Message-ID: <1374958658.73932.YahooMailNeo@web142606.mail.bf1.yahoo.com>
Date: 2013-07-27T20:57:38Z
From: arun
Subject: linear fit function with NA values
In-Reply-To: <45272633-bd71b012a593bdb925f36b87b43e3eaa@pmq5.m5r2.onet>
HI,
I couldn't get any error message with the data you provided.
return<- read.table(text="
????? ATI??????? AMU
-1? 0.734??? 9.003
0??? 0.999??? 2.001
1??? 3.097??? -1.003
2??????? NA??????? NA
3??????? NA??? 3.541
",sep="",header=TRUE)
median<- read.table(text="
????? 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
",sep="",header=TRUE)
?lapply(seq_len(ncol(return)),function(i) {lm(return[,i]~median[,i])})
[[1]]
Call:
lm(formula = return[, i] ~ median[, i])
Coefficients:
(Intercept)? median[, i]?
????? 4.696?????? -1.231?
[[2]]
Call:
lm(formula = return[, i] ~ median[, i])
Coefficients:
(Intercept)? median[, i]?
???? 3.3937????? -0.1607?
lapply(seq_len(ncol(return)),function(i) {lm(return[,i]~median[,i],na.action=na.omit)}) #same as above.
?sessionInfo()
R version 3.0.1 (2013-05-16)
Platform: x86_64-unknown-linux-gnu (64-bit)
locale:
?[1] LC_CTYPE=en_CA.UTF-8?????? LC_NUMERIC=C?????????????
?[3] LC_TIME=en_CA.UTF-8??????? LC_COLLATE=en_CA.UTF-8???
?[5] LC_MONETARY=en_CA.UTF-8??? LC_MESSAGES=en_CA.UTF-8??
?[7] LC_PAPER=C???????????????? LC_NAME=C????????????????
?[9] LC_ADDRESS=C?????????????? LC_TELEPHONE=C???????????
[11] LC_MEASUREMENT=en_CA.UTF-8 LC_IDENTIFICATION=C??????
attached base packages:
[1] stats???? graphics? grDevices utils???? datasets? methods?? base????
other attached packages:
[1] stringr_0.6.2? reshape2_1.2.2
loaded via a namespace (and not attached):
[1] plyr_1.8??? tools_3.0.1
BTW, It is better to ?dput() the example dataset.
A.K.
----- Original Message -----
From: iza.ch1 <iza.ch1 at op.pl>
To: arun <smartpink111 at yahoo.com>
Cc: R help <r-help at r-project.org>
Sent: Saturday, July 27, 2013 4:46 PM
Subject: Re: Re: [R] linear fit function with NA values
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 u?ytkownik arun <smartpink111 at yahoo.com> napisa?:
>
>
> 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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>
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