search for: lazert

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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
...vianet.on.ca (8.9.3/8.9.3) with SMTP id LAA24197 for <samba@lists.samba.org>; Wed, 12 Sep 2001 11:51:12 -0400 (EDT) Received: from ASUSTWO (ASUSTwo [10.1.1.5]) by TYAN586 (Weasel v1.23) for samba@lists.samba.org; 12 Sep 2001 11:47:42 To: samba <samba@lists.samba.org> From: Bruce LaZerte <mail@fwr.on.ca> Reply-To: Bruce LaZerte <mail@fwr.on.ca> Subject: Samba and smbfs do not report all files in a directory X-Mailer: The Polarbar Mailer; version=1.21; build=646 X-Mailer-Platform: Linux; architecture=x86; version=2.4.9muy00 X-Mailer-Java-VM: IBM Corporation; version=J2R...