similar to: analysis of life tables

Displaying 20 results from an estimated 2000 matches similar to: "analysis of life tables"

2005 Nov 17
1
Mean survival times
Dear list, I have data on insect survival in different cages; these have the following structure: deathtime status id cage S F G L S 1.5 1 1 C1 8 2 1 1 1 1.5 1 2 C1 8 2 1 1 1 11.5 1 3 C1 8 2 1 1 1 11.5 1 4 C1 8 2 1 1 1 There are 81 cages and
2004 Sep 19
1
Namespace problem
Now I try to add some C and Fortan code to my package, so the NAMESPACE file is useDynLib(eha) importFrom(survival, Surv) export(mlreg.fit, risksets) but I get ..... * checking R files for library.dynam ... OK * checking S3 generic/method consistency ... WARNING Error in .try_quietly({ : Error in library(package, lib.loc = lib.loc, character.only = TRUE, verbose = FALSE) :
2004 Sep 19
1
NAMESPACE Warning
I'm trying to learn how to use namespaces, so I have created a small package with two functions. My NAMESPACE file is importFrom(survival, Surv) export(mlreg.fit, risksets) because 'mlreg.fit' uses 'Surv' from 'survival'. However I get * checking package dependencies ... WARNING Namespace dependencies not required: survival Why? (If I remove 'importFrom' I
2007 Dec 04
2
weighted Cox proportional hazards regression
I'm getting unexpected results from the coxph function when using weights from counter-matching. For example, the following code produces a parameter estimate of -1.59 where I expect 0.63: d2 = structure(list(x = c(1, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 0, 0, 1, 0, 1, 0, 1, 0, 1), wt = c(5, 42, 40, 4, 43, 4, 42, 4, 44, 5, 38, 4, 39, 4, 4, 37, 40, 4, 44, 5, 45, 5, 44, 5), riskset =
2012 Oct 16
2
R Kaplan-Meier plotting quirks?
Hello. I apologize in advance for the VERY lengthy e-mail. I endeavor to include enough detail. I have a question about survival curves I have been battling off and on for a few months. No one local seems to be able to help, so I turn here. The issue seems to either be how R calculates Kaplan-Meier Plots, or something with the underlying statistic itself that I am misunderstanding. Basically,
2006 Jul 02
1
Calculation of lags
Hi, If I have the follow situation: A dependent variable (i.e. number of insects) that is affected by an independent variable (i.e. rain). The problem is that the measure of rain affect the population in other moment. So there exit a lag between the rain and the number of insects. Exist in R any tool to find what is this lag? Explain better. Suppose that I have a linear relationship
2008 Nov 18
2
matrix for diversity functions?
Hi, I have a small simple data frame (attached) - to compare diversity of insects encountered in disturbed and unditurbed site. What i have is the count of insects - the total number of times they were encountered over 30 monitoring slots. Can someone please check for me to make sure how the 'community data matrix' for the diversity function needs to be oriented so that i'm
2008 Aug 15
3
ylab with an exponent
plot(1,2, ylab= paste("insects", expression(m^2), sep=" ")) I get insects m^2 I would like m to the 2 what is the problem? -- Let's not spend our time and resources thinking about things that are so little or so large that all they really do for us is puff us up and make us feel like gods. We are mammals, and have not exhausted the annoying little problems of being
2007 Nov 08
6
Extract correlations from a matrix
Dear R users, suppose I have a matrix of observations for which I calculate all pair-wise correlations: m=matrix(sample(1:100,replace=T),10,10) w=cor(m,use="pairwise.complete.obs") How do I extract only those correlations that are >0.6? w[w>0.6] #obviously doesn?t work, and I can?t find a way around it. I would very much appreciate any help! Best wishes Christoph (using R
2005 Jan 07
6
coercing columns
Dear all, I have a data frame that looks like this: c1 c2 c3 A B C B C A A A B and so on; I?d like to produce one single vector consisting of the columns c1,c2, c3, such that vector=("A","B","A","B","C","A","C","A","B") I guess it?s easy to do but I don?t know how...Can anyone
2006 Sep 13
3
unexpected result in glm (family=poisson) for data with an only zero response in one factor
