[R] draw a circle with a gradient fill
Hi all,I would like to draw a simple circle where the color gradient follows
the rule color = 1/(r^2) where r is the distance from the circle. I would also
like to add a color bar with values going from -40 to -110 (and associate those
with the color gradient that fills the circle).
So far I experiemented with draw circle
install.packages('plotrix')require(plotrix)colors<-c('#fef0d9','#fdcc8a','#fc8d59','#e34a33','#b3')plot(c(-15,15),c(-15,15),type='n')draw.circle(0,
0, 10, nv = 1000, border = NULL, col = colors[1], lty = 1, lwd =
1)draw.circle(0, 0, 8, nv = 1000, border = NULL, col = colors[2], lty = 1, lwd
= 1)draw.circle(0, 0, 6, nv = 1000, border = NULL, col = colors[3], lty = 1,
lwd = 1)draw.circle(0, 0, 4, nv = 1000, border = NULL, col = colors[4], lty =
1, lwd = 1)draw.circle(0, 0, 2, nv = 1000, border = NULL, col = colors[5], lty
= 1, lwd = 1)
but this package does not give easily color gradients and so my solutions
contains 5 same colors filled rings.
Will there be any suggested improvements on my code above?
Thanks a lotAlex
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[R] add a color bar
Hi all,the code you will find at the bottom of the screen creates a 3d diagram of antenna measurements. I am adding also to this Figure a color bar,and I wanted to ask you if I can add a color bar (which package?) that will scale as the windows is maximized. My current color bar is a bit too rought and gets pixelized each time I maximize the window. Any suggestions?I would like to thank you for your replyRegardsAlex [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] scalable and dynamic color bar
Hi,I am using rgl to plot 3d graphics. You can find below some executable code. I would like to add a color bar that scales as the window size scale. The solution I currently have gives a color bar that gets pixelated once you maximize the window. Can you suggest of alternatives? ThanksAlex [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Writing my 3D plot function
Thanks a lot for this. I never realized that my yahoo mail does not send plain
text. So this is the code I have
require("rgls")
degreeToRadian<-function(degree){
return (0.01745329252*degree)
}
turnPolarToX<-function(Amplitude,Coordinate){
return (Amplitude*cos(degreeToRadian(Coordinate)))
}
turnPolarToY<-function(Amplitude,Coordinate){
return (Amplitude*sin(degreeToRadian(Coordinate)))
}
X1<-turnPolarToX(1,1:360)
Y1<-turnPolarToY(1,1:360)
Z1<-rep(0,360)
X2<-turnPolarToX(1,1:360)
Y2<-rep(0,360)
Z2<-turnPolarToY(1,1:360)
test3<-runif(360)
X3<-rep(0,360)
Y3<-turnPolarToX(1,1:360)
Z3<-turnPolarToY(1,1:360)
Min<-min(X1,Y1,Z1,X2,Y2,Z2,X3,Y3,Z3)
Max<-max(X1,Y1,Z1,X2,Y2,Z2,X3,Y3,Z3)
plot3d(X1,Y1,Z1,xlim=c(Min,Max),ylim=c(Min,Max),zlim=c(Min,Max),box=TRUE,axe=FALSE,add=TRUE,col="red",type="l")
plot3d(X2,Y2,Z2,xlim=c(Min,Max),ylim=c(Min,Max),zlim=c(Min,Max),box=TRUE,axe=FALSE,add=TRUE,col="green",type="l")
plot3d(X3,Y3,Z3,xlim=c(Min,Max),ylim=c(Min,Max),zlim=c(Min,Max),box=TRUE,axe=TRUE,add=FALSE,col="blue",type="l")
I hope this helps
Regards
Alex
On Monday, June 26, 2017 1:46 AM, Jeff Newmiller <jdnew...@dcn.davis.ca.us>
wrote:
Please look at what I see in your code below (run-on code mush) to understand
part of why it is important for you to send your email as plain text as the
Posting Guide indicates. You might find [1] helpful.
[1] https://wiki.openstack.org/wiki/MailingListEtiquette
--
Sent from my phone. Please excuse my brevity.
On June 25, 2017 2:42:26 PM EDT, Alaios via R-help <r-help@r-project.org> wrote:
>Hi all,I had a question last week on asking for a function that will
>help me draw three different circles on x,y,z axis based on polar
>coordinates (Each X,Y,Z circle are coming from three independent
>measurements of 1-360 polar coordinates). It turned out that there is
>no such function in R and thus I am trying to write my own piece of
>code that hopefully I will be able to share. I have spent some time to
>write some code based on the rgl library (Still not 100% sure that this
>was the best option).
>My input are three polar circles X,Y,Z with a good example being the
>image
>belowhttps://www.mathworks.com/help/examples/antenna/win64/xxpolarpattern_helix.png
>
>So for X axis my input is a 2D matrix [360,2] including a single
>measurement per each polar coordinate. The first thing I tried was to
>turn my polar coordinates to cartesian ones by writing two simple
>functions. This works so far and I was able to print three simple
>circles on 3d spaceb but the problem now are the legends I need to put
>that remain on cartesian coordinates. As you can see from the code
>below all circles should have radius 1 (in terms of simplicity) but
>unfortunately I have the cartesian coordinates legends that do not help
>on reading my Figure. You can help me by executing the code below
>
>require("rgls")
>degreeToRadian<-function(degree){ return (0.01745329252*degree)}
>turnPolarToX<-function(Amplitude,Coordinate){ return
>(Amplitude*cos(degreeToRadian(Coordinate)))}
>turnPolarToY<-function(Amplitude,Coordinate){ return
>(Amplitude*sin(degreeToRadian(Coordinate)))}
># Putting the first circle on 3d space. Circle of radius
>1X1<-turnPolarToX(1,1:360)Y1<-turnPolarToY(1,1:360)Z1<-rep(0,360)
># Putting the second circle on 3d space. Circle of radius
>1X2<-turnPolarToX(1,1:360)Y2<-rep(0,360)Z2<-turnPolarToY(1,1:360)
># Putting the third circle on 3d space. Circle of radius
>1X3<-rep(0,360)Y3<-turnPolarToX(1,1:360)Z3<-turnPolarToY(1,1:360)
>Min<-min(X1,Y1,Z1,X2,Y2,Z2,X3,Y3,Z3)Max<-max(X1,Y1,Z1,X2,Y2,Z2,X3,Y3,Z3)plot3d(X1,Y1,Z1,xlim=c(Min,Max),ylim=c(Min,Max),zlim=c(Min,Max),box=TRUE,axe=FALSE,add=TRUE,col="red",type="l")plot3d(X2,Y2,Z2,xlim=c(Min,Max),ylim=c(Min,Max),zlim=c(Min,Max),box=TRUE,axe=FALSE,add=TRUE,col="green",type="l")plot3d(X3,Y3,Z3,xlim=c(Min,Max),ylim=c(Min,Max),zlim=c(Min,Max),box=TRUE,axe=TRUE,add=FALSE,col="blue",type="l")
>Would it be also possible to include an jpeg image file on a rgls plot?
>Thanks a lotRegardsAlex
>[[alternative HTML version deleted]]
>
>__
>R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
>https://stat.ethz.ch/mailman/listinfo/r-help
>PLEASE do read the posting guide
>http://www.R-project.org/posting-guide.html
>and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Writing my 3D plot function
Hi all,I had a question last week on asking for a function that will help me
draw three different circles on x,y,z axis based on polar coordinates (Each
X,Y,Z circle are coming from three independent measurements of 1-360 polar
coordinates). It turned out that there is no such function in R and thus I am
trying to write my own piece of code that hopefully I will be able to share. I
have spent some time to write some code based on the rgl library (Still not
100% sure that this was the best option).
My input are three polar circles X,Y,Z with a good example being the image
belowhttps://www.mathworks.com/help/examples/antenna/win64/xxpolarpattern_helix.png
So for X axis my input is a 2D matrix [360,2] including a single measurement
per each polar coordinate. The first thing I tried was to turn my polar
coordinates to cartesian ones by writing two simple functions. This works so
far and I was able to print three simple circles on 3d spaceb but the problem
now are the legends I need to put that remain on cartesian coordinates. As you
can see from the code below all circles should have radius 1 (in terms of
simplicity) but unfortunately I have the cartesian coordinates legends that do
not help on reading my Figure. You can help me by executing the code below
require("rgls")
degreeToRadian<-function(degree){ return (0.01745329252*degree)}
turnPolarToX<-function(Amplitude,Coordinate){ return
(Amplitude*cos(degreeToRadian(Coordinate)))}
turnPolarToY<-function(Amplitude,Coordinate){ return
(Amplitude*sin(degreeToRadian(Coordinate)))}
# Putting the first circle on 3d space. Circle of radius
1X1<-turnPolarToX(1,1:360)Y1<-turnPolarToY(1,1:360)Z1<-rep(0,360)
# Putting the second circle on 3d space. Circle of radius
1X2<-turnPolarToX(1,1:360)Y2<-rep(0,360)Z2<-turnPolarToY(1,1:360)
# Putting the third circle on 3d space. Circle of radius
1X3<-rep(0,360)Y3<-turnPolarToX(1,1:360)Z3<-turnPolarToY(1,1:360)
Min<-min(X1,Y1,Z1,X2,Y2,Z2,X3,Y3,Z3)Max<-max(X1,Y1,Z1,X2,Y2,Z2,X3,Y3,Z3)plot3d(X1,Y1,Z1,xlim=c(Min,Max),ylim=c(Min,Max),zlim=c(Min,Max),box=TRUE,axe=FALSE,add=TRUE,col="red",type="l")plot3d(X2,Y2,Z2,xlim=c(Min,Max),ylim=c(Min,Max),zlim=c(Min,Max),box=TRUE,axe=FALSE,add=TRUE,col="green",type="l")plot3d(X3,Y3,Z3,xlim=c(Min,Max),ylim=c(Min,Max),zlim=c(Min,Max),box=TRUE,axe=TRUE,add=FALSE,col="blue",type="l")
Would it be also possible to include an jpeg image file on a rgls plot?
Thanks a lotRegardsAlex
[[alternative HTML version deleted]]
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] 3D plot with coordinates
Thanks. So after searching 4 hours last night it looks like that there is no R package that can do this right now. Any other ideas or suggestions might be helpful.RegardsAlex On Wednesday, June 21, 2017 3:21 PM, Alaios via R-help <r-help@r-project.org> wrote: Thanks Duncan for the replyI can not suppress anything these are radiation pattern measurements that are typically are taken at X,Y and Z planes. See an example here, where I want to plot the measurements for the red, green and blue planes (so the image below withouth the 3d green structure inside)https://www.researchgate.net/publication/258391165/figure/fig7/AS:322947316240401@1454008048835/Radiation-pattern-of-Archimedean-spiral-antenna-a-3D-and-b-elevation-cuts-at-phi.png I am quite confident that there is a tool in R to help me do this 3D plot, and even better rotatable. Thanks for the reply to allAlex On Wednesday, June 21, 2017 1:07 PM, Duncan Murdoch <murdoch.dun...@gmail.com> wrote: On 21/06/2017 5:23 AM, Alaios via R-help wrote: > Thanks a lot for the reply.After looking at different parts of the code > today I was able to start with simple 2D polar plots as the attached pdf > file. In case the attachment is not visible I used the plot.polar function > to create something like > that.https://vijaybarve.files.wordpress.com/2013/04/polarplot-05.png > Now the idea now will be to put three of those (for X,Y,Z) in a 3d rotatable > plane. I tried the rgl function but is not clear how I can use directly polar > coordinates to draw the points at the three different planes. > Any ideas on that? You can't easily do what you're trying to do. You have 6 coordinates to display: the 3 angles and values corresponding to each of them. You need to suppress something. If the values for matching angles correspond to each other (e.g. x=23 degrees and y=23 degrees and z=23 degrees all correspond to the same observation), then I'd suggest suppressing the angles. Just do a scatterplot of the 3 corresponding values. It might make sense to join them (to make a path as the angles change), and perhaps to colour the path to indicate the angle (or plot text along the path to show it). Duncan Murdoch > Thanks a lot.RegardsAlex > > On Tuesday, June 20, 2017 9:49 PM, Uwe Ligges ><lig...@statistik.tu-dortmund.de> wrote: > > > 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 >> plot those on the xyz axes that have degress. >> Is there a function or library to look at R cran? The ideal will be that >> after plotting I will be able to rotate the shape. >> I would like to thank you in advance for your helpRegardsAlex >> [[alternative HTML version deleted]] >> >> __ >> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see >> https://stat.ethz.ch/mailman/listinfo/r-help >> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html >> and provide commented, minimal, self-contained, reproducible code. >> > > > > > > > __ > R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] 3D plot with coordinates
Thanks Duncan for the replyI can not suppress anything these are radiation pattern measurements that are typically are taken at X,Y and Z planes. See an example here, where I want to plot the measurements for the red, green and blue planes (so the image below withouth the 3d green structure inside)https://www.researchgate.net/publication/258391165/figure/fig7/AS:322947316240401@1454008048835/Radiation-pattern-of-Archimedean-spiral-antenna-a-3D-and-b-elevation-cuts-at-phi.png I am quite confident that there is a tool in R to help me do this 3D plot, and even better rotatable. Thanks for the reply to allAlex On Wednesday, June 21, 2017 1:07 PM, Duncan Murdoch <murdoch.dun...@gmail.com> wrote: On 21/06/2017 5:23 AM, Alaios via R-help wrote: > Thanks a lot for the reply.After looking at different parts of the code > today I was able to start with simple 2D polar plots as the attached pdf > file. In case the attachment is not visible I used the plot.polar function > to create something like > that.https://vijaybarve.files.wordpress.com/2013/04/polarplot-05.png > Now the idea now will be to put three of those (for X,Y,Z) in a 3d rotatable > plane. I tried the rgl function but is not clear how I can use directly polar > coordinates to draw the points at the three different planes. > Any ideas on that? You can't easily do what you're trying to do. You have 6 coordinates to display: the 3 angles and values corresponding to each of them. You need to suppress something. If the values for matching angles correspond to each other (e.g. x=23 degrees and y=23 degrees and z=23 degrees all correspond to the same observation), then I'd suggest suppressing the angles. Just do a scatterplot of the 3 corresponding values. It might make sense to join them (to make a path as the angles change), and perhaps to colour the path to indicate the angle (or plot text along the path to show it). Duncan Murdoch > Thanks a lot.RegardsAlex > > On Tuesday, June 20, 2017 9:49 PM, Uwe Ligges ><lig...@statistik.tu-dortmund.de> wrote: > > > 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 >> plot those on the xyz axes that have degress. >> Is there a function or library to look at R cran? The ideal will be that >> after plotting I will be able to rotate the shape. >> I would like to thank you in advance for your helpRegardsAlex >> [[alternative HTML version deleted]] >> >> __ >> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see >> https://stat.ethz.ch/mailman/listinfo/r-help >> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html >> and provide commented, minimal, self-contained, reproducible code. >> > > > > > > > __ > R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] 3D plot with coordinates
Thanks a lot for the reply.After looking at different parts of the code today I was able to start with simple 2D polar plots as the attached pdf file. In case the attachment is not visible I used the plot.polar function to create something like that.https://vijaybarve.files.wordpress.com/2013/04/polarplot-05.png Now the idea now will be to put three of those (for X,Y,Z) in a 3d rotatable plane. I tried the rgl function but is not clear how I can use directly polar coordinates to draw the points at the three different planes. Any ideas on that? Thanks a lot.RegardsAlex On Tuesday, June 20, 2017 9:49 PM, Uwe Ligges <lig...@statistik.tu-dortmund.de> wrote: 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 plot those on > the xyz axes that have degress. > Is there a function or library to look at R cran? The ideal will be that > after plotting I will be able to rotate the shape. > I would like to thank you in advance for your helpRegardsAlex > [[alternative HTML version deleted]] > > __ > R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > XNoSink.pdf Description: Adobe PDF document __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] 3D plot with coordinates
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 look at R cran? The ideal will be that after plotting I will be able to rotate the shape. I would like to thank you in advance for your helpRegardsAlex [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How these Plots are called? Which package
can you see it now? I have uploaded it on my dropbox https://www.dropbox.com/s/9eikpabu6xflasa/Figure.jpg?dl=0 On Saturday, January 14, 2017 12:57 PM, John Kane <jrkrid...@yahoo.ca> wrote: No sign of attachment. On Saturday, January 14, 2017 5:42 AM, Alaios via R-help <r-help@r-project.org> wrote: Hello,how I can try something like that in R (in the attachment I am providing a sketch).Which packages would you try to use?I would like to thank you in advance for your helpRegardsAlex __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] How these Plots are called? Which package
Hello,how I can try something like that in R (in the attachment I am providing a sketch).Which packages would you try to use?I would like to thank you in advance for your helpRegardsAlex __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] why data.frame, mutate package and not lists
thanks for all the answers. I think also ggplot2 requires data.frames.If you want to add variable to data.frame you have to use attach, detach. Right?Any more links that discuss thoe two different approaches?Alex On Wednesday, September 14, 2016 5:34 PM, Bert Gunter <bgunter.4...@gmail.com> wrote: This is partially a matter of subjectve opinion, and so pointless; but I would point out that data frames are the canonical structure for a great many of R's modeling and graphics functions, e.g. lm, xyplot, etc. As for mutate() etc., that's about UI's and user friendliness, and imho my ho is meaningless. Best, Bert 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 Wed, Sep 14, 2016 at 6:01 AM, Alaios via R-help <r-help@r-project.org> wrote: > Hi all,I have seen data.frames and operations from the mutate package getting > really popular. In the last years I have been using extensively lists, is > there any reason to not use lists and use other data types for data > manipulation and storage? > Any article that describe their differences? I would like to thank you for > your replyRegardsAlex > [[alternative HTML version deleted]] > > __ > R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] why data.frame, mutate package and not lists
Hi all,I have seen data.frames and operations from the mutate package getting really popular. In the last years I have been using extensively lists, is there any reason to not use lists and use other data types for data manipulation and storage? Any article that describe their differences? I would like to thank you for your replyRegardsAlex [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] moran's I visualization example
Hi there,in case one has found a nice and easy reproducible example of a Morans'I example where neighborhoods are depicted and their calculated correlations are visible as well.Point is to make some examples that I can share with students that want to understand fast what is the notion about. RegardsAlex [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] greek characters in Figures
> install.packages("latex2exp")Installiere Paket nach
> ‘/home/apa/R/x86_64-pc-linux-gnu-library/3.2’(da ‘lib’ nicht
> spezifiziert)Warnung: kann nicht auf den Index für das Repository
> https://cran.cnr.Berkeley.edu/src/contrib zugreifen: nicht unterstütztes URL
> SchemaWarnmeldung:Paket ‘latex2exp’ ist nicht verfügbar (for R version 3.2.4
> Revised) > install.packages("latex2expr")Installiere Paket nach
> ‘/home/apa/R/x86_64-pc-linux-gnu-library/3.2’(da ‘lib’ nicht
> spezifiziert)Warnung: kann nicht auf den Index für das Repository
> https://cran.cnr.Berkeley.edu/src/contrib zugreifen: nicht unterstütztes URL
> SchemaWarnmeldung:Paket ‘latex2expr’ ist nicht verfügbar (for R version 3.2.4
> Revised)
On Monday, May 2, 2016 9:39 PM, David Winsemius <dwinsem...@comcast.net>
wrote:
> On May 2, 2016, at 10:32 AM, Alaios via R-help <r-help@r-project.org> wrote:
>
> Dear all,I am trying to write in my Figure labels short equations that
> contain greek characters
>
> For example: C(h) = sigma^2 * rho(h).
