linear regression by column

Hi everyone.

I have a DF with the first column being my independant variable and  
all
other columns the dependent variables.

Something like:

x	y1	y2	y3
...	...	...	...
...	...	...	...

What I'm trying to do is to perform a linear model for each of my  
"y". It is
pretty simple with loops, but I'm trying to vectorize it using  
*apply*.

For instance, I tried something like:

apply(DF, 1, function(DF){lm(DF[,1] ~ Band1[,2:5])})

But apparently it does not work.

.

On Feb 29, 2012, at 1:53 PM, Filoche wrote:

Hi everyone.

I have a DF with the first column being my independant variable and  
all
other columns the dependent variables.

Something like:

x	y1	y2	y3
...	...	...	...
...	...	...	...

What I'm trying to do is to perform a linear model for each of my  
"y". It is
pretty simple with loops, but I'm trying to vectorize it using  
*apply*.

For instance, I tried something like:

apply(DF, 1, function(DF){lm(DF[,1] ~ Band1[,2:5])})

apply( DF[2:5], 2, function(x){lm(DF[,1] ~ x)})

You need to use the variable name that you created in the function  
call and loop over columns, not rows.

But apparently it does not work.

For about four or five reasons.

.

David Winsemius, MD
West Hartford, CT

On Feb 29, 2012, at 6:39 PM, David Winsemius wrote:

On Feb 29, 2012, at 1:53 PM, Filoche wrote:

Hi everyone.

I have a DF with the first column being my independant variable and
all
other columns the dependent variables.

Something like:

x	y1	y2	y3
...	...	...	...
...	...	...	...

What I'm trying to do is to perform a linear model for each of my
"y". It is
pretty simple with loops, but I'm trying to vectorize it using
*apply*.

For instance, I tried something like:

apply(DF, 1, function(DF){lm(DF[,1] ~ Band1[,2:5])})

apply( DF[2:5], 2, function(x){lm(DF[,1] ~ x)})

You need to use the variable name that you created in the function
call and loop over columns, not rows.

I read the request wrong. It would be:

apply( DF[2:5], 2, function(y){y ~ DF$x)})

But apparently it does not work.

For about four or five reasons.

.

David Winsemius, MD
West Hartford, CT

David Winsemius, MD
West Hartford, CT