On Feb 14, 2013, at 1:10 AM, Kategoricus wrote:
Hello,
probably a newbie question, but i didnt find any information on
plotting/regressing w.r.t. a date data type. My trials were
unfruitful. Can
anyone help ? Thanks in advance!
Here is my interaction with R:
tabelle
date number date2
1 2009-01-1 1673 2009-01-01
2 2009-12-1 2111 2009-12-01
3 2010-7-1 2487 2010-07-01
4 2013-2-1 4301 2013-02-01
regression.punkte<-lm(tabelle$number ~ tabelle$date2)
Fehler in model.frame.default(formula = tabelle$number ~ tabelle
$date2, :
ungültiger Typ (list) für die Variable 'tabelle$date2'
I'm guessing that you used as.POSIXlt to create that date2 variable.
POSIXlt classed variables are lists and as such create all sorts of
problems for functions that are expecting atomic vectors. If I'm
right, you will ahve better chances of success by converting to
POSIXct class.
> tabelle <- read.table(text="date number date2
+ 1 2009-01-1 1673 2009-01-01
+ 2 2009-12-1 2111 2009-12-01
+ 3 2010-7-1 2487 2010-07-01
+ 4 2013-2-1 4301 2013-02-01", header=TRUE, stringsAsFactors=FALSE)
> tabelle$date2 <- as.POSIXlt(tabelle$date2)
> regression.punkte<-lm(tabelle$number ~ tabelle$date2)
Error in model.frame.default(formula = tabelle$number ~ tabelle
$date2, :
invalid type (list) for variable 'tabelle$date2'
> tabelle$date2 <- as.POSIXct(tabelle$date2)
> regression.punkte<-lm(tabelle$number ~ tabelle$date2)
> regression.punkte
Call:
lm(formula = tabelle$number ~ tabelle$date2)
Coefficients:
(Intercept) tabelle$date2
-2.405e+04 2.082e-05
regression.punkte<-lm(tabelle$number ~ tabelle$date)
plot(tabelle$date2,tabelle$number, pch=19, xlab="date of retrieval",
ylab="number of animals") ***R DRAWS THIS***
regression.punkte
Call:
lm(formula = tabelle$number ~ tabelle$date)
Coefficients:
(Intercept) tabelle$date2009-12-1 tabelle$date2010-7-1
1673 438 814
tabelle$date2013-2-1
2628
abline(regression.punkte, lwd=2) ***R DOES NOT DRAW LINE***
Warnmeldung:
In abline(regression.punkte, lwd = 2) :
nutze nur die ersten beiden von 4 Regressionskoeffizienten
summary(regression.punkte)
Call:
lm(formula = tabelle$number ~ tabelle$date)
Residuals:
ALL 4 residuals are 0: no residual degrees of freedom!
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 1673 NA NA NA
tabelle$date2009-12-1 438 NA NA NA
tabelle$date2010-7-1 814 NA NA NA
tabelle$date2013-2-1 2628 NA NA NA
Residual standard error: NaN on 0 degrees of freedom
Multiple R-squared: 1, Adjusted R-squared: NaN
F-statistic: NaN on 3 and 0 DF, p-value: NA
David Winsemius, MD
Alameda, CA, USA
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