>
> I am googling it and there are many packages available but unfortunately they
> do not look available for my 3.2.4 latex version
>
> install.packages("latex2expr")
My efforts found a 'latex2exp' but no 'latex2expr'. Are you sure you are not
missplelling the package name?
> Installiere Paket nach ‘/home/apa/R/x86_64-pc-linux-gnu-library/3.2’(da ‘lib’
> nicht spezifiziert)Warnung: kann nicht auf den Index für das Repository
> https://cran.cnr.Berkeley.edu/src/contrib zugreifen: nicht unterstütztes URL
> SchemaWarnmeldung:Paket ‘latex2expr’ ist nicht verfügbar (for R version 3.2.4
> Revised)
>
> Any ideas what else I can try?I would like to thank you in advance for your
> replyRegardsAlex
> [[alternative HTML version deleted]]
>
> __
> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
David Winsemius
Alameda, CA, USA
[[alternative HTML version deleted]]
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[R] greek characters in Figures
Dear all,I am trying to write in my Figure labels short equations that contain
greek characters
For example: C(h) = sigma^2 * rho(h).
I am googling it and there are many packages available but unfortunately they
do not look available for my 3.2.4 latex version
install.packages("latex2expr")Installiere Paket nach
‘/home/apa/R/x86_64-pc-linux-gnu-library/3.2’(da ‘lib’ nicht
spezifiziert)Warnung: kann nicht auf den Index für das Repository
https://cran.cnr.Berkeley.edu/src/contrib zugreifen: nicht unterstütztes URL
SchemaWarnmeldung:Paket ‘latex2expr’ ist nicht verfügbar (for R version 3.2.4
Revised)
Any ideas what else I can try?I would like to thank you in advance for your
replyRegardsAlex
[[alternative HTML version deleted]]
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Data structure to hold the following
The tables and vectors storing the data will be used for accessing the data (sequentially is also fine) to do calculations as needed. RegardsAlex On Monday, February 15, 2016 7:17 PM, Bert Gunter <bgunter.4...@gmail.com> wrote: I would say that it depends on what you want to do with the data. Bert On Monday, February 15, 2016, Alaios via R-help <r-help@r-project.org> wrote: Dear all,I am using R to emulate radio propagation dynamics. I have 90 antennas in a region and each of these 90 antennas hold information about 36 points (these are all exactly the same and there is no need to differentiate them further) Each of these antennas now should keep information about the distances from the 36 points (each of the 90 antennas have a different distance for each of the unique 36 points), which gives use in total 90 times a 36 elements vector. Each antenna should also keep a [36,600] matrix (each matrix describes in details the relation between one of the 90 antennas and one of the 36 points and 600 elements are needed for doing that). In total that mines 90 times [36,600] matrices What is an appropriate data structure for doing that in R ? I am giving fixed numbers here but in reality the 90,36 and 600 are just examples. Variable should be used here and thus we do not talk about fixed data structures before hand. I would like to thank you for your replyRegardsAlex [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- 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 ) [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Data structure to hold the following
Dear all,I am using R to emulate radio propagation dynamics. I have 90 antennas in a region and each of these 90 antennas hold information about 36 points (these are all exactly the same and there is no need to differentiate them further) Each of these antennas now should keep information about the distances from the 36 points (each of the 90 antennas have a different distance for each of the unique 36 points), which gives use in total 90 times a 36 elements vector. Each antenna should also keep a [36,600] matrix (each matrix describes in details the relation between one of the 90 antennas and one of the 36 points and 600 elements are needed for doing that). In total that mines 90 times [36,600] matrices What is an appropriate data structure for doing that in R ? I am giving fixed numbers here but in reality the 90,36 and 600 are just examples. Variable should be used here and thus we do not talk about fixed data structures before hand. I would like to thank you for your replyRegardsAlex [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] [FORGED] find numbers that fall in a region or the next available.
Thanks. I am using distm of the geoshere package.I still wonder if there is a package that can tell me if a gps coordinate or not falls inside my area that is defined as: bbox <- c(min(PlotPoints[, 1])-0.001, min(PlotPoints[, 2])-0.001, max(PlotPoints[, 1])+0.001, max(PlotPoints[, 2])+0.001) PlotPoints are gps coordinates. That would make it sure that I have no mistakes in my code. Any ideas?Alex On Tuesday, February 2, 2016 11:33 PM, Rolf Turner <r.tur...@auckland.ac.nz> wrote: On 03/02/16 11:04, Alaios via R-help wrote: > Dear all,I have GPS coordinates (one vector for longitude and one for > latitude: GPSLong and GPSLat) of small are that is around 300meters X > 300 meters (location falls inside UK).At the same time I have two > more vectors (Longitude and Latitude) that include position of food > stores again the UK I would like to find within my 300x300 square > area which as the food stores that fall inside.I thought to try to > find which of the Longitude of the food stores fall inside my area. I > tried something the below > > Longitude[Longitude>(min(GPSLong)-0.001)&<(max(GPSLong)+0.001)] > but this returned me zero results.The next option would be the code > to return me at least the place that falls outside but still is close > to that region.'Do you have any idea how to do that and not fall back > in the time consuming look at each element iteration? > I would like to thank you for your reply You could make use of the distfun() function from the spatstat package. Represent your "small area" as an object of class "owin". The longitude and latitude coordinates will be treated as if they were Euclidean coordinates, but over distances of the order of 300 metres this should not matter much. You could of course convert your long and lat coordinates to metres, using some appropriate projection, which might make more sense in your context. cheers, Rolf Turner -- Technical Editor ANZJS Department of Statistics University of Auckland Phone: +64-9-373-7599 ext. 88276 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] [FORGED] find numbers that fall in a region or the next available.
That is a great tip thanks.That would indeed bring me points that are the closes to my area.. but if I am not wront that returns points that are part of a circle surface. It might be that I get a point that is just 50 meters outside of my map area. Is not that true? I would need after I find closest point to confirm which ones fall inside my map region. Alex On Wednesday, February 3, 2016 8:36 PM, David L Carlson <dcarl...@tamu.edu> wrote: Look at the point.in.polygon() and over() functions in package sp. - David L Carlson Department of Anthropology Texas A University College Station, TX 77840-4352 -Original Message- From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Alaios via R-help Sent: Wednesday, February 3, 2016 2:42 AM To: Rolf Turner; R-help Mailing List Subject: Re: [R] [FORGED] find numbers that fall in a region or the next available. Thanks. I am using distm of the geoshere package.I still wonder if there is a package that can tell me if a gps coordinate or not fal ls inside my area that is defined as: bbox <- c(min(PlotPoints[, 1])-0.001, min(PlotPoints[, 2])-0.001, max(PlotPoints[, 1])+0.001, max(PlotPoints[, 2])+0.001) PlotPoints are gps coordinates. That would make it sure that I have no mistakes in my code. Any ideas?Alex On Tuesday, February 2, 2016 11:33 PM, Rolf Turner <r.tur...@auckland.ac.nz> wrote: On 03/02/16 11:04, Alaios via R-help wrote: > Dear all,I have GPS coordinates (one vector for longitude and one for > latitude: GPSLong and GPSLat) of small are that is around 300meters X > 300 meters (location falls inside UK).At the same time I have two > more vectors (Longitude and Latitude) that include position of food > stores again the UK I would like to find within my 300x300 square > area which as the food stores that fall inside.I thought to try to > find which of the Longitude of the food stores fall inside my area. I > tried something the below > > Longitude[Longitude>(min(GPSLong)-0.001)&<(max(GPSLong)+0.001)] > but this returned me zero results.The next option would be the code > to return me at least the place that falls outside but still is close > to that region.'Do you have any idea how to do that and not fall back > in the time consuming look at each element iteration? > I would like to thank you for your reply You could make use of the distfun() function from the spatstat package. Represent your "small area" as an object of class "owin". The longitude and latitude coordinates will be treated as if they were Euclidean coordinates, but over distances of the order of 300 metres this should not matter much. You could of course convert your long and lat coordinates to metres, using some appropriate projection, which might make more sense in your context. cheers, Rolf Turner -- Technical Editor ANZJS Department of Statistics University of Auckland Phone: +64-9-373-7599 ext. 88276 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] find numbers that fall in a region or the next available.
Dear all,I have GPS coordinates (one vector for longitude and one for latitude: GPSLong and GPSLat) of small are that is around 300meters X 300 meters (location falls inside UK).At the same time I have two more vectors (Longitude and Latitude) that include position of food stores again the UK I would like to find within my 300x300 square area which as the food stores that fall inside.I thought to try to find which of the Longitude of the food stores fall inside my area. I tried something the below Longitude[Longitude>(min(GPSLong)-0.001)&<(max(GPSLong)+0.001)] but this returned me zero results.The next option would be the code to return me at least the place that falls outside but still is close to that region.'Do you have any idea how to do that and not fall back in the time consuming look at each element iteration? I would like to thank you for your replyRegardsAlex [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] getting values from php or javascript
Dear all,I would like to execute some php or javascripts I found on the web. see at middle of this page towards bottom http://www.howtocreate.co.uk/php/gridref.php#samples Is there any way I can call the php function for example directly from R? I would like to thank you in advance for your replyRegardsAlex [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] improve my ggplot look
I am giving to the community two examples
My code currently look like:
breaks<-c(0,0.01,0.05,0.08,0.1,0.15,0.2,0.3,0.5,0.7,0.9,1)
ggmap(mp, darken = 0) + geom_point(aes(Longitude, Latitude, colour
=Divergence), data = PlotPoints, size =10)+
scale_colour_gradientn(colours=rainbow(9),breaks=breaks)+
geom_point(aes(Longitude,Latitude),size=15,data=stationaryPoint,colour="Red",shape="X",size=15)
where I added the gradientn in attempt to specify specific limits with a
specific color for a a given range.
I am attaching two examples where you would seethat the values on the colorramp
are not the same. In the first case is the 0.20 that gets the purple number and
in the second case the 1 (the latter is not printed in the color ramp)
Can someone explain me in what small change Ineed here?
I would like to thank you for your replyRegardsAlex
On Tuesday, November 24, 2015 12:43 PM, Alaios via R-help
<r-help@r-project.org> wrote:
Dear Dennis, it would be better if not plotting the lon and lat. Keeping it
blank is better for the aesthetics of my map.
I am not sure how I can give a reproducible example herebut I want to
ggmap(mp, darken = 0) + geom_point(aes(Longitude, Latitude,
colour=Divergence), data = PlotPoints, size =10)+
scale_colour_gradient2(low=muted("red"),mid="green", high=muted("blue"),trans
="log")+geom_point(aes(Longitude,Latitude),size=15,data=stationaryPoint,colour="Red",shape="X",size=15)
log10(seq(1,20,length.out=100))
to make the color scale with fixed colors (0.3 for green for example) and
increase it in size so the numbers in the bar are still visible when the figure
is getting smaller in size
RegardsAlex
On Friday, November 20, 2015 11:02 PM, Dennis Murphy <djmu...@gmail.com>
wrote:
Hi Alex:
The documentation for ggmap tells you that its x-y coordinates are
lat-long. What did you want to plot instead?
There are several theme options for legends, as well as a few more in
the guide_legend() function. As far as the color scale, you should be
able to set it with hex codes in each map.
Since there is no reproducible example with which to work, I can't
really help/comment much further. If you come up with one, please post
it back to the group so that others can see it. Some of them have more
experience with mapping in ggplot2 than I do.
Dennis
On Thu, Nov 19, 2015 at 8:45 PM, Alaios <ala...@yahoo.com> wrote:
> dear Dennis,
> thanks for your answers.
> I have changed my code to look like
>
> ggmap(mp, darken = 0) + geom_point(aes(Longitude, Latitude, colour
> =Divergence), data = PlotPoints, size =10)+
> scale_colour_gradient2(low=muted("red"),mid="green",
> high=muted("blue"),trans =
> "log")+geom_point(aes(Longitude,Latitude),size=15,data=stationaryPoint,colour="Red",shape="X",size=15)#
> log10(seq(1,20,length.out=100))
>
>
> current issues:
> -ggmap : plots the coordinates (longitudes and latitudes like x and y
> labels), which I do not want to do
> -Increase font size in the color bar
> -the color scale as it is now is totally fine, it is more that I want to
> make the colors there fixed, since different maps have slightly different
> numbers printed there (not much but that hinders comparisons between the
> maps).
>
> I would like to thank you for your support
> Regards
> Alex
>
>
>
> On Thursday, November 19, 2015 7:13 PM, Dennis Murphy <djmu...@gmail.com>
> wrote:
>
>
> Let's try this againsorry for hitting Send inadvertently.
>
> On Thu, Nov 19, 2015 at 5:09 AM, Alaios via R-help <r-help@r-project.org>
> wrote:
>> Dear all,the following line of code
>> print me a map of an area with the points I need. I only new two minor
>> adjustments
>> ggmap(mp, darken = 0) + geom_point(aes(Longitude, Latitude, colour
>> =Error), data = PlotPoints, size = 6)+
>> scale_colour_gradient2(low=muted("red"),mid="green",
>> high=muted("blue"),trans =
>> "log")+geom_point(aes(Longitude,Latitude),data=stationaryPoint,colour="Red",shape="s",size=12)
>>
>> I want to specify my color ramp to have the specific scale
>> 0,10,040,130,4
>
> You can specify the hex codes for colors in any of the scale_colour*()
> functions.
>>
>> I also want to plot in a way where the fonts will be binger and the
>> legends as well.Any ideas how I can do that?
>
> See ?theme.
>
> Dennis
>
>
>
>> I would like to thank you for your reply
>> RegardsAlex
>>
>>
>> [[alternative HTML version deleted]]
>>
>> __
>> R-help@r-project.org mailing list
Re: [R] improve my ggplot look
Dear Dennis, it would be better if not plotting the lon and lat. Keeping it
blank is better for the aesthetics of my map.
I am not sure how I can give a reproducible example herebut I want to
ggmap(mp, darken = 0) + geom_point(aes(Longitude, Latitude,
colour=Divergence), data = PlotPoints, size =10)+
scale_colour_gradient2(low=muted("red"),mid="green", high=muted("blue"),trans
="log")+geom_point(aes(Longitude,Latitude),size=15,data=stationaryPoint,colour="Red",shape="X",size=15)
log10(seq(1,20,length.out=100))
to make the color scale with fixed colors (0.3 for green for example) and
increase it in size so the numbers in the bar are still visible when the figure
is getting smaller in size
RegardsAlex
On Friday, November 20, 2015 11:02 PM, Dennis Murphy <djmu...@gmail.com>
wrote:
Hi Alex:
The documentation for ggmap tells you that its x-y coordinates are
lat-long. What did you want to plot instead?
There are several theme options for legends, as well as a few more in
the guide_legend() function. As far as the color scale, you should be
able to set it with hex codes in each map.
Since there is no reproducible example with which to work, I can't
really help/comment much further. If you come up with one, please post
it back to the group so that others can see it. Some of them have more
experience with mapping in ggplot2 than I do.
Dennis
On Thu, Nov 19, 2015 at 8:45 PM, Alaios <ala...@yahoo.com> wrote:
> dear Dennis,
> thanks for your answers.
> I have changed my code to look like
>
> ggmap(mp, darken = 0) + geom_point(aes(Longitude, Latitude, colour
> =Divergence), data = PlotPoints, size =10)+
> scale_colour_gradient2(low=muted("red"),mid="green",
> high=muted("blue"),trans =
> "log")+geom_point(aes(Longitude,Latitude),size=15,data=stationaryPoint,colour="Red",shape="X",size=15)#
> log10(seq(1,20,length.out=100))
>
>
> current issues:
> -ggmap : plots the coordinates (longitudes and latitudes like x and y
> labels), which I do not want to do
> -Increase font size in the color bar
> -the color scale as it is now is totally fine, it is more that I want to
> make the colors there fixed, since different maps have slightly different
> numbers printed there (not much but that hinders comparisons between the
> maps).
>
> I would like to thank you for your support
> Regards
> Alex
>
>
>
> On Thursday, November 19, 2015 7:13 PM, Dennis Murphy <djmu...@gmail.com>
> wrote:
>
>
> Let's try this againsorry for hitting Send inadvertently.
>
> On Thu, Nov 19, 2015 at 5:09 AM, Alaios via R-help <r-help@r-project.org>
> wrote:
>> Dear all,the following line of code
>> print me a map of an area with the points I need. I only new two minor
>> adjustments
>> ggmap(mp, darken = 0) + geom_point(aes(Longitude, Latitude, colour
>> =Error), data = PlotPoints, size = 6)+
>> scale_colour_gradient2(low=muted("red"),mid="green",
>> high=muted("blue"),trans =
>> "log")+geom_point(aes(Longitude,Latitude),data=stationaryPoint,colour="Red",shape="s",size=12)
>>
>> I want to specify my color ramp to have the specific scale
>> 0,10,040,130,4
>
> You can specify the hex codes for colors in any of the scale_colour*()
> functions.
>>
>> I also want to plot in a way where the fonts will be binger and the
>> legends as well.Any ideas how I can do that?
>
> See ?theme.
>
> Dennis
>
>
>
>> I would like to thank you for your reply
>> RegardsAlex
>>
>>
>> [[alternative HTML version deleted]]
>>
>> __
>> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide
>> http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>
>
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] improve my ggplot look
Dear all,the following line of code
print me a map of an area with the points I need. I only new two minor
adjustments
ggmap(mp, darken = 0) + geom_point(aes(Longitude, Latitude, colour =Error),
data = PlotPoints, size = 6)+
scale_colour_gradient2(low=muted("red"),mid="green", high=muted("blue"),trans =
"log")+geom_point(aes(Longitude,Latitude),data=stationaryPoint,colour="Red",shape="s",size=12)
I want to specify my color ramp to have the specific scale
0,10,040,130,4
I also want to plot in a way where the fonts will be binger and the legends as
well.Any ideas how I can do that?
