Re: [R] subset arg in (modified) evalq
Try this:
with(subset(data, x 0), summary(y + z))
On 5/18/07, Vadim Ogranovich [EMAIL PROTECTED] wrote:
Hi,
When using evalq to evaluate expressions within a say data.frame context I
often wish there was a 'subset' argument, much like in lm() or any ather
advanced regression model. I would be grateful for a tip how to do this.
Here is an illustration of what I want:
n - 100
data - data.frame(x=rnorm(n), y=rnorm(y), z=rnorm(z))
# this works
evalq({ i - 0x; summary(y[i] + z[i]) }, data)
# I want to do the above w/o explicit subscripting, e.g.
myevalq(summary(y + z), subset=0x, data)
Thanks,
Vadim
[[alternative HTML version deleted]]
__
R-help@stat.math.ethz.ch 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-help@stat.math.ethz.ch 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.
Re: [R] subset arg in (modified) evalq
I would check your performance assumption with an actual test before
concluding such but at any rate subset does have a select argument. See
?subset
On 5/18/07, Vadim Ogranovich [EMAIL PROTECTED] wrote:
Thanks Gabor! This does exactly what I wanted.
One follow-up question, how to extract the var names, in this case y, z,
from the expression? The subset function creates a new object and this may
be expensive when the data has a lot of irrelevant collumns. So I thougth
that I could reduce this to the columns I actually need.
Thanks,
Vadim
- Original Message -
From: Gabor Grothendieck [EMAIL PROTECTED]
To: Vadim Ogranovich [EMAIL PROTECTED]
Cc: r-help@stat.math.ethz.ch
Sent: Friday, May 18, 2007 9:19:49 AM (GMT-0600) America/Chicago
Subject: Re: [R] subset arg in (modified) evalq
Try this:
with(subset(data, x 0), summary(y + z))
On 5/18/07, Vadim Ogranovich [EMAIL PROTECTED] wrote:
Hi,
When using evalq to evaluate expressions within a say data.frame context I
often wish there was a 'subset' argument, much like in lm() or any ather
advanced regression model. I would be grateful for a tip how to do this.
Here is an illustration of what I want:
n - 100
data - data.frame(x=rnorm(n), y=rnorm(y), z=rnorm(z))
# this works
evalq({ i - 0x; summary(y[i] + z[i]) }, data)
# I want to do the above w/o explicit subscripting, e.g.
myevalq(summary(y + z), subset=0x, data)
Thanks,
Vadim
[[alternative HTML version deleted]]
__
R-help@stat.math.ethz.ch 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-help@stat.math.ethz.ch 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.
Re: [R] subset arg in (modified) evalq
Thanks Gabor! This does exactly what I wanted.
One follow-up question, how to extract the var names, in this case y, z, from
the expression? The subset function creates a new object and this may be
expensive when the data has a lot of irrelevant collumns. So I thougth that I
could reduce this to the columns I actually need.
Thanks,
Vadim
- Original Message -
From: Gabor Grothendieck [EMAIL PROTECTED]
To: Vadim Ogranovich [EMAIL PROTECTED]
Cc: r-help@stat.math.ethz.ch
Sent: Friday, May 18, 2007 9:19:49 AM (GMT-0600) America/Chicago
Subject: Re: [R] subset arg in (modified) evalq
Try this:
with(subset(data, x 0), summary(y + z))
On 5/18/07, Vadim Ogranovich [EMAIL PROTECTED] wrote:
Hi,
When using evalq to evaluate expressions within a say data.frame context I
often wish there was a 'subset' argument, much like in lm() or any ather
advanced regression model. I would be grateful for a tip how to do this.
Here is an illustration of what I want:
n - 100
data - data.frame(x=rnorm(n), y=rnorm(y), z=rnorm(z))
# this works
evalq({ i - 0x; summary(y[i] + z[i]) }, data)
# I want to do the above w/o explicit subscripting, e.g.
myevalq(summary(y + z), subset=0x, data)
Thanks,
Vadim
[[alternative HTML version deleted]]
__
R-help@stat.math.ethz.ch 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.
[[alternative HTML version deleted]]
__
R-help@stat.math.ethz.ch 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.
Re: [R] subset arg in (modified) evalq
Sorry, I didn't explain myself clear enough. I knew about the select arg in
subset(). My question was, given the expression expression(summary(x+y)), how
to extract all names that will be looked up during its evaluation.
As to checking performance assumptions, you are right, in most cases the
overhead is negligible, but sometimes I work with really big data sets.
