search for: bleu

...a is not indexed to give: dd.deriv2 <- function (Blev, beta, gamm, GL) { .expr1 <- GL^gamm .value <- Blev + rep(beta, each = 17) * .expr1 .grad <- array(0, c(length(.value), 5), list(NULL, c("Blev", "beta.rouge", "beta.vert", "beta.bleu", "gamm" ))) .grad[, "Blev"] <- 1 .grad[1:17, "beta.rouge"] <- .expr1[1:17] .grad[18:34, "beta.vert"] <- .expr1[1:17] .grad[35:51, "beta.bleu"] <- .expr1[1:17] .grad[, "gamm"] <- ifelse(GL == 0, 0,...

...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 idea ? Thanks in advance, Ptit Bleu (with R 2.9.0 under Windows ). -- View this message in context: http://r.789695.n4.nabble.com/Problem-with-as-POSIXlt-hour-tp2255758p2255758.html Sent from the R help mailing list archive at Nabble.com.

...les of the directory to the database. Is it a problem with the way I save tfichiers or is it a problem with the condition if (nfichiers!="0") ? Could you please give me some advices to correct my script (written with Tinn-R) ? I thank you in advance for your help. Have a nice week, Ptit Bleu. PS : Ptit Bleu means something like "Full Newbye" in french. So thanks to be patient :-) PPS : I hope you understand my french english ------------------------------------------ # Connexion a la base de donnees database de MySQL library(DBI) library(RMySQL) drv<-dbDriver("My...

...t;) and it will change namevar by Device2 into the file myfile.rnw before creating the .tex file ? Or is it possible to do it via another language, that is: loading the rnw file, modifying it, saving it and finally calling Sweave ? Whatever the solution, I need you help. In advance thank you. Ptit Bleu. -- View this message in context: http://www.nabble.com/Sweave-%3A-change-value-in-rnw-file-to-generate-multiple-%22single%22-reports---tp14256204p14256204.html Sent from the R help mailing list archive at Nabble.com.

...re 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 change the structure of the function). I hope you understand my french explainations ... Thanks in advance for your help, Have a nice week-end, Ptit Bleu. df1<-data.frame(c(1,2,3,4), c(9,6,5,8), c(5,4,8,6)) names(df1)<-c("C1", "C2", "C3") fcttest<-function(a1,a2,a3) { v<-a1*a2 w<-a2+a3 return(c(v,w)) } # The following for loop works result<-data.frame() for (i in 1:length(df1[,1])) { result&l...

...polynom. But can I do this and, if yes, how can I detect and automatically replace these "NA" ? To conclude, I must say that I read some posts on NA but I must confess that it is still not clear to me (maybe because I'm french and R-newbie ...) Thanks in advance for your help, Ptit Bleu. pfit Call: lm(formula = pfitmax ~ poly(vfitmax, 8, raw = T)) Coefficients: (Intercept) poly(vfitmax, 8, raw = T)1 poly(vfitmax, 8, raw = T)2 poly(vfitmax, 8, raw = T)3 poly(vfitmax, 8, raw = T)4 -2.790e+04 6.276e+03 -5.645...

...these numbers, I tried to convert them using as.numeric. But then I get as.numeric("095842087016731010") 95842087016731008 I understand the disappearing of the first 0 but I don't understand the substitution of the last 0 by 8. Is there a way to avoid that ? Thanks in advance, Ptit Bleu. -- View this message in context: http://r.789695.n4.nabble.com/Problem-with-long-number-from-character-to-numeric-class-tp3297121p3297121.html Sent from the R help mailing list archive at Nabble.com.

...d 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 welcome. Thanks in advance, Ptit Bleu. # aaa : x-axis # bbb : y-axis # ddd : group 1 or group 2 # fff : 2 different dates aaa<-rep(seq(1:100),2) bbb<-aaa*aaa*c(rep(1,50),rep(0.5,50)) ddd<-c(rep(1,50),rep(2,50)) fff<-rep(c(rep("2010-11-01",25),rep("2010-11-10",25)),2) library(lattice) # bbb as a funct...

...ppez.net/forums/d740403/autres-langages/algorithmes/contribuez/image-geometrie-projective-homography/ I've searched with these keywords: R-project homography warping, but with no success. I'm not very good at R and I don't know java. So any help is welcome. Have a nice week-end, Ptit Bleu. -- View this message in context: http://r.789695.n4.nabble.com/Homography-with-R-tp3837111p3837111.html Sent from the R help mailing list archive at Nabble.com.

...t 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 english), it seems that it doesn't work with values in data frame. Could you please help me to do this ? Thanks in advance, Have a nice week, Ptit Bleu. -- View this message in context: http://www.nabble.com/Must-be-easy%2C-but-haven%27t-found-the-function-%28numerical-integration%29-tf4465684.html#a12732936 Sent from the R help mailing list archive at Nabble.com.