I would like to thank you for your reply
RegardsAlex
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] [FORGED] Re: merging-binning data
Thanks.That is what I want. It is more that I do not know how to read factors that these two functions return Browse[1]> y $`13.6954016405008` [1] (13.2,115] Levels: (13.2,115] (115,217] (217,318] (318,420] (420,522] $`88.5092280867206` [1] (13.2,115] Levels: (13.2,115] (115,217] (217,318] (318,420] (420,522] $`137.931810364616` [1] (115,217] Levels: (13.2,115] (115,217] (217,318] (318,420] (420,522] str(y) List of 30 $ 13.6954016405008: Factor w/ 5 levels "(13.2,115]","(115,217]",..: 1 $ 88.5092280867206: Factor w/ 5 levels "(13.2,115]","(115,217]",..: 1 $ 137.931810364616: Factor w/ 5 levels "(13.2,115]","(115,217]",..: 2 $ 138.559590072838: Factor w/ 5 levels "(13.2,115]","(115,217]",..: 2 $ 143.085897171535: Factor w/ 5 levels "(13.2,115]","(115,217]",..: 2 $ 177.678839068735: Factor w/ 5 levels "(13.2,115]","(115,217]",..: 2 $ 177.819693807561: Factor w/ 5 levels "(13.2,115]","(115,217]",..: 2 $ 184.752000138622: Factor w/ 5 levels "(13.2,115]","(115,217]",..: 2 $ 211.255591076421: Factor w/ 5 levels "(13.2,115]","(115,217]",..: 2 $ 238.951618624679: Factor w/ 5 levels "(13.2,115]","(115,217]",..: 3 $ 241.609050762905: Factor w/ 5 levels "(13.2,115]","(115,217]",..: 3 $ 252.528297510773: Factor w/ 5 levels "(13.2,115]","(115,217]",..: 3 3 3 $ 254.643586371518: Factor w/ 5 levels "(13.2,115]","(115,217]",..: 3 I need to be able to keep the items within their groups and at the same time to keep the label of the group so to be able to use it for plotting purposes. How I can do that?RegardsAlex On Wednesday, November 4, 2015 11:20 PM, Rolf Turner <r.tur...@auckland.ac.nz> wrote: I have been vaguely following this thread and have become very confused given the complications that seem to have appeared. The original question was: >>>>> On Tue, Nov 3, 2015 at 10:47 AM, Alaios via R-help <r-help@r-project.org> >>>>> wrote: >>>>>> Dear all,I am not exactly sure on what is the proper name of what I am >>>>>> trying to do. >>>>>> I have a vector that looks like Actually you appear to have a 32 x 1 *matrix* (NOT the same thing!) that looks like: >>>>>> binDistance >>>>>> [,1] >>>>>> [1,] 238.95162 >>>>>> [2,] 143.08590 >>>>>> [3,] 88.50923 >>>>>> [4,] 177.67884 >>>>>> [5,] 277.54116 >>>>>> [6,] 342.94689 >>>>>> [7,] 241.60905 >>>>>> [8,] 177.81969 >>>>>> [9,] 211.25559 >>>>>> [10,] 279.72702 >>>>>> [11,] 381.95738 >>>>>> [12,] 483.76363 >>>>>> [13,] 480.98841 >>>>>> [14,] 369.75241 >>>>>> [15,] 267.73650 >>>>>> [16,] 138.55959 >>>>>> [17,] 137.93181 >>>>>> [18,] 184.75200 >>>>>> [19,] 254.64359 >>>>>> [20,] 328.87785 >>>>>> [21,] 273.15577 >>>>>> [22,] 252.52830 >>>>>> [23,] 252.52830 >>>>>> [24,] 252.52830 >>>>>> [25,] 262.20084 >>>>>> [26,] 314.93064 >>>>>> [27,] 366.02996 >>>>>> [28,] 442.77467 >>>>>> [29,] 521.20323 >>>>>> [30,] 465.33071 >>>>>> [31,] 366.60582 >>>>>> [32,] 13.69540 A later addendum to the question indicated that the OP wanted labels for the result consisting of the endpoints of the intervals into which the data were subdivided. Unless I am misunderstanding, this is trivial to accomplish using cut() and split(): x <- c(238.95162, 143.0859, 88.50923, 177.67884, 277.54116, 342.94689, 241.60905, 177.81969, 211.25559, 279.72702, 381.95738, 483.76363, 480.98841, 369.75241, 267.7365, 138.55959, 137.93181, 184.752, 254.64359, 328.87785, 273.15577, 252.5283, 252.5283, 252.5283, 262.20084, 314.93064, 366.02996, 442.77467, 521.20323, 465.33071, 366.60582, 13.6954) f <- cut(x,5) y <- split(x,f) y $`(13.2,115]` [1] 88.50923 13.69540 $`(115,217]` [1] 143.0859 177.6788 177.8197 211.2556 138.5596 137.9318 184.7520 $`(217,318]` [1] 238.9516 277.5412 241.6090 279.7270 267.7365 254.6436 273.1558 252.5283 [9] 252.5283 252.5283 262.2008 314.9306 $`(318,420]` [1] 342.9469 381.9574 369.7524 328.8779 366.0300 366.6058 $`(420,522]` [1] 483.7636 480.9884 442.7747 521.2032 465.3307 Is this not the result that you want? If not, what *is* the result that you want? cheers, Rolf Turner -- Technical Editor ANZJS Department of Statistics University of Auckland Phone: +64-9-373-7599 ext. 88276 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] merging-binning data
Thanks it works great and gives me group numbers as integers and thus I can with which group the elements as needed (which (groups== 2)) Question though is how to keep also the labels for each group. For example that my first group is the [13,206) RegardsAlex On Wednesday, November 4, 2015 1:00 PM, Boris Steipe <boris.ste...@utoronto.ca> wrote: I would transform the original numbers into integers which you can use as group labels. The row numbers of the group labels are the indexes of your values. Example: assume your input vector is dBin nGroups <- 5 # number of groups groups <- (dBin - min(dBin)) / (max(dBin) - min(dBin)) # rescale to the range [0,1] groups <- floor(groups * nGroups) + 1 # discretize to nGroups integers Now you can eg. get the indices for group 2 groups[groups == 2] Depending on the nature of your input data, it may be better to keep these groups in a column adjacent to your values, rather than in a separate vector, or even better to just calculate the groups on the fly in your downstream analysis with the approach given above in a function, rather than storing them at all. These are simple operations that should not add perceptibly to execution time. Cheers, Boris On Nov 4, 2015, at 6:40 AM, Alaios via R-help <r-help@r-project.org> wrote: > Thanks for the answer. Split does not give me the indexes though but only in > which group they fall in. I also need the index of the group. Is the first, > the second .. group?Alex > > > > On Tuesday, November 3, 2015 5:05 PM, Ista Zahn <istaz...@gmail.com> wrote: > > > Probably > > split(binDistance, test). > > Best, > Ista > > On Tue, Nov 3, 2015 at 10:47 AM, Alaios via R-help <r-help@r-project.org> > wrote: >> Dear all,I am not exactly sure on what is the proper name of what I am >> trying to do. >> I have a vector that looks like >> binDistance >> [,1] >> [1,] 238.95162 >> [2,] 143.08590 >> [3,] 88.50923 >> [4,] 177.67884 >> [5,] 277.54116 >> [6,] 342.94689 >> [7,] 241.60905 >> [8,] 177.81969 >> [9,] 211.25559 >> [10,] 279.72702 >> [11,] 381.95738 >> [12,] 483.76363 >> [13,] 480.98841 >> [14,] 369.75241 >> [15,] 267.73650 >> [16,] 138.55959 >> [17,] 137.93181 >> [18,] 184.75200 >> [19,] 254.64359 >> [20,] 328.87785 >> [21,] 273.15577 >> [22,] 252.52830 >> [23,] 252.52830 >> [24,] 252.52830 >> [25,] 262.20084 >> [26,] 314.93064 >> [27,] 366.02996 >> [28,] 442.77467 >> [29,] 521.20323 >> [30,] 465.33071 >> [31,] 366.60582 >> [32,] 13.69540 >> so numbers that start from 13 and go up to maximum 522 (I have also many >> other similar sets).I want to put these numbers into 5 categories and thus I >> have tried cut >> >> >> Browse[2]> >> test<-cut(binDistance,seq(min(binDistance)-0.1,max(binDistance),length.out=scaleLength+1)) >> Browse[2]> test >> [1] (217,318] (115,217] (13.7,115] (115,217] (217,318] (318,420] >> [7] (217,318] (115,217] (115,217] (217,318] (318,420] (420,521] >> [13] (420,521] (318,420] (217,318] (115,217] (115,217] (115,217] >> [19] (217,318] (318,420] (217,318] (217,318] (217,318] (217,318] >> [25] (217,318] (217,318] (318,420] (420,521] (420,521] (420,521] >> [31] (318,420] (13.7,115] >> Levels: (13.7,115] (115,217] (217,318] (318,420] (420,521] >> >> >> I want then for the numbers of my initial vector that fall within the same >> "category" lets say the (318,420] to be collected on a vector.I rephrase it >> the indexes of my initial vector that have a value between 318 to 420 to be >> put in a same vector that I can process then as I want. >> How I can do that effectively in R? >> I would like to thank you for your replyRegardsAlex >> >> [[alternative HTML version deleted]] >> >> __ >> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see >> https://stat.ethz.ch/mailman/listinfo/r-help >> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html >> and provide commented, minimal, self-contained, reproducible code. > > > [[alternative HTML version deleted]] > > __ > R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] merging-binning data
you are right.by labels I mean the "categories", "breaks" that my data fall in.To be part of group 2 for example you have to be in the range of [110,223) I need to keep those for my plots. Did I describe it more precisely now?Alex On Wednesday, November 4, 2015 2:09 PM, Boris Steipe <boris.ste...@utoronto.ca> wrote: I don't understand: - where does the "label" come from? (It's not an element of your data that I see.) - what do you want to do with this "label" i.e. how does it need to be associated with the data? B. On Nov 4, 2015, at 7:57 AM, Alaios <ala...@yahoo.com> wrote: > Thanks it works great and gives me group numbers as integers and thus I can > with which group the elements as needed (which (groups== 2)) > > Question though is how to keep also the labels for each group. For example > that my first group is the [13,206) > > Regards > Alex > > > > On Wednesday, November 4, 2015 1:00 PM, Boris Steipe > <boris.ste...@utoronto.ca> wrote: > > > I would transform the original numbers into integers which you can use as > group labels. The row numbers of the group labels are the indexes of your > values. > > Example: assume your input vector is dBin > > nGroups <- 5 # number of groups > groups <- (dBin - min(dBin)) / (max(dBin) - min(dBin)) # rescale to the range > [0,1] > groups <- floor(groups * nGroups) + 1 # discretize to nGroups integers > > Now you can eg. get the indices for group 2 > > groups[groups == 2] > > Depending on the nature of your input data, it may be better to keep these > groups in a column adjacent to your values, rather than in a separate vector, > or even better to just calculate the groups on the fly in your downstream > analysis with the approach given above in a function, rather than storing > them at all. These are simple operations that should not add perceptibly to > execution time. > > Cheers, > Boris > > > > > > > On Nov 4, 2015, at 6:40 AM, Alaios via R-help <r-help@r-project.org> wrote: > > > Thanks for the answer. Split does not give me the indexes though but only > > in which group they fall in. I also need the index of the group. Is the > > first, the second .. group?Alex > > > > > > > > On Tuesday, November 3, 2015 5:05 PM, Ista Zahn <istaz...@gmail.com> > >wrote: > > > > > > Probably > > > > split(binDistance, test). > > > > Best, > > Ista > > > > On Tue, Nov 3, 2015 at 10:47 AM, Alaios via R-help <r-help@r-project.org> > > wrote: > >> Dear all,I am not exactly sure on what is the proper name of what I am > >> trying to do. > >> I have a vector that looks like > >> binDistance > >> [,1] > >> [1,] 238.95162 > >> [2,] 143.08590 > >> [3,] 88.50923 > >> [4,] 177.67884 > >> [5,] 277.54116 > >> [6,] 342.94689 > >> [7,] 241.60905 > >> [8,] 177.81969 > >> [9,] 211.25559 > >> [10,] 279.72702 > >> [11,] 381.95738 > >> [12,] 483.76363 > >> [13,] 480.98841 > >> [14,] 369.75241 > >> [15,] 267.73650 > >> [16,] 138.55959 > >> [17,] 137.93181 > >> [18,] 184.75200 > >> [19,] 254.64359 > >> [20,] 328.87785 > >> [21,] 273.15577 > >> [22,] 252.52830 > >> [23,] 252.52830 > >> [24,] 252.52830 > >> [25,] 262.20084 > >> [26,] 314.93064 > >> [27,] 366.02996 > >> [28,] 442.77467 > >> [29,] 521.20323 > >> [30,] 465.33071 > >> [31,] 366.60582 > >> [32,] 13.69540 > >> so numbers that start from 13 and go up to maximum 522 (I have also many > >> other similar sets).I want to put these numbers into 5 categories and thus > >> I have tried cut > >> > >> > >> Browse[2]> > >> test<-cut(binDistance,seq(min(binDistance)-0.1,max(binDistance),length.out=scaleLength+1)) > >> Browse[2]> test > >> [1] (217,318] (115,217] (13.7,115] (115,217] (217,318] (318,420] > >> [7] (217,318] (115,217] (115,217] (217,318] (318,420] (420,521] > >> [13] (420,521] (318,420] (217,318] (115,217] (115,217] (115,217] > >> [19] (217,318] (318,420] (217,318] (217,318] (217,318] (217,318] > >> [25] (217,318] (217,318] (318,420] (420,521] (420,521] (420,521] > >> [31] (318,420] (13.7,115] > >> Levels: (13.7,115] (115,217] (217,318] (318,420] (420,521] > >> > >> > >>
Re: [R] merging-binning data
Thanks for the answer. Split does not give me the indexes though but only in which group they fall in. I also need the index of the group. Is the first, the second .. group?Alex On Tuesday, November 3, 2015 5:05 PM, Ista Zahn <istaz...@gmail.com> wrote: Probably split(binDistance, test). Best, Ista On Tue, Nov 3, 2015 at 10:47 AM, Alaios via R-help <r-help@r-project.org> wrote: > Dear all,I am not exactly sure on what is the proper name of what I am trying > to do. > I have a vector that looks like > binDistance > [,1] > [1,] 238.95162 > [2,] 143.08590 > [3,] 88.50923 > [4,] 177.67884 > [5,] 277.54116 > [6,] 342.94689 > [7,] 241.60905 > [8,] 177.81969 > [9,] 211.25559 > [10,] 279.72702 > [11,] 381.95738 > [12,] 483.76363 > [13,] 480.98841 > [14,] 369.75241 > [15,] 267.73650 > [16,] 138.55959 > [17,] 137.93181 > [18,] 184.75200 > [19,] 254.64359 > [20,] 328.87785 > [21,] 273.15577 > [22,] 252.52830 > [23,] 252.52830 > [24,] 252.52830 > [25,] 262.20084 > [26,] 314.93064 > [27,] 366.02996 > [28,] 442.77467 > [29,] 521.20323 > [30,] 465.33071 > [31,] 366.60582 > [32,] 13.69540 > so numbers that start from 13 and go up to maximum 522 (I have also many > other similar sets).I want to put these numbers into 5 categories and thus I > have tried cut > > > Browse[2]> > test<-cut(binDistance,seq(min(binDistance)-0.1,max(binDistance),length.out=scaleLength+1)) > Browse[2]> test > [1] (217,318] (115,217] (13.7,115] (115,217] (217,318] (318,420] > [7] (217,318] (115,217] (115,217] (217,318] (318,420] (420,521] > [13] (420,521] (318,420] (217,318] (115,217] (115,217] (115,217] > [19] (217,318] (318,420] (217,318] (217,318] (217,318] (217,318] > [25] (217,318] (217,318] (318,420] (420,521] (420,521] (420,521] > [31] (318,420] (13.7,115] > Levels: (13.7,115] (115,217] (217,318] (318,420] (420,521] > > > I want then for the numbers of my initial vector that fall within the same > "category" lets say the (318,420] to be collected on a vector.I rephrase it > the indexes of my initial vector that have a value between 318 to 420 to be > put in a same vector that I can process then as I want. > How I can do that effectively in R? > I would like to thank you for your replyRegardsAlex > > [[alternative HTML version deleted]] > > __ > R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] merging-binning data
Thanks for your comments. Actually only the last group has a single element.
The first group is always "full" of members and as that it works fine. Some
constant spacing between the groups would be good as well and thus I will check
quantiles.
Thanks for the great support and time invested on thisRegardsAlex
On Wednesday, November 4, 2015 3:34 PM, Boris Steipe
<boris.ste...@utoronto.ca> wrote:
Whatever approach is "best" to define subsets depends completely on the
semantics of the data. Your approach (a fixed number of equally spaced breaks)
is the right one if the absolute ranges of the data is important. It should be
obvious that either the top or the bottom group could contain only a single
element, and also that any or all of the intermediate groups could be empty.
If you want to control the number of elements in your groups, use quantiles
instead.
Your application may require to define the breaks in other ways. The code I
have given you doesn't generalize well, as it depends on the equal spacing of
breaks. As I mentioned earlier, I would not store the groups at all - but would
define a function that returns a vector of elements in the group, and in the
function body I would clearly and explicitly define the conditions for group
membership (and comment it). That is how you make code for a task like this
explicit and _maintainable_.
Cheers,
Boris
On Nov 4, 2015, at 9:19 AM, Alaios <ala...@yahoo.com> wrote:
> Thanks everything is solved and I was even able to plot boxplots as needed.
> The only minor is that the max element falls in the last category and is only
> the single one element. Perhaps this can be from the way my data look like.
> Retgards
> Alex
>
>
>
> On Wednesday, November 4, 2015 3:06 PM, Boris Steipe
> <boris.ste...@utoronto.ca> wrote:
>
>
> The breaks are just the min() and max() in your groups. Something like
>
> sprintf("[%5.2f,%5.2f]", min(dBin[groups==2]), max(dBin[groups==2]))
>
> ... should achieve what you need.
>
>
> B.
>
>
>
> On Nov 4, 2015, at 8:45 AM, Alaios <ala...@yahoo.com> wrote:
>
> > you are right.
> > by labels I mean the "categories", "breaks" that my data fall in.
> > To be part of group 2 for example you have to be in the range of [110,223)
> > I need to keep those for my plots.
> >
> > Did I describe it more precisely now?
> > Alex
> >
> >
> >
> > On Wednesday, November 4, 2015 2:09 PM, Boris Steipe
> > <boris.ste...@utoronto.ca> wrote:
> >
> >
> > I don't understand:
> > - where does the "label" come from? (It's not an element of your data that
> > I see.)
> > - what do you want to do with this "label" i.e. how does it need to be
> > associated with the data?
> >
> >
> > B.
> >
> >
> >
> > On Nov 4, 2015, at 7:57 AM, Alaios <ala...@yahoo.com> wrote:
> >
> > > Thanks it works great and gives me group numbers as integers and thus I
> > > can with which group the elements as needed (which (groups== 2))
> > >
> > > Question though is how to keep also the labels for each group. For
> > > example that my first group is the [13,206)
> > >
> > > Regards
> > > Alex
> > >
> > >
> > >
> > > On Wednesday, November 4, 2015 1:00 PM, Boris Steipe
> > > <boris.ste...@utoronto.ca> wrote:
> > >
> > >
> > > I would transform the original numbers into integers which you can use as
> > > group labels. The row numbers of the group labels are the indexes of your
> > > values.
