Hi all,
I want to define a function such that one of its argument if passed do one
thing and if not passed do the second thing. So basically, I have to check
whether the argument is passed or not inside the function. I am trying to
use 'exists' function to do this.
f =
Thanks David. Big help, problem solved.
Regards,
Utkarsh
Original Message
Subject: Re: [R] calling exists function inside
another function is not working
From: David Winsemius dwinsem...@comcast.net
To: utkarshsinghal
utkarsh.sing...@global-analytics.com
Cc: r help r
Hi all,
Is there any way I can add more columns to an existing filebacked big.matrix
object.
In general, I want a way to modify an existing big.matrix object, i.e., add
rows/columns, rename colnames, etc.
I tried the following:
library(bigmemory)
x =
Dec 2010 18:29:38 +0530
From: utkarshsinghal [1]utkarsh.sing...@global-analytics.com
To: r help [2]r-h...@stat.math.ethz.ch
Hi all,
Is there any way I can add more columns to an existing filebacked big.matrix
object.
In general, I want a way to modify an existing big.matrix
A H24.6221 9
4B L28.2254 9
5B M28.8259 9
6B H18.8169 9
On Mon, Nov 23, 2009 at 3:15 AM, utkarshsinghal
utkarsh.sing...@global-analytics.com wrote
Hi All,
I am currently doing the following to compute summary statistics of
aggregated data:
a = aggregate(warpbreaks$breaks, warpbreaks[,-1], mean)
b = aggregate(warpbreaks$breaks, warpbreaks[,-1], sum)
c = aggregate(warpbreaks$breaks, warpbreaks[,-1], length)
ans = cbind(a, b[,3], c[,3])
Hi All,
I have recently *re*-installed R-2.9.1 in my Linux machine. Since then,
I am unable to plot using the usual interactive device.
plot(1:10)
This plots in a pdf file Rplots.pdf in my working directory.
sessionInfo()
R version 2.9.1 (2009-06-26)
i686-pc-linux-gnu
locale:
Hi All,
I am using rpart function to model my data:
library(rpart)
set.seed(1)
x1 = sample(c(a,b),100,T)
x2 = runif(100)
y = ifelse((x1==a x2=0.5)|(x1==b x20.5), 0, 1)
data = data.frame(x1,x2,y)
fit = rpart(y~x1+x2)
plot(fit); text(fit, use.n=TRUE)
Now I want to use variable x1 for the first
Hi All,
d = data.frame(a=1:10,b=1:10)
by1 = rep(c(a,b),5)
by(d, by1, function(z) z[,,drop=F])
by1: a
a b
1 1 1
3 3 3
5 5 5
7 7 7
9 9 9
by1: b
a b
2 2 2
4 4 4
6 6 6
8 8 8
10 10 10
by(d, by1, function(z)
...@imail.org
801.408.8111
*From:* utkarshsinghal [mailto:utkarsh.sing...@global-analytics.com]
*Sent:* Wednesday, July 08, 2009 2:24 AM
*To:* Greg Snow
*Cc:* Thomas Lumley; r help
*Subject:* Re: [R] bigglm() results different from glm()+Another question
Hi Greg,
Many thanks for your
.
Hope this helps,
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
*From:* utkarshsinghal [mailto:utkarsh.sing...@global-analytics.com]
*Sent:* Tuesday, July 07, 2009 12:10 AM
*To:* Greg Snow
*Cc:* Thomas
801.408.8111
*From:* utkarshsinghal [mailto:utkarsh.sing...@global-analytics.com]
*Sent:* Monday, July 06, 2009 8:58 AM
*To:* Thomas Lumley; Greg Snow
*Cc:* r help
*Subject:* Re: [R] bigglm() results different from glm()+Another question
The AIC of the biglm models is highly dependent
Thank you Mr. Lumley and Mr. Greg. That was helpful.
Regards
Utkarsh
Thomas Lumley wrote:
On Fri, 3 Jul 2009, utkarshsinghal wrote:
Hi Sir,
Thanks for making package available to us. I am facing few problems
if you can give some hints:
Problem-1:
The model summary and residual
26647.79
500021647.79
1 11647.79
Regards
Utkarsh
utkarshsinghal wrote:
Thank you Mr. Lumley and Mr. Greg. That was helpful.
Regards
Utkarsh
Thomas Lumley wrote:
On Fri, 3 Jul 2009, utkarshsinghal wrote:
Hi Sir,
Thanks for making package available to us. I am facing
Hi Sir,
Thanks for making package available to us. I am facing few problems if
you can give some hints:
Problem-1:
The model summary and residual deviance matched (in the mail below) but
I didn't understand why AIC is still different.
