similar to: executable R script under xp (to avoid migration toward Matlab or C++)

Displaying 20 results from an estimated 6000 matches similar to: "executable R script under xp (to avoid migration toward Matlab or C++)"

2010 Nov 24
2
looking for the RMySQL package for R 2.12.0 under XP
Hello, I convinced our new Ph student to work with R and I managed to convince the guy in charge of the computer networks to install R 2.12.0 (we haven't the admin rights). I wanted to install on his computer the same packages I use (but I have R 2.11) and especially RMySQL. Problem : there is no more RMySQL package in the list pf the packages. I've tried to install my version but it is
2011 Sep 23
2
Homography with R
Hello, I would like to know if it exists a package including something equivalent to this page: http://www.developpez.net/forums/d740403/autres-langages/algorithmes/contribuez/image-geometrie-projective-homography/ http://www.developpez.net/forums/d740403/autres-langages/algorithmes/contribuez/image-geometrie-projective-homography/ I've searched with these keywords: R-project homography
2011 Feb 09
5
Problem with long number (from character to numeric class)
Hello, I have a text file with one column containing long number but stored as string. I download the file with read.table (and colClass) and the first row of this column is : "095842087016731010" As I need to make some calculations with these numbers, I tried to convert them using as.numeric. But then I get as.numeric("095842087016731010") 95842087016731008 I understand
2010 Jun 15
3
Problem with as.POSIXlt$hour
Hello, In one of my functions, I need to extract the hour from a date. For example: as.POSIXlt("2010-03-27 02:00:00")$hour gives 2 as expected. It works for all the dates I've tested except the following one which is in my list of dates: as.POSIXlt("2010-03-28 02:00:00")$hour which gives 0. I don't understand why it doesn't work with this particular date. Any
2010 Mar 12
2
looking for a package to solve current-voltage equation I=f(I)
Hello, I'm looking for a R-package to solve such an equation : http://n4.nabble.com/file/n1590462/iv.png which is one current-voltage equation of a solar cell. Thanks in advance, Ptit Bleu. -- View this message in context: http://n4.nabble.com/looking-for-a-package-to-solve-current-voltage-equation-I-f-I-tp1590462p1590462.html Sent from the R help mailing list archive at Nabble.com.
2010 Nov 30
2
xyplot : superimposed 2 groups in different panels
Hello, I would like to plot the following xyplot : for each date of fff (1 date per panel), bbb=f(aaa) for the two groups (ddd=1 and ddd=2) superimposed. I can do it by group (see below) but not together. I looked at http://lmdvr.r-forge.r-project.org/figures/figures.html but I haven't found what I was looking for (to be honest, I haven't understood all the examples). Any help will be
2010 Nov 19
3
Still confused with lapply
Hello, I would like to replace a for loop with lapply in order to speed up the treatment of my data (I've read lapply can help to do it). At the end of the message, you will find a simple example (function is more complex and data frames is more than 1 million of rows) of what I would like to do, that is applying the same calculation to all the rows of the data frame (knowing that I cannot
2012 Jun 13
2
add horizontal reference lines in lattice bwplot
I'm plotting a lattice bwplot from a data.frame and would like to add some horizontal lines indicating some reference values (preferably with text annotation) as well as a grid of regular log tics. The following illustrates the plot: data(ToothGrowth) require("lattice") p = bwplot(exp(len) ~ supp, ToothGrowth, panel = function(..., box.ratio) { panel.grid(h=-1,
2011 Dec 15
2
lm and R-squared (newbie)
Hello, I've two data.frames (data1 and data4), dec="." and sep=";". http://r.789695.n4.nabble.com/file/n4199964/data1.txt data1.txt http://r.789695.n4.nabble.com/file/n4199964/data4.txt data4.txt When I do plot(data1$nx,data1$ny, col="red") points(data4$nx,data4$ny, col="blue") , results seem very similar (at least to me) but the R-squared of
2009 Dec 04
1
Problem with date x-axis
Hello, I have a "friday" problem. I'm trying to plot the number of measures by day with this command between the 2008-07-31 and the 2009-11-13 (it works) : plot(unique(as.Date(strptime(data$date, format="%Y-%m-%d %H:%M:%S))), measuresbyday) then I would like to plot vertical lines for each month. But it doesn't work because par()$usr gives 14071.16 14579.84 -7.16 213.16
2009 Dec 11
1
lattice - 2 graphs per panel with 2 differents colours
Hello, I would like to get a lattice plot of 8 panels (unique(df$fact)=8) with 2 graphs (df$y1 and df$y2 as a function of df$x) and 1 red point at (500, ymax) per panel. The script below is quite ok but I'm not able to define two different colors for the two graphs. If you have an idea how to use the "col" function in order to attribute the colors, it will be very kind of you to
2012 Jun 28
1
add constraints to nls or use another function
Hello, I'm trying to fit experimental data with a model and nls. For some experiments, I have data with x from 0 to 1.2 and the fit is quite good. But it can happen that I have data only the [0,0.8] range (see the example below) and, then, the fit is not correct. I would like to add a constraint, for example : the second derivative must be positive. But I don't know how to add this to
2007 Dec 10
7
Sweave : change value in rnw file to generate multiple "single" reports ?