> > >
> > > Example: assume your input vector is dBin
> > >
> > > nGroups <- 5 # number of groups
> > > groups <- (dBin - min(dBin)) / (max(dBin) - min(dBin)) # rescale to the
> > > range [0,1]
> > > groups <- floor(groups * nGroups) + 1 # discretize to nGroups integers
> > >
> > > Now you can eg. get the indices for group 2
> > >
> > > groups[groups == 2]
> > >
> > > Depending on the nature of your input data, it may be better to keep
> > > these groups in a column adjacent to your values, rather than in a
> > > separate vector, or even better to just calculate the groups on the fly
> > > in your downstream analysis with the approach given above in a function,
> > > rather than storing them at all. These are simple operations that should
> > > not add perceptibly to execution time.
> > >
> > > Cheers,
> > > Bori
[R] merging-binning data
Dear all,I am not exactly sure on what is the proper name of what I am trying to do. I have a vector that looks like binDistance [,1] [1,] 238.95162 [2,] 143.08590 [3,] 88.50923 [4,] 177.67884 [5,] 277.54116 [6,] 342.94689 [7,] 241.60905 [8,] 177.81969 [9,] 211.25559 [10,] 279.72702 [11,] 381.95738 [12,] 483.76363 [13,] 480.98841 [14,] 369.75241 [15,] 267.73650 [16,] 138.55959 [17,] 137.93181 [18,] 184.75200 [19,] 254.64359 [20,] 328.87785 [21,] 273.15577 [22,] 252.52830 [23,] 252.52830 [24,] 252.52830 [25,] 262.20084 [26,] 314.93064 [27,] 366.02996 [28,] 442.77467 [29,] 521.20323 [30,] 465.33071 [31,] 366.60582 [32,] 13.69540 so numbers that start from 13 and go up to maximum 522 (I have also many other similar sets).I want to put these numbers into 5 categories and thus I have tried cut Browse[2]> test<-cut(binDistance,seq(min(binDistance)-0.1,max(binDistance),length.out=scaleLength+1)) Browse[2]> test [1] (217,318] (115,217] (13.7,115] (115,217] (217,318] (318,420] [7] (217,318] (115,217] (115,217] (217,318] (318,420] (420,521] [13] (420,521] (318,420] (217,318] (115,217] (115,217] (115,217] [19] (217,318] (318,420] (217,318] (217,318] (217,318] (217,318] [25] (217,318] (217,318] (318,420] (420,521] (420,521] (420,521] [31] (318,420] (13.7,115] Levels: (13.7,115] (115,217] (217,318] (318,420] (420,521] I want then for the numbers of my initial vector that fall within the same "category" lets say the (318,420] to be collected on a vector.I rephrase it the indexes of my initial vector that have a value between 318 to 420 to be put in a same vector that I can process then as I want. How I can do that effectively in R? I would like to thank you for your replyRegardsAlex [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] threshold and replace values in a matrix
Thanks. replace also looks to work okay. On Wednesday, October 21, 2015 3:43 PM, PIKAL Petr <petr.pi...@precheza.cz> wrote: Hi I wonder why you are asking that after quite a long use of R. test[test < (-.5)] <- (-1) Double loops seems to me the last resort in R if any other approach fails. Cheers Petr > -Original Message- > From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Alaios > via R-help > Sent: Wednesday, October 21, 2015 3:35 PM > To: R-help Mailing List > Subject: [R] threshold and replace values in a matrix > > Dear all I have a table as that. > test<-matrix(data=rnorm(100),ncol=10) > and I want to find the values that are below my thresholdthreshold<- - > 0.5and replace them with a -1 instead. > > I can of course write a double nested for loop to check one by one > elementif (test[i,j]<= threshold) test[i,j]<- -1 but that would be > rather inneficient sice I have very large tables. > Does R offer any "automation" for matrix data types? > I would like to thank you in advance for your replyRegardsAlex > > [[alternative HTML version deleted]] > > __ > R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting- > guide.html > and provide commented, minimal, self-contained, reproducible code. Tento e-mail a jakékoliv k němu připojené dokumenty jsou důvěrné a jsou určeny pouze jeho adresátům. Jestliže jste obdržel(a) tento e-mail omylem, informujte laskavě neprodleně jeho odesílatele. Obsah tohoto emailu i s přílohami a jeho kopie vymažte ze svého systému. Nejste-li zamýšleným adresátem tohoto emailu, nejste oprávněni tento email jakkoliv užívat, rozšiřovat, kopírovat či zveřejňovat. Odesílatel e-mailu neodpovídá za eventuální škodu způsobenou modifikacemi či zpožděním přenosu e-mailu. V případě, že je tento e-mail součástí obchodního jednání: - vyhrazuje si odesílatel právo ukončit kdykoliv jednání o uzavření smlouvy, a to z jakéhokoliv důvodu i bez uvedení důvodu. - a obsahuje-li nabídku, je adresát oprávněn nabídku bezodkladně přijmout; Odesílatel tohoto e-mailu (nabídky) vylučuje přijetí nabídky ze strany příjemce s dodatkem či odchylkou. - trvá odesílatel na tom, že příslušná smlouva je uzavřena teprve výslovným dosažením shody na všech jejích náležitostech. - odesílatel tohoto emailu informuje, že není oprávněn uzavírat za společnost žádné smlouvy s výjimkou případů, kdy k tomu byl písemně zmocněn nebo písemně pověřen a takové pověření nebo plná moc byly adresátovi tohoto emailu případně osobě, kterou adresát zastupuje, předloženy nebo jejich existence je adresátovi či osobě jím zastoupené známá. This e-mail and any documents attached to it may be confidential and are intended only for its intended recipients. If you received this e-mail by mistake, please immediately inform its sender. Delete the contents of this e-mail with all attachments and its copies from your system. If you are not the intended recipient of this e-mail, you are not authorized to use, disseminate, copy or disclose this e-mail in any manner. The sender of this e-mail shall not be liable for any possible damage caused by modifications of the e-mail or by delay with transfer of the email. In case that this e-mail forms part of business dealings: - the sender reserves the right to end negotiations about entering into a contract in any time, for any reason, and without stating any reasoning. - if the e-mail contains an offer, the recipient is entitled to immediately accept such offer; The sender of this e-mail (offer) excludes any acceptance of the offer on the part of the recipient containing any amendment or variation. - the sender insists on that the respective contract is concluded only upon an express mutual agreement on all its aspects. - the sender of this e-mail informs that he/she is not authorized to enter into any contracts on behalf of the company except for cases in which he/she is expressly authorized to do so in writing, and such authorization or power of attorney is submitted to the recipient or the person represented by the recipient, or the existence of such authorization is known to the recipient of the person represented by the recipient. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] threshold and replace values in a matrix
Dear all I have a table as that. test<-matrix(data=rnorm(100),ncol=10) and I want to find the values that are below my thresholdthreshold<- -0.5and replace them with a -1 instead. I can of course write a double nested for loop to check one by one elementif (test[i,j]<= threshold) test[i,j]<- -1 but that would be rather inneficient sice I have very large tables. Does R offer any "automation" for matrix data types? I would like to thank you in advance for your replyRegardsAlex [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Multinomial Fitting Distrbution
Dear all,I am trying to fit a heavy tailed distribution and I have tried working with the mix function of the mixdist package.It looks like that this package allows fitting two distributions (or move) of the same family and not combining different distributions (so mixing a geometric with a normal and so on). After trying with the mix tool all the different combinations have failed. (I am not know if images are allowed here as attachments and thus I am sharing a link with the uploaded image ) http://alexpal.smugmug.com/photos/i-N76qWsM/0/O/i-N76qWsM.jpg You can see that the three different families of distribution failed to capture correct the heavy tail at the end. Can you please sugest me which functions (package) I could try in R for combining two different distributions for a fit? I would like to thank you for your reply | | | | | | | | | | | | | | | View on alexpal.smugmug.com | Preview by Yahoo | | | | | [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] From replicate to accesing sublists
Dear all,I have a R structure that was created with replicate.The data sets looks to be a matrix with each cell being a list. str(error_suburban_0[,1],max.level=1) List of 4 $ vaR :List of 20 ..- attr(*, class)= chr variogram $ Shadowing:List of 2 ..- attr(*, class)= chr geodata $ FIT :List of 1 $ propmodel:List of 12 ..- attr(*, class)= chr lm The error_suburban is a matrix that each field so error_suburban_0[,1], error_suburban_0[,2], error_suburban_0[,3], error_suburban_0[,4],... and so on, contains the four sublists $ vaR :List of 20 ..- attr(*, class)= chr variogram $ Shadowing:List of 2 ..- attr(*, class)= chr geodata $ FIT :List of 1 $ propmodel:List of 12 ..- attr(*, class)= chr lm I would like to pick for each of these matrix elements to collect only the $Shadowing sublist error_suburban_0[,1], error_suburban_0[,2], error_suburban_0[,3], error_suburban_0[,4]... and so on Right now I am implementing this by a for loop that access each matrix element sequentially. Can you please advice me if there is a better approach to do that in R? Regards Alex [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] change language at console
Hi all,I am a linux R user and my default R environemnt (after writing R in linux console) returns the error messages in German (I am not the system adminitstrator and I can not change system settings). I know that the English package is also installed so I guess I need to set some environmental variable for this.Any idea how I am doing that? RegardsAlex [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] From replicate to accesing sublists
Thanks.The code you gave me at the end works correctly.. I was wondering if there is more efficient way to access each element withouth this classic for loop. RegardsAlex On Wednesday, April 1, 2015 5:54 PM, David Winsemius dwinsem...@comcast.net wrote: On Apr 1, 2015, at 3:34 AM, Alaios via R-help wrote: Dear all,I have a R structure that was created with replicate.The data sets looks to be a matrix with each cell being a list. str(error_suburban_0[,1],max.level=1) List of 4 $ vaR :List of 20 ..- attr(*, class)= chr variogram $ Shadowing:List of 2 ..- attr(*, class)= chr geodata $ FIT :List of 1 $ propmodel:List of 12 ..- attr(*, class)= chr lm The error_suburban is a matrix that each field so error_suburban_0[,1], error_suburban_0[,2], error_suburban_0[,3], error_suburban_0[,4],... and so on, contains the four sublists $ vaR :List of 20 ..- attr(*, class)= chr variogram $ Shadowing:List of 2 ..- attr(*, class)= chr geodata $ FIT :List of 1 $ propmodel:List of 12 ..- attr(*, class)= chr lm I would like to pick for each of these matrix elements to collect only the $Shadowing sublist error_suburban_0[,1], error_suburban_0[,2], error_suburban_0[,3], error_suburban_0[,4]... and so on Right now I am implementing this by a for loop that access each matrix element sequentially. It would have been better to show the results of dim() or dput(). Matrix objects (which are capable of holding lists) should be accessible with either a single or double argument to [. This should deliver contents: for (i in 1:4) print( error_suburban_0[i]$Shadowing ) If the matrix has 4 or more rows, then that would be accessing only from the first column. If fewer than 4 rows, you would be wrapping around to later columns. -- David. Can you please advice me if there is a better approach to do that in R? Regards Alex [[alternative HTML version deleted]] This is a plain text mailing list. Please reconfigure your email client to sent in plain text. -- David Winsemius Alameda, CA, USA [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] check a list that a sublist exists
Hi all,I have a list that has the following fields.
$`80`
[1] Error in if (fitcass1[[2]] == \Error\) { : \n Fehlender Wert, wo
TRUE/FALSE nötig ist\n
attr(,class)
[1] try-error
attr(,condition)
simpleError in if (fitcass1[[2]] == Error) { print(sprintf(error at
fitting gamma distribution with %s periods. Mean %f %f Sd %f %f, flag,
mean1, mean2, sd1, sd2))} else { return(fitcass1)}: Fehlender Wert, wo
TRUE/FALSE nötig ist
$`81`
[1] 0
$`9`
[1] 0
$`79`
$parameters
pi mu sigma
1 0.9996796725 1.654832 127.6542
2 0.0003203275 17183.001125 302.8063
$se
pi.se mu.se sigma.se
1 2.113882e-05 0.1439152 14.22274
2 2.113882e-05 38.3582148 NaN
$distribution
[1] gamma
and so one. The content of each first level sublist are never the same. I want
for each first-level sublist my list has to check fast that the current element
, lets say the 79th has the $parameters.then I would keep only the numbers from
the sublists that have this $parameters inside them and skip all the rest.
I tried something like exists(Mylist[[i]]$parameters) but it does not workI
also tried the is.numeric(Mylist[[i]]$parameters)) but this line fails when the
current sublist does not contain the $parameters field.
Can you please help me sort this out?
RegardsAlex
[[alternative HTML version deleted]]
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Re: [R] Save a list of list and search for values
Hi,thanks all for the answer.I am using mclapply to call the lapply many times
as needed. My function returns only a value if the fit is succesful.For testing
if the fit is sucessfuly my code works like that
fitcass1-tryCatch(mix(mixdat=mydataOnVector,mixpar=params,dist=distribution),error=function(e)
list(e,Error))
if (fitcass1[[2]]==Error){
print(sprintf(error at fitting
gamma distribution with %s periods. Mean %f %f Sd %f
%f,flag,mean1,mean2,sd1,sd2))
}else{
(code trunctated)... where I do some plots
so at the end of the function the code looks like
if (fitcass1[[2]]!=Error)
return(fitcass1)
then I am calling the function above with
keeptheBigListAsJimSuggested-mclapply(expandMeanSigmaOn_list,callFunctionAbove,mydataOnVector=mydataOnVector,filename=filename,mc.cores=64)
If I am not wrong that would work. I will try later after my code stops
executing.
Any more comments on this?
RegardsAlex
On Thursday, February 26, 2015 3:39 PM, jim holtman jholt...@gmail.com
wrote:
You store it as a list of lists and can then use the lapply function
to navigate for values.
result - lapply(1:1, function(x){
mix(param[x]) # whatever your call to 'mix' is with some data
})
Jim Holtman
Data Munger Guru
What is the problem that you are trying to solve?
Tell me what you want to do, not how you want to do it.
On Thu, Feb 26, 2015 at 9:27 AM, Alaios via R-help r-help@r-project.org wrote:
Dear all,in my code I am using the mix() function that returns results in a
list. The result looks like
List of 10
$ parameters :'data.frame': 2 obs. of 3 variables:
..$ pi : num [1:2] 0.77 0.23
..$ mu : num [1:2] -7034 162783
..$ sigma: num [1:2] 20235 95261
$ se :'data.frame': 2 obs. of 3 variables:
..$ pi.se : num [1:2] 0.0423 0.0423
..$ mu.se : num [1:2] 177 12422
..$ sigma.se: num [1:2] 1067 65551
$ distribution: chr norm
$ constraint :List of 8
..$ conpi : chr NONE
..$ conmu : chr NONE
..$ consigma: chr NONE
..$ fixpi : NULL
..$ fixmu : NULL
..$ fixsigma: NULL
..$ cov : NULL
..$ size : NULL
$ chisq : num 28
$ df : num 5
$ P : num 3.67e-05
$ vmat : num [1:5, 1:5] 1.79e-03 -3.69e-01 -1.17e+02 2.95e+01
-2.63e+03 ...
$ mixdata :Classes ‘mixdata’ and 'data.frame': 11 obs. of 2 variables:
..$ X : num [1:11] 1e+04 2e+04 3e+04 4e+04 5e+04 6e+04 7e+04 8e+04 9e+04
1e+05 ...
..$ count: int [1:11] 993 137 82 30 21 5 7 14 21 2 ...
$ usecondit : logi FALSE
- attr(*, class)= chr mix
In my code I am trying around 10.000 fit (and each of these fits returns the
list above) and I want to keep those in a way that later on I would be able
to search inside all the lists.For example I would like to find inside those
10.000 lists which one has the smallest $chisq value. What would be a
suitable way to implement that in R? Luckily I am working in a computer with
a lot of ram so storing 10.000 lists temporary in memory before saving to
disk would not be a problem.
What would you suggest me?
RegardsAlex
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[[alternative HTML version deleted]]
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Save a list of list and search for values
Sorry Jim I forgot to reply on what I am trying to do.I have many data sets
that contain some numbers: I am trying to fit those with a mixture
distribution. For that I am using the mix function that returns me back the
fitted parameters and the chi square (which is a first performance
indicator).After all the fits are completed and gathered on the big list, I
would like to see which are the fits with the better chi square numbers and
what are their typical numbers. After I see that results I would have to think
more on how to proceed
RegardsAlex
On Friday, February 27, 2015 1:58 PM, Alaios ala...@yahoo.com wrote:
Hi,thanks all for the answer.I am using mclapply to call the lapply many times
as needed. My function returns only a value if the fit is succesful.For testing
if the fit is sucessfuly my code works like that
fitcass1-tryCatch(mix(mixdat=mydataOnVector,mixpar=params,dist=distribution),error=function(e)
list(e,Error))
if (fitcass1[[2]]==Error){
print(sprintf(error at fitting
gamma distribution with %s periods. Mean %f %f Sd %f
%f,flag,mean1,mean2,sd1,sd2))
}else{
(code trunctated)... where I do some plots
so at the end of the function the code looks like
if (fitcass1[[2]]!=Error)
return(fitcass1)
then I am calling the function above with
keeptheBigListAsJimSuggested-mclapply(expandMeanSigmaOn_list,callFunctionAbove,mydataOnVector=mydataOnVector,filename=filename,mc.cores=64)
If I am not wrong that would work. I will try later after my code stops
executing.
Any more comments on this?
RegardsAlex
On Thursday, February 26, 2015 3:39 PM, jim holtman jholt...@gmail.com
wrote:
You store it as a list of lists and can then use the lapply function
to navigate for values.
result - lapply(1:1, function(x){
mix(param[x]) # whatever your call to 'mix' is with some data
})
Jim Holtman
Data Munger Guru
What is the problem that you are trying to solve?
Tell me what you want to do, not how you want to do it.
On Thu, Feb 26, 2015 at 9:27 AM, Alaios via R-help r-help@r-project.org wrote:
Dear all,in my code I am using the mix() function that returns results in a
list. The result looks like
List of 10
$ parameters :'data.frame': 2 obs. of 3 variables:
..$ pi : num [1:2] 0.77 0.23
..$ mu : num [1:2] -7034 162783
..$ sigma: num [1:2] 20235 95261
$ se :'data.frame': 2 obs. of 3 variables:
..$ pi.se : num [1:2] 0.0423 0.0423
..$ mu.se : num [1:2] 177 12422
..$ sigma.se: num [1:2] 1067 65551
$ distribution: chr norm
$ constraint :List of 8
..$ conpi : chr NONE
..$ conmu : chr NONE
..$ consigma: chr NONE
..$ fixpi : NULL
..$ fixmu : NULL
..$ fixsigma: NULL
..$ cov : NULL
..$ size : NULL
$ chisq : num 28
$ df : num 5
$ P : num 3.67e-05
$ vmat : num [1:5, 1:5] 1.79e-03 -3.69e-01 -1.17e+02 2.95e+01
-2.63e+03 ...
