]*,
and (2) read_sas *[from library 'haven']. *But couldn't find what I am
looking for.
Best regards,
Utkarsh Singhal
91.96508.54333
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list -- To UNSUBS
not be quite
>> as hard as it sounds.
>>
>> Incidentally, do teach your mailer to not send plain text. It is not much
>> of a problem this time, but HTML mails can become quite unreadable on the
>> list.
>>
>>
>> -pd
>>
>>
>> > On 10
# Model-1
library(lmer)
coef(lmer(Reaction ~ Days + (1| Subject), sleepstudy))
# Model-2
coef(lm(Reaction ~ Days + Subject, sleepstudy))
Can somebody tell me the reason? Are the above formulations actually
different or is it due to different optimization method used?
Thank you.
Utkar
this, can you alternately answer the
following: "Is it possible to define the 'lm' and 'lmer' models above so
they produce the same results (at least in terms of predictions)?"
Thanks again.
Utkarsh Singhal
91.96508.54333
On 12 July 2016 at 19:15, Thierry Onkelinx wrote
123456
295.0310 305.4983 315.9656 326.4329 336.9002 347.3675
And these are quite different from the mixedmodel:
> fit_lmer = lmer(Reaction ~ Days + (1| Subject), sleepstudy)
> head(predict(fit_lmer))
12345
lly go away with
large data. So, I am not too concerned about that.
I can sleep peacefully if you just say that the difference is due to
optimization method used or distributional assumptions :).
But if it is anything more too it, then how do I decide which of the above
two models to choose in what scenari
Hi All,
I want to "merge" two datasets by column "ID" and I don't want the result to
be sorted by "ID". I am doing the following:
> z = merge(x, y, by = "ID", sort=F)
The result is not sorted by "ID". But (as oppose to what I expected) it is
not even in the original order of eith
Thanks, but I kind of new this way already and it doesn't seem an optimal
thing to do.
What I was looking for is to pass some argument in 'merge' itself
which doesn't change the ordering of 'x'. Or, more than that, I am
interested in knowing that why is it changing the o
Hi All,
I ran the following lines in R:
print(object.size(a <- rep(1,10^6)),units="Mb")
print(object.size(a <- rep(3.542,10^6)),units="Mb")
print(object.size(b <- rep("x",10^6)),units="Mb")
print(object.size(b <- rep("xyzxyz xyz",10^6)),units="Mb")
print(object.size(b <- 1:10^6),units="Mb")
prin
Hi R,
I have an excel file in which the third column is "date" and others are
"character" and "numeric".
Number of columns are 12
If I use this to read the file in R: x = read.xls("D:\\file.xls")
The problem is that my date column is read in julian dates.
So I am using:
Sorry but I was interested in reading as date format from the excel
itself. Is there any way of doing this?
Regards
Utkarsh
-Original Message-
From: Gabor Grothendieck [mailto:[EMAIL PROTECTED]
Sent: Tuesday, March 25, 2008 8:43 PM
To: Utkarsh Singhal
Cc: [EMAIL
,
Regards
Utkarsh Singhal | Amba Research
Ph +91 80 3980 8017 | Mob +91 99 0295 8815
Bangalore * Colombo * London * New York * San José * Singapore *
www.ambaresearch.com <mailto:utkar...@ambaresearch.com>
This e-mail may contain confidential and/or privileged i...{{drop
his?
Thank you,
Regards
Utkarsh Singhal | Amba Research
Ph +91 80 3980 8017 | Mob +91 99 0295 8815
Bangalore * Colombo * London * New York * San José * Singapore *
www.ambaresearch.com <mailto:utkar...@ambaresearch.com>
This e-mail may contain confidential and/or privileged i...{{d
: Utkarsh Singhal
Cc: R-help Forum
Subject: Re: [R] how to label the branches of a tree
You may have to change/scale the sizes of the font by using "cex" and then to
keep all labels within the plotting window, use "xpd=TRUE". Like in
text(fit, use.n=TRUE, cex=0.8, xpd=TRUE)
Hi R,
I am getting this error while trying to use 'lrm' function with nine
independent variables:
> res =
lrm(y1994~WC08301+WC08376+WC08316+WC08311+WC01001+WC08221+WC08106+WC0810
1+WC08231,data=y)
singular information matrix in lrm.fit (rank= 8 ). Offending
variable(s):
WC08101 WC0822
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