similar to: calculating power of log rank test

my program given below can some one make it presentable. I trying to simulate survival data and calculate the power. I think i could have done better. s=10 number=0 count1=0;count2=0;count3=0;count4=0;count5=0;count6=0;count7=0;count8=0; count9=0; count11=0;count22=0;count33=0;count44=0;count55=0;count66=0;count77=0; count88=0;count99=0; while(s!=0){ n=100 n1=n/2 n2=n/4

hie i'm tryimg to generate two survival data using the following code (I know its ugly ) but it seems to repeat two of the variables can any one tell me whats the porblem. n=20 n1=n/2 n2=n/4 a11=1 ;a12=1.4 ;a21=16 ;a22=a12 * a21 t1<-array(1,c(n1)) t2<-array(2,c(n1)) treatgrp=matrix(c(t1,t2))

hie when i use plot.survfit to plot more than one graph why I only see the last graph how do i see the other graphs.for example n=20 n1=n/2 n2=n/4 a11=4;a12=4 ;a21=4 ;a22=4 t1<-array(1,c(n1)) t2<-array(2,c(n1)) treatgrp=matrix(c(t1,t2))

Hi All I have a dataset of 200 columns and 1000 rows , there are 3 repeated values under each column (7,8,10). I wanted to calculate the frequency of each value under each column and then apply the function maf () given that the frequency of each value is known. I can do the analysis step by step like this :- > Values A B C ... 200 1 7 10 7 2

Hi, To clarify the default behavior that Boris is referencing below, note the definition of the 'bin' argument to the tabulate() function: bin: a numeric vector ***(of positive integers)***, or a factor. Long vectors are supported. I added the asterisks for emphasis. This is also noted in the examples used for the function in ?tabulate at the bottom of the help page. The second

Thank you for your effort Bert.., I knew what is the problem now, the values (1,2,3) were only an example. The values I have are 0 , 1, 2 . Tabulate () function seem to ignore calculating the frequency of 0 values and this is my exact problem as the frequency of 0 values should also be calculated for the maf to be calculated correctly. ________________________________ From: Bert Gunter

Always reply to the list. I am not a free, private consultant! "For example, if I have the values : 1 , 2 , 3 in each column, applying Tabulate () would calculate the frequency of 1 and 2 without 3" Huh?? > x <- sample(1:3,10,TRUE) > x [1] 1 3 1 1 1 3 2 3 2 1 > tabulate(x) [1] 5 2 3 Cheers, Bert Bert Gunter "The trouble with having an open mind is that people

|> x <- sample(0:2, 10, replace = TRUE) |> x [1] 1 0 2 1 0 2 2 0 2 1 |> tabulate(x) [1] 3 4 |> table(x) x 0 1 2 3 3 4 B. > On Nov 10, 2017, at 4:32 AM, Allaisone 1 <allaisone1 at hotmail.com> wrote: > > > > Thank you for your effort Bert.., > > > I knew what is the problem now, the values (1,2,3) were only an example. The values I have are

Hello, consider the following IR code : %count4 = getelementptr inbounds %struct.r32, %struct.r32* %cur.087, i64 0, i32 4 How to check in the instruction, whether the operand is a structure or not if(isa<GetElementPtrInst>(instruction)) { GetElementPtrInst *getElementPtrInst=dyn_cast<GetElementPtrInst>(&instruction); //check if getElemetPtrInst operands structure or array. }

Hi, I have a question about some code in ScalarEvolution.cpp, in the function HowManyLessThans. Specifically, this function returns CouldNotCompute if the iteration step is greater than one even when the NoWrap flags are set (if the step goes past the limit and wraps). Here's the comment: } else if (isKnownPositive(Step)) { // Test whether a positive iteration can step

I am trying to compare Levene,bartlett, count5 tests for a random normal sample for the significance level and power, the problem seems bug free but with obvious wrong result, anyone can help, I send the R code as attachment. Thanks http://www.nabble.com/file/p16992744/758%2Bproject 758+project -- View this message in context:

Thanks Ryan, I'm able to retrieved the type using the following code: Type *type=getElementPtrInst->getSourceElementType(); On Fri, Aug 12, 2016 at 7:00 PM, Ryan Taylor <ryta1203 at gmail.com> wrote: > Take a look at visitGetElementPtrInst in InstructionCombining.cpp for some > examples about how to iterate over GEP and check for type. > >

I have struggled with compiling 2.0.5a on Solaris 2.6 with gcc version 2.8.1 for the last couple of weeks (not a high priority project) and decided to try a precompiled version just in case I was being stupid. Well, perhaps I'm stupider (is that a word) than I thought because I installed the package and I still cannot connect. I have read through the docs and tried more than a few

Ok, I assumed you wanted to compute a matrix M for all states such that M[i][j]= transition from state i to state j / number of transition from state i but from what you just answered it looks like you want to compute a matrix M for a set of states S such that: M[S_i][S_j]= transition from state S_i to state S_j / number of transition from state S_i where S_i is in the set S ? 2017-08-28

Hi, I think you overthought this one a little bit, I don't know if this is the kind of code you are expecting but I came up with something like that: generate_transition_matrix <- function(data, n_states) { #To be sure I imagine you should check n_states is right at this point transitions <- matrix(0, n_states, n_states) #we could improve a little bit here because at

Chuck (Is it fine to call you Chuck?) has far more R jutsu than I do obviously. I don't know much about pmin and factor but it might worth looking into if you want to manipulate states by names (I assume this is why one might want to use it?) generate_transition_matrix <- function(data, states) prop.table(table(head(data, -1), tail(data, -1)), 1)[states,]

Hello, I am trying to implement a formula aij= transition from state S_i to S_j/no of transition at state S_i Code I have written is working with three state {1,2,3 }, but if the number of states become={1,2,3,4,......n} then the code will not work, so can some help me with this. For and some rows of my data frame look like

#I have a dataset with two factor. I want to combine those factors into a single factor and count the number of data values for each new factor. The following gives a comparable dataframe: a <- rep(c("a", "b"), c(6,6)) b <- rep(c("c", "d"), c(6,6)) df <- data.frame(f1=a, f2=b, d=rnorm(12)) df # I use the 'interaction' function to combine

Hi all, today I tried to compile R in 64-bit mode on Linux on Power4. I used the following parameter with configure. ./configure -without-x CC='xlC' F77='xlf' CXX='xlC' CFLAGS='-q64 -g -qfullpath' FFLAGS='-q64 -g -qfullpath' CXXFLAGS='-q64 -g -qfullpath' MAIN_LDFLAGS='-q64 -g -qfullpath' LDFLAGS='-q64' SHLIB_CXXLDFLAGS='-q64

I have a macro and when I call it I have something like this: exten => s,1,NoOp(Group Count: ${GROUP_COUNT(MYGROUP)}) exten => s,n,Set(GROUP()=MYGROUP) ;Set Group exten => s,n,NoOp(Group List: ${GROUP_LIST()}) exten => s,n,NoOp(Group Count: ${GROUP_COUNT(MYGROUP)}) The GROUP_COUNT returns zero before the call to GROUP but also returns 0 after the call to GROUP. If I