$ mixdata :Classes ‘mixdata’ and 'data.frame': 11 obs. of 2 variables:
..$ X : num [1:11] 1e+04 2e+04 3e+04 4e+04 5e+04 6e+04 7e+04 8e+04 9e+04
1e+05 ...
..$ count: int [1:11] 993 137 82 30 21 5 7 14 21 2 ...
$ usecondit : logi FALSE
- attr(*, class)= chr mix
In my code I am trying around 10.000 fit (and each of these fits returns the
list above) and I want to keep those in a way that later on I would be able
to search inside all the lists.For example I would like to find inside those
10.000 lists which one has the smallest $chisq value. What would be a
suitable way to implement that in R? Luckily I am working in a computer with
a lot of ram so storing 10.000 lists temporary in memory before saving to
disk would not be a problem.
What would you suggest me?
RegardsAlex
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[[alternative HTML version deleted]]
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Save a list of list and search for values
Dear all,in my code I am using the mix() function that returns results in a list. The result looks like List of 10 $ parameters :'data.frame': 2 obs. of 3 variables: ..$ pi : num [1:2] 0.77 0.23 ..$ mu : num [1:2] -7034 162783 ..$ sigma: num [1:2] 20235 95261 $ se :'data.frame': 2 obs. of 3 variables: ..$ pi.se : num [1:2] 0.0423 0.0423 ..$ mu.se : num [1:2] 177 12422 ..$ sigma.se: num [1:2] 1067 65551 $ distribution: chr norm $ constraint :List of 8 ..$ conpi : chr NONE ..$ conmu : chr NONE ..$ consigma: chr NONE ..$ fixpi : NULL ..$ fixmu : NULL ..$ fixsigma: NULL ..$ cov : NULL ..$ size : NULL $ chisq : num 28 $ df : num 5 $ P : num 3.67e-05 $ vmat : num [1:5, 1:5] 1.79e-03 -3.69e-01 -1.17e+02 2.95e+01 -2.63e+03 ... $ mixdata :Classes ‘mixdata’ and 'data.frame': 11 obs. of 2 variables: ..$ X : num [1:11] 1e+04 2e+04 3e+04 4e+04 5e+04 6e+04 7e+04 8e+04 9e+04 1e+05 ... ..$ count: int [1:11] 993 137 82 30 21 5 7 14 21 2 ... $ usecondit : logi FALSE - attr(*, class)= chr mix In my code I am trying around 10.000 fit (and each of these fits returns the list above) and I want to keep those in a way that later on I would be able to search inside all the lists.For example I would like to find inside those 10.000 lists which one has the smallest $chisq value. What would be a suitable way to implement that in R? Luckily I am working in a computer with a lot of ram so storing 10.000 lists temporary in memory before saving to disk would not be a problem. What would you suggest me? RegardsAlex [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] from expand.grid to a list (lapply)
Hi all,I would like to use an expand.grid functionality that would give me at the end a listso far my code looks like: sigma_max_On_seq-seq(0.05,100,length.out=1) mean_max_On_seq-seq(2,1),length.out=1) expandMeanSigmaOn-expand.grid(mean_max_On_seq,mean_max_On_seq,sigma_max_On_seq,sigma_max_On_seq) that gives me at the end all the possible combinations as a data frame.Browse[1] str((expandMeanSigmaOn)) 'data.frame': XXX obs. of 4 variables: $ Var1: num 1 11432 22864 34295 45727 ... $ Var2: num 1 1 1 1 1 1 1 1 1 1 ... $ Var3: num 0.05 0.05 0.05 0.05 0.05 0.05 0.05 0.05 0.05 0.05 ... $ Var4: num 0.05 0.05 0.05 0.05 0.05 0.05 0.05 0.05 0.05 0.05 ... - attr(*, out.attrs)=List of 2 ..$ dim : int 10 10 10 10 ..$ dimnames:List of 4 .. ..$ Var1: chr Var1= 1.00 Var1= 11432.44 Var1= 22863.89 Var1= 34295.33 ... .. ..$ Var2: chr Var2= 1.00 Var2= 11432.44 Var2= 22863.89 Var2= 34295.33 ... .. ..$ Var3: chr Var3= 0.05000 Var3= 31.48065 Var3= 62.91131 Var3= 94.34196 ... .. ..$ Var4: chr Var4= 0.05000 Var4= 31.48065 Var4= 62.91131 Var4= 94.34196 ... but I want to have a list that I would be able to use it as input into a lapply function. The lapply should use then each list entry independently giving as input variables only the current $Var1,$Var2,$Var3,$Var4 How I can convert in R the data.frame into a list? I would like to thank you in advance for your replyRegardsAlex [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Parallel for that does not keep the results
Hi all,I am working in a multi core machine and I am trying to make some
parallel code to speed up the process.
I have seen already the foreach packet but it looks like that it always combine
the results on a list. My case though is simpler since I am plotting and saving
in external files, inside the loop, and thus I do not need to keep anything
from the loop.My code looks like
expandMeanSigmaOn-expand.grid(1:100,100:200,5:10,5000:6000)
for (i in 1:length(expandMeanSigmaOn$Var1)){
mean1-expandMeanSigmaOn$Var1[i]
mean2-expandMeanSigmaOn$Var2[i]
sd1-expandMeanSigmaOn$Var3[i]
sd2-expandMeanSigmaOn$Var4[i]
fitcass1-mix(mydata,mixparam(c(mean1,mean2),(c(sd1,sd2)),gamma)))
pdf(file=paste(filename,ON.pdf,sep=));plot(fitcass1);dev.off() # plotting
and saving
save(OnFitList,file=paste(filename,ON.Rdata,sep=)) # plotting and saving
}
}
}
as you can see from the code above, given some input values I am trying some
fits, which then I am saving the results. Do you think that the foreach
packaget would be suitable (since the returning list can grow very large and
eat up memory) or should I try some alternative package?
I would like to thank you in advance for your replyRegardsAlex
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[R] save structure to be accesible later
Dear all,I have a function that returns the following list. At the end I will call my function 1000 times and I want to keep for each of these 1000 results (the structure as given below)in an order to be accessible later (Load the 1000 results and access them within a for loop for example) How should I approach the issue?I would like to thank you for your exampleRegardsAlex P,S The result of my function str(fitcass1) List of 10 $ parameters :'data.frame': 2 obs. of 3 variables: ..$ pi : num [1:2] 0.833 0.167 ..$ mu : num [1:2] 8828 11 ..$ sigma: num [1:2] 18085 1543 $ se :'data.frame': 2 obs. of 3 variables: ..$ pi.se : num [1:2] NA NA ..$ mu.se : num [1:2] NA NA ..$ sigma.se: num [1:2] NA NA $ distribution: chr gamma $ constraint :List of 8 ..$ conpi : chr NONE ..$ conmu : chr NONE ..$ consigma: chr NONE ..$ fixpi : NULL ..$ fixmu : NULL ..$ fixsigma: NULL ..$ cov : NULL ..$ size : NULL $ chisq : num 52.4 $ df : num 5 $ P : num 4.57e-10 $ vmat : num [1:5, 1:5] NA NA NA NA NA NA NA NA NA NA ... $ mixdata :Classes ‘mixdata’ and 'data.frame': 11 obs. of 2 variables: ..$ X : num [1:11] 1e+04 2e+04 3e+04 4e+04 5e+04 6e+04 7e+04 8e+04 9e+04 1e+05 ... ..$ count: int [1:11] 993 137 82 30 21 5 7 14 21 2 ... $ usecondit : logi FALSE - attr(*, class)= chr mix [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] combinations between two vectors
Hi all,I am looking for a function that would give me all the combinations between two vectors.Lets take as example the test-seq(1,3,by=5000) Browse[2] test [1] 1 5001 10001 15001 20001 25001 I want all the combinations between two times the test... I think this is called permutation so a function that could do permutation(test,test)and produce the following 1,11,50011,100011,15001 3,13,5001...25001,20001,25001,25001 is there such a function ? RegardsAlex [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] combining two distributions
(I am sorry if you have received this email twice but it does not look sent on my client) Hi all,I am having some heavy tailed data and I am trying to think of the more appropriate package for the fitting.The canonical try should be something like exponential and pareto or exponential + gamma (or gamma + gamma with different shape parameters). I am trying to have one distribution that is classical thin tailed one and another one which is more like a power law or close (pareto,gamma,weibull, log-normal). Do you have any recommendation with which packages I can start on and if the functions you have in mind they can also provide information like AIC or BIC so to help me choose the best combinations? I would like to thank you in advance for your help RegardsAlex [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] combining two distributions
Hi all,I am having some heavy tailed data and I am trying to think of the more appropriate package for the fitting.The canonical try should be something like exponential and pareto or exponential + gamma (or gamma + gamma with different shape parameters). I am trying to have one distribution that is classical thin tailed one and another one which is more like a power law or close (pareto,gamma,weibull, log-normal). Do you have any recommendation with which packages I can start on and if the functions you have in mind they can also provide information like AIC or BIC so to help me choose the best combinations? I would like to thank you in advance for your help RegardsAlex [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] combining two distributions
Hi,thanks a lot for the answer. I think that my problem is before packages. I have a large amount of different datasets and I had a random look on their histograms. I know that some of them look like the combination I have explained (exponential+pareto, exponential+gamma) but I am not sure still.I wanted to know if there is a preliminary approach where a fitting algorithms tries some of the suggested combinations and give me hints where to focus my future research.I will have a look on the package you suggested. RegardsAlex On Monday, December 8, 2014 11:23 PM, Spencer Graves spencer.gra...@prodsyse.com wrote: Have you considered distr and related packages? If this does not solve your problem, have you considered searching with findFn in the sos package? If that still does not produce sufficient enlightenment, please try this list again with commented, minimal, self-contained, reproducible code describing what you want in a bit more detail (as indicated at the end of emails on this list). Spencer On 12/8/2014 10:45 AM, Alaios via R-help wrote: (I am sorry if you have received this email twice but it does not look sent on my client) Hi all,I am having some heavy tailed data and I am trying to think of the more appropriate package for the fitting.The canonical try should be something like exponential and pareto or exponential + gamma (or gamma + gamma with different shape parameters). I am trying to have one distribution that is classical thin tailed one and another one which is more like a power law or close (pareto,gamma,weibull, log-normal). Do you have any recommendation with which packages I can start on and if the functions you have in mind they can also provide information like AIC or BIC so to help me choose the best combinations? I would like to thank you in advance for your help RegardsAlex [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] starting with mixture distribution
Hi.I was using the last night the fitdistr package to start with some fitting. I have some data sets that even though have a very gaussian distribution it looks like that also it has some very heavy tails, that can not be accurately be modelled by a gaussian distribution.Where should I give a try with mixture of distribution? RegardsAlex [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] From Strings to Variable name. When get does not work
Hi all, I would like to turn some long strings like MyString$Myfield$MySubfield into variables but it looks like that the get does not like lists so for example: test-list(a=2) test $a [1] 2 get(test) $a [1] 2 get(test$a) Fehler in get(test$a) : Objekt 'test$a' nicht gefunden lapply(test,function(x) return (x$a)) Fehler in x$a : $ operator is invalid for atomic vectors What should I do to read those lists I have as strings? I would like to thank you in advance for your reply regards A [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Read text file
Hi all I am trying to read some text files with the following format: 1377262633.948000 $GPRMC,125708.00,A,5047.66107,N,00603.65528,E,0.203,247.36,230813,,,A*60 1377262633.958000 $GPVTG,247.36,T,,M,0.203,N,0.377,K,A*3B 1377262633.968000 $GPGGA,125708.00,5047.66107,N,00603.65528,E,1,09,0.85,169.3,M,46.5,M,,*52 1377262633.978000 $GPGSA,A,3,29,21,31,25,16,05,06,13,271.78,0.85,1.57*0C 1377262633.998000 $GPGSV,3,1,12,03,01,266,,05,16,043,39,06,21,263,43,13,07,330,43*70 1377262634.008000 $GPGSV,3,2,12,16,37,302,45,18,03,149,,21,59,166,33,23,04,304,16*75 1377262634.028000 $GPGSV,3,3,12,25,18,129,21,27,11,260,39,29,45,071,47,31,35,211,47*7C but this returns me the following: read.csv(sensor_0.log,sep=,) Error in read.table(file = file, header = header, sep = sep, quote = quote, : more columns than column names I guess the problem is that the columns are not consistent on a per row basis. What I am trying to do though is to read only the lines that contain the $GPGLL or the $GPGLA entries (in the example they corresponds to 3rd and 4th line). How can I do this in R? Regards A [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Ignore errors and proceed to next
Hi all,
I have a very large number of vectors that I want first to look fast which
distribution might be considered candidate for fitting.
I made a simple loop that checks for all vector (the code below is for one
vector and being called for each vector separately). If a good fit is found
this is dumped to a txt file so to allow me later on, see results
distList-c(norm,exp,gamma,lnorm)
for (dist in distList) {
if (gofstat(fitdist(onVector,distr=dist))$kstest ==not rejected){
# keep it
out-capture.output(gofstat(fitdist(onVector,distr=distr))$ks)
print(sprintf(Saving to file %s ,filename))
cat(out,file=paste(filename,.txt,sep=),sep=\n,append=TRUE)
}
}
the major problem is that these for loops return errors (sometimes for a given
vector a specific distribution does not make sense or the vector might be zero
or containing only many times the same element) . The easiest would be in
errors and warning just move to the next element of the for loop.
Something like that
for ()
{
if error==skip to next element
else do normal stuff
}
how I can do that in R?
Regards
Alex
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Ignore errors and proceed to next
Hi thanks for the answer.
I have so vast majority of vectors that some distributions will not work for
some cases but they work for others pretty well. This is the reason I store
only those that worked to helpe get a coarse understanding how to proceed. If
you have a better idea let me know
On Friday, February 14, 2014 10:35 AM, Frede Aakmann Tøgersen
fr...@vestas.com wrote:
Hi
See
?try
?tryCatch
So you still want to do these distribution tests even though you have been
warned not to?
Yours sincerely / Med venlig hilsen
Frede Aakmann Tøgersen
Specialist, M.Sc., Ph.D.
Plant Performance Modeling
Technology Service Solutions
T +45 9730 5135
M +45 2547 6050
fr...@vestas.com
http://www.vestas.com
Company reg. name: Vestas Wind Systems A/S
This e-mail is subject to our e-mail disclaimer statement.
Please refer to www.vestas.com/legal/notice
If you have received this e-mail in error please contact the sender.
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of Alaios
Sent: 14. februar 2014 10:14
To: R-help@r-project.org
Subject: [R] Ignore errors and proceed to next
Hi all,
I have a very large number of vectors that I want first to look fast which
distribution might be considered candidate for fitting.
I made a simple loop that checks for all vector (the code below is for one
vector and being called for each vector separately). If a good fit is found
this
is dumped to a txt file so to allow me later on, see results
distList-c(norm,exp,gamma,lnorm)
for (dist in distList) {
if (gofstat(fitdist(onVector,distr=dist))$kstest ==not rejected){
# keep it
out-capture.output(gofstat(fitdist(onVector,distr=distr))$ks)
print(sprintf(Saving to file %s ,filename))
cat(out,file=paste(filename,.txt,sep=),sep=\n,append=TRUE)
}
}
the major problem is that these for loops return errors (sometimes for a
given vector a specific distribution does not make sense or the vector might
be zero or containing only many times the same element) . The easiest
would be in errors and warning just move to the next element of the for
loop.
Something like that
for ()
{
if error==skip to next element
else do normal stuff
}
how I can do that in R?
Regards
Alex
[[alternative HTML version deleted]]
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Which distribution to select (massive fitting)
Hi all, I have a large number of measurements from which I select a large number of unique vectors. For each vectors I would like to test which distribution might be a candidate for fitting. It is impossible to look on each vector separately but I can inside a for loop test different models and based on their goodness of fit to make offline decisions (I will be saving goodness of fits results on a text file). Do you know given a vector how I can get the goodness of fit for the basic distributions : norm, lnorm, pois, exp, gamma, nbinom, geom, beta, unif and logis Is it possible to try many of those (or at least some of the above) and try to get these results? Regards Alex [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Density or Boxplot with median and mean
Hi there, I would like to be able to draw a density plot or a box plot where the median and the median and the mean would be visible. If I decide a density plot I need to put two big marks one for the median and one for the mean, which I do not know how I can achieve to put marks in a density plot. For that I am using plot(density(myVector)) while on the boxplot median is already visible but mean not. To have the mean there I would have to add one more line on each boxplot, perhaps of different color but I am not sure if that is possible in R. boxplot(myVector) I am using where myVector can be something like myVector-seq(1,200) I would like to thank you all in advance Regards Alex [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Density or Boxplot with median and mean
Thanks Jim.. once again your rock On Wednesday, January 22, 2014 9:51 AM, Jim Lemon j...@bitwrit.com.au wrote: On 01/22/2014 07:37 PM, Alaios wrote: Hi there, I would like to be able to draw a density plot or a box plot where the median and the median and the mean would be visible. If I decide a density plot I need to put two big marks one for the median and one for the mean, which I do not know how I can achieve to put marks in a density plot. For that I am using plot(density(myVector)) while on the boxplot median is already visible but mean not. To have the mean there I would have to add one more line on each boxplot, perhaps of different color but I am not sure if that is possible in R. boxplot(myVector) I am using where myVector can be something like myVector-seq(1,200) Hi Alex, On a density plot you can use abline: abline(v=mean(myVector),col=red) abline(v=median(myVector),col=green) I don't know of any boxplot function that will plot two measures of central tendency, but box.heresy in plotrix will plot any one measure that you like. Jim [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Select 5 identical numbers
Dear all, I would like to select from a vector of 10 elements, 5 of those that are identical. How can I do that in R? I guess one way would be to taking random numbers and see if that appeared again. Is though there a more straightforward approach? Regards Alex [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Select 5 identical numbers
Hi, I want from a vector containing has like 1000 elements to select X of it randomly but with never selecting the same element again. Each one should be unique element of the vector. Is this more precise now? Regards Alex On Monday, January 20, 2014 7:54 PM, Alaios ala...@yahoo.com wrote: Dear all, I would like to select from a vector of 10 elements, 5 of those that are identical. How can I do that in R? I guess one way would be to taking random numbers and see if that appeared again. Is though there a more straightforward approach? Regards Alex [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] save table to txt file in a printable form
Hi there, I would like to save tabular (in R just matrices) in a txt file but in a way that they would be somehow readable. That means keeping columns aligned and rows so one can read easily for example the 2,3 element. IS there a way to do that in R? capture.output for example does not produce tables in the way I want to have those. Regards Alex [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Grid type of sampling in geodata
Hi all, I have spatial field created with grf and I was wondering if I can sample in lines, something that can resemble sampling outdoors at the streets. Random sampling looks to far of what I want to have. there is in geodata package the sample.geodata but this looks like to be random samples. I would like to thank you in advance for your help Regards A. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Colour Legend text, in a print color2D.matplot
Hi Jim, it worked nicely. par(mar) looks to control the space at the bottom left right and top.. What If I want to increase a bit the distance between the plotted matrix and the added color legend ? They look to be at my version slightly packed together. Regards Alex On Monday, November 11, 2013 3:07 AM, Alaios ala...@yahoo.com wrote: Thanks! It worked. Regards Alex On Monday, November 11, 2013 12:02 AM, Jim Lemon j...@bitwrit.com.au wrote: On 11/11/2013 04:57 AM, Alaios wrote: Hi all, I am plotting very nice looking mattrices with plotrin... so far so good, I would like though to ask you if it would be possible to add at the bottom of the color.legend (this lovely color bar that maps colors to numbers). Would that be possible to do that? Hi Alex, It is not too difficult: # first allow printing outside the plot par(xpd=TRUE) # get the x component of the label # from the values passed to color.legend # see the help page labelx-(xl+xr)/2 # get the y coordinate from the y value labely-yb-strheight(M) # display the label text(labelx,labely,mylabel) # restore the clipping par(xpd=FALSE) Jim [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Handle Gps coordinates
Dear all, I would like to ask you if there are any gps libraries. I would like to be able to handle them, -like calculate distances in meters between gps locations, -or find which gps location is closer to a list of gps locations. Is there something like that in R? I would like to tthank you in advance for your reply Regards Alex [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Colour Legend text, in a print color2D.matplot
Hi all, I am plotting very nice looking mattrices with plotrin... so far so good, I would like though to ask you if it would be possible to add at the bottom of the color.legend (this lovely color bar that maps colors to numbers). Would that be possible to do that? I would like to thank you in advance for your reply Regards Alex [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Plotrix: Add text below the color- legend.
