search for: idy

Hi, I am trying to calculate net reclassification improvement (NRI) and Inegrated Discrimination Improvement (IDI) for a survival dataset to compare 2 risk models. It seems that the improveProb() in Hmisc package does this only for binary outcome, while rcorrp.cens() does take survival object, but doesn't output NRI or IDI. Can anyone suggest any other packages that can calculate NRI and IDI

...s i1 and therefore either and'ing or xor'ing the results of my comparison with the value 1. This is causing the IR after running opt to be correct, but when it hits my backend, there are some changes that generate something weird. int gID = width; const int idx = gID % 64; const int idy = gID / 64; pv[gID] = 0; if( idy > 63 && idx > 0 ) { pv[gID]=1; } generates conditionals : ult r1043.x, 63, r1026.x ilt r1041.x, r1039.x, 1 and r1041.x, r1041.x, r1043.x if_logicalz r1041.x //w...

...em. > I have no clue how the 1 gets put in there (!(idx > 0)) == (idx <= 0) == (idx < 1). Most likely the DAGCombiner is doing this. What should be happening is something like the following: Branch lowering transforms the code into something like the following to save a jump: if (!(idy > 63 && idx > 0)) goto afterif; <code in if> afterif: <code after if> Then, !(idy > 63 && idx > 0) gets combined to (idy < 64 || idx < 1). Then, your legalization transforms the branch condition into (!(idy < 64 || idx < 1) == 0). Then, combine...

...s put in there > > (!(idx > 0)) == (idx <= 0) == (idx < 1). Most likely the DAGCombiner > is doing this. > > What should be happening is something like the following: > Branch lowering transforms the code into something like the following > to save a jump: > if (!(idy > 63 && idx > 0)) goto afterif; > <code in if> > afterif: > <code after if> > > Then, !(idy > 63 && idx > 0) gets combined to (idy < 64 || idx < 1). > > Then, your legalization transforms the branch condition into (!(idy < > 6...

given what ?identical says, i find the following odd: x = 1:10 y = 1:10 all.equal(x,y) [1] TRUE identical(x,y) [1] TRUE y[11] = 11 y = y[1:10] all.equal(x,y) [1] TRUE identical(x,y) [1] FALSE y [1] 1 2 3 4 5 6 7 8 9 10 length(y) [1] 10 looks like a bug. platform i686-pc-linux-gnu arch i686 os linux-gnu system

Dear friends, suppose my dataset is the following data: id<-1:9 x<-c(1,2,3,1,2,3,1,2,3) y<-c(1,1,1,2,2,2,3,3,3) data<-data.frame(id,x,y) id x y 1 1 1 1 2 2 2 1 3 3 3 1 4 4 1 2 5 5 2 2 6 6 3 2 7 7 1 3 8 8 2 3 9 9 3 3 i want to do sampling like this:say the sample size is 3. First: random sampling from x; Next ,random sampling from

Dear all, I am interested to evaluate reclassification using net reclassification improvement and Integrated Discrimination Index IDI after survival analysis (Cox proportional hazards using stcox). I search a R package or a R code that specifically addresses the categorical NRI for time-to-event data in the presence of censored observation and, if possible, at different follow-up time points. I

Hi, We've had some issues with auth. /var/log/secure is full of 1000s of these lines: Jun 4 19:12:08 khan dovecot-auth: pam_unix(dovecot:auth): authentication failure; logname= uid=0 euid=0 tty=dovecot ruser= rhost=127.0.0.1 user=user123 Users can usually login OK with their ldap credentials, but occasionally logins slow to a crawl if not outright fail, esp people checking

Following Duncan's suggestion, I forward the below to R-devel. vQ -------- Original Message -------- Subject: Re: [R] Randomly remove condition-selected rows from a matrix Date: Fri, 02 Jan 2009 10:34:52 -0500 From: Duncan Murdoch <murdoch at stats.uwo.ca> To: Wacek Kusnierczyk <Waclaw.Marcin.Kusnierczyk at idi.ntnu.no> CC: R help <R-help at stat.math.ethz.ch>

