Displaying 5 results from an estimated 5 matches for "manuelperera".
2005 Oct 28
3
splitting a character field in R
Dear R users,
I have a dataframe with one character field, and I would like to create two
new fields (columns) in my dataset, by spliting the existing character
field into two using an existing substring.
... something that in SAS I could solve e.g. combining substr(which I am
aware exist in R) and "index" for determining the position of the pattern
within the string.
e.g. if my
2005 Nov 07
2
lattice chart: different definitions for series
Hi enthusiasts,
Trying to create a single chart in lattice with different plotting
definitions for the different series (two series should be drawn with lines
and the other without them)
I am using a dataset, which includes a grouping variable e.g. clinic with
three levels, the variable "year" and a continous variable: "mct".
In the graph the variable "year" is in
2006 May 26
1
query: sample size calculation
Dear all,
I am doing something wrong.
I am trying to apply a formula for sample size calculation as in the book
"Design and Analysis of Clinical Trials", from Chow et all.
There a suggested sample size strategy uses the formula
0.30/0.45=F(0.80,2n,2n)/F(0.025,2n,2n)
which gives n=96, as in the book.
(here F(alfa,k,n) is the upper (alfa)th quantile of an F distribution with
k,n
2005 Oct 24
0
query on xtable(Sweave)
Dear all,
I am exporting to latex a dataframe via xtable(Sweave). My dataframe
includes 4 rows and 4 columns. The first two rows containing real
values(e.g. a laboratory parameter's mean value), and the two rows at the
bottom containing only integers (number of cases).
using display=c('fg','fg','fg','fg') as an argument to xtable I tried to
reach my goal, of
2005 Oct 25
0
query on xtable output
Dear all,
I am exporting to latex a matrix via xtable. My matrix includes e.g. 4 rows
and 4 columns. The first two rows containing real values(e.g. a laboratory
parameter's mean value), and the two rows at the bottom containing only
integers (number of cases).
>
mymat<-matrix(c(c(10.52,2.52,12.35,3.63),c(3.52,16.25,13.62,6.36),c(11,12,15,16),c(14,15,16,18)),4,4,byrow=TRUE)
> mymat