Hi all, I am plotting very nice and sexy images with plotrix :) so far so good, I would like though to ask you if it would be possible to add at the bottom of the color.legend (this lovely color bar that maps colors to numbers). Would that be possible to do that? I would like to thank you in advance for your reply Regards Alex [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Colour Legend text, in a print color2D.matplot
Thanks! It worked. Regards Alex On Monday, November 11, 2013 12:02 AM, Jim Lemon j...@bitwrit.com.au wrote: On 11/11/2013 04:57 AM, Alaios wrote: Hi all, I am plotting very nice looking mattrices with plotrin... so far so good, I would like though to ask you if it would be possible to add at the bottom of the color.legend (this lovely color bar that maps colors to numbers). Would that be possible to do that? Hi Alex, It is not too difficult: # first allow printing outside the plot par(xpd=TRUE) # get the x component of the label # from the values passed to color.legend # see the help page labelx-(xl+xr)/2 # get the y coordinate from the y value labely-yb-strheight(M) # display the label text(labelx,labely,mylabel) # restore the clipping par(xpd=FALSE) Jim [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] convert one digit numbers to two digits one
Hi all, the following returns the hour and the minutes paste(DataSet$TimeStamps[selectedInterval$start,4], DataSet$TimeStamps[selectedInterval$start,5],sep=:) [1] 12:3 the problem is that from these two I want to create a time stamp so 12:03. The problem is that the number 3 is not converted to 03. Is there an easy way when I have one digit integer to add a zero in the front? Two digits integers are working fine so far, 12:19, or 12:45 would appear correctly I would like to thank you in advance for your help Regards Alex [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] add a color band
Hi Jim Lemon, thanks for the help, I appreciate this. right now my code looks like. par(mar=c(5,4,4,5)) color2D.matplot(data,1,c(0,1),0,xlab=,ylab=Spans, main=color.scale,xrange=c(-110,-50),border=NA,axes=F) color.legend(357,30,370,100,seq(-110,-50,length.out=13), align=rb,rect.col=color.scale(1:13,1,c(0,1),0), gradient=y) my major problem now is that the a. text in the color bar is squeezed so -50 overlaps with -60 and so on b. for some reason the color bar sometimes (the same code is called for all the data matrices I have) is misaligned in different positions each time Could you please also help me with those two? Regards Alex On Monday, October 28, 2013 9:00 AM, Alaios ala...@yahoo.com wrote: Hi Jim and thanks for your answer... I might be too tired with my new born or just exhausted. I am sharing for everyone a small data snipset that you can load https://www.dropbox.com/s/fh8jhwujgunmtrb/DataToPlotAsImage.Rdata load(DataToPlotAsImage.Rdata) require(plotrix) browser() test-data # this transforms the values of test into red-yellow color2D.matplot(test,axes=F,xlab=,ylab=,main=color.scale, extremes=c(#FF,#00),show.legend=FALSE) axis(1,at=seq(1,ncol(test),length.out=10),labels=seq(201,300,length.out=10)) color.legend(104,30,112,70,seq(-110,-30,length=11), align=rb,rect.col=color.scale(1:30,1,c(0,1),0),gradient=y) as you can see I have problems where the legend appears. My par(usr returned me par(usr) # [1] 0 351 0 200 but I am not sure how to read that to place the legend at a useful place. second I am not sure why the image is so full with black rows.. What I want is to have the legend visible and later on customize the x axis to write custom string of different size... First I need though to fix the more severe problems as I have described RegardsAlex On Sunday, October 27, 2013 12:25 PM, Jim Lemon j...@bitwrit.com.au wrote: On 10/27/2013 08:39 AM, Alaios wrote: Hi Jim and thanks for your answer... I might be too tired with my new born or just exhausted. I am attaching for everyone a small data snipset that you can load load(DataToPlotAsImage.Rdata) require(plotrix) browser() test-data # this transforms the values of test into red-yellow color2D.matplot(test,axes=F,xlab=,ylab=,main=color.scale, extremes=c(#FF,#00),show.legend=FALSE) axis(1,at=seq(1,ncol(test),length.out=10),labels=seq(201,300,length.out=10)) color.legend(104,30,112,70,seq(-110,-30,length=11), align=rb,rect.col=color.scale(1:30,1,c(0,1),0),gradient=y) as you can see I have problems where the legend appears. My par(usr returned me par(usr) # [1] 0 351 0 200 but I am not sure how to read that to place the legend at a useful place. second I am not sure why the image is so full with black rows.. What I want is to have the legend visible and later on customize the x axis to write custom string of different size... First I need though to fix the more severe problems as I have described Hi Alex, I'm not sure why you have created a copy of data to plot it. I can get quite a sensible plot using this: par(mar=c(5,4,4,5)) color2D.matplot(data,1,c(0,1),0,xlab=,ylab=, main=color.scale,xrange=c(-110,-50),border=NA) color.legend(357,30,370,100,seq(-110,-50,length.out=6), align=rb,rect.col=color.scale(1:6,1,c(0,1),0), gradient=y) Notice several things. First, when you have a large number of cells in a plot like this, setting the border to NA means that you don't get mostly borders (default = black) in the plot. The second thing is that your data range is -107.18150 to -54.07662. In order to get rounded numbers in your legend, I have set the xrange argument to -110 to -50. This gives a neat looking legend that spans your data, a bit like the pretty function would do. It also means that the color mapping is to that range and is the same in the legend as in the plot. I have left enough space on the right of the plot to fit in the legend, as that was where you said you wanted it in your last email. What par(usr) tells you is the dimensions of the plot in user units. Here it is x=0 at the left, x=351 at the right, y=0 at the bottom and y=200 at the top. Jim [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] add a color band
Hi Jim, thanks for the answer! I toyed around as you said and now they look cool and sexy! many thanks! Do you know if there is a way to compress a bit the output? The image type of functions have the useRaster parameter to set to have a compressed output. Right now the produced plots are around 400kBytes which is a lot and I am looking for ways to reduce image size. I would like to thank you for you reply Alex On Monday, November 4, 2013 11:59 AM, Jim Lemon j...@bitwrit.com.au wrote: On 11/04/2013 08:09 PM, Alaios wrote: Hi Jim Lemon, thanks for the help, I appreciate this. right now my code looks like. par(mar=c(5,4,4,5)) color2D.matplot(data,1,c(0,1),0,xlab=,ylab=Spans, main=color.scale,xrange=c(-110,-50),border=NA,axes=F) color.legend(357,30,370,100,seq(-110,-50,length.out=13), align=rb,rect.col=color.scale(1:13,1,c(0,1),0), gradient=y) my major problem now is that the a. text in the color bar is squeezed so -50 overlaps with -60 and so on b. for some reason the color bar sometimes (the same code is called for all the data matrices I have) is misaligned in different positions each time Could you please also help me with those two? Hi Alex, For your first question, I would simply extend the color legend vertically: color.legend(357,30,370,150,seq(-110,-50,length.out=13), align=rb,rect.col=color.scale(1:13,1,c(0,1),0), gradient=y) For the second one, you obviously have different dimensions for the data matrices. So, let's step through a method for getting the legend position and size from the plot itself. As I have written a few times previously, par(usr) gives you the dimensions of the plot in user units. For the example above, the dimensions were: x - 0-351 y - 0-200 With a bit of arithmetic, you can work out that the legend positions in the above are: xylim-par(usr) # x position of lower left corner of legend xl-xylim[2]+diff(xylim[1:2])*0.017 # y position of lower left corner yb-xylim[3]+diff(xylim[3:4])*0.15 # x position of upper right corner of legend xr-xylim[2]+diff(xylim[1:2])*0.054 # y position of upper right corner yt-xylim[3]+diff(xylim[3:4])*0.75 Having these lines means that you can get the position and size of the legend about right from the information provided by par(usr) even if you change the number of cells in the matrix passed to color2D.matplot. Then you would call: color.legend(xl,yb,xr,yt,seq(-110,-50,length.out=13), align=rb,rect.col=color.scale(1:13,1,c(0,1),0), gradient=y) unless of course you wanted to change the values in the scale markings. I'll leave that for you to work out. Jim [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] remap values from one vector to the other
Hi everyone, I am plotting some legend and I am using the axis(at=..) to specify the place to plot the marks I want. My plotted data have ncol(x) so the at places have values that span from 1 to ncol(x) there I would like to be able to map values that go from 880e6 to 1020e6. so 880e6 remaps to 1 1020e6 repams to ncol(x) what I do not know how to do is to remap the values that are between 880e6 and 1020e6 between then 1 and ncol(x). Can someone help me find an appropriate function for that? I would like to thank you in advance for your help Regards Alex [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Select fixed number of elements
Hi all, I have in my code some vectors that are not of equal size. I would like to be able for each of these vectors select 6 elements that are (almost) equally spaced. So the first one would be at (or close) to the beginning the last one at (or close) to the end and the other 4 equally spaced between first and last element. How can I do something like that on a vector of not known size? I would like to thank you in advance for your help Regards Alex [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] help me align the legend bar
Hi, I have some code that you can simply execute: require(plotrix) test-matrix(data=rnorm(1,-100,5),nrow=100) color2D.matplot(test,axes=F,xlab=,ylab=,main=color.scale, extremes=c(#FF,#00),show.legend=FALSE) axis(1,at=seq(1,ncol(test),length.out=10),labels=seq(201,300,length.out=10)) color.legend(84,30,125,70,seq(-110,-30,length=11), align=rb,rect.col=color.scale(1:30,1,c(0,1),0),gradient=y) What I would like to do is to make space at the right for the color band . The band should have one color from -110 to -30 with scales of 10. Can someone help me with that? Regards Alex [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] TelosB external modules, guide
Hi all, I would like to ask your help regarding connecting external modules to telosb. I have found that tiny os offers many possibilities for that as ADC, GPIOs, SPI, UART, I2C I have never learned anything regarding those. Can someone please let me know if there is any simple guide on these interfaces to show with a primitive example how to start with? I would like to thank you in advance for your help Regards Alex [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] add a color bar in a plot
Hi all, I am trying to add a color legend to my plot. As an example I am giving you a bit of code that you can run. I am sharing for everyone a small data snipset that you can load https://www.dropbox.com/s/fh8jhwujgunmtrb/DataToPlotAsImage.Rdata load(DataToPlotAsImage.Rdata) require(plotrix) browser() test-data # this transforms the values of test into red-yellow color2D.matplot(test,axes=F,xlab=,ylab=,main=color.scale, extremes=c(#FF,#00),show.legend=FALSE) axis(1,at=seq(1,ncol(test),length.out=10),labels=seq(201,300,length.out=10)) color.legend(104,30,112,70,seq(-110,-30,length=11), align=rb,rect.col=color.scale(1:30,1,c(0,1),0),gradient=y) as you can see I have problems where the legend appears. My par(usr returned me par(usr) # [1] 0 351 0 200 but I am not sure how to read that to place the legend at a useful place. second I am not sure why the image is so full with black rows.. What I want is to have the legend visible and later on customize the x axis to write custom string of different size... First I need though to fix the more severe problems as I have described RegardsAlex [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] add a color band
Hi Jim and thanks for your answer... I might be too tired with my new born or just exhausted. I am sharing for everyone a small data snipset that you can load https://www.dropbox.com/s/fh8jhwujgunmtrb/DataToPlotAsImage.Rdata load(DataToPlotAsImage.Rdata) require(plotrix) browser() test-data # this transforms the values of test into red-yellow color2D.matplot(test,axes=F,xlab=,ylab=,main=color.scale, extremes=c(#FF,#00),show.legend=FALSE) axis(1,at=seq(1,ncol(test),length.out=10),labels=seq(201,300,length.out=10)) color.legend(104,30,112,70,seq(-110,-30,length=11), align=rb,rect.col=color.scale(1:30,1,c(0,1),0),gradient=y) as you can see I have problems where the legend appears. My par(usr returned me par(usr) # [1] 0 351 0 200 but I am not sure how to read that to place the legend at a useful place. second I am not sure why the image is so full with black rows.. What I want is to have the legend visible and later on customize the x axis to write custom string of different size... First I need though to fix the more severe problems as I have described RegardsAlex On Sunday, October 27, 2013 12:25 PM, Jim Lemon j...@bitwrit.com.au wrote: On 10/27/2013 08:39 AM, Alaios wrote: Hi Jim and thanks for your answer... I might be too tired with my new born or just exhausted. I am attaching for everyone a small data snipset that you can load load(DataToPlotAsImage.Rdata) require(plotrix) browser() test-data # this transforms the values of test into red-yellow color2D.matplot(test,axes=F,xlab=,ylab=,main=color.scale, extremes=c(#FF,#00),show.legend=FALSE) axis(1,at=seq(1,ncol(test),length.out=10),labels=seq(201,300,length.out=10)) color.legend(104,30,112,70,seq(-110,-30,length=11), align=rb,rect.col=color.scale(1:30,1,c(0,1),0),gradient=y) as you can see I have problems where the legend appears. My par(usr returned me par(usr) # [1] 0 351 0 200 but I am not sure how to read that to place the legend at a useful place. second I am not sure why the image is so full with black rows.. What I want is to have the legend visible and later on customize the x axis to write custom string of different size... First I need though to fix the more severe problems as I have described Hi Alex, I'm not sure why you have created a copy of data to plot it. I can get quite a sensible plot using this: par(mar=c(5,4,4,5)) color2D.matplot(data,1,c(0,1),0,xlab=,ylab=, main=color.scale,xrange=c(-110,-50),border=NA) color.legend(357,30,370,100,seq(-110,-50,length.out=6), align=rb,rect.col=color.scale(1:6,1,c(0,1),0), gradient=y) Notice several things. First, when you have a large number of cells in a plot like this, setting the border to NA means that you don't get mostly borders (default = black) in the plot. The second thing is that your data range is -107.18150 to -54.07662. In order to get rounded numbers in your legend, I have set the xrange argument to -110 to -50. This gives a neat looking legend that spans your data, a bit like the pretty function would do. It also means that the color mapping is to that range and is the same in the legend as in the plot. I have left enough space on the right of the plot to fit in the legend, as that was where you said you wanted it in your last email. What par(usr) tells you is the dimensions of the plot in user units. Here it is x=0 at the left, x=351 at the right, y=0 at the bottom and y=200 at the top. Jim [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Create Time Lists with a for loop
Dear all, in my code I have written the following list TimeFramesShort -list(c(strptime(2011-10-12 10:59:00,%Y-%m-%d %H:%M:%S),strptime(2011-10-13 11:02:00,%Y-%m-%d %H:%M:%S)) # rest items were truncated I was wondering if could somehow take the first element of the list and create element that are only 5 minutes apart something like list(c(strptime(2011-10-12 10:59:00,%Y-%m-%d %H:%M:%S),strptime(2011-10-12 11:04:00,%Y-%m-%d %H:%M:%S), c(strptime(2011-10-12 11:04:00,%Y-%m-%d %H:%M:%S),strptime(2011-10-12 11:09:00,%Y-%m-%d %H:%M:%S), c(strptime(2011-10-12 11:09:00,%Y-%m-%d %H:%M:%S),strptime(2011-10-12 11:13:00,%Y-%m-%d %H:%M:%S) ... and so on. The problem I see is that pure string manipulation will not work as after one hour that should also make changes in the hours. Regards Alex [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] How to avoid this warning message
Dear all, I would like to ask you how I can avoid this warning I am getting when I am writing some results in a text file 5: In write.table(x = cor(t(collectMean_UL), t(collectMean_DL)), ... : appending column names to file the code that saves to the file look like: write(x=TEMPERATURE,file=paste(Folder,/Corr_,.txt,sep=),append=TRUE) write.table(x=cor(t(Data)),file=paste(Folder,/Cor_,.txt,sep=),eol = \r\n,append=TRUE) if someone can help on that... I would like to thank you in advance for your reply Regards Alex [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Compare two lists, with their sublists that have same structure
Dear all, I would like to ask your help concering two R lists. If I did everything should have the same structure (that means the same number of sublists, and their sublists also the same number of sublists). What would change between the two lists is the contents of each element in the lists. Could you please help me understand how I can do that in R? I would like to thank you in advance for your help Regards Alex [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] add a color band
Hi Jim and thanks for your answer... I might be too tired with my new born or just exhausted. I am attaching for everyone a small data snipset that you can load load(DataToPlotAsImage.Rdata) require(plotrix) browser() test-data # this transforms the values of test into red-yellow color2D.matplot(test,axes=F,xlab=,ylab=,main=color.scale, extremes=c(#FF,#00),show.legend=FALSE) axis(1,at=seq(1,ncol(test),length.out=10),labels=seq(201,300,length.out=10)) color.legend(104,30,112,70,seq(-110,-30,length=11), align=rb,rect.col=color.scale(1:30,1,c(0,1),0),gradient=y) as you can see I have problems where the legend appears. My par(usr returned me par(usr) # [1] 0 351 0 200 but I am not sure how to read that to place the legend at a useful place. second I am not sure why the image is so full with black rows.. What I want is to have the legend visible and later on customize the x axis to write custom string of different size... First I need though to fix the more severe problems as I have described Regards Alex On Friday, October 25, 2013 11:45 PM, Jim Lemon j...@bitwrit.com.au wrote: On 10/25/2013 11:16 PM, Alaios wrote: Hi Jim and thanks for help I will need some help to make -the legend visible (probably at the right side of the window) -split the legend at 0.1 steps -1,-0.9,-0.8.1 and -assign a color heat.color(30). I have selected orange and reds -show me the way to change the size of legend if it would be needed par(mar=c(7,4,4,6)) test-matrix(data=runif(1),nrow=100) color2D.matplot(test,axes=F,xlab=,ylab=,show.legend=FALSE) #,show.legend=TRUE axis(1,at=seq(1:nrow(test)),labels=seq(201,300)) # color.legend(11,6,11.8,9,seq(-1,1,length=10),rect.col=heat.colors(30),gradient=y) Hi Alex, You can do this in at least two ways. The first example uses the color.scale function to assign colors close to those of heat.colors. The second shows how to use heat.colors if you want that. library(plotrix) par(mar=c(7,4,4,6)) test-matrix(data=runif(1),nrow=100) # this transforms the values of test into red-yellow color2D.matplot(test,axes=F,xlab=,ylab=,main=color.scale, extremes=c(#FF,#00),show.legend=FALSE) axis(1,at=seq(1:nrow(test)),labels=seq(201,300)) # use par(usr) to find out the plot dimensions, # then place the legend where you want it color.legend(104,30,112,70,seq(-1,1,length=11), align=rb,rect.col=color.scale(1:30,1,c(0,1),0),gradient=y) # now try it with heat.colors test.cut-matrix(as.numeric(cut(test,breaks=seq(0,1,length.out=30))),nrow=100) color2D.matplot(test.cut,axes=F,xlab=,ylab=,main=heat.colors, cellcolors=heat.colors(30)[test.cut],show.legend=FALSE) axis(1,at=seq(1:nrow(test)),labels=seq(201,300)) color.legend(104,30,112,70,seq(-1,1,length=11), align=rb,rect.col=heat.colors(30),gradient=y) Jim__ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] add a color band
Hi all, I would like to ask your help to add a color band (Πam not sure regarding the right term, this color band at the right of the plot describing values with their corresponding color. For now I have only this code test-matrix(data=runif(1),nrow=100) plot(test,axes=FALSE) axis(1,at=c(0,1),labels=c(a,b)) # but I would like to add at the right side the color legend. How can I do that in R? I would like to thank you in advance for your help R. Alex [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] add a color band
Hi Jim and thanks for help I will need some help to make -the legend visible (probably at the right side of the window) -split the legend at 0.1 steps -1,-0.9,-0.8.1 and -assign a color heat.color(30). I have selected orange and reds -show me the way to change the size of legend if it would be needed par(mar=c(7,4,4,6)) test-matrix(data=runif(1),nrow=100) color2D.matplot(test,axes=F,xlab=,ylab=,show.legend=FALSE) #,show.legend=TRUE axis(1,at=seq(1:nrow(test)),labels=seq(201,300)) # color.legend(11,6,11.8,9,seq(-1,1,length=10),rect.col=heat.colors(30),gradient=y) I would like to thank you in advance for your help Regards Alex On Friday, October 25, 2013 11:50 AM, Jim Lemon j...@bitwrit.com.au wrote: On 10/25/2013 08:38 PM, Alaios wrote: Hi all, I would like to ask your help to add a color band (Ã⢠am not sure regarding the right term, this color band at the right of the plot describing values with their corresponding color. For now I have only this code test-matrix(data=runif(1),nrow=100) plot(test,axes=FALSE) axis(1,at=c(0,1),labels=c(a,b)) # butàI would like to add at the rightàside the color legend. How can I do that in R? I would like to thank you in advance for your help Hi Alex, If all else fails, look at color.legend (plotrix). Jim [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Plot.raster hides the axis layer
Hi all, I am trying to plot a raster object (I can explain why but the point is that it would be a raster objeçt).. I have selected a small code to show you exactly the problem require(raster) test-matrix(data=runif(1),nrow=100) m-raster(test) plot(m,axes=FALSE) axis(1,at=c(0,1),labels=c(a,b)) # THIS DOES NOT CHANGE THE INVISIBLE AXIS WHILE plot(test,axes=FALSE) axis(1,at=c(0,1),labels=c(a,b)) # WHEN I AM PLOTTING A NON RASTER OBJECT THE AXIS APPEARS. Can someone explain me what I can do so a raster layer to show correctly the axis? Please remember that I have to do that on a raster object so coercing it will not help .. I would like to thank you in advance for your help Regards Alex [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Where is element 30?
