On Apr 25, 2012, at 6:02 PM, Greg Snow wrote:
I believe that fortune(312) applies here. As my current version of
fortunes does not show this I am guessing that it is in the
development version and so here is what fortune(312) will eventually
print (unless something changes or I got something wrong):
The problem here is that the $ notation is a magical shortcut and like
any other magic if used incorrectly is likely to do the programmatic
equivalent of turning
yourself into a toad.
—Greg Snow (in response to a user that wanted to access a column
whose name is
stored in y via x$y rather than x[[y]])
R-help (February 2012)
I second this self-numeration, ... errr, self-nomination. I thought
it was by the maintainer of the mind reading machine, but I see on
investigation that was Dirk E. I see that 291 is the highest current
fortune but ... hey ... what the heck ... why not have some NULL
fortunes hanging around?
--
David
On Tue, Apr 24, 2012 at 9:42 PM, Jim Silverton <jim.silver...@gmail.com
> wrote:
Hi,
I have the following problem- I want to access a list whose
elements are
imp1, imp2, imp3 etc I tried theusing the paste comand in a for
loop see
the last for loop below. But I keep calling it df but df = imp1
(for the
first run). Any ideas on how I can access the elements of the list?
Isaac
require(Amelia)
library(Amelia)
data.use <- read.csv("multiplecarol.CSV", header=T)
names(data.use) = c("year", "dischargex1", "y", "pressurex2" ,
"windx3")
ts <- c (c(1:12), c(1:12), c(1:12), c(1:12), c(1:12), c(1:12),
c(1:12),
c(1:6) )
length(ts)
data.use = cbind(ts, data.use)
#a.out2 <- amelia(data.use, m = 1000, idvars = "year")
n.times = 100
a.out.time <- amelia(data.use, m = n.times, ts="ts", idvars="year",
polytime=2)
constant.col = dischargex1.col = pressurex2.col = windx3.col =
rep(0,n.times)
for (i in 1: n.times)
{
x = c("imp",i)
df = paste(x, collapse = "")
data1 = a.out.time[[1]]$df
attach(data1)
y = as.numeric(y)
dischargex1 = as.numeric(dischargex1)
pressurex2 = as.numeric(pressurex2)
windx3 = as.numeric(windx3)
multi.regress = lm(y~ dischargex1 + pressurex2 + windx3)
constant.col[i] = as.numeric(multi.regress[[1]][1])
dischargex1.col[i] = as.numeric(multi.regress[[1]][2])
pressurex2.col[i] = as.numeric(multi.regress[[1]][3])
windx3.col[i] = as.numeric(multi.regress[[1]][4])
}
--
Thanks,
Jim.
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--
Gregory (Greg) L. Snow Ph.D.
538...@gmail.com
______________________________________________
R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
David Winsemius, MD
West Hartford, CT
______________________________________________
R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
