search for: z1

...this string into function ( R function ). thanks for any tips. cleber #e.g. fun_String = "-100*x1 + 0*x2 + 100*x3" fun <- function(x1,x2,x3){ return( ############ evaluation( fun_String ) ############ ) True String mathematical function :-( :-( > nomes [1] "8.49*x1*z1 + 6.13*x1*z2 + 6.4*x1*z3 + 6.9*x2*z1 + 4.54*x2*z2 + 3.99*x2*z3 + 19.31*x3*z1 + 12.49*x3*z2 + 3.86*x3*z3 + 5.25*x1*z1*z2 + -6.2*x1*z1*z3 + 9.07*x1*z2*z3 + 10.87*x2*z1*z2 + 9.78*x2*z1*z3 + 49.05*x2*z2*z3 + 4.56*x1*x2*z1 + -4.9*x1*x2*z2 + 4.01*x1*x2*z3 + -0.39*x3*z1*z2 + 14.34*x3*z1*z3 + 0.7*x3*z2*z3...

Hi, R users, I'm trying to transform a matrix A into B (see below). Anyone knows how to do it in R? Thanks. Matrix A (zone to zone travel time) zone z1 z2 z3 z1 0 2.9 4.3 z2 2.9 0 2.5 z3 4.3 2.5 0 B: from to time z1 z1 0 z1 z2 2.9 z1 z3 4.3 z2 z1 2.9 z2 z2 0 z2 z3 2.5 z3 z1 4.3 z3 z2 2.5 z3 z3 0 The real matrix I have is much larger, with more than 2000 zones. But I think it should be the same thing if I can transform A into B. Thanks. Gar...

...;fontsize fontsize$default<-16 trellis.par.set("fontsize",fontsize) a<-c(1,2,4,5,4,3,3,3) b<-c(2,5,1,1,1,3,3,3) c<-c(3,5,1,2,2,5,5,5) as.data.frame(cbind(a,b,c))->q21 varnames<-colnames(q21) i<-9999 for (i in 1:3) { round(table(q21[,i])/sum(table(q21[,i])),3)*100->z1 as.data.frame.table(table(q21[,i]))->z1 round(z1[,2]/sum(z1[,2]),3)*100->z1[,2] colnames(z1)<-c(varnames[i],"y") xnames<-levels(z1[,1]) paste(xnames[1]," - low",sep="")->xnames[1] paste(xnames[length(xnames)]," - high",sep=&qu...

...attice_0.20-6 foreign_0.8-50 [9] rgdal_0.7-18 sp_0.9-99 loaded via a namespace (and not attached): [1] RCurl_1.91-1 grid_2.15.1 tools_2.15.1 twitteR_0.99.19 It's very simple but I am not able to plot it with following commands for (i in 1:9) {ifelse(z1$e[[i]] > 0 , plot(z1$x,z1$y),plot(z1$x,z1$e))} Or ifelse(z1$e > 0,plot(z1$x,z1$y),plot(z1$x,z1$e)) Bhupendrasinh Thakre [[alternative HTML version deleted]]

...not a single suggestion. I believe I have provided sufficient information for at least some help. Here I repost the question with several modifications. I want to run nested fixed-factor Anova in R on different experiments. I have 48 levels of the main factor x1 and 242 levels of the nested factor z1, and continuous response variable y1 with around 15.000 data points. There is no interaction between specification of the main factor and nested factor levels. When I run lm on using nested design anov1=lm(c(as.vector(y1))~as.factor(x1)+ as.factor(x1)/as.factor(z1)) I get a warning summary(anov...

Hello again, after obtaining the keytab file I tried to use kinit keytab.file followed by the spn $ samba-tool spn list z1 z1 User CN=z1,CN=Users,DC=home,DC=lan has the following servicePrincipalName: zookeeper/ap42.home.lan $ samba-tool domain exportkeytab z1.ktab --principal=z1 $ samba-tool domain exportkeytab z1.ktab --principal=zookeeper/ap42.home.lan $ kinit -V -k -t z1.ktab zookeeper/ap42.home.lan Usin...

