similar to: Bold in expression in Y label

Hi all: I am trying to use the BiodiversityR package in Rcommander since I need to box-cox transform some data. However, even when trying with the example files, I get this error just after selecting the box cox transformation in the Menu, and Rcommander and Rstudio just shut down. What is it? What could I do? Thanks a lot The instruction at "0x00000001" referenced memory at

in the tutorial of official llvm doc, chapter 3, it deals with arguement of function as follow: for (Function::arg_iterator AI = F->arg_begin(); Idx != Args.size(); ++AI, ++Idx) { AI->setName(Args[Idx]); // NamedValues is map<string, Value*> NamedValues[Args[Idx]] = AI; and when it try to get the value of arguement, it simply does: Value

Ahh yes, sorry about that. Here's the corrected snippet: # Create an Example Data Frame Containing Car x Color data carnames <- c("bmw","renault","mercedes","seat") carcolors <- c("red","white","silver","green") datavals <- round(rnorm(16, mean=10, sd=4),1) data <- data.frame(Car=rep(carnames,4),

hi, I have my data time expressed in character string exple "5:20" (hour:min) and i want to convert these in times recognized by R. I have tried the POSIXct function: balise07$Hour <- as.POSXIct(balise07$Hour) but it didn't work. Do you know why? Do you know how i can convert my string character in a real time? Thanks Karine HEERAH Master 2 mention

Hi I have some data that i want to plot but i don't find how to do it. I have car types (bmw,renault,mercedes,seat ...), colors and a number for each car type-color relation.I want to come up with a matrix representation of cars vs colors where in each intersection i could set a dot proportional in size to my third variable. Can anybody give me a clue of hoe to come up with such

hello, every one, My guest os''s kernel version is suse10''s kernel-2.6.13.4-xen. Now, I want to change kernel version to kernel-2.6.13.4-smp. what should I do? Need I modify vm machine configue ? thanks a lot. Kathy _________________________________________________________________ 免费下载 MSN Explorer: http://explorer.msn.com/lccn

I read the tutorial doc and some info of SSA, finally understand it. Thanks for your help. And the segmentation error of loading arguementation occurs, (gdb error info) Program received signal SIGSEGV, Segmentation fault. 0x0855bb68 in llvm::LoadInst::LoadInst(llvm::Value*, char const*, bool, llvm::Instruction*) () code is like follows: //#include necessary header files int main(){

Earlier today I sent a question to Frank Harrell as an R developer with whom I am most familiar. He suggested also that I put my questions to the list for additional responses. Next month I'll be giving a talk on R as an example of high quality open source software. I think there is much to learn from R as a high quality extensible product that (at least as far as I can tell) has never

Dear R forum Suppose I have a data.frame df = data.frame(id = c(1:6), x = c(15, 21, 14, 21, 14, 38), y = c(36, 38, 55, 11, 5, 18), x.1 = c(15, 21, 14, 21, 14, 38), z = c("D", "B", "A", "F", "H", "P")) > df   id  x  y    x.1 z 1  1 15 36  15 D 2  2 21 38  21 B 3  3 14 55  14 A 4  4 21 11  21 F 5  5 14  5  14 H 6  6 38 18  38 P

Hello, I am working with a customer that has SSH F-Secure (NT windows). We currently have OPENSSH installed in our unix server. We are trying to copy a file from our unix box to an NT windows and received the following error: Remote: Failed to lauch child process! How can we resolve this problem. Thanks, Kathy Tablan Confidentiality Notice: This e-mail message, including any

Dear R forum I have one stupid question, but I have no other solution to it in sight? Suppose some R process creates graphs etc alongwith main output as data.frame e.g output1 = data.frame(bands = c("A", "B", "C"), results = c(74, 108,  65)) I normally save this output as some csv file. But I need to save this output as some image (I understand this is weird,

Hi everyone, Can anyone give me more details about the 'lp', 'risk', 'terms', and 'expected' options from the predict.coxph function? I actually found that risk is obtained from exp(lp). I can't find information on 'expected' type however. Is anybody can expand the information on this function? Thank you in advance, I will appreciate it.

hi On Sun, Apr 22, 2012 at 8:36 PM, Jianfei Hu <hujianfei258 at gmail.com> wrote: > in the  tutorial of official llvm doc, chapter 3, it deals with arguement of > function as follow: > > for (Function::arg_iterator AI = F->arg_begin(); Idx != Args.size(); >        ++AI, ++Idx) { >     AI->setName(Args[Idx]); > >     // NamedValues is map<string, Value*>

Dear R forum I have a data.frame df = data.frame(dates = c("4/15/2013", "4/14/2013", "4/13/2013", "4/12/2013"), values = c(47, 38, 56, 92)) I need to to create a vector by repeating the dates as "Current_date", 4/15/2013, 4/14/2013, 4/13/2013, 4/12/2013,  "Current_date", 4/15/2013, 4/14/2013, 4/13/2013, 4/12/2013, Current_date,

hello, I have a problem to open an hdf file. i have downloaded the package 'hdf5' as it was advised on R seek. But when i try to load the file, the R console sends me an eror message: setwd("C:/Documents and Settings/Karine/Bureau/data/") #install.packages('hdf5') library("hdf5") sea_ice <- hdf5load("asi-s6250-20090704-v5i.hdf", load =

Apologies if this has been asked before. I am having trouble understanding the R mailing list never mind R! I am relatively new to R having migrated from Minitab and SPSS. I have managed to do some more complicated statistics such as hierarchical partitioning of variance on an 80,000 record dataset but have to admit that drawing a simple bar plot I could do by hand is proving extremely

Dear R forum, I have a data.frame as defied below - df = data.frame(names = c("C", "A", "A", "B", "C", "B", "A", "B", "C"), dates = c("4/15/2013", "4/13/2013", "4/15/2013", "4/13/2013", "4/13/2013", "4/15/2013", "4/14/2013",

Dear R forum I have a data.frame df = data.frame(currency_type = c("EURO_o_n", "EURO_o_n", "EURO_1w", "EURO_1w", "USD_o_n", "USD_o_n", "USD_1w", "USD_1w"), rates = c(0.47, 0.475, 0.461, 0.464, 1.21, 1.19, 1.41, 1.43))   currency_type     rates 1      EURO_o_n   0.470 2      EURO_o_n   0.475 3       EURO_1w   0.461

Dear Forum, I have a data frame as  mydat = data.frame(basel_asset_class = c(2, 8, 8 ,8), defa_frequency = c(0.15, 0.07, 0.03, 0.001)) > mydat   basel_asset_class defa_frequency 1                 2          0.150 2                 8          0.070 3                 8          0.030 4                 8          0.001 I need to get the subset of this data.frame where no of records for the

Dear Forum, I have a data frame as  mydat = data.frame(basel_asset_class = c(2, 8, 8 ,8), defa_frequency = c(0.15, 0.07, 0.03, 0.001)) > mydat   basel_asset_class defa_frequency 1                 2          0.150 2                 8          0.070 3                 8          0.030 4                 8          0.001 I need to get the subset of this data.frame where no of records for the