AIC(m1)
[1] 532965
AIC(m1big_longer)
[1] 101442.9
Hi All,
I am currently using R version 2.8.1 on linux cent os 4.4 (i386) and
want to upgrade to version 2.9.1. It seems to me that version-2.9.1 is
it not for my OS.
Am I right?
Regards
Utkarsh
[[alternative HTML version deleted]]
__
values,
but not necessarily true.
On Fri, Jun 19, 2009 at 8:18 AM, Petr PIKAL petr.pi...@precheza.cz
mailto:petr.pi...@precheza.cz wrote:
Hi
utkarshsinghal utkarsh.sing...@global-analytics.com
mailto:utkarsh.sing...@global-analytics.com napsal dne
17.06.2009 15:29:34
, utkarshsinghal
utkarsh.sing...@global-analytics.com wrote:
Hi Jim,
What you are saying is correct. Although, my computer might not
have same
speed and I am getting the following for 10M entries:
user system elapsed
0.559 0.038 0.607
Moreover, in the case of character
Hi All,
There are several replies to the question below, but I think there must
exist a better way of doing so.
I just want to check whether all the elements of a vector are same. My
vector has one million elements and it is highly likely that there are
distinct elements in the first few
] == x)))
[1] FALSE
user system elapsed
0.180.000.19
This was for 10M entries.
On Tue, Jun 16, 2009 at 7:42 AM, utkarshsinghal
utkarsh.sing...@global-analytics.com
mailto:utkarsh.sing...@global-analytics.com wrote:
Hi All,
There are several replies
(1,1000)) # 10,000,000
system.time(print(all(x[1] == x)))
[1] FALSE
user system elapsed
0.180.000.19
This was for 10M entries.
On Tue, Jun 16, 2009 at 7:42 AM, utkarshsinghal
utkarsh.sing...@global-analytics.com
mailto:utkarsh.sing...@global-analytics.com
Hi all,
I am using glm function with family binomial(logit) to fit logistic
regression model. My data is very big and the algorithm is such that it
has to run glm function hundreds of times. Now *I need only the
**estimates of the coefficients and std. error in my output, *but
apparently
I define the following function:
(Please don't wonder about the use of this function, this is just a
simplified version of my actual function. And please don't spend your
time in finding an alternate way of doing the same as the following does
not exactly represent my function. I am only
Yeah, seems so obvious now. What a blunder, poor me.
Perfect explanation. Thanks
Thomas Lumley wrote:
On Wed, 27 May 2009, utkarshsinghal wrote:
I define the following function:
(Please don't wonder about the use of this function, this is just a
simplified version of my actual function
Hi all,
I am performing a stepwise regression by running the step function on
an lm object. Now I want to save the intermediate iterations. I know
the argument trace=T will print it on the console, but I rather want to
assign it to some R object or may be output it in a CSV or text file.
can have all the intermediate
iterations also as R objects.
Thanks in advance.
Gabor Grothendieck wrote:
Try this:
out - capture.output(example(step))
On Mon, May 18, 2009 at 9:11 AM, utkarshsinghal
utkarsh.sing...@global-analytics.com wrote:
Hi all,
I am performing a stepwise
Hi All,
Can anybody explain why the following three ways of extracting residuals
from a glm object are giving me different outputs:
idv = runif(1000,0,1)
dv = rbinom(1000,1,0.5)
d = data.frame(idv,dv)
fit = glm(dv~idv, data=d, family=binomial)
head(residuals(fit))
1 2
Hi All,
Is there any way in R to calculate the difference between two dates in
years, months days.
date1 = as.Date(1993-11-23)
date2 = as.Date(2009-04-15)
I want my output to be:15 years 4 months 27 days
Any hint would be helpful.
Regards
Utkarsh
[[alternative HTML
Hi All,
I have vector of length 52, say, x=sample(30,52,replace=T). I want to
sort x and split into five *nearly equal groups*. Note that the
observations are repeated in x so in case of a tie I want both the
observations to fall in same group.
This seems a very common task to do, but still I
I never understood that why is the value returned by as.date function in
the library(survival) never matches with the description given in the
help file:
Following is the extract from ?as.date
Description:
Converts any of the following character forms to a Julian date:
8/31/56, 8-31-1956,
__
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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.
Hi R,
I have explored R archives a lot but couldn't find an efficient way of
doing the following:
I want to split a vector into sets of equal sizes. Is there any inbuilt
function of doing so with the option of specifying how to treat the
remaining observations. For example: suppose I want
__
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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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