similar to: random value changes in a vector

I need help with (hopefully) just one more thing. I have been fussing with this for quite some time and have decided just to give up and ask! I want to match a column in a matrix to a vector. I found a "which" command that I thought would be helpful as it does the following: > g=c(1,5,3,2,7) > which(g==5) [1] 2 As the above gave which placement in the g vector corresponded to

I need help determining the unique columns of a matrix and the numbers of each unique column. For example, let's say I have a matrix with 6 columns, 2 of these are filled with the value of 1 and the other 4 are filled with the value of 0. I would then like to have a command that tells me what the unique columns are (so columns with 1s and columns with 0s) and the number of each type of

Hello, I'm having two similar problems with for loop and I would appreciate any help or comment. I want to use "for loop" to calculate series of initial values for an optimization problem. But some initial values have my function quit due to problems like calculating the inverse of singular matrices. I don't want to make my program check determinants and skip it if it is

Dear List, Im new to R - making a transition from SAS. I have a space delimited file with the following structure. Each line in the datafile is identified by the first letter. A = Inventory (Inventory) X = Stratum (Stratum_no Total Ye=year established) P = Plot (Plot_no age slope= species) T = Tree (tree_no frequency) L = Leader (leader diameter height) F = Feature (start_height finish_height

Hi, Using openSUSE 13.1 on s390x machine I get two new regressions with llvm 3.4rc1: [ 1412s] /home/abuild/rpmbuild/BUILD/llvm/unittests/ADT/BitVectorTest.cpp:39: Failure [ 1412s] Value of: Vec.count() [ 1412s] Actual: 1 [ 1412s] Expected: 5U [ 1412s] Which is: 5 [ 1412s] /home/abuild/rpmbuild/BUILD/llvm/unittests/ADT/BitVectorTest.cpp:42: Failure [ 1412s] Value of: Vec.all() [ 1412s]

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'm a week-old R user, and have become stuck trying to create usable CSV outputs for post-processing. I am using the package Rioja, which provides small datasets of results. I am running several analyses 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

Dear All, I'm trying to use waldtest to test poolability (parameter stability) between two logistic regressions. Because I need to use robust standard errors (using sandwich), I cannot use anova. anova has no problems running the test, but waldtest does, indipendently of specifying vcov or not. waldtest does not appear to see that my models are nested. H0 in my case is the the vector of

Dear Spencer and Prof. Fox, Thank you for your replies. 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

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

Caching computation in 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 =

[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{} section

Hi, I want to do a meta-analysis with count data for treatement/control cases. Mi problem is that I need to use zero values (an informative zero value) for the mean and standard deviation for one of the treatement, but R has a problem: "Studies with zero values for sd.e or sd.c get no weight in meta-analysis". I can agroup the case by Family (byvar=Family). ¿Can you help me? Thanks!

Hi All, I *think* it's ddply because 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,

I'm new to R 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",

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

Hello, everyone! I picked up R several months ago and have adopted it as my choice for statistical programming. Coming from a Java background, I can honestly say that R is not only free, it is better tha S-plus: the lexical scope in R makes it very simple to simulate Java's object model. For this, I encourage everyone to read the artcle: Robert Gentleman and Ross Ihaka (2000) "Lexical

Dear helpers, I wrote a code which estimates a multi-equation model with generalized least squares (GLS). I can use GLS because I know the covariance matrix of the residuals a priori. However, it is a bit slow and I wonder if anybody would be able to point out a way to make it faster (it is part of a bigger code and needs to run several times). Any suggestion would be greatly appreciated. Carlo

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Dear all, I would be very grateful if you 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