:
> Shorter/simpler alternative for adding a alpha channel
>
> adjustcolor("lightblue", alpha = 0.5)
>
>
> So I would use something like:
>
>
>
Shorter/simpler alternative for adding a alpha channel
adjustcolor("lightblue", alpha = 0.5)
So I would use something like:
# Open new plot; make sure limits are ok; but don't plot
plot(0, 0, xlim=c(1,20),
ylim = range(c(mean1+sd1, mean2+sd2, mean1-sd1, mean2-sd2)),
type="n", las=1,
xla
Thank you very much for the help. Turn out I can use 'density' to
differentiate the overlaid area.
On Tue, Sep 19, 2023, 16:16 Ivan Krylov wrote:
> В Tue, 19 Sep 2023 13:21:08 +0900
> ani jaya пишет:
>
> >
> polygon(c(1:20,20:1),c(mean1[1:20]+sd1[1:20],mean1[20:1]),col="lightblue")
> >
> polygo
В Tue, 19 Sep 2023 13:21:08 +0900
ani jaya пишет:
> polygon(c(1:20,20:1),c(mean1[1:20]+sd1[1:20],mean1[20:1]),col="lightblue")
> polygon(c(1:20,20:1),c(mean1[1:20]-sd1[1:20],mean1[20:1]),col="lightblue")
> polygon(c(1:20,20:1),c(mean2[1:20]+sd2[1:20],mean2[20:1]),col="lightyellow")
> polygon(c(1:
),c(mean2[1:20]+sd2[1:20],mean2[20:1]),col="lightyellow")
polygon(c(1:20,20:1),c(mean2[1:20]-sd2[1:20],mean2[20:1]),col="lightyellow")
lines(mean1,lty=1,lwd=2,col="blue")
lines(mean2,lty=1,lwd=2,col="yellow")
what i want is like in this source, but using
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:
>
>
>
Hello, All:
I want to plot level, velocity, and acceleration in three panels with
only one x axis. The code below does this using "layout". However, I
want the three plot areas to be of equal size, and this won't do that:
If I stretch the plot vertically, the relative sizes of the three
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.
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.
I would like a setup where code is run in one wi
Of Sorkin, John
Sent: Thursday, December 8, 2022 8:06 PM
To: r-help@r-project.org (r-help@r-project.org)
Subject: [R] Plot a line using ggplot2
[External Email]
Colleagues,
I am trying to plot a simple line using ggplot2. I get the axes, but I don't
get the line. Please let me know what
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
)
Colleagues,
I am trying to plot a simple line using ggplot2. I get the axes, but I don't
get the line. Please let me know what my error I am making.
Thank you,
John
# Define x and y values
PointEstx <- Estx+1.96*SE
PointEsty <- 1
row2 <- cbind(PointEstx,PointEsty)
linedata<- data_frame(rbind(r
Try group_by() function
> Try fill= or color= in the aes() statement in ggplot.
>
> What is the goal?
> Do you want lines, bars, or dots?
>
> Is this a class?
>
> Tim
>
> -Original Message-
> From: R-help On Behalf Of ektaraful
> Sent: Saturday, February
Thank you!
On Wed, Oct 13, 2021 at 12:00 PM Deepayan Sarkar
wrote:
>
> On Wed, Oct 13, 2021 at 12:13 PM Luigi Marongiu
> wrote:
> >
> > I have seen that the only package that easily rotate the plot is
> > ggplot, so I ran:
> > ```
> > library(ggplot2)
> > df = data.frame(MR = c(negative_mr, unce
On Wed, Oct 13, 2021 at 12:13 PM Luigi Marongiu
wrote:
>
> I have seen that the only package that easily rotate the plot is
> ggplot, so I ran:
> ```
> library(ggplot2)
> df = data.frame(MR = c(negative_mr, uncertain_mr, positive_mr),
> FCN = c(negative_fcn, uncertain_fcn, positive_fcn))
> p <- gg
I have seen that the only package that easily rotate the plot is
ggplot, so I ran:
```
library(ggplot2)
df = data.frame(MR = c(negative_mr, uncertain_mr, positive_mr),
FCN = c(negative_fcn, uncertain_fcn, positive_fcn))
p <- ggplot(df, aes(x=MR)) +
geom_density()
p + coord_flip()
```
Even in this c
Hi
I don't think you need (or want) the 'gridGraphics' package for this.