Hello, I'm still trying to make the life of my colleagues easier. Nice, isn't it ? At the moment, I'm looking for a way to generate multiple "single report". In fact I have a .rnw file which send a query to a MySQL database (rs<-dbSendQuery(con, statement="select * from treatdata where name='Device1'") But of course my colleagues have many devices and
2007 Aug 27
4
Problem with save or/and if (I think but maybe not ...)
Hi, I recently discovered the R program and I thought it could be useful to me. I have to analyse data saved as .Px file (x between 0 and 8 - .P0 files have 18 lines at the beginning that I have to skip). New files are generated everyday. This is my strategy : In order to analyse the data, I first want to copy the new data in a database in MySQL (which already contains the previous data). So
2007 Nov 07
3
Can I replace NA by 0 (if yes, how) ?
Hello, I'm trying to fit some points with a 8-degrees polynom (result of lm is stored in pfit). In most of the case, it is ok but for some others, some coefficients are "NA". I don't really understand the meaning of these "NA". And the problem is that I can't perform a derivation (pderiv<-as.function((deriv(polynomial(pfit$coefficients))))) on pfit due to the
2007 Sep 17
4
Must be easy, but haven't found the function (numerical integration)
Hi, I have a data frame of 2 columns with the following types : data$day char data$value num And I plot my data with : plot(strptime(donnees$day,format="%Y-%m-%d %H:%M:%S"),donnees$value, type="l") And I'd just like to get the numerical value of the integration of this graph. I looked at ?integrate but, as far as I understood (that is, not very much, due to my poor
2007 Nov 20
2
Friendly way to link R - MySQL and non-(R and Mysql) users ?
Thanks to your help, I managed to link R and a Mysql Database, send queries, plot the results and put everything in a pdf document (with Sweave). My co-workers find the job not bad but they would like to have a friendly interface to send queries and see the graphs (the pdf document is not necessary). Something like this coul be a good starting point : --------------------------------------- |
2007 Dec 17
2
Must be obvious but not to me : problem with regular expression
Hi, I have a vector called nfichiers of 138 names of file whose extension is .P0 or P1 ... to P8. The script is not the same when the extension is P0 or P(1 to 8). Examples of file names : [128] "Output0.P0" [129] "Output0.P1" [130] "Output0.P2" [131] "Output01102007.P0" [132] "Output01102007.P1" [133]
2009 Jun 05
2
find a sequence of characters in a vector
Hello, I'm just looking for an easy way to find the positions of a complete sequence in a bigger vector. For example : c("a","z","e") in c("a","z","e","r","t","a","z","a","z","e","c") and the result should be 1 8 that is the positions of the beginning of
2007 Nov 06
1
How to find the zero (only the real solution) with the package polynom ?
Hello, I have 3 columns : a, b and a*b I would like to find the pair (a,b) so that a*b is the minimum but not from the points I measured but from the fit of the curve (I have more points that the ones given below but I fit only on this part because I know that the minimum a*b is in this interval). I thought doing it this way : - to fit a*b=f(a) abfit<-lm(ab ~ poly(a,8,raw=T)) - to use the