search for: guazzetti

...to be able to stop a stupid thought in its pathway from the brain to the keyboard". :-) Sorry once again and thank for your patience. Stefano > -----Messaggio originale----- > Da: Prof Brian Ripley [mailto:ripley at stats.ox.ac.uk] > Inviato: venerd?? 23 luglio 2004 16.30 > A: Guazzetti Stefano > Cc: Luis Rideau Cruz; r-help at stat.math.ethz.ch > Oggetto: Re: R: [R] retrieve rows from frame assuming criterion > [corrected] > > > On Fri, 23 Jul 2004, Guazzetti Stefano wrote: > > > sorry for my previus (WRONG) > > answer, as someone already point...

...a "pu" object is returned by curve > pu [1] -0.1802445 1.1802445 The behaviour is reproducible with both R 2.0.0 and R 2.0.1 I can see that a promise of evaluation of "pu" is made within curve but I cannot understand completely what happens. Thanks in advance, Stefano Guazzetti platform i386-pc-mingw32 arch i386 os mingw32 system i386, mingw32 status major 2 minor 0.0 year 2004 month 10 day 04 language R

sorry for my previus (WRONG) answer, as someone already pointed out a solution could be subset(data, PUNTAR==c("IX49","IX48")) > -----Messaggio originale----- > Da: Guazzetti Stefano > Inviato: venerd?? 23 luglio 2004 15.55 > A: 'Luis Rideau Cruz'; r-help at stat.math.ethz.ch > Oggetto: R: [R] retrieve rows from frame assuming criterion > > > Assuming your data.frame is called "data" > > data[data$PUNTAR==c("IX49",...

...c("b", "c")) a b 3 b 3 7 b 7 9 b 9 10 c 10 > subset(x, a %in% c("b", "c")) a b 3 b 3 4 b 4 7 b 7 8 b 8 9 b 9 10 c 10 What matters is the use of "%in%" instead of "==", not the use of subset(). Andy > From: Guazzetti Stefano > > sorry for my previus (WRONG) > answer, as someone already pointed out > a solution could be > > subset(data, PUNTAR==c("IX49","IX48")) > > > > > -----Messaggio originale----- > > Da: Guazzetti Stefano > > Inviato: ve...

It seems that Access needs that you surround the dates with a # symbol. You probably need something like. res <- sqlQuery(channel, "select * from test_table where market = 'atl-bos' and competitor = 'delta' and dd = #2007-11-20#") Hope this helps, Stefano -----Messaggio originale----- Da: r-help-bounces at r-project.org [mailto:r-help-bounces at r-project.org]Per

a closer look to the help on predict.glm will reveal that the function accepts a 'type' argument. In you case 'type = response' will give you the results in probabilities (that it seems to be what you are looking for). There also is an example on use of the 'type' argument at the end of the page. Stefano -----Messaggio originale----- Da: r-help-bounces at r-project.org

One possibility ist boxplot(sapply(ListOfNames, get, env = .GlobalEnv)) Hope that this helps Thomas -----Urspr??ngliche Nachricht----- Von: Lefebure Tristan [mailto:Tristan.Lefebure at univ-lyon1.fr] Gesendet: Mittwoch, 7. Juli 2004 10:34 An: r-help at stat.math.ethz.ch Betreff: [R] boxplot a list of objects Hi list, #Imagine we have vectors of different length (in practice 100 vectors):

Dear listers, Let use say that I want to display the Pixel dataset (in the NLME library) xyplot(pixel~day|Dog, groups=Side, data=Pixel, panel=panel.superpose, panel.groups=panel.xyplot, type="b", pch=16) (I know, there are better examples, ...) Now, how could I change the colours of the symbols accordingly to another factor changing within subject and side form time to time? (let us

...the function itself > replace function (x, list, values) { x[list] <- values x } <environment: namespace:base> Stefano >-----Messaggio originale----- >Da: r-help-bounces at stat.math.ethz.ch >[mailto:r-help-bounces at stat.math.ethz.ch]Per conto di >Guazzetti Stefano >Inviato: Thursday, May 04, 2006 07:55 AM >A: isaia at econ.unito.it; ruser2006 at yahoo.com >Cc: r-help at stat.math.ethz.ch >Oggetto: [R] R: [Re:] function to replace missing values with median >value?]] > > >there is also a replace function...