Hi I have a vector like that readCsvFile$V1 [1] 30 31 32 33 34 35 36 37 38 39 310 311 312 313 314 315 316 317 318 [20] 319 320 321 20 21 22 23 24 25 26 27 28 29 210 211 212 213 214 215 [39] 216 217 218 219 220 221 222 223 40 41 42 43 44 45 46 47 48 49 410 [58] 411 412 413 414 415 416 417 418 419 420 421 and I am looking to find where the number 31 is located. So I need one function to return me the index of where the number 31 is. Is there a function for that in R? Regards Alex [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] plot correlation matrix
Hi all, I am having 4 vectors like Data: num [1:4, 1:32] -82.8 -81.8 -75.5 -107.6 -87.6 ... and I want to calculate the correlation between those. Is there a graphical way in R to plot the correlations or not? I would like to thank you in advance for your help Regards Alex [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Lists with numbers lists and strings
Dear all, I have a list that is created like that Spans-list( c(837e6,842e6), c(832e6,837e6), c(930.1e6,935.1e6) ) I would like to include a second list that will contain the string that would correspond to the numbers at the left side. I would like thus inside my list to have two sublists. The first onew would be the Spans and the second one the Caption list. Could you please help me understand how I can do that? Regards Alex [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Averaging Out many rows from a column AND funtion to string
Dear all,
1) I have a very large matrix of
str(keep)
num [1:153899, 1:3415] -98.6 -95.8 -96.4 -95.8 -98 ...
that I would like to reduce its size to something like
str(keep)
num [1:1000, 1:3415] -98.6 -95.8 -96.4 -95.8 -98 ...
or anything similar in size as this is a matrix that needs plotting (so is ok
if it is 1000 row, 995, or 1123)
I think what I need here is a way of selecting multiple rows and averaging per
column (notice that the column number should stay the same)
b. I would like to be able to convert strings that are function names to real
function calls. For example I have something like
LogFunction- function(){}
FunctionIndex- rbind (c(1,LogFunction),
c(2,TakeFunction)
)
print(sprintf('Using the function %s',FunctionIndex[1,1]))
# call the FunctionIndex[1,1] somehow
I would like to thank you in advance for your help
Regards
Alex
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Averaging Out many rows from a column AND funtion to string
see inline
From: PIKAL Petr petr.pi...@precheza.cz
Sent: Wednesday, March 27, 2013 11:24 AM
Subject: RE: [R] Averaging Out many rows from a column AND funtion to string
Hi
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of Alaios
Sent: Wednesday, March 27, 2013 9:13 AM
To: R help
Subject: [R] Averaging Out many rows from a column AND funtion to
string
Dear all,
1) I have a very large matrix of
str(keep)
num [1:153899, 1:3415] -98.6 -95.8 -96.4 -95.8 -98 ...
that I would like to reduce its size to something like
str(keep)
num [1:1000, 1:3415] -98.6 -95.8 -96.4 -95.8 -98 ... or anything
similar in size as this is a matrix that needs plotting (so is ok if it
is 1000 row, 995, or 1123)
I think what I need here is a way of selecting multiple rows and
averaging per column (notice that the column number should stay the
same)
Make an index variable and aggregate values according it
Something like
idx-cut(1:153899, 153)
keep.ag-aggregate(keep, list(idx), mean)
1) Thanks that returned a data frame.. How I can have a matrix at the end?
2) I want to have a string that can be used also for calling a function with
the same name. Think of using mean to call mean. I need to have R interpret
the string in different ways.
Regards
Alex
b. I would like to be able to convert strings that are function names
to real function calls. For example I have something like
LogFunction- function(){}
FunctionIndex- rbind (c(1,LogFunction),
c(2,TakeFunction)
)
print(sprintf('Using the function %s',FunctionIndex[1,1])) # call the
FunctionIndex[1,1] somehow
I am not sure if I understand correctly.
Is this what you want?
myf -function(fun=mean, arg) eval(call(fun, arg))
Regards
Petr
I would like to thank you in advance for your help
Regards
Alex
[[alternative HTML version deleted]]
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Averaging Out many rows from a column AND funtion to string
see inline
From: PIKAL Petr petr.pi...@precheza.cz
Sent: Wednesday, March 27, 2013 1:50 PM
Subject: RE: [R] Averaging Out many rows from a column AND funtion to string
Hi
Â
Sent: Wednesday, March 27, 2013 11:46 AM
To: PIKAL Petr; R help
Subject: Re: [R] Averaging Out many rows from a column AND funtion to string
Â
see inline
Â
From:PIKAL Petr petr.pi...@precheza.cz
Sent: Wednesday, March 27, 2013 11:24 AM
Subject: RE: [R] Averaging Out many rows from a column AND funtion to string
Hi
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of Alaios
Sent: Wednesday, March 27, 2013 9:13 AM
To: R help
Subject: [R] Averaging Out many rows from a column AND funtion to
string
Dear all,
1) I have a very large matrix of
str(keep)
 num [1:153899, 1:3415] -98.6 -95.8 -96.4 -95.8 -98 ...
that I would like to reduce its size to something like
str(keep)
 num [1:1000, 1:3415] -98.6 -95.8 -96.4 -95.8 -98 ... or anything
similar in size as this is a matrix that needs plotting (so is ok if it
is 1000 row, 995, or 1123)
I think what I need here is a way of selecting multiple rows and
averaging per column (notice that the column number should stay the
same)
Make an index variable and aggregate values according it
Something like
idx-cut(1:153899, 153)
keep.ag-aggregate(keep, list(idx), mean)
1) Thanks that returned a data frame.. How I can have a matrix at the end?
use as.matrix(data.frame) on numeric part
a bit of my code that you can re run. I convert a 30,30 matrix to a 10,30.Â
but it looks at the end that I do not get the Data part of the data.fram
correctly:
Data-matrix(data=rnorm(900,80,20),nrow=30,ncol=30)
idx-cut(1:30, 10)
keep.ag-aggregate(Data, list(idx), mean)
str(as.matrix(keep.ag) ) # it does not look like integers
2) I want to have a string that can be used also for calling a function with
the same name. Think of using mean to call mean. I need to have R interpret
the string in different ways.
To call âmeanâ of what? I am sure I do not understand what is your
intention.
LogFunction- function(){}
FunctionIndex- rbind (c(1,LogFunction),
              c(2,TakeFunction))
print(sprintf('Using the function %s',FunctionIndex[1,1]))
This does not contain much clue.
Regards
Petr
PS. Do not use HTML mail messages.
Regards
Alex
b. I would like to be able to convert strings that are function names
to real function calls. For example I have something like
LogFunction- function(){}
FunctionIndex- rbind (c(1,LogFunction),
            c(2,TakeFunction)
           )
print(sprintf('Using the function %s',FunctionIndex[1,1])) # call the
FunctionIndex[1,1] somehow
I am not sure if I understand correctly.
Is this what you want?
myf -function(fun=mean, arg) eval(call(fun, arg))
Regards
Petr
I would like to thank you in advance for your help
Regards
Alex
   [[alternative HTML version deleted]]
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Averaging Out many rows from a column AND funtion to string
I have fixed it like that:
keep-matrix(data=rnorm(900,80,20),nrow=30,ncol=30)
ShrinkTo-500
idx-cut(1:nrow(keep), ShrinkTo)
keep.ag-aggregate(keep, list(idx), mean)
PlotMe-data.matrix(keep.ag[2:ShrinkTo])
The only problem is that takes ages to finish. Would it be possible to
convert it to something like lapply?
Regards
Alex
From: PIKAL Petr petr.pi...@precheza.cz
Sent: Wednesday, March 27, 2013 4:02 PM
Subject: RE: [R] Averaging Out many rows from a column AND funtion to string
Hi
Â
str(as.matrix(keep.ag[,-1]) ) Â # does look like numeric
num [1:10, 1:30] 75.1 93 79 81.7 76.3 ...
- attr(*, dimnames)=List of 2
 ..$ : NULL
 ..$ : chr [1:30] V1 V2 V3 V4 ...
Â
Please read and follow what was recommended.
Â
quote
Â
use as.matrix(data.frame) on numeric part
                                       Â
                 ^^^
Â
aggregate produces data frame with its first column being your idx variable,
which is factor. Trying to convert it whole to matrix results in character
matrix. You need to exclude first column from conversion
Â
And please can you explain how mean(rnorm(whatever)) shall be integer?
Â
Regards
Petr
Â
Sent: Wednesday, March 27, 2013 3:01 PM
To: PIKAL Petr; R help
Subject: Re: [R] Averaging Out many rows from a column AND funtion to string
Â
see inline
Â
Â
From:PIKAL Petr petr.pi...@precheza.cz
Sent: Wednesday, March 27, 2013 1:50 PM
Subject: RE: [R] Averaging Out many rows from a column AND funtion to string
Hi
Â
Sent: Wednesday, March 27, 2013 11:46 AM
To: PIKAL Petr; R help
Subject: Re: [R] Averaging Out many rows from a column AND funtion to string
Â
see inline
Â
From:PIKAL Petr petr.pi...@precheza.cz
Sent: Wednesday, March 27, 2013 11:24 AM
Subject: RE: [R] Averaging Out many rows from a column AND funtion to string
Hi
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of Alaios
Sent: Wednesday, March 27, 2013 9:13 AM
To: R help
Subject: [R] Averaging Out many rows from a column AND funtion to
string
Dear all,
1) I have a very large matrix of
str(keep)
 num [1:153899, 1:3415] -98.6 -95.8 -96.4 -95.8 -98 ...
that I would like to reduce its size to something like
str(keep)
 num [1:1000, 1:3415] -98.6 -95.8 -96.4 -95.8 -98 ... or anything
similar in size as this is a matrix that needs plotting (so is ok if it
is 1000 row, 995, or 1123)
I think what I need here is a way of selecting multiple rows and
averaging per column (notice that the column number should stay the
same)
Make an index variable and aggregate values according it
Something like
idx-cut(1:153899, 153)
keep.ag-aggregate(keep, list(idx), mean)
1) Thanks that returned a data frame.. How I can have a matrix at the end?
use as.matrix(data.frame) on numeric part
Â
a bit of my code that you can re run. I convert a 30,30 matrix to a 10,30.Â
but it looks at the end that I do not get the Data part of the data.fram
correctly:
Data-matrix(data=rnorm(900,80,20),nrow=30,ncol=30)
idx-cut(1:30, 10)
keep.ag-aggregate(Data, list(idx), mean)
str(as.matrix(keep.ag) ) # it does not look like integers
Â
Â
Â
Â
Â
Â
Â
2) I want to have a string that can be used also for calling a function with
the same name. Think of using mean to call mean. I need to have R interpret
the string in different ways.
To call âmeanâ of what? I am sure I do not understand what is your
intention.
LogFunction- function(){}
FunctionIndex- rbind (c(1,LogFunction),
              c(2,TakeFunction))
print(sprintf('Using the function %s',FunctionIndex[1,1]))
This does not contain much clue.
Regards
Petr
PS. Do not use HTML mail messages.
Regards
Alex
b. I would like to be able to convert strings that are function names
to real function calls. For example I have something like
LogFunction- function(){}
FunctionIndex- rbind (c(1,LogFunction),
            c(2,TakeFunction)
           )
print(sprintf('Using the function %s',FunctionIndex[1,1])) # call the
FunctionIndex[1,1] somehow
I am not sure if I understand correctly.
Is this what you want?
myf -function(fun=mean, arg) eval(call(fun, arg))
Regards
Petr
I would like to thank you in advance for your help
Regards
Alex
   [[alternative HTML version deleted]]
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Controlling Raster plots
Hi,
I have a few raster layers and I would like to customized their x and y axis.
I have tried already something like:
require('raster')
keep-matrix(data=rnorm(900,80,20),nrow=30,ncol=30)
xlab-seq(100e6,200e6,length.out=5)
test-raster(keep)
plot(test,ylab=,xaxt=n,yaxt=n)
axis(1, at=seq(0,1,length.out=5), xlab)
that normally would work in any other plot function.
Could you please help me add the xaxis ?
I would like to thank you in advance for your reply
Regards
Alex
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Averaging Out many rows from a column AND funtion to string
Well my true matrix is
of
num [1:153899, 1:3415]
that I want to convert to something like
num [1:1000, 1:3415] (keeping column number the same).
I only give subsets here to allow others to run the code at their computer
Thanks a lot
Regards
Alex
From: PIKAL Petr petr.pi...@precheza.cz
Sent: Wednesday, March 27, 2013 4:45 PM
Subject: RE: [R] Averaging Out many rows from a column AND funtion to string
Hi
Â
system.time({
+ keep-matrix(data=rnorm(900,80,20),nrow=30,ncol=30)
+ ShrinkTo-500
+ idx-cut(1:nrow(keep), ShrinkTo)
+ keep.ag-aggregate(keep, list(idx), mean)
+ PlotMe-as.matrix(keep.ag[,-1])
+ })
  user system elapsed
  29.23   0.14  29.37
Â
Â
It takes 30 seconds when using 30 rows. It is not enough time to get a cup
of tee, which I do not consider ages. Maybe split lapply approach or
data.matrix or  **ply could be quicker but I do not consider worth spending
hours to elaborate some solution which will spare 20 sec computing time.
Â
Regards
Petr
Â
Sent: Wednesday, March 27, 2013 4:12 PM
To: PIKAL Petr; R help
Subject: Re: [R] Averaging Out many rows from a column AND funtion to string
Â
I have fixed it like that:
keep-matrix(data=rnorm(900,80,20),nrow=30,ncol=30)
ShrinkTo-500
idx-cut(1:nrow(keep), ShrinkTo)
keep.ag-aggregate(keep, list(idx), mean)
PlotMe-data.matrix(keep.ag[2:ShrinkTo])
The only problem is that takes ages to finish. Would it be possible to
convert it to something like lapply?
Regards
Alex
Â
Â
From:PIKAL Petr petr.pi...@precheza.cz
Sent: Wednesday, March 27, 2013 4:02 PM
Subject: RE: [R] Averaging Out many rows from a column AND funtion to string
Hi
Â
str(as.matrix(keep.ag[,-1]) ) Â # does look like numeric
num [1:10, 1:30] 75.1 93 79 81.7 76.3 ...
- attr(*, dimnames)=List of 2
 ..$ : NULL
 ..$ : chr [1:30] V1 V2 V3 V4 ...
Â
Please read and follow what was recommended.
Â
quote
Â
use as.matrix(data.frame) on numeric part
                                       Â
                 ^^^
Â
aggregate produces data frame with its first column being your idx variable,
which is factor. Trying to convert it whole to matrix results in character
matrix. You need to exclude first column from conversion
Â
And please can you explain how mean(rnorm(whatever)) shall be integer?
