search for: inv

...lt;- 1 D1<-data.frame() for(t in tim){ D1<-rbind(D1,I_N) } D1<-data.matrix(D1[as.vector(table(i.ind,t.ind))==1,]) D2<-data.frame() for(i in nam){ D2<-rbind(D2,I_T) } D2<-data.matrix(D2[as.vector(table(t.ind,i.ind))==1,]) D<-data.matrix(cbind(D1,D2)) Q<-I-D%*%ginv(crossprod(D))%*%t(D)#fQ(D1)-fQ(D1)%*%D2%*%ginv(t(D2)%*%fQ(D1)%*%D2)%*%t(D2)%*%fQ(D1) Q } ############################## library(plm) library(lmtest) data(Grunfeld) y_i<-Grunfeld$inv-ave(Grunfeld$inv,index=Grunfeld$firm) y_t<-Grunfeld$inv-ave(Grunfeld$inv,index=Grunfeld$year) y_it<-(...

...dtest recognize the nesting. Here's the lines I run: (BTW, I try to use robust standard errors because what I normally use (glm.binomial.disp) to correct for overdispersion does not converge for the unpooled model. But this is another story....) # poolability for leva.fin03.d # pooled model inv.log.leva.base = glm(mix.au.bin ~ cat.gap.tot + leva.fin03.d + ... + sud0nord1, data = inv.sub.au, family = binomial, maxit = 1000) # I deleted almost all variables to make the line more readable # overdispersed pooled model - NOT THE PROBLEM NOW inv.log.leva.base.disp = glm.binomial.disp(inv.log...

[Hope this is the right list where to send...] An attempt to update package 'mnormt' involves the addition of a small new function called 'pd.solve'. When I come to the package checking stage, an error occurs in parsing pd.solve.Rd. The full transcript of the outcome is copied below (it includes details on my installation) but the critical point is where the \examples{} secti...

...ies. I'll very appreciate, if you have any ideas concerning the problem described below. First, I'd like to describe the model in brief. In general I consider a model with three equations. First one is for annual GRP growth - in general it looks like: 1) GRP growth per capita = G(investment, migration, initial GRP per capita, spatial lag on GRP growth + some additional explanatory variables to control for regional disparities), where spatial lag on GRP growth in year t is simply spatial weights matrix W(n*n) (weights - are inverse square distances between regional centers;...

...n rails? Simple example: factorial modulus a large number input: integer x output: factorial( x ) % 12345678901234567 I want it so that if it computes factorial of N once, it will not have to compute for N again. code: class SiteController < ApplicationController caches_action :factorial, :inv def examine @inv = @params[''inv''] @outv = factorial @inv.to_i end def factorial( inv ) y=1 while (inv > 0) do y*=inv inv-=1 end (y % 12345678901234567) end end I am running in production mode so caching should be enabled.

...s in a loop and iteratively adding the results to a *list* ("combined"). Within each iteration I use the following: > combined[[i]] <- performance(fit) With two iterations I get the following list, as output to a CSV: object.RMSE object.R2 object.RMSE.1object.R2.1 WA.inv 0.321543059 0.86388897 0.350494954 0.882600618 WA.cla 0.345947482 0.86388897 0.373078015 0.882600618 WA.inv.tol 0.308649941 0.874585589 0.336975196 0.89148291 WA.cla.tol 0.330038831 0.874585589 0.356895789 0.89148291 Obviously if I run thousands of iterations this continues on to the...

Hello all, How do I actually use the output of predict.lm(..., type="terms") to predict new term values from new response values? I'm a chromatographer trying to use R (2.15.1) for one of the most common calculations in that business: - Given several chromatographic peak areas measured for control samples containing a molecule at known (increasing) concentrations, first

I submit the following matrix to both MATLAB and R x= 0.133 0.254 -0.214 0.116 0.254 0.623 -0.674 0.139 -0.214 -0.674 0.910 0.011 0.116 0.139 0.011 0.180 MATLAB's inv(x) provides the following 137.21 -50.68 -4.70 -46.42 -120.71 27.28 -8.94 62.19 -58.15 6.93 -7.89 36.94 8.35 11.17 10.42 -14.82 R's solve(x) provides: 261.94 116.22 150.92 -267.78 116.22 344.30 286.68 -358.30 150.92 286.68 252.96 -334.09 -267.78 =358.30 -334.09 475.22 inv(x)*x = I(4)...

...up the case by Family (byvar=Family). ¿Can you help me? Thanks! > datos<-read.table(file="datos_totales.csv",head=TRUE,sep=",",dec=",") > datos[10:20,] Study ControlN ControlMean ControlSD TreatedN TreatedMean TreatedSD Family 10 Ali.Gmb.S1.Inv 3 893.3333 180.36999 3 1213.33333 410.52812 Capitellidae 11 Ali.Gmb.S2.Inv 3 640.0000 348.71192 3 666.66667 220.30282 Capitellidae 12 Ali.Gmb.S3.Inv 3 426.6667 184.75209 3 920.00000 628.64935 Capitellidae 13 Ali.Cul.S1.Ver 3 213.3333...

