similar to: Best way to select good points in a noisy signal ?

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

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

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

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 : --------------------------------------- |

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]

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

Hi, I read the posts for 2 hours and ?list and tried many comninations but I haven't found the answer to this basic question. So I decided to post my question even if it is a silly one ... What is the instruction to retrieve, for example, the "D" of the first list ? Thanks in advance, Ptit Bleu. > x<-list(LETTERS[1:5], LETTERS[10:20]) > x [[1]] [1] "A"

Hello, I would like to fit data with the following formula : y=V*(1+alpha*(x-25)) where y and x are my data, V is a constant and alpha is the slope I'm looking for. How to translate this into R-language ? At the moment, I only know : lm(y ~ x) Sorry for such a basic question. I thought I could find the solution in a post but I have to confess that, up to know, I'm not able to understand

Hi, I keep on trying to write some small scripts in order to learn R but even with basic scripts I have problems ... I start with the name of a file which is in fact the time the file has been generated (I cannot change the format). Then I convert namefile with strptime. The problem occurs when I add another time from another file with append. It displays some informations I don't want. I

Hello, I have a list of data.frame rowsplit : List of 15 $ (0,0.025] :'data.frame': 169 obs. of 7 variables: $ (0.025,0.05]:'data.frame': 174 obs. of 7 variables: $ (0.05,0.075]:'data.frame': 92 obs. of 7 variables: $ (0.075,0.1] :'data.frame': 76 obs. of 7 variables: $ (0.1,0.125] :'data.frame': 37 obs. of 7 variables: $

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

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

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

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

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

Hello, I'm trying to produce graphs automatically from data stored in database. Before saving the graphs, I would like to maximize the size of the graphs. The best would be to directly open maximized windows with x11() but up to now I failed doing it. I tried different widths and heighs but I never managed to obtain a full screen window. Is there a command to do it ? Thanks in advance, Ptit

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

Hello, I discovered R two years ago and thanks to the R-community I managed to write some scripts to analyze my data stored in mysql databases. The only problem is that I am the only one using R in the lab. Colleagues mainly use Matlab (but not with mysql, only with text files) but regularly come to me to get data treated with R-scripts !!!. To allow the use of my scripts by other people, my

Hello, I'm trying to plot dayly evolution of the temperature over France from Global Forecast System files ("I'm trying" is the right expression...). akilonlat03 is the temperature for different latitudes and longitudes ? 3 o'clock. akilonlat06 is the temperature for different latitudes and longitudes ? 6 o'clock. I would like to plot akilonlat03 and then akilonlat06

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