Thanks a lot for your help,
Vadim
- Original Message -
From: Gabor Grothendieck ggrothendieck @ gmail .com
To: Vadim Ogranovich vogranovich @ jumptrading .com
Cc: r-help @stat.math. ethz .ch
Sent: Friday, May 18, 2007 9:53:26 AM ( GMT-0600 ) America/Chicago
Subject: Re: [R] subset arg in (modified) evalq
I would check your performance assumption with an actual test before
concluding such but at any rate subset does have a select argument. See
?subset
On 5/18/07, Vadim Ogranovich vogranovich @ jumptrading .com wrote:
Thanks Gabor ! This does exactly what I wanted.
One follow-up question, how to extract the var names, in this case y, z,
from the expression? The subset function creates a new object and this may
be expensive when the data has a lot of irrelevant collumns . So I thougth
that I could reduce this to the columns I actually need.
Thanks,
Vadim
- Original Message -
From: Gabor Grothendieck ggrothendieck @ gmail .com
To: Vadim Ogranovich vogranovich @ jumptrading .com
Cc: r-help @stat.math. ethz .ch
Sent: Friday, May 18, 2007 9:19:49 AM ( GMT-0600 ) America/Chicago
Subject: Re: [R] subset arg in (modified) evalq
Try this:
with(subset(data, x 0), summary(y + z))
On 5/18/07, Vadim Ogranovich vogranovich @ jumptrading .com wrote:
Hi,
When using evalq to evaluate expressions within a say data.frame context I
often wish there was a 'subset' argument, much like in lm () or any ather
advanced regression model. I would be grateful for a tip how to do this.
Here is an illustration of what I want:
n - 100
data - data.frame(x= rnorm (n), y= rnorm (y), z= rnorm (z))
# this works
evalq ({ i - 0x; summary(y[i] + z[i]) }, data)
# I want to do the above w/o explicit subscripting , e.g.
myevalq (summary(y + z), subset=0x, data)
Thanks,
Vadim
[[alternative HTML version deleted]]
__
R-help @stat.math. ethz .ch 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.
[[alternative HTML version deleted]]
__
R-help@stat.math.ethz.ch 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.
Re: [R] subset arg in (modified) evalq
Try this:
e - quote(summary(y + z))
all.vars(e)
On 5/18/07, Vadim Ogranovich [EMAIL PROTECTED] wrote:
Sorry, I didn't explain myself clear enough. I knew about the select arg in
subset(). My question was, given the expression expression(summary(x+y)),
how to extract all names that will be looked up during its evaluation.
As to checking performance assumptions, you are right, in most cases the
overhead is negligible, but sometimes I work with really big data sets.
Thanks a lot for your help,
Vadim
- Original Message -
From: Gabor Grothendieck [EMAIL PROTECTED]
To: Vadim Ogranovich [EMAIL PROTECTED]
Cc: r-help@stat.math.ethz.ch
Sent: Friday, May 18, 2007 9:53:26 AM (GMT-0600) America/Chicago
Subject: Re: [R] subset arg in (modified) evalq
I would check your performance assumption with an actual test before
concluding such but at any rate subset does have a select argument. See
?subset
On 5/18/07, Vadim Ogranovich [EMAIL PROTECTED] wrote:
Thanks Gabor! This does exactly what I wanted.
One follow-up question, how to extract the var names, in this case y, z,
from the expression? The subset function creates a new object and this may
be expensive when the data has a lot of irrelevant collumns. So I thougth
that I could reduce this to the columns I actually need.
Thanks,
Vadim
- Original Message -
From: Gabor Grothendieck [EMAIL PROTECTED]
To: Vadim Ogranovich [EMAIL PROTECTED]
Cc: r-help@stat.math.ethz.ch
Sent: Friday, May 18, 2007 9:19:49 AM (GMT-0600) America/Chicago
Subject: Re: [R] subset arg in (modified) evalq
Try this:
with(subset(data, x 0), summary(y + z))
On 5/18/07, Vadim Ogranovich [EMAIL PROTECTED] wrote:
Hi,
When using evalq to evaluate expressions within a say data.frame context
I
often wish there was a 'subset' argument, much like in lm() or any ather
advanced regression model. I would be grateful for a tip how to do this.
Here is an illustration of what I want:
n - 100
data - data.frame(x=rnorm(n), y=rnorm(y), z=rnorm(z))
# this works
evalq({ i - 0x; summary(y[i] + z[i]) }, data)
# I want to do the above w/o explicit subscripting, e.g.
myevalq(summary(y + z), subset=0x, data)
Thanks,
Vadim
[[alternative HTML version deleted]]
__
R-help@stat.math.ethz.ch 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-help@stat.math.ethz.ch 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.
