# [R] linear fit function with NA values

iza.ch1 iza.ch1 at op.pl
Sat Jul 27 22:46:59 CEST 2013

```Hi

Thanks for your hints. I would like to describe my problem better and give an examle of the data that I use.

I conduct the event study and I need to create abnormal returns for the daily stock prices. I have for each stock returns from time period of 8 years. For some days I don't have the data for many reasons. in excel file they are just empty cells but I convert my data into 'zoo' and then it is transformed into NA. I get something like this

return

ATI        AMU
-1   0.734     9.003
0    0.999     2.001
1    3.097     -1.003
2        NA        NA
3        NA     3.541

median
ATI        AMU
-1   3.224     -2.003
0    2.999     -1.301
1    1.3        -1.003
2    4.000     2.442
3       -10     4.511

I want to regress first column return with first column median and second column return with second column median. when I do
OLS<-lapply(seq_len(ncol(return)),function(i) {lm(return[,i]~median[,i])})
I get an error message. I would like my function to omit the NAs and for example for ATI returns to take into account only the values for -1,0,1 and regress it against the same values from ATI in median which means it would also take only (3.224, 2.999, 1.3)

Is it possible to do it?

Thanks a lot

W dniu 2013-07-27 17:33:30 użytkownik arun <smartpink111 at yahoo.com> napisał:
>
>
> HI,
> set.seed(28)
> dat1<- as.data.frame(matrix(sample(c(NA,1:20),100,replace=TRUE),ncol=10))
>
> set.seed(49)
> dat2<- as.data.frame(matrix(sample(c(NA,40:80),100,replace=TRUE),ncol=10))
>  lapply(seq_len(ncol(dat1)),function(i) {lm(dat2[,i]~dat1[,i])}) #works bcz the default setting removes NA
> Regarding the options:
> ?lm()
> na.action: a function which indicates what should happen when the data
>           contain ‘NA’s.  The default is set by the ‘na.action’ setting
>           of ‘options’, and is ‘na.fail’ if that is unset.  The
>           ‘factory-fresh’ default is ‘na.omit’.  Another possible value
>           is ‘NULL’, no action.  Value ‘na.exclude’ can be useful.
>
>  lapply(seq_len(ncol(dat1)),function(i) {lm(dat2[,i]~dat1[,i],na.action=na.exclude)})
> #or
>  lapply(seq_len(ncol(dat1)),function(i) {lm(dat2[,i]~dat1[,i],na.action=na.omit)})
>
> lapply(seq_len(ncol(dat1)),function(i) {lm(dat2[,i]~dat1[,i],na.action=na.fail)})
> #Error in na.fail.default(list(`dat2[, i]` = c(54L, 59L, 50L, 64L, 40L,  :
>  # missing values in object
>
> In your case, the error is different.  It could be something similar to the below case:
> dat1[,1]<- NA
>
> lapply(seq_len(ncol(dat1)),function(i) {lm(dat2[,i]~dat1[,i],na.action=na.omit)})
> #Error in lm.fit(x, y, offset = offset, singular.ok = singular.ok, ...) :
>  # 0 (non-NA) cases # here it is different
>
>  lapply(seq_len(ncol(dat1)),function(i) {try(lm(dat2[,i]~dat1[,i]))}) #works in the above case.  It may not work in your case.
>
> You need to provide a reproducible example to understand the situation better.
> A.K.
>
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> ----- Original Message -----
> From: iza.ch1 <iza.ch1 at op.pl>
> To: r-help at r-project.org
> Cc:
> Sent: Saturday, July 27, 2013 8:47 AM
> Subject: [R] linear fit function with NA values
>
> Hi
>
> Quick question. I am running a multiple regression function for each column of two data sets. That means as a result I get several coefficients. I have a problem because data that I use for regression contains NA. How can I ignore NA in lm function. I use the following code for regression:
> OLS<-lapply(seq_len(ncol(es.w)),function(i) {lm(es.w[,i]~es.median[,i])})
> as response I get
> Error in lm.fit(x, y, offset = offset, singular.ok = singular.ok, ...) :
>   all values NA
>
> thanks for help :)
>
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