similar to: is there any multiple log-rank tests function with adjusted P-value implemented ?

How can I get the Log - Rank p value to be output? The chi square value can be output, so I was thinking if I can also have the degrees of freedom output I could generate the p value, but can't see how to find df either. > (survtest <- survdiff(Surv(time, cens) ~ group, data = surv,rho=0)) Call: survdiff(formula = Surv(time, cens) ~ group, data = surv, rho = 0) N Observed

> On Feb 13, 2018, at 4:02 PM, array chip via R-help <r-help at r-project.org> wrote: > > Hi all, > > The survdiff() from survival package has an argument "rho" that implements Fleming-Harrington weighted long rank test. > > But according to several sources including "survminer" package

> On Feb 14, 2018, at 5:26 PM, David Winsemius <dwinsemius at comcast.net> wrote: > >> >> On Feb 13, 2018, at 4:02 PM, array chip via R-help <r-help at r-project.org> wrote: >> >> Hi all, >> >> The survdiff() from survival package has an argument "rho" that implements Fleming-Harrington weighted long rank test. >>

hie Im trying to calculate the power of the logrank test for different values of rho .I was just wandering whether the following programme would do it. any suggestions are welcome s=50 number=1 count1=0;count2=0;count3=0;count4=0;count5=0;count6=0;count7=0;count7=0; count8=0;count9=0 while(s!=0){ n=20 n1=n/2

Hi all,? The survdiff() from survival package has an argument "rho" that implements Fleming-Harrington weighted long rank test.? But according to several sources including "survminer" package (https://cran.r-project.org/web/packages/survminer/vignettes/Specifiying_weights_in_log-rank_comparisons.html), Fleming-Harrington weighted log-rank test should have 2 parameters

DeaR experts, I was asked for a log-rank pairwise survival comparison. I've a straightforward way to do this using the SAS system: http://support.sas.com/documentation/cdl/en/statug/63033/HTML/default/viewer.htm#/documentation/cdl/en/statug/63033/HTML/default/statug_lifetest_sect019.htm What I've found in R is shown below, but it's not a logrank test, I suppose. (The documentation

how do I check the rank of a matrix ? say A= 1 0 0 0 1 0 then rank(A)=2 what is this function? thanks I did try help.search("rank"), but all the returned help information seem irrelevant to what I want. I would like to know how people search for help information like this. rank(base) Sample Ranks SignRank(stats) Distribution of the

This is more of a general question without data. After doing 'survdiff', from the 'survival' package, on strata including four groups (so 4 curves on a Kaplan Meier curve) you get a chi squared p-value whether to reject the null hypothesis or not. Is there a method to followup with pairwise testing on the respective groups? I have searched the library but have come up with

Hello all- I am doing a survival analysis for two species of invasive plants I outplanted to edges and interiors of island and mainland sites in a local reservoir. I am using the KM estimate and had no problem doing survdiff for my data using the following code: S4<-Surv(outplant$SurvTime, outplant$StatusD6) diff4=survdiff(S4 ~ outplant$Species+outplant$SiteType+outplant$EdgInt) diff4

Hi list-members can anybody tell me why > pairwise.t.test(val, fac) produces an empty p-table. As shown below: Pairwise comparisons using t tests with pooled SD data: val and fac AS AT Fhh Fm Fmk Fmu GBS Gf HFS Hn jAL Kol R_Fill AT - - - - - - - - - - - - - Fhh - - - - - - - - - - - - - Fm - - - - - - -

Thank you Terry. Right now I can use comp() from survMisc package to do the 2-parameter version of F-H weighting. I think both SAS and stata offer the 2-parameter version, so just?thought it would be nice if survdiff() can have that option given it's standard package in R.? Thanks! John On Friday, February 16, 2018, 7:08:46 AM PST, Therneau, Terry M., Ph.D. <therneau at mayo.edu>

Dear all, Does anyone knows how I could extract the p-value in: > survdiff(Surv(tempo,status) ~ grupo,data=dados1,rho=1) Call: survdiff(formula = Surv(tempo, status) ~ grupo, data = dados1, rho = 1) N Observed Expected (O-E)^2/E (O-E)^2/V grupo=1 21 5.12 12.00 3.94 14.5 grupo=2 21 14.55 7.68 6.16 14.5 Chisq= 14.5 on 1 degrees of freedom,

Hi list, I want to use the p-value from the survdiff function (package survival) to reuse within a function in a Kaplan-Meier plot. The p-value is somehow not a component of the value list ?! Thanks in advance -- A. Goralczyk G?ttingen, Ger.

Dear R-help, I have a question about pairwise.t.test and adjustment for multiple comparisons for paired data points. I have the following data: n=c("x", "x", "x", "x", "x", "x", "x", "x", "x", "x", "y", "y", "y", "y", "y", "y",

Dear all, I am perplexed when trying to get the same results using pairwise.t.test and t.test. I'm using examples in the ISwR library, >attach(red.cell.folate) I can get the same result for pairwise.t.test and t.test when I set the variances to be non-equal, but not when they are assumed to be equal. Can anyone explain the differences, or what I'm doing wrong? Here's an example

Dear SIr, In case of looking at the codes of the fuction, "cov", we find all the codings below. But, incase of "mean", we don't find the contents. Please show me the way to look at all the codings of any functions. Best regards, Kei ----------------------- > cov function (x, y = NULL, use = "all.obs", method = c("pearson",

Hello, I would like to work with Orange as part of GSoC 2012(and continue henceforth). Apologies for joining in a bit late- i was waiting to get a proper grasp of things before discussing it here. Currently I am a Masters students in Mathematics with my bachelors in Computer Science[integrated dual degree]. Over the last year and a half, I have worked on a few ML projects and have a couple of

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

Hi, I have just encountered a strange behaviour from 'cor' with regards to the treatment of NAs when calculating Spearman correlations. I guess it is a subtle bug. If I understand the help page correctly, the two modes 'complete.obs' and 'pairwise.complete.obs' specify how to deal with correlation coefficients when calculating a correlation _matrix_. When calculating

All, Given the following commands: > ag.m35<-read.table("m35.txt",header=TRUE,sep=",") > ag.m35.lp<-subset(ag.m35, race=="lp") > aov.m35.lp=aov(year~time,data=ag.m35.lp) > anova.m35.lp=anova(aov.m35.lp) > anova.m35.lp Analysis of Variance Table Response: year Df Sum Sq Mean Sq F value Pr(>F) time 1 11.56 11.5649 1.7282