You just need to flip the x-/y-data that you get from density() (and
make sure that you align the y-axes of the two plots).
The following code shows an example (and adds a bit of par() control to
eliminate white space).
I don't know the gridGraphics package, and I haven't looked closely at what
you are trying to do. But note that lattice functions construct grid
"grobs" that can be saved and plotted in arbitrary, including rotated,
viewports, using the print.trellis function. I frankly am pretty ignorant
about suc
Hello,
I would like to show a density plot of the Y axis. To do that, I would
like to split the plot into a panel 2/3 long and a density plot 1/3
long. The problem is that, since the density is on the Y axis, the
density plot should be rotated byb90 degrees. I tried with the package
gridGraphics bu
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 )
Dear R-help ML,
I have downloaded the NUTS shapefiles from here:
https://ec.europa.eu/eurostat/web/gisco/geodata/reference-data/administrative-units-statistical-units/nuts
I can load them with the following code:
library(rgdal)
shp_bdir <- [PATH TO SHAPE FILE]
layername <- "NUTS_RG_
Hello,
I have just released to CRAN the precisePlacement package which provides a
selection of tools that make it easier to place elements onto a (base R)
plot exactly where you want them. It allows users to identify points and
distances on a plot in terms of inches, pixels, margin lines, data
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
Dear all,
I want to plot a Generalized Estimating Equation (GEE) model, with
interactions and the confidence interval using a similar style to ggplot2
. I have tried several packages but all of them produce errors. My idea
I wanted to know if anyone knows a graphical package which works with GEE
sday, May 19, 2021 3:14 PM
To: r-help@r-project.org
Subject: [R] Plot with some countries in red
Dear R-experts,
Here below a toy R code example. I would like some countries (not all of them)
"Italy", "Canada", "Greece" and "Norway" to appear in red co
Dear R-experts,
Here below a toy R code example. I would like some countries (not all of them)
"Italy", "Canada", "Greece" and "Norway" to appear in red color. The others
remaining black. How can I do that without big changes in my R code ? Indeed, I
would like my R code to remain like this as
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
Hello
is it possible to color the data of a dataframe according to the
values of one column?
I have a dataframe that OI have subdivided into X and Y, with a third
Z with the sample number. I would like to plot Y~X but coloring using
Z. But I would like to use base R and not lines.
Is that possible?
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
Hi,
I would like to plot the histogram of data and fit it with a lognormal
distribution.
The ideal, would be to superimpose the fit on the histogram and write
the results of the fit on the figure.
Right now, I was able to plot the histogram and fit the density with a
lognormal, but I can't
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
f$y)
>
> gives you points. You need to set labels to x axis though.
>
> Cheers
> Petr
>
> > -Original Message-
> > From: R-help On Behalf Of Luigi Marongiu
> > Sent: Thursday, August 27, 2020 2:16 PM
> > To: r-help
> > Subject: [R
-help On Behalf Of Luigi Marongiu
> Sent: Thursday, August 27, 2020 2:16 PM
> To: r-help
> Subject: [R] plot factors with dots in R
>
> Hello,
> I have a dataframe as follows:
> ```
> x = c("0 pmol", "10 pmol", "100 pmol", "1000 pmol")
&
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" "10 pmol" "100 pmol" "1000 pmol"
$ y: num 0.931 1.891 2.24 2.792
```
I wou
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
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 plots. The first title doesn't show sigma as a math symbol, while the
sec
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
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
base is 100 and each day calvulate the cumulat
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
1 - 100 of 1969 matches
Mail list logo