...1.5377 23.1632 0.066 0.947107 Mutant.IDS 14.6333 23.1632 0.632 0.527934 Mutant.IDT 48.8900 23.1632 2.111 0.035458 * Mutant.IDU 58.9597 23.1632 2.545 0.011313 * Mutant.IDV 81.7657 23.1632 3.530 0.000467 *** Mutant.IDW 82.9576 23.1632 3.581 0.000386 *** Mutant.IDY 49.1926 23.1632 2.124 0.034343 * Mutant.IDZ 51.0381 23.1632 2.203 0.028170 * Mutant.IDZA 116.0487 23.1632 5.010 8.38e-07 *** Mutant.IDZB 56.4402 23.1632 2.437 0.015287 * Mutant.IDZC -14.5305 23.1632 -0.627 0.530838 Mutant.IDZD -5.0069 23.1632 -0.216 0.828983 Mutant...

On Nov 18, 2008, at 11:24 AM, Villmow, Micah wrote: > Is there a way to tell LLVM to treat Boolean results as 32bit values > instead of 1 bit values? LLVM IR doesn't have a concept of C level booleans. What problem are you trying to solve? -Chris > > Thanks, > > Micah Villmow > Systems Engineer > Advanced Technology & Performance > Advanced Micro Devices

Hello, I am wondering about the behaviour of strsplit. When the pattern matches the beginning of the search string, the mepty string is added to the result, but that's not the case when the pattern matches the end of the search string: strsplit(" hello dolly ") [1] "" "hello" "dolly" The man for strsplit explains the algorithm: " The algorithm

Is there a way to tell LLVM to treat Boolean results as 32bit values instead of 1 bit values? Thanks, Micah Villmow Systems Engineer Advanced Technology & Performance Advanced Micro Devices Inc. 4555 Great America Pkwy, Santa Clara, CA. 95054 P: 408-572-6219 F: 408-572-6596 -------------- next part -------------- An HTML attachment was scrubbed... URL:

First; is this a Wine error number? 11001 The dialog box has the Wine logo in the upper left corner. Jim -------------- next part -------------- An HTML attachment was scrubbed... URL: http://www.winehq.org/pipermail/wine-users/attachments/20080106/569d6998/attachment.htm

Full_Name: Wacek Kusnierczyk Version: 2.8.0 and 2.10.0 r48242 OS: Ubuntu 8.04 Linux 32 bit Submission from: (NULL) (129.241.110.161) In the following code: duplicated(data.frame(), incomparables=NA) # Error in if (!is.logical(incomparables) || incomparables) .NotYetUsed("incomparables != FALSE") : # missing value where TRUE/FALSE needed the raised error is clearly not the

In the example: x = rep(c(0,0,1,1),4) y = rep(c(0,1,0,1),4) z = c(1,0,1,0,0,0,1,1,0,1,0,0,1,1,1,1) f = as.factor(c(rep("a",4),rep("b",4),rep("c",4),rep("d",4))) levelplot(z~x+y|f,data.frame(x=x,y=y,z=z,f=f)) I noted that the last ("d") plot remains empty. I guess the reason for this is that the values are constant (1), but I consider it more

Attached is a suggested patch for src/main/character.c. It does not fix a bug, but rather provides an improvement on the recent extension of do_grep. In essence, instead of four occurences of 'invert ^ LOGICAL(ind)[i]' that accommodate for the option 'invert' added to grep, there is one occurence of 'LOGICAL(ind)[i] ^= invert' that modifies in-place the logical vector of

The attached patch provides a modification to the recent fix/improvement to do_grep already included in the most recent development version. The original fix added new functionality to the grep function by adding a new parameter, 'invert'. In the source code for the underlying do_grep, the value of the parameter is used to invert the logical match-no match flag vector ind. The

the following is a trivialized version of some functional code i tried to use in r: (funcs = lapply(1:5, function(i) function() i)) # a list of no-parameter functions, each with its own closure environment, # each supposed to return the corresponding index when applied to no arguments sapply(funcs, function(func) func()) # supposed to return c(1,2,3,4,5) there is absolutely nothing unusual in

Hello! I have the problem that in my function the passed variable is not used, but the variable name of the dataframe itself?- difficult to explain, but an easy example: TestFunc<-function(df, group) { ??? print(names(subset(df, select=group))) } df1<-data.frame(group="G1", visit="V1", value=0.9) TestFunc(df1, c("group", "visit")) Result: [1]