R help - Apr 2010 - Problem calculating multiple regressions on a data frame.

Hi there,
I am stuck trying to solve what should be a fairly easy problem.
I have a data frame that essentially consists of (ID, time as seqMonth,
variable, value) and i want to find the regression coefficient of value vs
time for each combination of ID and Variable.
I have tried several approaches and none of them seems to work as i
expected.
For example, i have tried:

theSplit<-split(theTestLineal, list(as.factor(theTestLineal $ids),
as.factor(theTestLineal $variable)))

I can then use 
lm(value~seqMonth,data=zongSplit[[1]])
...
lm(value~seqMonth,data=zongSplit[[4]])

that works well, (it fails for some combinations of ID and variable where
there is one datapoint)

however when i try to use an lapply:
lapply(zongSplit,function(x)lm(value~seqMonth,data=x,na.action=na.exclude))

it fails, with error message:
Error in lm.fit(x, y, offset = offset, singular.ok = singular.ok, ...) :
 0 (non-NA) cases

I have tried to change the na.action with no success (na.pass, na.fail,
na.exclude... all give the same error message)


I have also tried to follow the approach suggested by Charles Sharpsteen
(http://www.mail-archive.com/r-help@r-project.org/msg74759.html) with
similar results.
The code is as follows:
theModels <- by( theTestLineal, list( theTestLineal$ids,
zongTestLineal$variable), function( dataSlice ){
linMod <- lm( value ~ seqMonth, data = dataSlice )

# Slope and intercept may be recovered from the output of the coef()
function:
intercept <- coef( linMod )[1]
slope <- coef( linMod )[2]

# The R-Squared value is returned by the summary() function:
rsq <- summary( linMod )[[ ''r.squared'' ]]

# The summary function also provides statistics for the F-distribution,
# extract them, reformat as a list, rename and feed to pf() using do.call()
# in order to get the p-value:
fStats <- as.list( summary( linMod )[[ ''fstatistic'' ]] )
names( fStats ) <- c( ''q'', ''df1'',
''df2'' )
fStats[[ ''lower.tail'' ]] <- FALSE

pVal <- do.call( pf, fStats )

return(data.frame( slope, intercept, rsq, pVal ))
})

Any help will be appreciated!

	[[alternative HTML version deleted]]

Hi

what about

fit <- lm(value~seqMonth+ids+varable, data=theTestLineal)

or similar approach using

?lme

See also

?interaction

Regards
Petr

r-help-bounces at r-project.org napsal dne 27.04.2010 13:48:32:
> Hi there,
> I am stuck trying to solve what should be a fairly easy problem.
> I have a data frame that essentially consists of (ID, time as seqMonth,
> variable, value) and i want to find the regression coefficient of value 
vs
> time for each combination of ID and Variable.
> I have tried several approaches and none of them seems to work as i
> expected.
> For example, i have tried:
> 
> theSplit<-split(theTestLineal, list(as.factor(theTestLineal $ids),
> as.factor(theTestLineal $variable)))
> 
> I can then use 
> lm(value~seqMonth,data=zongSplit[[1]])
> ...
> lm(value~seqMonth,data=zongSplit[[4]])
> 
> that works well, (it fails for some combinations of ID and variable 
where
> there is one datapoint)
> 
> however when i try to use an lapply:
> 
lapply(zongSplit,function(x)lm(value~seqMonth,data=x,na.action=na.exclude))
> 
> it fails, with error message:
> Error in lm.fit(x, y, offset = offset, singular.ok = singular.ok, ...) :
>  0 (non-NA) cases
> 
> I have tried to change the na.action with no success (na.pass, na.fail,
> na.exclude... all give the same error message)
> 
> 
> I have also tried to follow the approach suggested by Charles Sharpsteen
> (http://www.mail-archive.com/r-help at r-project.org/msg74759.html) with
> similar results.
> The code is as follows:
> theModels <- by( theTestLineal, list( theTestLineal$ids,
> zongTestLineal$variable), function( dataSlice ){
> linMod <- lm( value ~ seqMonth, data = dataSlice )
> 
> # Slope and intercept may be recovered from the output of the coef()
> function:
> intercept <- coef( linMod )[1]
> slope <- coef( linMod )[2]
> 
> # The R-Squared value is returned by the summary() function:
> rsq <- summary( linMod )[[ 'r.squared' ]]
> 
> # The summary function also provides statistics for the F-distribution,
> # extract them, reformat as a list, rename and feed to pf() using 
do.call()
> # in order to get the p-value:
> fStats <- as.list( summary( linMod )[[ 'fstatistic' ]] )
> names( fStats ) <- c( 'q', 'df1', 'df2' )
> fStats[[ 'lower.tail' ]] <- FALSE
> 
> pVal <- do.call( pf, fStats )
> 
> return(data.frame( slope, intercept, rsq, pVal ))
> })
> 
> Any help will be appreciated!
> 
>    [[alternative HTML version deleted]]
> 
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide 
http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.

