Hello,
Maybe instead of a loop, vectorize with logical indices.
i1 <- is.na(jindex)
i2 <- is.numeric(jindex)
if(any(!i1)){
if(any(!i2)){
words <- jindex[!i1 & !i2]
pattern <- paste(words, collapse = "|")
jindex <- grep(pattern = pattern, x.label, value = FALSE)
}
jj <- jindex[
I am an author of the paper behind the fad package. I suspect that the call is
not correct. Actually, fad does not quite account for time series or other
structured data and you have to enter it, as in all general EFA packages as a n
x p matrix, with n the number of observations and p the number
Thank to Rui, Jeff, and Bert. They are all very useful.
Somewhat related is the following, in which jindex is a numeric or
alphanumeric vector in a function that starts with
try<-function(, jindex=NA)
In the if loop, in the first line I am trying to determine whether the
vector jindex is
This not being a question about R, but rather about statistics, or possibly
about a contributed package, means (per the Posting Guide) that you should be
asking in a statistics forum like stats.stackexchange.com or corresponding with
the author of the package in question. If you are lucky someon
Dear R users,
I want to find the latent factors from a kind of time-series data
describing temporal changes of concentration using a factor analysis
technique called 'factor analysis of dynamic structure (FADS).' I learned
how to form the data for the analysis using a proper package embedding
FADS
Hello,
The pattern can be assembled with paste(., collapse = "|").
With the same vector of names, nms:
words <- c("black","conserv")
pattern <- paste(words, collapse = "|")
grep(pattern = pattern, nms, value = TRUE)
#[1] "x1.black" "x1.conserv" "x2.black" "x2.conserv"
Hope this helps,
Ru
Regular expression patterns are not vectorized... only the data to be searched
are. Use one of the many websites dedicated to tutoring regular expressions to
learn how they work. (Using function names like "names" as data names is bad
practice.)
nms <- c( "x1.one", "x1.black", "x1.othrrace", "x
On 5/8/21 10:00 AM, Steven Yen wrote:
Below, the first command simply creates a list of 16 names (labels)
which can be ignore.
In the 2nd and 3rd commands, I am able to identify names containing
"black".
In line 4, I am trying to identify names containing "black" or
"conserv" but obviousl
Below, the first command simply creates a list of 16 names (labels)
which can be ignore.
In the 2nd and 3rd commands, I am able to identify names containing "black".
In line 4, I am trying to identify names containing "black" or "conserv"
but obviously it does not work. Can someone help? Thank
9 matches
Mail list logo
