Displaying 7 results from an estimated 7 matches for "lazert".
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lazerte
2003 Mar 31
4
Convert char vector to numeric table
I'm a great fan of read.table(), but this time the data had a lot of cruft. So I used readLines() and editted the char vector to eventually get something like this:
" 23.4 1.5 4.2"
" 19.1 2.2 4.1"
and so on. To get that into a 3 col numeric table, I first just used:
writeLines(data,"tempfile")
2006 Sep 26
2
Sort problem with merge (again)
# R version 2.3.1 (2006-06-01) Debian Linux "testing"
# Is the following behaviour a bug, feature or just a lack of
# understanding on my part? I see that this was discussed here
# last March with no apparent resolution.
d <- as.factor(c("1970-04-04","1970-08-11","1970-10-18"))
x <- c(9,10,11)
ch <- data.frame(Date=d,X=x)
d <-
2003 Mar 24
1
using tapply with a matrix?
I've successfully created my own function (topbot) that takes a vector as input and used it with tapply() and an input vector (data$depth) and classification factor (data$profile):
vols <- tapply(data$depth, data$profile, topbot)
works great.
But now I want to do something similar, except my function will take a 3 column matrix with nrows() equal to the factor's length. tapply
2003 May 15
1
Manly's randomization analysis of multiple regression
My wife has been using a diagnostic from Manley (1991; "Randomization and MonteCarlo Methods in Biology") that compares a normal multiple regression's performance with that using random predicted variables.
Is there something like this already available in R?
If not, the "boot" package looks like a good place to start looking for methods, no?
Thanks in advance
Bruce
2003 Aug 07
0
predict(lm(etc.), some_data) -> "numeric envir arg not of length one " ???
...arg not of length one
???
> df$flow[p]<- predict(tlo,df$flow1[p])
This loess prediction works however.
Could someone explain the linear model's "numeric envir arg not of length one" error for me?
Thanks in advance ...
Bruce L.
__________________________________________
Bruce LaZerte
Grandview Lake in Muskoka
Baysville, Ontario, Canada
2003 Aug 09
1
lm(), na.exclude and predict()
I'm trying to get predict.lm() to return an NA for each NA row in it's input vector, so the output is the same length as the input.
I thought that using na.action=na.exclude with lm() would do that. But apparently not ??
df <- data.frame(x=c(NA,1,2,3,NA),y=c(0,2,3,4,0))
tl <- lm(y~x,df,na.action=na.exclude)
predict.lm(tl,data.frame(x=c(2.5,NA,3,4,5)))
1 3 4 5
3.5 4.0 5.0
2003 Dec 01
0
No subject
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