search for: afit

...restrictions on abrreviated names. The anova.coxph command, in a slavish copy of anova.lm etc, returns a data frame with column labels of loglik Chisq Df Pr(>|Chi|) If one tries to extract the final column of the table errors result since it is not a standard R variable name. > afit <- anova(lm(conc ~ uptake, CO2)) > afit$P [1] 2.905607e-06 NA Warning message: In `$.data.frame`(afit, P) : Name partially matched in data frame > afit$Pr(>F) Error: unexpected '>' in "afit$Pr(>" The second is a result of allowing TAB completion of t...

...rnative hypothesis: true difference in means is not equal to 0 95 percent confidence interval: -0.2156863 1.6584968 sample estimates: mean of x mean of y 0.3589240 -0.3624813 > > A = c(rep('x',N1),rep('y',N2)) > Y = c(x,y) > fr = data.frame(Y=Y,A=as.factor(A)) > afit=aov(Y ~ A,fr) > > X=model.matrix(afit) > B=afit$coefficients > V=solve(t(X) %*% X) > > mse=tail(summary(afit)[[1]]$'Mean Sq',1) > df=tail(summary(afit)[[1]]$'Df',1) > t_statisitic=(B/(mse * sqrt(diag(V))))[[2]] > 2*(1-pt(abs(t_statisitic),df))#the p-value...

...ar2) in defining dt (and dt2) in one part of predict.coxph: # #if (ncol(y) == 2) { # if (se.fit) { # dt <- (chaz * newx[indx2, ]) - xbar[indx2, ] # *** SHOULD BE JUST xbar # se[indx2] <- sqrt(varh + rowSums((dt %*% object$var) * dt)) * newrisk[indx2] # } #} #else { # j2 <- approx(afit$time, 1:afit.n, newy[indx2, 2], method = "constant", f = 0, yleft = 0, yright = afit.n)$y # chaz2 <- approx(-afit$time, afit$cumhaz, -newy[indx2, 2], method = "constant", rule = 2, f = 0)$y # chaz2 <- c(0, afit$cumhaz)[j2 + 1] # pred[indx2] <- (chaz2 - chaz) * newris...

I have the R code at the end. The last command gives me "1 observation deleted due to missingness". I don't understand what this error message. Could somebody help me understand it and how to fix the problem? > summary(afit) Df Sum Sq Mean Sq F value Pr(>F) A 2 0.328 0.16382 0.1899 0.82727 B 3 2.882 0.96057 1.1136 0.34644 C 4 1.676 0.41897 0.4857 0.74617 A:B 6 5.133 0.85554 0.9919 0.43404 A:C 8 14.193 1.77406 2.0568 0.04543 * B:C...

...] "contr.treatment" Regards, Peng > a=2 > b=3 > n=4 > A = rep(sapply(1:a,function(x){rep(x,n)}),b) > B = as.vector(sapply(sapply(1:b, function(x){rep(x,n)}), function(x){rep(x,a)})) > Y = A + B + rnorm(a*b*n) > fr = data.frame(Y=Y,A=as.factor(A),B=as.factor(B)) > afit=aov(Y ~ A + B,fr) > model.matrix(afit) (Intercept) A2 B2 B3 1 1 0 0 0 2 1 0 0 0 3 1 0 0 0 4 1 0 0 0 5 1 1 0 0 6 1 1 0 0 7 1 1 0 0 8 1 1 0 0 9 1 0 1 0 10 1 0...

I was wondering if there is a way of editting strings in R. I have a set of strings and each set is a row of numbers and paranthesis. For example the first row is: (0 2)(3 4)(7 9)(5 9)(1 5) and I have a thousand or so such rows. I was wondering how I could get the corresponding string obtained by adding 1 to all the numbers in the string above. Dursun [[alternative HTML version deleted]]

...cients (A1, A2, A3, B1, B2, B3, B4)? > a=3 > b=4 > n=1000 > A = rep(sapply(1:a,function(x){rep(x,n)}),b) > B = as.vector(sapply(sapply(1:b, function(x){rep(x,n)}), function(x){rep(x,a)})) > Y = A + B + rnorm(a*b*n) > > fr = data.frame(Y=Y,A=as.factor(A),B=as.factor(B)) > afit=aov(Y ~ A + B - 1,fr) > afit$coefficients A1 A2 A3 B2 B3 B4 1.993067 2.969252 3.965888 1.005445 2.020717 3.023940 Regards, Peng

...1 1 [4,] 1 1 [5,] 1 2 [6,] 1 2 [7,] 1 2 [8,] 1 2 [9,] 1 3 [10,] 1 3 [11,] 1 3 [12,] 1 3 [13,] 2 4 [14,] 2 4 [15,] 2 4 [16,] 2 4 [17,] 2 5 [18,] 2 5 [19,] 2 5 [20,] 2 5 [21,] 2 6 [22,] 2 6 [23,] 2 6 [24,] 2 6 > Y = A + B + rnorm(a*b*n) > fr = data.frame(Y=Y,A=as.factor(A),B=as.factor(B)) > afit=aov(Y ~ A / B - 1,fr) > X=model.matrix(afit) > solve(t(X) %*% X) Error in solve.default(t(X) %*% X) : Lapack routine dgesv: system is exactly singular

...fb +fa*fc +fb*fc +fa*fe +fc*fe +fa*fb*fc +fa*fc*fe + rnorm(1) + } > > aframe = data.frame( + A=as.factor(X[,'A']) + , B=as.factor(X[,'B']) + , C=as.factor(X[,'C']) + , D=as.factor(X[,'D']) + , E=as.factor(X[,'E']) + ,Y) > > afit=aov(Y ~ A * B * C * D * E, aframe) > > summary(afit) Df Sum Sq Mean Sq A 2 1512240 756120 B 3 453324 151108 C 4 2549895 637474 D 5 2 0.3693 E 6 1451057 241843 A:B 6 33875 5646 A:C 8...

...ontrast between '3' and '4'). Could somebody let me know what is the correct way to compute the contrast in the following example? library(contrast) a=3 n=4 A = as.vector(sapply(1:a,function(x){rep(x,n)})) Y = A + rnorm(a*n) aframe = data.frame(Y=Y, A=as.factor(A)) aframe afit=aov(Y ~ A - 1, aframe) contrast(afit, list=(A='3'), list=(A='4'))

Hi; I am having problems inverting matrices using the function solve() For example R can not invert the following matrix [,1] [,2] [,3] [,4] [,5] [1,] 25 500 11250 275000 7.106250e+06 [2,] 500 11250 275000 7106250 1.906250e+08

...other panel function) through each of multiple data series plotted on the same graph? Specifically, while one can do something like xyplot(a+b+c~x) which plots three series, a,b & c, but can one automatically fit lines through each of them? I suppose one could generate three more variables afit, bfit, and cfit with a model & predict and then plot them, but wondered if there was an easier way. Thank you for any advice. Here is an example: # use an example panel function using smooth.spline; however, the issue relates to all panel functions # a panel function to fit smoothed lines t...

Hi, I don't quite understand what are the return values of aov. I know that it has 'coefficients'. But I need to know what all the other return values are. Can somebody let me know how to figure them? Value: An object of class 'c("aov", "lm")' or for multiple responses of class 'c("maov", "aov", "mlm",