you could try mapply mydata2<-mapply("[", mydata, lapply(mydata, function(x) !x %in% A)) mydata2[[1]]<-A #to replace the obviously deleted elements of "A" mydata2 mydata2[[1]] mydata2[[2]] mydata2[[3]] mydata2[[4]] Stefano -----Messaggio originale----- Da: r-help-bounces at stat.math.ethz.ch [mailto:r-help-bounces at stat.math.ethz.ch]Per conto di jastar

trellis.device(bg="white", color=F) before your call to splom could make what you want but take also a look at ?trellis.par.set Stefano > -----Messaggio originale----- > Da: v.demartino2 at virgilio.it [mailto:v.demartino2 at virgilio.it] > Inviato: mercoled? 3 marzo 2004 12.10 > A: r-help > Oggetto: [R] Changing background in splom et al. > > > Context:

You probably mean something like: ti <- 1:1000 e1 <- rnorm(1000) e2 <- rnorm(1000) x <- 0.0001*ti+e1 y2 <- -2+x+e2 y <- ifelse(y2>0,1,0) plot(x, y, pch = 16, col = "darkblue", main = expression(paste("Scatter diagram of ", italic(y[t]), "against ", italic(x[t]))), xlab = expression(italic(x[t])), ylab =

Take a look at the "scales" argument in dotpolot. Maybe you need something like: position<-exp(-5:0)/(1+exp(-5:0)) dotplot(type~freq/(3027-freq), scales=list(x=list(log=T, at=position, lab=round(position, 3))) ) Stefano >-----Messaggio originale----- >Da: r-help-bounces at stat.math.ethz.ch >[mailto:r-help-bounces at stat.math.ethz.ch]Per

assuming that the rows are sorted correctly dat id x y 1 50 1647685 4815259 2 50 1647546 4815196 3 50 1647454 4815294 4 50 1647405 4815347 5 50 1647292 4815552 6 50 1647737 4815410 7 74 1647555 4815201 8 74 1647464 4815023 9 74 1646970 4815129 10 74 1646895 4815264 11 74 1646762 4815513 > list.dat<-split(dat, dat$id) > > closed.polygons<-lapply(list.dat,

Perhaps using 'ave' and 'cut': df <- data.frame(x=runif(100, 0.1, 1), y=rnorm(100, 0.2, 0.6)) df$xcut<-cut(df$x, seq(0, 1, 0.1)) df$z<-ave(df$y, df$xcut) df[order(df$x),] Stefano -----Messaggio originale----- Da: r-help-bounces at stat.math.ethz.ch [mailto:r-help-bounces at stat.math.ethz.ch]Per conto di e.rapsomaniki at mail.cryst.bbk.ac.uk Inviato: venerd? 15

Oats[Oats$Variety %in% c("Victory", "Golden Rain"),] or subset(Oats, Variety %in% c("Victory", "Golden Rain")) Stefano -----Messaggio originale----- Da: r-help-bounces at stat.math.ethz.ch [mailto:r-help-bounces at stat.math.ethz.ch]Per conto di elyakhlifi mustapha Inviato: luned? 23 aprile 2007 9.56 A: R-help at stat.math.ethz.ch Oggetto: [R] extract from

try this > a<-c(1, 6, 8, 9) > 1*(1:10 %in% a) [1] 1 0 0 0 0 1 0 1 1 0 Stefano -----Messaggio originale----- Da: r-help-bounces at r-project.org [mailto:r-help-bounces at r-project.org]Per conto di njhuang86 Inviato: Friday, June 12, 2009 4:45 PM A: r-help at r-project.org Oggetto: [R] Creating this vector, any suggetions? Suppose I have the first vector: c(1, 6, 8, 9) I will

?unique as an example > mat<-matrix(c(1,2,3,1,1,2,1,2,3,4,7,5), ncol=3, byrow=T) > mat #rows 1 and 3 are identical [,1] [,2] [,3] [1,] 1 2 3 [2,] 1 1 2 [3,] 1 2 3 [4,] 4 7 5 > unique(mat) [,1] [,2] [,3] [1,] 1 2 3 [2,] 1 1 2 [3,] 4 7 5 Stefano -----Messaggio originale----- Da: r-help-bounces at r-project.org

It seems to me that you are using polygon in a wrong way. What you probably need could be something like: polygon(c(rev(t$z), t$z), c(rep(0, nrow(t)), t$ht), col=2, border=NA) Stefano -----Messaggio originale----- Da: r-help-bounces at stat.math.ethz.ch [mailto:r-help-bounces at stat.math.ethz.ch]Per conto di LL Inviato: sabato 26 maggio 2007 12.34 A: r-help at stat.math.ethz.ch

?any any(x==2) Stefano -----Messaggio originale----- Da: r-help-bounces at r-project.org [mailto:r-help-bounces at r-project.org]Per conto di Dimitri Liakhovitski Inviato: mercoled? 23 settembre 2009 17.38 A: R-Help List Oggetto: [R] Function to check if a vector contains a given value? Dear R'rs, is there a function that checks if a given vector contains a certain value. E.g.,