...tive 6 0 1 0 0 1 1 1 1 Negative 7 1 0 1 1 1 1 1 1 Negative 8 0 0 0 0 1 0 0 1 Negative 9 0 1 1 1 1 0 0 1 Negative 10 0 0 0 1 1 0 1 0 Positive 11 0 0 0 0 1 0 0 1 Negative 12 0 0 1 1 1 1 1 0 Positive 13 0 1 1 0 1 1 1 1 Negative" z1 <- read.table(textConnection(Zero1), header=TRUE) z1 str(z1) A simple way I can use mosaic plot mosaicplot(table(z1)) library(vcd) mosaic(table(z1)) I have tried to learn ?xtabs ?table and ?ftable but I can't figure out. I need a barplot for all variables and the result maybe like | |...

..._excl_5_282_559.txt', header=TRUE, sep="\t") daten<-as.matrix(daten) #2) create empty matrix: indxind<-matrix(nrow=617, ncol=617) indxind[1:20,1:19] #3) compare cells to each other, score: for (s in 3:34) { #walks though the matrix colum by colum, starting at colum 3 for (z1 in 1:617) { #for each current colum, take one row (z1)... for (z2 in 1:617) { #...and compare it to another row (z2) of the current colum if (z1!=z2) {topf<-indxind[z1,z2] if (daten[2*z1-1,s]==daten[2*z2-1,s]) topf<-topf+1 #actually, 2 rows make up 1 individua...

Dear Rusers, I would like some comments about the following results (under R-2.2.0) > m = matrix(1:6 , 2 , 3) > m [,1] [,2] [,3] [1,] 1 3 5 [2,] 2 4 6 > z1 = m[(m[,1]==2),] > z1 [1] 2 4 6 > is.matrix(z1) [1] FALSE > z2 = m[(m[,1]==0),] > z2 [,1] [,2] [,3] > is.matrix(z2) [1] TRUE Considered together, I'm a bit surprised about the returned types from z1 and z2. I would not have been surprised if z1 would still have been...

...posted elsewhere I cannot find it. I need to perform multiple integration where some of the variables are in the bounds of the other variables. I was trying to use R2Cuba function but cannot set the upper and lower bounds. My code so far is : int <- function(y){ u2 = y[1] z2 = y[2] u1 =y[3] z1 = y[4] ff <- u1*(z1-u1)*u2*(z2-u2)*exp(-0.027*(12-z2)) return(ff) } cuhre(4,1,int,rel.tol=1e-3,lower=c(y[4],y[4],4,4),upper=c(12,y[4],12,y[2]),abs.tol= 1e-12,flags= list(verbose=2, final=0)) I know that code is wrong but it shows that variables are in both the upper and lower bounds. Thanks...

...  I use fCopulae package to draw different graphs of univariate and bivariate skew t.  But the plots titles overlap.  I tried using cex.main, font.main to adjust the size but they still overlaps.  Here is my code: par(mfrow = c(3, 1)) mu = 0 Omega = 1 alpha1 = 0 alpha2 = 1.5 alpha3 = 2 alpha4 = 0.5 Z1 = matrix(dmvst(x, 1, mu, Omega, alpha1, df = Inf), length(x)) Z2 = matrix(dmvst(x, 1, mu, Omega, alpha2, df = 5), length(x)) Z3 = matrix(dmvst(x, 1, mu, Omega, alpha3, df = 5), length(x)) Z4 = matrix(dmvst(x, 1, mu, Omega, alpha4, df = Inf), length(x)) plot(x,Z1, ylim=range(Z1,Z2), type='l'...

...6, p60, p65, pc, ph, pd, pb50, pb60) { k50 <- exp(p50) k56 <- exp(p56) k60 <- exp(p60) k65 <- exp(p65) c <- exp(pc) h <- exp(ph) d <- exp(pd) b50 <- exp(pb50) b60 <- exp(pb60) a <- (k50+k56) b <- k65 e <- k56 f <- (k60+k65) z1 <- (-(a+f)/2 - sqrt((a+f)^2/4 - a*f + b*e)) z2 <- (-(a+f)/2 + sqrt((a+f)^2/4 - a*f + b*e)) K <- ((z1+a)/(z2-z1)) B1 <- (b/(z2-z1)*b60 - K*b50) A1 <- (b50-B1) X1 <- (b*d/(z2-z1)-K*(c*b4i+h)) X2 <- (K*c*(b4i-b40)) X3 <- (c*b4i + h - X1) X4 <- (c*(b40-b4i)-...