...cause the function recognizes vr1, etc, in other parts of the function. Here's some code: # create dataset PROV.PM.FBCTS <- c(0.00 ,0.00, 33205.19, 25994.56, 23351.37, 26959.56 ,27632.58, 26076.24, 0.00, 0.00 , 6741.42, 18665.09 ,18129.59 ,21468.39 ,21294.60 ,22764.82, 26076.73) FBCTS.INV.TOT <- c(0 , 0, 958612, 487990, 413344, 573347, 870307, 552681 , 0, 0 , 163831 , 400161 , 353000, 358322 , 489969, 147379, 1022769) FBCTS.REC.TOT <- c(0 , 0, 1638818 ,297119 , 299436 ,414164 , 515735, 529001 , 0, 0 , 427341 , 625893 ,437854 , 407091, 425119 ,...

Full_Name: Ole Christensen Version: 1.6.1 OS: linux-gnu Submission from: (NULL) (130.225.18.176) In package boot : > inv.logit(800) [1] NaN where it should have been 1. The problem is caused by exp(x) returning Inf when x is large. One way of fixing the problem [there may be better ways] would be to include the line out[x > 709] <- 1 in inv.logit() Cheers Ole > version _ platform i686-...

Dear R users this may be a simple question - but i would appreciate any thoughts does anyone know how you would get one lower and one upper confidence interval for a set of data that consists of proportions. i.e. taking a usual confidence interval for normal data would result in the lower confidence interval being negative - which is not possible given the data (which is constrained between

...and i'm having some trouble with a bubble chart. Basically I have 3 series (a,b,c), but the third one is a binnary variable (assumes only 0 or 1 to the entire data). How can I use these binnary information to make 2 different colours in a bubble chart?. I.e., I'm using this code: symbols(inv$a, inv$b, circles=radius, inches=0.35, fg="white", bg="red", xlab="aa", ylab="bb", add=TRUE) and i want the bg color to be red if c=1 and blue if c=0. thanks, Murilo -- View this message in context: http://r.789695.n4.nabble.com/Simple-question-about-symb...

...# matrix of exogenous #### MODEL ESTIMATION ### # build the big X matrix needed for GLS estimation: X <- kronecker(diag(1,3),x) Y <- c(y) # stack the y in a vector # residual covariance matrix for each observation covar <- kronecker(sigma,diag(1,N)) # GLS betas covariance matrix inv.sigma <- solve(covar) betav <- solve(t(X)%*%inv.sigma%*%X) # GLS mean parameter estimates betam <- betav%*%t(X)%*%inv.sigma%*%Y ----- end of code ----

...with R from 'iris' data? Well, I think that the basic soluction to calculate this distances is: # # --- Begin R script 1 --- # x = as.matrix(iris[,1:4]) tra = iris[,5] man = manova(x ~ tra) # Mahalanobis E = summary(man)$SS[2] #Matrix E S = as.matrix(E$Residuals)/man$df.residual InvS = solve(S) ms = matrix(unlist(by(x, tra, mean)), byrow=T, ncol=ncol(x)) colnames(ms) = names(iris[1:4]) rownames(ms) = c('Set', 'Ver', 'Vir') D2.12 = (ms[1,] - ms[2,])%*%InvS%*%(ms[1,] - ms[2,]) print(D2.12) D2.13 = (ms[1,] - ms[3,])%*%InvS%*%(ms[1,] - ms[3,]) print(D2.13)...

...nd Graphical Statistics, 9, 491-508. I have coded a simulation program which runs at 1/5 the speed of minimum usability on a Pentium 700 PC (windows me). I have spent quite some time trying to optimize the program. So my question is: on which platform does R run the fastest? I would be willing to invest on a faster platform. Thanks in advance. My program follows in case someone can help me improve it. Thanks. Jason # Liao and Lipsitz (2001) # A REML type estimate of variance components in generalized linear mixed models # This program runs on R. It does not run on S-plus. rm(list=ls(all=...

Hi, I am working on a dataset in which a number of venture capitalists invest in a number of firms. What I am creating is an asymmetric matrix M in which m(ij) is the volume (sum) of coinvestments of VC i with VC j (i.e., how much has VC i invested in companies that VC j also has investments in). The output should look like the "coinvestments" matrix produced wi...

Dear All, My question is simple but I need someone to help me out. Suppose I have a positive definite matrix A. The funtion chol() gives matrix L, such that A = L'L. The inverse of A, say A.inv, is also positive definite and can be factorized as A.inv = M'M. Then A = inverse of (A.inv) = inverse of (M'M) = (inverse of M) %*% (inverse of M)' = ((inverse of M)')'%*% (inverse of M)', i.e. if define B = transpose of (inverse of M...

...ou could help me with: Given the regularized gamma function Reg=int_0^r (x^(k-1)e^(-x))dx/int_0^Inf (x^(k-1)e^(-x))dx ; 0<r<Inf (which is eventually the ratio of the Incomplete gamma function by the gamma function), does anyone know of a package in R that would evaluate the derivative of the inverse of Reg with respect to k? I am aware that the function "Rgamma.inv" of the package "Zipfr" evaluates the inverse of Reg and I'm wondering wether there is a function that would evaluate the derivative of the inverse.. Alternatively, a good numerical integration package/...

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