search for: vumc

Displaying 9 results from an estimated 9 matches for "vumc".

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2006 Oct 29
2
match lists
...of list B in such a way that element 1 of list A corresponds to element 1 of list B (and element 2 of A with 2 of B). So, in total there will be 4 new matrices, 1 for each column of list A. For example, for column a of list A the new matrix will be: 2 40 50 3 60 70 3 60 70 thanks, Martijn VUmc Amsterdam [[alternative HTML version deleted]]
2009 Jan 20
0
vacancy postdoc computational systems biology - Amsterdam
...Clinical Genetics VU University Medical Centre Phone: +31 20 5987460 Contact details: Dr JHGM van Beek Mail address: van der Boechorststraat 7 1081 BT Amsterdam The Netherlands Visiting address: Faculty of Earth and Life Sciences (W&N building) De Boelelaan 1085 Amsterdam Room B-423 http://vumc.nl; http://ibivu.nl; http://home.hccnet.nl/jhgm.van.beek e-mail: hans.van.beek at falw.vu.nl secondary e-mail: hans.vanbeek at vumc.nl (checked infrequently) FOR YOUR INTEREST freely downloable scientific article on dynamics of energy conversion in the heart : Click http://home.hccnet.nl/jhgm.van....
2013 Apr 17
1
simulation\bootstrap of list factors
Dear R experts, I am trying to simulate a list containing data matrices. Unfortunately, I don't manage to get it to work. A small example: n=5 nbootstrap=2 subsets<-list() for (i in 1:n){ subsets[[i]] <- rnorm(5, mean=80, sd=1) for (j in 1:nbootstrap){ test<-list() test[[j]]<-subsets[[i]] } } How can I get test to be 2 simulation rounds with
2011 Oct 18
1
Function in nested loop
Hi all, I would like to run a function with several nested conditions, which are completely factorial. The input data (x1) has two different sample sizes, so: x1 <- dat1 x1 <- dat2 Then a can have 3 different values: a <- 0.15 a <- 0.35 a <- 0.50 Then b can have 2 different values: b <- data.matrix (c(0.5,0.5,0.5)) b <- data.matrix (c(0.2,0.4,0.6)) Then d can have 5
2008 Dec 01
1
gee + rcs
Hi all, I have fitted a gee model with the gee package and included restricted cubic spline functions. Here is the model: chol.g <- gee(SKIN ~ rcs(CHOLT, 3), id=ID, data=chol, family=binomial(link="logit"), corstr="exchangeable") To extract the log odds I use: predict.glm(chol.g, type = "link") Now I want to compute the logg odds for specific CHOLT values
2013 Mar 29
2
changing y-axis intervals in a boxplot
Dear R-users, I'm a bit of a rooky in R and do not know how to change the y-axis intervals of a boxplot. The y-axis in my boxplot ranges from 0 to 100 which is fine. The problem is the interval level in between. Now it increases with steps of 20 but I'd like to have it changed by steps of 5. Anyone knows how to do this in R? I used the following code for producing this boxplot:
2017 Sep 29
0
Error in Lordif: slope is missing or negative
Hi all I am not an experienced user of R. I am trying to perform DIF analysis using Lordif and I get the follow error: > GroupDIF <- lordif(resp.data=Resp, group=Group, criterion="R2", pseudo.R2="McFadden", R2.change=0.02) Iteration: 500, Log-Lik: -137340.437, Max-Change: 0.00119 EM cycles terminated after 500 iterations. (mirt) | Iteration: 1, 14 items flagged for DIF
2011 Sep 19
3
Replace a for loop with a function
Hi all, I would like to replace the for loop in the code below with a function to improve the speed and to make the script more efficient. The loop creates a vector of integers (x) with the probability of f for each integer. The length of f is variable, but sums to 1. I tried to use a function with optional arguments which did not work. Here is the code: f <- data.matrix(c(0.5,0.15,0.35))
2012 Sep 07
3
error: in catg (xi, name=nam, label=lab): "LO2" has <2 category levels
Dear R-users, During a fit procedure in a Logistic prediction model I encounter the following problem: error: in catg (xi, name=nam, label=lab: X has <2 category levels The following code is used: fit <-lrm(MRI_Diag_RC ~ factor(O4_1r) + N6_1r + leeftijd + LO1 + LO2 + LO3+ LO4+ LO5+ LO6+ LO7+ LO8+ LO9+ LO10+ LO11+ LO12+ LO13 + LO14+ LO15+ LO16+ LO17+ LO18+ LO19+ LO20+ LO21+ LO22+ LO23+