...posts about this topic but all need a server (if I read well). I'd just want to use it on a local machine. Is there a way to realize such an interface and to link it directly to an R-script (which will send the query to MySQL and will plot the result of the query) ? Thanks for your help, Ptit Bleu. -- View this message in context: http://www.nabble.com/Friendly-way-to--link-R---MySQL-and-non-%28R-and-Mysql%29-users---tf4844081.html#a13858847 Sent from the R help mailing list archive at Nabble.com.

...I tried "manually" with a small vector : > s [1] "aa.P0" "bb.P0" "cc.P1" "dd.P2" > s[grep(".P[^0]", s)] [1] "cc.P1" "dd.P2" It works !!! Has someone an idea to solve this small problem ? Thanks in advance, Ptit Bleu. -- View this message in context: http://www.nabble.com/Must-be-obvious-but-not-to-me-%3A-problem-with-regular-expression-tp14370723p14370723.html Sent from the R help mailing list archive at Nabble.com.

...ot;,"r","t","a","z","a","z","e","c")) 1 2 3 meaning that each character is in the bigger vector. It must be easy, except for me. Sorry. If you have a solution, thanks in advance to share it. Have a good week-end, Ptit Bleu. -- View this message in context: http://www.nabble.com/find-a-sequence-of-characters-in-a-vector-tp23888063p23888063.html Sent from the R help mailing list archive at Nabble.com.

...sage : uniroot(deriv(polynomial(pfit$coefficients)), c(0.8,0.2)) Error in uniroot(deriv(polynomial(abfit$coefficients)), c(0.8, 0.2)) : Impossible to find the function "f" Is there a way to extract this real solution ? I thank you in advance for your help, Have a nice day, Ptit Bleu. ------------------------------------------------- Set of points a b a*b 0.8 0.033320357 0.026656286 0.76 0.023930636 0.018187283 0.72 0.0153903 0.011081016 0.68 0.007844914 0.005334542 0.64 0.001598132 0.001022804 0.6 -0.00315939 -0.001895634 0.56 -0.006402796 -0.003585...

...site a possible solution using rtools but it was very unclear to me (I'm only an end-user and French ...) and with no admin rights, I think I won't be able to follow the proposed procedure. Do you know where I can find a zip file of the RMySQL package for R 2.12 ? Thanks in advance, Ptit Bleu. PS : everybody around me is using matlab. If I say that I have a problem to connect R to MySQL, I will have to move to matlab. So please, help !!! :-) -- View this message in context: http://r.789695.n4.nabble.com/looking-for-the-RMySQL-package-for-R-2-12-0-under-XP-tp3057537p3057537.html Se...

...t) to translate them into matlab langage (and then into .exe) or into C++ meaning abandoning R. And I don't want to. So is it possible to make executables from R scripts ? It seems it is not the case but I hope I missed a way to do it. Thanks in advance for your answers, Have a nice day, Ptit Bleu. -- View this message in context: http://n4.nabble.com/executable-R-script-under-xp-to-avoid-migration-toward-Matlab-or-C-tp1555965p1555965.html Sent from the R help mailing list archive at Nabble.com.

...ile xy.txt ( http://www.nabble.com/file/p18773387/xy.txt xy.txt ), I can clearly see a trend. Is there a function or a package able to take the median value of y for an interval of x (x +/- a defined value) to plot nice graph (at least a better one) ? Thanks in advance, Have a nice week-end, Ptit Bleu. -- View this message in context: http://www.nabble.com/Best-way-to-select-good-points-in-a-noisy-signal---tp18773387p18773387.html Sent from the R help mailing list archive at Nabble.com.

...c(as.POSIXct(unique(as.Date(strptime(data$date, format="%Y-%m-%d %H:%M:%S)))[1])) corresponding to "2008-08-01" gives 1217376000 I think the last value is ok but I don't understand why the par()$usr values are so low. Any idea ? Thanks in advance and have a nice week-end, Ptit Bleu. -- View this message in context: http://n4.nabble.com/Problem-with-date-x-axis-tp948642p948642.html Sent from the R help mailing list archive at Nabble.com.

...al to the array extent If I do the same with a table with only 44 columns, without the autoincremental column, it works ... but I need this last column. Could you please help me to modify my commands so that I can copy 44-columns data.frame to the 45-columns MySQL table ? Thanks in advance, Ptit Bleu. -- View this message in context: http://r.789695.n4.nabble.com/sqlsave-and-mysql-database-with-autoincremental-column-tp3251425p3251425.html Sent from the R help mailing list archive at Nabble.com.

...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 the posts I read concerning lm (the level of the questions are too high for me and my english quite poor as well). Have a nice evening, Ptit Bleu. -- View this message in context: http://www.nabble.com/Newbie%27s-question-about-lm-tp18511262p18511262.html Sent from the R help mailing list archive at Nabble.com.