search for: nom

Hi, Thanks for showing me how to use RODBC and odbcConnect. Now works nicely. The question I've got now is: noms <- list.files(pattern=".DBF") # removing extension names: noms <- sapply(noms, function(x) as.character(strsplit(x,".DBF")) , USE.NAMES=F) for (i in 1:length(noms)) { s <- sqlFetch(bdades, noms[i]) # etc. } But it seems that sqlFetch() doesn't accept the...

...([0-9]){4}"? subgrupo_rg <- "_([0-9]){1,3}."? mon_rg <- "SMGM|UMAM|MN"? ? ? ruta <- 'C:/Users/bdominguez/Documents/H0711/Bond/1909/'? archivos<-list.files(path=ruta,pattern = '*.pdf')? ? ? imagen <- image_read_pdf(path=paste(ruta,"/",nombre,".pdf",sep=""))? prueba <-image_ocr(imagen, language = 'eng')? lineas<-unlist(str_split(prueba,pattern = "\n"))? lineasp<-unlist(str_split(prueba[2],pattern = "\r\n"))? ? newnom <- NULL? renglones <- NULL? for (nombre in archivos){?...

Let me present to you my problem : I have a character vector x and I would like to obtain the indices of the elements of this vector that yielded exactly a match. For example, x=nom, pattern="b", I would to obtain 2 because "b" is on the second position. First program : nom <- c("a","b","ab") grep("b",nom) 2 3 Then I try the option extended =FALSE (instead of TRUE by default) and I obtain '2 3' a second...

Hello I have the following problem : The dataframe TEST has multiple lines for a same person because : there are differents values of Nom or differents values of Prenom but the values of Matricule or Sexe or Date.de.naissance are the same. TEST <- structure(list(Matricule = c(66L, 67L, 67L, 68L, 89L, 90L, 90L, 91L, 108L, 108L, 108L), Nom = structure(c(1L, 2L, 2L, 4L, 8L, 5L, 6L, 9L, 3L, 3L, 7L), .Label = c("CHICHE", &qu...

...t;_([0-9]){1,3}."? > mon_rg <- "SMGM|UMAM|MN"? > ? > ? > ruta <- 'C:/Users/bdominguez/Documents/H0711/Bond/1909/'? > archivos<-list.files(path=ruta,pattern = '*.pdf')? > ? > ? > imagen <- image_read_pdf(path=paste(ruta,"/",nombre,".pdf",sep=""))? > prueba <-image_ocr(imagen, language = 'eng')? > lineas<-unlist(str_split(prueba,pattern = "\n"))? > lineasp<-unlist(str_split(prueba[2],pattern = "\r\n"))? > ? > newnom <- NULL? > renglones <- NUL...

...expression(italic(phantom(0)^{137}*Cs)), expression(italic(phantom(0)^{129*m}*Te))), cex=3 ) But, since I have a lot of isotopes to write on the plot, I would like to construct automatically the labels. So I wrote the code below which works fine. listenoms <- list(nom=c("Ge","Cs","Te"),num=c("78","137","129*m")) n <- length(listenoms$nom) resu <- "c(" for( i in 1:(n-1)) { resu <- paste(resu,paste("expression(italic(phantom(0)^{",...

...expression(italic(phantom(0)^{137}*Cs)), expression(italic(phantom(0)^{129*m}*Te))), cex=3 ) But, since I have a lot of isotopes to write on the plot, I would like to construct automatically the labels. So I wrote the code below which works fine. listenoms <- list(nom=c("Ge","Cs","Te"),num=c("78","137","129*m")) n <- length(listenoms$nom) resu <- "c(" for( i in 1:(n-1)) { resu <- paste(resu,paste("expression(italic(phantom(0)^{",...

...onf driver = sql } protocol imap { mail_plugins = quota imap_quota mail_log notify } protocol pop3 { mail_plugins = quota } protocol lda { mail_plugins = quota sieve } dovecot-sql.conf password_query = SELECT \ concat(s_courriel_utilisateur.utilisateur, '@', s_courriel_domaine.nom) AS user,s_courriel_utilisateur.password as password \ FROM s_courriel_utilisateur left join s_courriel_alias on s_courriel_utilisateur.id=s_courriel_alias.utilisateur and s_courriel_alias.archive !=1 ,s_courriel_domaine \ WHERE s_courriel_utilisateur.archive !=1 \...

Dear R community, I'm trying to remove a loop from my code but I'm stock and I can't find a good way to do it. Hopefully one of you will have something clever to propose. Here is a simplified example: I have a squared matrix: > nom.plac2 <- c("102", "103", "301", "303","304", "403") > poids2 <- matrix(NA, 6,6, dimnames=list(nom.plac2,nom.plac2)) > poids2 102 103 301 303 304 403 102 NA NA NA NA NA NA 103 NA NA NA NA NA NA 301 NA NA NA N...

Hy all, I wish to use a variable as rownames for a set of datas. By example : > nom<-"toto" > prenom<-"tutu" > res<-c(1,2) > res<-t(res) > res [,1] [,2] [1,] 1 2 > colnames(res)<-c(nom,prenom) > res toto tutu [1,] 1 2 > nom [1] "toto" > I wish to call the rowname by the var...

