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...fferent directory. I would like to migrate to samba4 AD-DC but i need to keep extra-domain users and non-samba attributes. I would like keep all users in LDAP and syncing their passwords from samba with "passwd program" but samba4 ignores these directives. This is my system Linux anubi.ausl.fe.it 2.6.32-358.6.1.el6.i686 #1 SMP Tue Apr 23 18:13:20 UTC 2013 i686 i686 i386 GNU/Linux My samba version (rpm from SOGo Repository but i tested also compiling from 4.0.5 sources) Version 4.0.1-4.centos6.1 My smb.conf # Global parameters [global] workgroup = AUSLFE realm =...

is there anyone who can explain me how to use "passwd program" in samba4 -- *Dr. Michael Cinti* *mi.cinti at ausl.fe.it <mailto:mi.cinti at ausl.fe.it>* U.O. Tecnologia della Comunicazione e della Informazione (I.C.T.) Azienda Usl Ferrara Ospedale del Delta - via Valle Oppio, 2 - 44023 Lagosanto (FE) Tel. +39-0533-723221 Tel. +39-0533-723163 ---------------------------------------------------------------...

Dear all, curve(x^3*(1-x)^7, from = 0, to = 1) works as expected but, omitting the "xlim" or the "to" and "from" arguments and calling "curve" more than once: par(mfrow = c(2,2)) for (i in 1:4) curve(x^3*(1-x)^7) gives an expected (al least to me) result. Note also that a "pu" object is returned by curve > pu [1] -0.1802445 1.1802445

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

...-- Matt --------------------------------------------------------------------------------------------- Thanks a lot for answering! Greetings, David __________________________________________________________________________ Verschicken Sie SMS direkt vom Postfach aus - in alle deutschen und viele ausl?ndische Netze zum gleichen Preis! https://produkte.web.de/webde_sms/sms

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

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 =

Yes, paraphrasing Murphy I can say of myself: "Nothing seems 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

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.,

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

what about using "mapply"? splitted.value<-with(x.1, split(VALUE, GROUP)) splitted.freq<-with(x.1, split(FREQUENCY, GROUP)) mapply(weighted.mean, splitted.value, w=splitted.freq) Stefano >-----Messaggio originale----- >Da: r-help-bounces at stat.math.ethz.ch >[mailto:r-help-bounces at stat.math.ethz.ch]Per conto di >james.holtman at convergys.com

oops!, I pressed the 'send' key too soon, ... see ?replace replace(x, is.na(x), median(x, na.rm=T)) take also a look at 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