similar to: Issue with c++ .C call

Dear group, Here is my df : df <- structure(list(DESCRIPTION = c("PRM HGH GD ALUMINIUM USD 09/07/10 ", "PRM HGH GD ALUMINIUM USD 09/07/10 ", "PRIMARY NICKEL USD 04/06/10 " ), CREATED.DATE = structure(c(18361, 18361, 18325), class = "Date"), QUANITY = c(-1L, 1L, 1L), CLOSING.PRICE = c("2,415.90", "2,415.90",

Hi, ? I'm trying to carry out Cramer von Mises tests between pairs of vectors belonging to a discrete distribution (concretely frequencies from 0 to 200). However, the program crashes in the attempt. The problem seems to be that these vectors only have positive integer numbers (+ zero). When I add a random very small positive decimal to the non-decimal part everything works fine (files prm1

Dear group, Here is my data.frame : df <- structure(list(DESCRIPTION = c("PRM HGH GD ALU", "PRM HGH GD ALU", "PRIMARY NICKEL", "PRIMARY NICKEL", "PRIMARY NICKEL", "PRIMARY NICKEL", "STANDARD LEAD ", "STANDARD LEAD ", "STANDARD LEAD ", "STANDARD LEAD ", "STANDARD LEAD ",

Is there a repository of Facter plugins that people have made? I''m working on a few that i wouldn''t mind sharing when complete.

Found this solution. It is maybe not the most elegant way, but it does the job. > a=as.data.frame(substr(lme$DESCRIPTION,1,14)) > colnames(a)=c("DESCRIPTION") > lme=as.data.frame(c(a,lme[,2:3])) > lme DESCRIPTION CLOSING.PRICE POSITION 1 PRIMARY NICKEL 25,755.7100 0 2 PRIMARY NICKEL 25,760.8600 0 3 PRM HGH GD ALU 2,415.9000 0

Dear group, I know this issue has been already covered, and before you reply I must say I have read the R-FAQ and search the mailing list archive. I still can't manage to change my factor to numeric as I couldn't find any clear answer. Here is my df : Pose1 <- structure(list(DESCRIPTION = structure(c(1L, 2L, 3L, 4L, 5L, 8L), .Label = c(" SUGAR NO.11 May/10 ", "COTTON

> Dear All > > I have data as follows. > > D T M L > 0.20 1 03 141 > 0.32 1 07 62 > 0.50 1 05 49 > 0.80 1 04 46 > 0.20 2 14 130 > 0.32 2 17 52 > 0.50 2 13 41 > 0.80 2 14 36 > 0.20 3 24 120 > 0.32

Dear List, I'm having a curious problem with lapply(). I've used it before to convert a subset of columns in my dataframe, to factors, and its worked. But now, on re-running the identical code as before it just doesn't convert the columns into factors at all. As far as I can see I've done nothing different, and its strange that it shouldn't do the action. Has anybody

This is in reference to David Schmitt''s writeup [1] I like that writeup a lot; I just had a few comments/clarifications: * Namespaces: For now, we just use a flat namespace for modules, i.e. you can''t have modules within modules, and you always know that in a fully qualified name ''foo::bar::baz'' that ''foo'' is the module name and

Dear List, I thought it would be much easier to put a second query into a second mail. I need to print 426*10000 blocks of variance components data, where 426 is the number of columns of data that have 10000 permutations of variance generated for each of them. I have tried printing out a smaller number of permutations for a smaller number of markers and that has worked. However, since a

Hallo, I own a NETUPS 450 (Model on the sticker: PRM 450i; upsstats reports "UPS 450 VA FW -0024") which perfectly works with the upscode2 driver. Maybe you want to add it to your list, even if the model is >10 years old. I suppose this is true for the whole series. According manual the come in the 450VA, 700VA, 1000VA and 1500VA strengths. They have been sold under different

I just installed R's RPM on a Linux box using red hat 8.0. I would like to use some graphical interface (e.g. gnome) but I have not managed to do it. Marcos [[alternative HTML version deleted]]

Dear R users, The package 'mfp' that fits fractional polynomial terms to predictors. Example: data(GBSG) f <- mfp(Surv(rfst, cens) ~ fp(age, df = 4, select = 0.05) + fp(prm, df = 4, select = 0.05), family = cox, data = GBSG) print(f) To describe the association between the original predictor, eg. age and risk for different values of age I can plot it the polynomials

Dear list, here are my two data frames: av <- structure(list(DESCRIPTION = c("COFFEE C Sep/10", "COPPER Sep/10", "CORN Dec/10", "CRUDE OIL miNY Sep/10", "GOLD Aug/10", "HENRY HUB NATURAL GAS Sep/10", "PALLADIUM Sep/10", "SILVER Sep/10", "SOYBEANS Nov/10", "SPCL HIGH GRADE ZINC USD", "SUGAR

Hello R community, I have been using R's inbuilt maximum likelihood functions, for the different methods (NR, BFGS, etc). I have figured out how to use all of them except the maxBHHH function. This one is different from the others as it requires an observation level gradient. I am using the following syntax: maxBHHH(logLik,grad=nuGradient,finalHessian="BHHH",start=prm,iterlim=2)

Hello..I am not very proficient in Linux kernel stuffs although I know my basics. I have a question and all suggestions/solutions will be highly appreciated... I got to know that RHEL 5 has inbuilt Xen Support...So I tried to install a RHEL 5.0 Server on my P4 machine. On top of it I tried to install Xen specific RPMs and some other RPMs needed by Xen. The additinal RPMs added after base

Hi there, I would like to find a more efficient way of permuting the rows and columns of a symmetrical matrix that represents ecological or actual distances between objects in space. The permutation is of the type used in a Mantel test. Specifically, the permutation has to accomplish something like this: Original matrix addresses: a11 a12 a13 a21 a22 a23 a31 a32 a33 Example

I''ve been thinking about how to make Puppet a killer app, and I think there is one specific thing that could be done to help make that happen: People using Puppet now to consistently state what they *really* wish Puppet could do for them -- the recent thread around Puppet on Windows is in that vein, but it''s not specific enough. For example, I''d like to know what

Dear group, I have these 2 following data frame: allcon <- structure(list(DESCRIPTION = structure(1:17, .Label = c("COFFEE C Jul/10", "COPPER May/10", "CORN Jul/10", "CORN May/10", "COTTON NO.2 Jul/10", "CRUDE OIL miNY May/10", "GOLD Jun/10", "HENRY HUB NATURAL GAS May/10", "ROBUSTA COFFEE (10)

Dear group, Here is my df : pose16 <- structure(list(DESCRIPTION = structure(c(1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 12L), .Label = c("COPPER May/10", "COTTON NO.2 Jul/10", "CRUDE OIL miNY May/10", "GOLD Jun/10", "ROBUSTA COFFEE (10) Jul/10", "SOYBEANS Jul/10", "SUGAR NO.11 Jul/10", "SUGAR NO.11 May/10",