[R] matrix with standard errors of lm model
Hi
Can someone give me a hint on how to create a matrix with standard errors from
lm model? I have already managed to get the matrix with coefficients:
coef-as.data.frame(sapply(seq_len(ncol(es.w)),function( i) {x1-
summary(lm(es.w[,i]~es.median[,i]));x1$coef[,1]}))
but I can't get the one like this for standard errors. I do regression for each
column.
Thanks a lot :)
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[R] t.test error
Hi I receive a very strange error message after trying to do t-test. When I write the code t.test(x) I get an error message: error in t.test(x) : function sqr not found I don't understand this problem. Can someone help me how to do it right? Thanks a lot :) __ R-help@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.
[R] replace Na values with the mean of the column which contains them
Hi everyone
I have a problem with replacing the NA values with the mean of the column which
contains them. If I replace Na with the means of the rest values in the column,
the mean of the whole column will be still the same as if I would have omitted
NA values. I have the following data
de
[,1][,2] [,3]
[1,] NA -0.26928087 -0.1192078
[2,] NA 1.20925752 0.9325334
[3,] NA 0.38012008 -1.8927164
[4,] NA -0.41778861 1.4330507
[5,] NA -0.49677462 0.2892706
[6,] NA -0.13248754 1.3976522
[7,] NA -0.54179054 0.2295291
[8,] NA 0.35788624 -0.5009389
[9,] 0.27500571 -0.41467591 -0.3426560
[10,] -3.07568579 -0.59234248 -0.8439027
[11,] -0.42240954 0.73642396 -0.4971999
[12,] -0.26901731 -0.06768044 -1.6127122
[13,] 0.01766284 -0.40321968 -0.6508823
[14,] -0.80999580 -1.52283305 1.4729576
[15,] 0.20805934 0.25974308 -1.6093478
[16,] 0.03036708 -0.04013730 0.1686006
and I wrote the code
de[which(is.na(de))]-sapply(seq_len(ncol(de)),function(i)
{mean(de[,i],na.rm=TRUE)})
I get as the result
[,1][,2] [,3]
[1,] -0.50575168 -0.26928087 -0.1192078
[2,] -0.1376 1.20925752 0.9325334
[3,] -0.13412312 0.38012008 -1.8927164
[4,] -0.50575168 -0.41778861 1.4330507
[5,] -0.1376 -0.49677462 0.2892706
[6,] -0.13412312 -0.13248754 1.3976522
[7,] -0.50575168 -0.54179054 0.2295291
[8,] -0.1376 0.35788624 -0.5009389
[9,] 0.27500571 -0.41467591 -0.3426560
[10,] -3.07568579 -0.59234248 -0.8439027
[11,] -0.42240954 0.73642396 -0.4971999
[12,] -0.26901731 -0.06768044 -1.6127122
[13,] 0.01766284 -0.40321968 -0.6508823
[14,] -0.80999580 -1.52283305 1.4729576
[15,] 0.20805934 0.25974308 -1.6093478
[16,] 0.03036708 -0.04013730 0.1686006
It has replaced the NA values in first column with mean of first column
-0.505... and second cell with mean of second column etc.
I want to have the result like this:
[,1][,2] [,3]
[1,] -0.50575168 -0.26928087 -0.1192078
[2,] -0.50575168 1.20925752 0.9325334
[3,] -0.50575168 0.38012008 -1.8927164
[4,] -0.50575168 -0.41778861 1.4330507
[5,] -0.50575168 -0.49677462 0.2892706
[6,] -0.50575168 -0.13248754 1.3976522
[7,] -0.50575168 -0.54179054 0.2295291
[8,] -0.50575168 0.35788624 -0.5009389
[9,] 0.27500571 -0.41467591 -0.3426560
[10,] -3.07568579 -0.59234248 -0.8439027
[11,] -0.42240954 0.73642396 -0.4971999
[12,] -0.26901731 -0.06768044 -1.6127122
[13,] 0.01766284 -0.40321968 -0.6508823
[14,] -0.80999580 -1.52283305 1.4729576
[15,] 0.20805934 0.25974308 -1.6093478
[16,] 0.03036708 -0.04013730 0.1686006
Thanks in advance
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[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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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
ATIAMU
-1 0.734 9.003
00.999 2.001
13.097 -1.003
2NANA
3NA 3.541
median
ATIAMU
-1 3.224 -2.003
02.999 -1.301
11.3-1.003
24.000 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 smartpink...@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.
- Original Message -
From: iza.ch1 iza@op.pl
To: r-help@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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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.
__
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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.
[R] do not perform function if the outcome is NA
Hi you all
I have a question regarding the function. In my function I divide the values by
the standard errors and sometimes the standard error is equal to zero and I get
the result NA. Can I write the function in the way that if the outcome of the
function is zero then the function is not conducted and it stays the value (not
divided by standard errors)? the code for my function is the following:
standardised.abnormal.returns-lapply(seq_len(ncol(abnormal.returns)),function(i)
{abnormal.returns[,i]/standard.deviation[i,1]})
Thank you all for the help
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Re: [R] do not perform function if the outcome is NA
Hi Rui
Thanks a lot. It works perfect.
Izzie
Hello,
Try the following.
standardised.abnormal.returns -
lapply(seq_len(ncol(abnormal.returns)),function(i) {
if(standard.deviation[i,1] == 0)
abnormal.returns[,i]
else
abnormal.returns[,i]/standard.deviation[i,1]
})
Hope this helps,
Rui Barradas
Em 23-07-2013 22:58, iza.ch1 escreveu:
Hi you all
I have a question regarding the function. In my function I divide the
values by the standard errors and sometimes the standard error is equal to
zero and I get the result NA. Can I write the function in the way that if
the outcome of the function is zero then the function is not conducted and
it stays the value (not divided by standard errors)? the code for my
function is the following:
standardised.abnormal.returns-lapply(seq_len(ncol(abnormal.returns)),function(i)
{abnormal.returns[,i]/standard.deviation[i,1]})
Thank you all for the help
__
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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.
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[R] create data frame with coefficients from many regressions
Hi !
I want to ask if somebody knows the way to create data frame with
coefficients from many regressions
I regress the first column from ret against the first columns from median,
then the second with the second and so on.
This is the code used for regression
i-1:6
lapply(seq_len(ncol(ret)),function(i) {lm(ret[,i]~median[,i])}
I get 6 results for each regression
[[1]]
Call:
lm(formula = ret[, i] ~ median[, i])
Coefficients:
(Intercept) median[, i]
01
[[2]]
Call:
lm(formula = ret[, i] ~ median[, i])
Coefficients:
(Intercept) median[, i]
-1.411e-171.000e+00
now I would like to create a data frame with intercepts which looks like it
[[1]] [[2]]
Intercept
median
I tried to use ddply command but it does not work. I will be very grateful
for the hint :)
Thank you in advance
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[R] Linear regression repeat for each column
Hi everyone I need to calculate abnormal returns for different events applying event study methodology. I must create a market model in order to perform the analysis. I apply regression analysis to get OLS estimators. I have a problem to create a linear regression which I could repeat for each column in two different data frames (one with explainatory and one with explaning variables). It means that I want to regress column 1 from first data frame with column 1 from second data frame,, clumn two with column two etc. I tried to use the following code:#x is matrix containing stock returns y is matrix containing market index i-1:length(x) t[i] - lapply(t[i], lm(x[,1]~y[,i])) but it is not working. Could anybody help me? Thanks a lot Iza __ R-help@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.