Dear members, here is my trouble: My data consists of counts of trapped insects in different attractive traps. I usually use GLMs with a poisson error distribution to find out the differences between my traitments (and to look at other factor effects). But for some dataset where one traitment contains only zeros, GLM with poisson family fail to find any difference between this particular traitment
2010 Jul 07
1
Appropriateness of survdiff {survival} for non-censored data
I read through Harrington and Fleming (1982) but it is beyond my statistical comprehension. I have survival data for insects that have a very finite expiration date. I'm trying to test for differences in survival distributions between different groups. I understand that the medical field is most often dealing with censored data and that survival analysis, at least in the package survival,
2008 Nov 06
2
replacing characters in formulae / models
Dear all, How can I replace text in objects that are of class "formula"? y="a * x + b" class(y)="formula" grep("x",y) y[1] Suppose I would like to replace the "x" by "w" in the formula object "y". How can this be done? Somehow, the methods that can be used in character objects do not work 1:1 in formula objects... Many
2008 Nov 25
4
glm or transformation of the response?
Dear all, For an introductory course on glm?s I would like to create an example to show the difference between glm and transformation of the response. For this, I tried to create a dataset where the variance increases with the mean (as is the case in many ecological datasets): poissondata=data.frame( response=rpois(40,1:40), explanatory=1:40) attach(poissondata) However, I have run into
2008 Feb 22
3
Simultaneously summarizing many models
Dear R users, Let?s say I have 10 models, each named m1,m2,m3..., and I would like to summarize them automatically and simultaneously - e.g., to extract parameter estimates later on from all models; how can I do that? I have tried: x=1:10 #this creates some example data y=rnorm(10) m1=lm(x~y) m2=lm(x~1) sum.lms=function(x)summary(paste("m",x,sep="")) sum.lms(1:2) but
2005 May 02
4
"apply" question
Dear R users, I??ve got a simple question but somehow I can??t find the solution: I have a data frame with columns 1-5 containing one set of integer values, and columns 6-10 containing another set of integer values. Columns 6-10 contain NA??s at some places. I now want to calculate (1) the number of values in each row of columns 6-10 that were NA??s (2) the sum of all values on columns 1-5
2009 Jul 08
2
Randomizing a dataframe
Hi R-helpers, I have a dataframe (called data) with trees in rows (n=100) and insect species (n=10) in columns. My tree IDs are in a column called TREE and each species has a column labeled SPEC1, SPEC2, SPEC3, etc... I wish to randomize the values in my dataframe such that row and column totals are held constant, i.e. in my randomized data each tree will have the same number of individual
2010 May 26
2
sequential treatment of a vector for formula
Please pardon the simplicity of this question of biological nature. I'm trying to calculate a statistic, px, the proportion of a cohort that survives through the interval x:x+1. I have the vector from which the calc is to be made but I can't figure out how to tell R to take the current value and divide it by the next value. The formula is P0=L1/LO The following is an example of the
2016 Aug 22
3
Dial and start music on hold after timeout
Sorry, I forgot to write that the SIP peer must keep ringing while the announcement is being played. Le 22/08/2016 ? 17:42, John Kiniston a ?crit : > This seems like the obvious answer but maybe I'm misunderstanding the > question. > > exten => s,1,Dial(SIP/alice,20) > same => n,Playback(myannouncement) > same => n,NoOP(Whatever else you want to do goes
2007 Jun 15
1
complex contrasts and logistic regression
Hi, I am doing a retrospective analysis on a cohort from a designed trial, and I am fitting the model fit<-glmD(survived ~ Covariate*Therapy + confounder,myDat,X=TRUE, Y=TRUE, family=binomial()) My covariate has three levels ("A","B" and "C") and therapy has two (treated and control), confounder is a continuous variable. Also patients were randomized to