search for: ex3

...", "f"), e = 30)) ex2 <- data.frame(v1b = log(abs(rt(18, 3)) + 1), v2 = rep(c("2007", "2006", "2005"), 6), z = rep(c("a", "b", "c", "d", "e", "f"), e = 3)) ex3 <- merge(ex, ex2, by=c("v2","z")) D2007 <- ex3[ex3$z=="d" & ex3$v2==2007, ] D2006 <- ex3[ex3$z=="d" & ex3$v2==2006, ] C2007 <- ex3[ex3$z=="c" & ex3$v2==2007, ] quantile(D2007$v1, probs = c(0.05, 0.95)) quantile(D2006$v1, prob...

...0)) # the individual to be marked ex2 <- data.frame(v1b = log(abs(rt(18, 3)) + 1), v2 = rep(c("2007", "2006", "2005"), 6), z = rep(c("a", "b", "c", "d", "e", "f"), e = 3)) ex3 <- merge(ex, ex2, by=c("v2","z")) bwplot(v2 ~ v1 | z, data = ex3, layout=c(3,2), pch = "|", par.settings = list( plot.symbol = list(alpha = 1, col = "transparent",cex = 1,pch = 20)), panel = function(x, y){ panel.bw...

...instead of tropical and temperate are three ontogenetic states Larva Met Juv CTMAX CTMIN SP SP SP This is my scrip: Ex = subset(Expr, Outlayer=="N") Ex2 = subset (Ex, S0 == 1) Ex3 = subset(Ex2, Experimento == "CTMIN") Ex4 = subset(Ex2, Experimento == "CTMAX") Ex3$Stage <- ordered(Ex3$Stage, levels=c("LARVA", "MET", "JUV")) Ex4$Stage <- ordered(Ex4$Stage, levels=c("LARVA", "MET", "JUV")) par...

I've got two tables.... first one(table1): ID chrom start end Ex1 2 152 180 Ex2 10 2000 2220 Ex3 15 3000 4000 second one ( table2): chrom location name 2 160 Alv 2 190 GNN 2 100 ARg 10 210 GGG 15...

...g): error:12800067:DSO support routines::could not load the shared library: filename(libz.so): libz.so: cannot open shared object file: No such file or directory, error:12800067:DSO support routines::could not load the shared library: user=<>, rip=192.168.117.5, lip=192.168.117.2, session=<eX3e+Uv3k8DAqHUF> Mar 20 13:49:30 imap-login: Info: Disconnected: TLS initialization failed. (no auth attempts in 0 secs): user=<>, rip=192.168.117.5, lip=192.168.117.2, session=<eX3e+Uv3k8DAqHUF The library files exist in the include directory [/opt] # find . -name libz* -print ./lib/opkg...

...Linux R-2.3.0-2.fc4 R version 2.3.0 (2006-04-24) Copyright (C) 2006 R Development Core Team $ wc -l dataset.010MM.txt 10000000 dataset.010MM.txt $ head -3 dataset.010MM.txt 15623 3845 22309 $ wc -l dataset.100MM.txt 100000000 dataset.100MM.txt $ head -3 dataset.100MM.txt 15623 3845 22309 $ cat ex3.r options(width=1000) foo <- read.delim("dataset.010MM.txt") summary(foo) foo <- read.delim("dataset.100MM.txt") summary(foo) $ R < ex3.r R > foo <- read.delim("dataset.010MM.txt") R > summary(foo) X15623 Min. : 1 1st Qu.: 8152 Median :1...

...gt; 3. Traversing a data structure, where the recursion is simply to get > > to the leaves which is where all the fun stuff happens. > > > > > > Points 2 and 3 both share a trait that they're dividing/fanning out > > at each step: > > > > > > void ex3(i) { > > if (is_base(i)) { > > f(); > > return; > > } > > g(i); > > ex3(i / 2); > > ex3(i / 2 + 1); > > } > > > > > > The access pattern of such a function is not linear. It is a > > pre-order walk of a (binary, in this case) t...

...ase... 2. Divide and conquer algorithms. Quicksort, FFT butterflies, etc. 3. Traversing a data structure, where the recursion is simply to get to the leaves which is where all the fun stuff happens. Points 2 and 3 both share a trait that they're dividing/fanning out at each step: void ex3(i) { if (is_base(i)) { f(); return; } g(i); ex3(i / 2); ex3(i / 2 + 1); } The access pattern of such a function is not linear. It is a pre-order walk of a (binary, in this case) tree. As such we can't use the two-loop transform above, but would...

...is_uninitialized(newext), 1724 <+++<+++<+++<+++ext4_ext_get_actual_len(newext), 1725 <+++<+++<+++<+++nearex, len, nearex + 1, nearex + 2); 1726 <+++<+++memmove(nearex + 1, nearex, len); 1727 <+++<+++path[depth].p_ext = nearex; 1728 <+++} 2678 <+++<+++ex3 = &newex; 2679 <+++<+++ex3->ee_block = cpu_to_le32(iblock + max_blocks); 2680 <+++<+++ext4_ext_store_pblock(ex3, newblock + max_blocks); 2681 <...