Â
Regards
Petr
Â
Sent: Wednesday, March 27, 2013 3:01 PM
To: PIKAL Petr; R help
Subject: Re: [R] Averaging Out many rows from a column AND funtion to string
Â
see inline
Â
Â
From:PIKAL Petr petr.pi...@precheza.cz
Sent: Wednesday, March 27, 2013 1:50 PM
Subject: RE: [R] Averaging Out many rows from a column AND funtion to string
Â
Hi
Â
Sent: Wednesday, March 27, 2013 11:46 AM
To: PIKAL Petr; R help
Subject: Re: [R] Averaging Out many rows from a column AND funtion to string
Â
see inline
Â
From:PIKAL Petr petr.pi...@precheza.cz
Sent: Wednesday, March 27, 2013 11:24 AM
Subject: RE: [R] Averaging Out many rows from a column AND funtion to string
Hi
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of Alaios
Sent: Wednesday, March 27, 2013 9:13 AM
To: R help
Subject: [R] Averaging Out many rows from a column AND funtion to
string
Dear all,
1) I have a very large matrix of
str(keep)
 num [1:153899, 1:3415] -98.6 -95.8 -96.4 -95.8 -98 ...
that I would like to reduce its size to something like
str(keep)
 num [1:1000, 1:3415] -98.6 -95.8 -96.4 -95.8 -98 ... or anything
similar in size as this is a matrix that needs plotting (so is ok if it
is 1000 row, 995, or 1123)
I think what I need here is a way of selecting multiple rows and
averaging per column (notice that the column number should stay the
same)
Make an index variable and aggregate values according it
Something like
idx-cut(1:153899, 153)
keep.ag-aggregate(keep, list(idx), mean)
1) Thanks that returned a data frame.. How I can have a matrix at the end?
use as.matrix(data.frame) on numeric part
Â
a bit of my code that you can re run. I convert a 30,30 matrix to a 10,30.Â
but it looks at the end that I do not get the Data part of the data.fram
correctly:
Data-matrix(data=rnorm(900,80,20),nrow=30,ncol=30)
idx-cut(1:30, 10)
keep.ag-aggregate(Data, list(idx), mean)
str(as.matrix(keep.ag) ) # it does not look like integers
Â
Â
Â
Â
Â
Â
Â
2) I want to have a string that can be used also for calling a function with
the same name. Think of using mean to call mean. I need to have R interpret
the string in different ways.
To call âmeanâ of what? I am sure I do not understand what is your
intention.
LogFunction- function(){}
FunctionIndex- rbind (c(1,LogFunction),
              c(2,TakeFunction))
print(sprintf('Using the function %s',FunctionIndex[1,1]))
This does
Re: [R] Plot Matrix with Data
Hi,
Thanks for the answer.
It looks like that the mutate what I was missing so long..
That's my current attempt
Data-matrix(data=rnorm(900,80,20),nrow=30,ncol=30)
Lengths- 15
library(reshape2)
library(ggplot2)
require('plyr')
tdm - melt(Data)
tdm - mutate(tdm, col = cut(value, seq(15, 90, by=5), labels = seq(20, 90,
by=5)) )
test-colorRampPalette(c(orange, red),space=Lab,interpolate=linear)
ggplot(tdm, aes(x = Var1, y = Var2, fill = col)) +
geom_raster() +
scale_x_continuous(expand = c(0, 0)) +
scale_y_continuous(expand = c(0, 0)) +
labs(x = MHz, y = Threshold, fill = Duty Cycle)
Two things.
a. I have tried to make a color pallete that goes from orange through all the
orangish and redish hues to the red but as you can see that failed.
b. One think is that the colors visible in the legend in the right part are not
always visible. If for example there is a 20 value the 20 will appear and if
there is not but there is a 25 the bar will start from the 25 onwards.
I would like to thank everyone for their on going support
Regards
Alex
From: Dennis Murphy djmu...@gmail.com
Sent: Monday, March 25, 2013 9:17 PM
Subject: Re: [R] Plot Matrix with Data
Hi Alex:
Here are a couple of raw examples, one with a 'discretized' color
gradient and another with ordered factor levels in groups of length
10. Starting with tdm,
library(plyr)
library(ggplot2)
tdm - mutate(tdm, col = cut(value, seq(0, 140, by = 10),
labels = seq(10, 140, by = 10)) )
## or equivalently,
# tdm$col - cut(col$value, seq(0, 140, by = 10),
# labels = seq(10, 140, by = 10)) )
# The numeric factor levels are 1 - 14, so if we convert col to numeric,
# should multiply the values by 10. This plot uses a continuous
# color gradient whose midpoint is near the median value.
ggplot(tdm, aes(x = Var1, y = Var2, fill = 10 * as.numeric(col))) +
geom_raster() +
scale_x_continuous(expand = c(0, 0)) +
scale_y_continuous(expand = c(0, 0)) +
scale_fill_gradient2(low = green, mid = orange,
high = yellow, midpoint = 85) +
labs(x = MHz, y = Threshold, fill = Duty Cycle)
# This one uses col as a factor. The labels are the correct
# upper limits of each value bin. Obviously you want to use
# a smarter mechanism for assigning colors than the
# default. The colorspace package or the colorRampPalette()
# function might be useful here.
ggplot(tdm, aes(x = Var1, y = Var2, fill = col)) +
geom_raster() +
scale_x_continuous(expand = c(0, 0)) +
scale_y_continuous(expand = c(0, 0)) +
labs(x = MHz, y = Threshold, fill = Duty Cycle)
The values on the x-y axes are the row and column numbers of the
reshaped matrix. You may need to map them to values of the MHz and
Threshold values; I doubt that [0, 1] is the range of each variable
(as you have in another post using base graphics).
Dennis
Hi ,
I would like to use ggplot2 to plot a matrix as an image.
You can copy paste the following
Data-matrix(data=rnorm(900,80,20),nrow=30,ncol=30)
lengthOut-5
Lengths- 15
library(reshape2)
library(ggplot2)
tdm - melt(Data)
ggplot(tdm, aes(x = Var2, y = Var1, fill =
factor(value)),levels=seq(0,1,by=0.1)) +
labs(x = MHz, y = Threshold, fill = Duty Cycle) +
geom_raster(alpha=1) +
scale_fill_discrete(h.start=1,breaks=c(20,30,40,50,60,70,80,90),labels=c(20,30,40,50,60,70,80,90),fill=red)
+
scale_x_continuous(expand = c(0, 0)) +
scale_y_continuous(expand = c(0, 0))
# End of code part
What I wanted to do is to print the values
below 20, with an x color
between 20-30, with a y clor
between 30-40, with a z color
and so on
I would not care now for the color palette.
Any would do.
Could you please help me with that?
Regards
A
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[R] Increase font size in plots
Hi
I am using violin plots (type of boxplots) and I am trying to increase the
font size in the plots.
It looks like that the violin plots do not work as normal plots as the cex
parameters are ignored.
You can have a loot at the code below
require('vioplot')
data1-rnorm(100)
data2-rnorm(10)
data3-rnorm(1000)
vioplot(data1,data2,data3,cex.lab=2,cex.axis=2)
Regards
Alex
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Re: [R] Plot Matrix with Data
Hi,
it was very kind of you to help me again.
Your code works, and the reason I was using the limits as that is that my
dataset is slightly different than the one I used for showing the problem here.
More specifically this is my dataset
Browse[1] str(keep)
num [1:153899, 1:3415] -98.6 -95.8 -96.4 -95.8 -98 ...
Browse[1] (density(keep))
Call:
density.default(x = keep)
Data: keep (525565085 obs.); Bandwidth 'bw' = 0.1695
x y
Min. :-106.94 Min. :0.000
1st Qu.: -87.13 1st Qu.:0.0003593
Median : -67.31 Median :0.0050385
Mean : -67.31 Mean :0.0126038
3rd Qu.: -47.49 3rd Qu.:0.0084189
Max. : -27.67 Max. :0.1792683
and as you can see is rather large... that makes it a marathon to print it.
Would it be possible to average some of the arrows in a column by column basis
so to turn the
str(keep)
num [1:153899, 1:3415] -98.6 -95.8 -96.4 -95.8 -98 ...
to something like
num [1:1000, 1:3415] . How can I do that efficiently in R?
I would like to thank everyone in advance.
Regards
Alex
From: Dennis Murphy djmu...@gmail.com
Sent: Tuesday, March 26, 2013 10:43 AM
Subject: Re: [R] Plot Matrix with Data
Hi:
I don't understand why you're truncating the duty cycle at 90 when the
data range from 15 to 150 - the effect of doing that is the set of
grey squares in the heatmap corresponding to missing values.
The reason your color scale didn't work is because you never used it
in the ggplot() call, but I think that it would have been difficult to
incorporate it anyway because ggplot2 has its own conventions. It
appears that what you want is ggplot2::scale_fill_continuous().
I chucked the color factor and tried the following instead. You can
edit it however you want, and you can extend the length/width of the
color bar - see the help page of guide_colourbar(), in particular the
barheight argument.
http://docs.ggplot2.org/current/guide_colourbar.html
My shot at your graphic:
ggplot(tdm, aes(x = Var1, y = Var2, fill = value)) +
geom_raster() +
scale_x_continuous(expand = c(0, 0)) +
scale_y_continuous(expand = c(0, 0)) +
scale_fill_continuous(limits = c(15, 150),
breaks = seq(15, 150, by = 15),
low = orange, high = red,
guide = guide_colourbar(nbin = 28)) +
labs(x = MHz, y = Threshold, fill = Duty Cycle)
This color bar has 28 bins, 5 units apart, but the labels are 15 apart
since otherwise they overwrite each other. You can reduce the crowding
by increasing the height of the color bar as indicated earlier. This
appears to be closer to your intent with a lot less bother.
Dennis
Hi,
Thanks for the answer.
It looks like that the mutate what I was missing so long..
That's my current attempt
Data-matrix(data=rnorm(900,80,20),nrow=30,ncol=30)
Lengths- 15
library(reshape2)
library(ggplot2)
require('plyr')
tdm - melt(Data)
tdm - mutate(tdm, col = cut(value, seq(15, 90, by=5), labels = seq(20, 90,
by=5)) )
test-colorRampPalette(c(orange, red),space=Lab,interpolate=linear)
ggplot(tdm, aes(x = Var1, y = Var2, fill = col)) +
geom_raster() +
scale_x_continuous(expand = c(0, 0)) +
scale_y_continuous(expand = c(0, 0)) +
labs(x = MHz, y = Threshold, fill = Duty Cycle)
Two things.
a. I have tried to make a color pallete that goes from orange through all
the orangish and redish hues to the red but as you can see that failed.
b. One think is that the colors visible in the legend in the right part are
not always visible. If for example there is a 20 value the 20 will appear
and if there is not but there is a 25 the bar will start from the 25
onwards.
I would like to thank everyone for their on going support
Regards
Alex
From: Dennis Murphy djmu...@gmail.com
Sent: Monday, March 25, 2013 9:17 PM
Subject: Re: [R] Plot Matrix with Data
Hi Alex:
Here are a couple of raw examples, one with a 'discretized' color
gradient and another with ordered factor levels in groups of length
10. Starting with tdm,
library(plyr)
library(ggplot2)
tdm - mutate(tdm, col = cut(value, seq(0, 140, by = 10),
labels = seq(10, 140, by = 10)) )
## or equivalently,
# tdm$col - cut(col$value, seq(0, 140, by = 10),
# labels = seq(10, 140, by = 10)) )
# The numeric factor levels are 1 - 14, so if we convert col to numeric,
# should multiply the values by 10. This plot uses a continuous
# color gradient whose midpoint is near the median value.
ggplot(tdm, aes(x = Var1, y = Var2, fill = 10 * as.numeric(col))) +
geom_raster() +
scale_x_continuous(expand = c(0, 0)) +
scale_y_continuous(expand = c(0, 0)) +
scale_fill_gradient2(low = green, mid = orange,
high = yellow, midpoint = 85) +
labs(x = MHz, y = Threshold, fill = Duty Cycle)
#
[R] plot Raster
Dear all,
I am trying to plot an image so I am trying this through raster layer.
You can copy paste the following
require('raster')
Data-matrix(data=rnorm(900,80,20),nrow=30,ncol=30)
rasterData-raster(Data)
lengthOut-5
xAxisFrequencies-seq(800,900,length.out=lengthOut)
plot(rasterData, ylab=,xaxt=n,yaxt=n)
axis(1, at=seq(0,1,length.out=lengthOut), xAxisFrequencies, col.axis = blue)
What I want is to add a customized x label, but the last line I gave above
axis(1,at.) does not work.
It would be also nice to add a bit of legend also under the color bar to
explain there what are the values.
Could you please help me with those two ?
Regards
Alex
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[R] Plot Matrix with Data
Hi , I would like to use ggplot2 to plot a matrix as an image. You can copy paste the following Data-matrix(data=rnorm(900,80,20),nrow=30,ncol=30) lengthOut-5 Lengths- 15 library(reshape2) library(ggplot2) tdm - melt(Data) ggplot(tdm, aes(x = Var2, y = Var1, fill = factor(value)),levels=seq(0,1,by=0.1)) + labs(x = MHz, y = Threshold, fill = Duty Cycle) + geom_raster(alpha=1) + scale_fill_discrete(h.start=1,breaks=c(20,30,40,50,60,70,80,90),labels=c(20,30,40,50,60,70,80,90),fill=red) + scale_x_continuous(expand = c(0, 0)) + scale_y_continuous(expand = c(0, 0)) # End of code part What I wanted to do is to print the values below 20, with an x color between 20-30, with a y clor between 30-40, with a z color and so on I would not care now for the color palette. Any would do. Could you please help me with that? Regards A [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] All unique combinations
Dear all, I would like to have all unique combinations in the following matrix TimeIndex- rbind (c(1,Week_of_21_07-29_03), c(2,Thursday_21_03), c(3,Friday_22_03), c(4,Saturday_23_03), c(5,Sunday_24_03), c(6,Monday_25_03), c(7,Tuesday_26_03), c(8,Wednesday_27_03), c(9,Thursday_28_03), c(10,Friday_ 29_03)) As have a plotting code that takes two input arguments plot_two_time_frames(time1,LabelTime1,time2,LabelTime2) these inputs are time1: integer from TimeIndex. TimeIndex[i,1] LabelTime1: Labels from TimeIndex[i,2] time2: integer from TimeIndex, TimeIndex[i,1] LabelTime2: Labels from TimeIndex[i,2] the times 1 and time 2 should be unique combinations and never the same number. I understand that my explanation was not the best possible. But I would like to thank you in advance for your support Regards Alex [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] lm and Formula tutorial
Dear all, I was reading last night the lm and the Formula manual page, and 'I have to admit that I had tough time to understand their syntax. Is there a simpler guide for the dummies like me to start with? I would like to thank you in advance for your help Regards Alex [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] parallel execution in R
Dear all,
I have a piece of code that I want to run in parallel (I am working in system
of 16 cores)
foreach (i=(seq(-93,-73,length.out=21))) %dopar%
{
threshold-i
print(i)
do_analysis1(i,path)
do_analysis2(i,path)
do_something_else_analysis1(i,path)
something_else_now(i,path)
}
as you can see I have already tried to make this run in parallel, meaning for
every i value each of the 16 processor shoule take a block of the body such
as:
threshold-i
print(i)
do_analysis1(i,path)
do_analysis2(i,path)
do_something_else_analysis1(i,path)
something_else_now(i,,path)
and execute it . Unfortunately this does not work and oonly one processor looks
utilized.
Alternatively, mclapply have worked well in the past, but in this case I am not
sure how to convert the serial execution of the body of the loop to a list that
would be compatible with the mclapply.
I would like to thank you in advance for your help
Regards
Alex
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Re: [R] ggplot2 Increase font size
Hi, I am not quite sure what you meanÎ. I give again reproducible code: require(ggplot2) require(reshape) DataToPlot-matrix(data=rnorm(9),nrow=3,dimnames=list(seq(1,3),seq(4,6))) tdm-melt(DataToPlot) p- ggplot(tdm, aes(x = X2, y = X1, fill = factor(value))) +                  labs(x = MHz, y = Threshold, fill = Duty Cycle) +                  geom_raster(alpha=1) +                  scale_fill_discrete(h.start=1,breaks=c(0,0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9,1),labels=c(0,0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9,1)) +                  scale_x_continuous(expand = c(0, 0)) +                  scale_y_continuous(expand = c(0, 0)) DataToPlot in my case contains something like that: DataToPlot           4         5         6 1 -0.4135124 0.4643110 -0.7530622 2 0.8827643 -0.1702428 0.4607671 3 0.7942167 -1.2450487 -0.9380290 what I would like to have is to have one specific color for each of the following cases c(0,0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9,1) so for example the -0.4 would be categorized under the 0 category the 0.46 would be categorized under the 0.5 category the -0.75 would be categorized under the 0 category the 0.88 would be categorized under the 0.9 category right now the code I gave prints no color bar. If I change it like that. p- ggplot(tdm, aes(x = X2, y = X1, fill = factor(value))) +                  labs(x = MHz, y = Threshold, fill = Duty Cycle) +                  geom_raster(alpha=1) +                  scale_fill_discrete(h.start=1) +                  scale_x_continuous(expand = c(0, 0)) +                  scale_y_continuous(expand = c(0, 0)) colorbar comes back but as you will see the defined values is not how I want to categorize my data. Could you please help me find what I do not understand here? I would like to thank you in advance for your reply Regards Alex From: Ista Zahn istaz...@gmail.com Cc: R help R-help@r-project.org Sent: Monday, February 25, 2013 3:54 PM Subject: Re: [R] ggplot2 Increase font size Hi Alex, See ?theme Best, Ista Dear all, I am using the code as below  tdm - melt(matrixToPlot)   p- ggplot(tdm, aes(x = Var2, y = Var1, fill = factor(value))) +          labs(x = Mz, y = T, fill = D) +          geom_raster(alpha=1) +          scale_fill_discrete(h.start=1) +          scale_x_continuous(expand = c(0, 0)) +          scale_y_continuous(expand = c(0, 0)) to plot an two dimensional image . I would like to ask your help to replace the gray border with white color and increase the font size of x and y axis as wells as the legend of the color bar. Could you please give me the function names to use? Regards Alex     [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] ggplot2 Increase font size
Dear all, I am using the code as below tdm - melt(matrixToPlot) p- ggplot(tdm, aes(x = Var2, y = Var1, fill = factor(value))) + labs(x = Mz, y = T, fill = D) + geom_raster(alpha=1) + scale_fill_discrete(h.start=1) + scale_x_continuous(expand = c(0, 0)) + scale_y_continuous(expand = c(0, 0)) to plot an two dimensional image . I would like to ask your help to replace the gray border with white color and increase the font size of x and y axis as wells as the legend of the color bar. Could you please give me the function names to use? Regards Alex [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