Dear R-List, I want to use the label function (from Hmisc library) to allow for the names of my isotopes. library(Hmisc) library(lattice) library(grid) num <- c("78","137","129m") nom <- c("Ge","Cs","Te") df <- data.frame(GE78=seq(nom),CS137=seq(nom),TE129m=seq(nom)) if I use this function to create the labels : lab <- function(i) as.expression(bquote(italic(phantom(0)^{.(num[i])}*.(nom[i])))) label(df$GE78) <- lab(1) label(df$CS13...

Dear all, this is my first message to this mailing list - please advise if it is not the right place for the subject I've been using R very intensively the last 3-4 years and one of the most tedious tasks is modification of lookup or conversion tables So far, I have not found functions that create the commands for creating objects (vectors, data frames) based on the objects themselves -

Dear all, I 'd like to keep the names of variables when calling them in a function. An example might help to understand my problem : The following function puts in a new data frame counts and percent of a data.frame called as "tablo" the step " nom.chiffr[1] <- names(vari) " is useless as names from the original data.frame aren't kept in the function environement. Hoping I use appropriate R-vocabulary, I thank you for your help David descriptif <- function (tablo) { descriptifvar <- function (vari) { table(vari) leng...

Hi, I used Dr. Harrell's rms package to make a nomogram. Below is my code for nomogram and calculate total points and probability *in original data set* used for building the nomogram. *My question is how I get the formula for calculating the survival probability for this nomogram. Then I can use this formula to do validation by using other data...

...e 0.00 0.00 0.00 > iVec2 <- index(Vec2, verbose=TRUE) finalized tree user.self sys.self elapsed sort 7.78 0 7.79 tree 0.00 0 0.00 > > timefactor(any(Vec1=="a999"), any(iVec1=="a999"), 10, 100) finalized tree nom denom factor user.self 0.041 0.0036 11.38889 sys.self 0.000 0.0000 NaN elapsed 0.041 0.0036 11.38889 > timefactor(any(Vec2=="a999"), any(iVec2=="a999"), 10, 100) nom denom factor user.self 0.040 0.0036 11.11111 sys.self 0.000 0.0000 NaN elapsed...

..."gross value added > cup", > "gross value added coppy", "GFCF cup", "GFCF vol", "empl jobs", > "empl FTE", "employees jobs", "employees FTE", "D1", "D11", "D29-D39", > "D29 nom", "D39 nom", "P51c nom", "B2n+B3n", "total hours worked", > "hours worked employees"), var = c("PROD", "PKPY", "VALU", "VKPY", > "GFCF", "GKPY", "EMPN", "FTEN&quot...

...L, 2L, 3L, 3L, 4L, 4L, 5L, 5L, 6L, 6L, 7L, 7L), .Label = c("St1", "St2", "St3", "St4", "St5", "St6", "St7"), class = "factor"), MFI = c(10755.5, 9839, 5142.5, 4857, 1510.5, 1505, 502.5, 451, 215, 195.5, 58, 57, 15, 15), nom = c(206, 206, 125, 125, 68, 68, 38, 38, 24, 24, 13, 13, 6.5, 6.5), weights = c(0.0013946353028683, 0.00152454517735542, 0.00291686922702965, 0.00308832612723904, 0.0099304865938431, 0.00996677740863787, 0.0298507462686567, 0.0332594235033259, 0.0697674418604651, 0.0767263427109974, 0.25862068965517...

...or NT. (eg using: "net use F: \\slackware\files /user:nigels nigel" ) Windows NT reports: The account is not authorised to connect from this station. Windows 2000: The network location cannot be reached. from both windows machines i can do this: C:\>nbtstat -a slackware Table de noms NetBIOS de la machine distante Nom Type Etat --------------------------------------------- SLACKWARE <00> UNIQUE Inscrit SLACKWARE <03> UNIQUE Inscrit SLACKWARE <20> UNIQUE Inscrit ..__MSBROWSE__.<01> GROUP...

M?XICO NOM-001-SEMARNAT-2021 Comunicado del 27 de Enero 2023 Actualizaci?n de la NOM-001-SEMARNAT-2021 Aplicaci?n Regulatoria Obligatoria a partir del 11 de Marzo 2023 Fecha del evento 24 de Agosto 2023 Online v?a Internet Capacitaci?n Indispensable para Negocios y Empresas que realicen Descarga de Aguas...

...;ai vu qu''on pouvait g?n?rer des ?l?ment d''un formulaire gr?ce des m?thodes. Apres quelque petite recherches j''ai trouv? comment g?n?rer un select <%= select(''category'', ''category.id'', Category.find_all.collect {|category| category.nom}) %> comment faire pour utiliser l''option selected ? Merci -- Bolo Michelin Blog [Flash & RoR] http://blog.developpez.com/?blog=30 -------------- next part -------------- An HTML attachment was scrubbed... URL: http://wrath.rubyonrails.org/pipermail/rails/attachments/20060425/7c0...