Replace lm(...) with try(lm(...))


On Tue, Apr 27, 2010 at 7:48 AM, Luis Sisam?n <luis.sisamon at gmail.com>
wrote:
> Hi there,
> I am stuck trying to solve what should be a fairly easy problem.
> I have a data frame that essentially consists of (ID, time as seqMonth,
> variable, value) and i want to find the regression coefficient of value vs
> time for each combination of ID and Variable.
> I have tried several approaches and none of them seems to work as i
> expected.
> For example, i have tried:
>
> theSplit<-split(theTestLineal, list(as.factor(theTestLineal $ids),
> as.factor(theTestLineal $variable)))
>
> I can then use
> lm(value~seqMonth,data=zongSplit[[1]])
> ...
> lm(value~seqMonth,data=zongSplit[[4]])
>
> that works well, (it fails for some combinations of ID and variable where
> there is one datapoint)
>
> however when i try to use an lapply:
> lapply(zongSplit,function(x)lm(value~seqMonth,data=x,na.action=na.exclude))
>
> it fails, with error message:
> Error in lm.fit(x, y, offset = offset, singular.ok = singular.ok, ...) :
> ?0 (non-NA) cases
>
> I have tried to change the na.action with no success (na.pass, na.fail,
> na.exclude... all give the same error message)
>
>
> I have also tried to follow the approach suggested by Charles Sharpsteen
> (http://www.mail-archive.com/r-help at r-project.org/msg74759.html) with
> similar results.
> The code is as follows:
> theModels <- by( theTestLineal, list( theTestLineal$ids,
> zongTestLineal$variable), function( dataSlice ){
> linMod <- lm( value ~ seqMonth, data = dataSlice )
>
> # Slope and intercept may be recovered from the output of the coef()
> function:
> intercept <- coef( linMod )[1]
> slope <- coef( linMod )[2]
>
> # The R-Squared value is returned by the summary() function:
> rsq <- summary( linMod )[[ 'r.squared' ]]
>
> # The summary function also provides statistics for the F-distribution,
> # extract them, reformat as a list, rename and feed to pf() using do.call()
> # in order to get the p-value:
> fStats <- as.list( summary( linMod )[[ 'fstatistic' ]] )
> names( fStats ) <- c( 'q', 'df1', 'df2' )
> fStats[[ 'lower.tail' ]] <- FALSE
>
> pVal <- do.call( pf, fStats )
>
> return(data.frame( slope, intercept, rsq, pVal ))
> })
>
> Any help will be appreciated!
>
> ? ? ? ?[[alternative HTML version deleted]]
>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>

Gabor Grothendieck <ggrothendieck <at> gmail.com>
writes:
> 
> Replace lm(...) with try(lm(...))
> 
Thanks for all the replies. I managed to make it work with the try() trick, 
I actually padded the lm() on 
two levels of try() and is working now Sharpsteen approach.