Dear R users, I want to run nested fixed-factor Anova in R on different experiments. In this toy example I have 3 levels of the main factor x1 and 7 levels of the nested factor z1 x1 and continuous response variable y1. x1 [1] 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 2 [38] 2 2 2 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 [75] 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 z1 [1] 1 1 1 1 1 1 1 1 1 1 1...

Dear useRs, I have the following data "z" of two variables "x"(z[,1]) and "y"(z[,2]). > dput(z) structure(c(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66,

Hi everybody. The question: I get two vectors 'iFalseFalse' and 'i2'. I think they should be the same but they are not. Is it because R does not handle complicated logical expressions in such cases or I do something wrong? > z1 = c(NA, "", 3, NA, "", 3) > z2 = c("", "", 3, NA, 3, NA) > cV = (as.character(z1)==as.character(z2)) > cV [1] NA TRUE TRUE NA FALSE NA > > iFalse = (c(1:(length(z1))))[(cV==FALSE)] > iNonNA = (c(1:(length(z1))))[(!(is.na(cV)))] &g...

I have the joint distribution of three discrete random variables z1, z2 and z3 which is captured by "z" and "prob" as described below. For example, the probability for z1=0.46667, z2=-1 and z3=-1 is 2.752e-13. Also, the probability adds up to 1. > head(z) z1 z2 z3 [1,] -0.46667 -1.0000 -1.0000 [2,] -0.33333 -0.9333 -0.93...

...<- 2.0165   rho  <- 0.868   # simplified expressions   a      <- alp-0.5   c1     <- sqrt(pi)/(gamma(alp)*(1-rho)^alp)   c2     <- sqrt(rho)/(1-rho)   t1     <- exp(-z/(1-rho))   t2     <- (z/(2*c2))^a   bes1   <- besselI(z*c2,a)   t1bes1 <- t1*bes1    c1*t1bes1*t2 }   z1  <- 20 cdf <- integrate(integrand, lower = 0, upper = z1,abs.tol = FALSE) r   <- runif(1)   ## Newton iteration z2  <- z1 - (cdf - r)/integrand(z1)   Output   > z1  <- 20   > cdf <- integrate(integrand, lower = 0, upper = z1)   > r   <- runif(1)   > z2  <- z1 - (...

...s. However my tests (shown below) indicate that lm() actually seems to perform separate regressions for each of the response variables without accounting for their possible correlation. ### multivariate analysis using lm() ### ######################################## > ex7.8 <- data.frame(z1 = c(0, 1, 2, 3, 4), y1 = c(1, 4, 3, 8, 9), y2 = c(-1, -1, 2, 3, 2)) > > f.mlm <- lm(cbind(y1, y2) ~ z1, data = ex7.8) > summary(f.mlm) Response y1 : Call: lm(formula = y1 ~ z1, data = ex7.8) Residuals: 1 2 3 4 5 0 1 -2 1 0 Coefficients: Estimate Std. Error t valu...

Is there an easy way to compare complex numbers? Here is a small example: > (z1=polyroot(c(1,-.4,-.45))) [1] 1.111111-0i -2.000000+0i > (z2=polyroot(c(1,1,.25))) [1] -2+0i -2+0i > x=0 > if(any(identical(z1,z2))) x=99 > x [1] 0 # real and imaginary parts: > Re(z1); Im(z1) [1] 1.111111 -2.000000 [1] -8.4968e-21 8.4968e-21 > Re(z2); Im...

...ormal dependence structure with unit Frechet margins. Sigma <- matrix(c(1,.5*sqrt(1),.5*sqrt(1),1),2,2) # generate y <- mvrnorm(Nsam, c(0,0), Sigma) # random v <- cbind(pnorm(y[,1],mean = 0, sd = 1), pnorm(y[,2],mean = 0, sd = 1)) z <- cbind(-1/log(v[,1]),-1/log(v[,2])) z1 <- z[,1] z2 <- z[,2] And to get the scatter plot by: plot(z1,z2) How can I get the contour densities plots for (z1,z2) at 10^(-j/2) for j=2,...,9? Waiting for your reply!Many thanks! Xiao [[alternative HTML version deleted]]