...t was somewhat tricky. For completeness, I just wanted to show the substitute construction, which actually doesn't seem all that tricky to me. Here is the complete code sequence to check: ex1 <- expression(x < a) ## original ex2 <- bquote( .(ex1[[1]]) & y < b) ## Using bquote ex3 <- substitute( z & y < b,list(z = ex1[[1]])) ## using substitute directly identical(ex2,ex3) ## TRUE Cheers to all, Bert Gunter Genentech Nonclinical Statistics

...(it would be so awesome to have them in the introductory manual!) #Try to change the placement of the keywords and you are dead! 4 examples Ex1: if (1==1){ ?print('if') ?print('if again') ?}else ?print('else') Ex2: if (2==2) print('if') else print('else') Ex3: if (2==2){ ?print('if') ?print('if again') ?}else ?{ ?print('else') ?print('else2') ?} Ex4: if (2==2){ ?print('if') ?print('if again') }else print('else') cheers, ------------------------------------- Daniela

Hi, I seem to have a key colour problem with a lattice barchart. The colours on my key rectangle don't match the colours on the barchart. Here is my data frame: "LandFill" "Ruminants" "United States (USA)" 7777.21428 5528.16 "France" 200.527083 1299.87 "Australia" 185.878368 2448.17 "Russian Federation" 1752.8334 2024.29

...gt; 3. Traversing a data structure, where the recursion is simply to get > > to the leaves which is where all the fun stuff happens. > > > > > > Points 2 and 3 both share a trait that they're dividing/fanning out > > at each step: > > > > > > void ex3(i) { > > if (is_base(i)) { > > f(); > > return; > > } > > g(i); > > ex3(i / 2); > > ex3(i / 2 + 1); > > } > > > > > > The access pattern of such a function is not linear. It is a > > pre-order walk of a (binary, in this case) t...

...:could not load the shared library: >> filename(libz.so): libz.so: cannot open shared object file: No such file >> or directory, error:12800067:DSO support routines::could not load the >> shared library: user=<>, rip=192.168.117.5, lip=192.168.117.2, >> session=<eX3e+Uv3k8DAqHUF> >> Mar 20 13:49:30 imap-login: Info: Disconnected: TLS initialization >> failed. (no auth attempts in 0 secs): user=<>, rip=192.168.117.5, >> lip=192.168.117.2, session=<eX3e+Uv3k8DAqHUF >> >> The library files exist in the include director...

...X2 1 5 45,98,78 2 7 45,98,78 3 9 45,98,78 However, if my dataframe does not already exist, then I cannot create a dataframe with vectors using the following syntax: data.frame(x1=list(c(5,6,7),c(8,9,10))) What I want: ex3 X1 1 5,6,7 2 8,9,10 What I get: x1.c.5..6..7. x1.c.8..9..10. 1 5 8 2 6 9 3 7 10 R seems to pull out each element in each list component and treat them as separate ele...

...colonn is for as.factor) ; 5 x-values. I fail to use 'abline' command (regression's line don't display) in this particular context, on a 'stripchart' display. Any idea ? Here are the few commands that I execute, the lattest being my problem : > tab <- read.csv2("ex3.csv") > sapply(tab,data.class)     sujet         x        xf        lx       lxf         y "numeric" "numeric" "numeric" "numeric" "numeric" "numeric" > attach(tab) > lxf <- as.factor(lxf) > model1 <- lm(y~lx) &gt...

Dear Mr. H. Peter Anvin, I am trying to figure out the physical address of where the PXE DOS image (downloaded in PXE) is stored by MEMDISK in the Extended Memeory. Is it at E98000h as specified in the PXE spec or the address to is random. What I am trying to figure out is a safe range of extended memory segment (100000h to E98000h), where HP tools can safely allocate extended memory in DOS

Hi, there, Is there any way to undelete my files on ex3 file system? For whatever reason, I made a stupid mistake and deleted some very important files. I search on Internet and it's said there is no way to undelete under ext3. Is that absolutely true? Here is my last hope. I unmount the partition immediately (by force, device always busy). In fac...

...ncomplete Data in Generalized Linear Models" by Joseph G Ibrahim. i have half way through developing the R code based on this this paper, I need little help in tweaking my code furthure. ---------------------------------------------------------------------------- a=read.table("C:/sudhi/ex3.csv",header=T,sep=",") #Set design matrix n=1558 X1=as.matrix(rep(1,n)) X2=as.matrix((a[,1])) X3=as.matrix((a[,2])) X4=as.matrix((a[,3])) X5=as.matrix((a[,4])) X6=as.matrix((a[,5])) X7=as.matrix((a[,6])) X8=as.matrix((a[,7])) X=cbind(X1,X2,X3,X4,X5,X6,X7,X8) #Start Newton-Raphson...

If you go to http://hpux.cs.utah.edu/hppd/hpux/Networking/Misc/samba-2.2.3a/ you can at least get the precompiled version of 2.2.3a for 10.20 if that will work for you. Otherwise roll your own. Thanks, Mike -----Original Message----- From: CHENEY,JOHN (HP-Australia,ex3) [mailto:john.cheney@hp.com] Sent: Wednesday, January 22, 2003 2:44 PM To: 'twilson@stl-inc.com' Cc: 'samba@lists.samba.org' Subject: [Samba] RE: samba digest, Vol 1 #2098 - 22 msgs Tim Looks like you got a binary compiled for HP-UX 11.x, as that version of HP-UX introduced libc....