That function is pure R code, so it might 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
>&g
) 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]]
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https
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 scenario.
Regards,
Utkarsh Singhal
91.9
6
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))
12 3 4 56
292.1888 302.6561 313.1234 323.5907 334.05
nately 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 <thierry.onkel...@inbo.be> wrote:
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.
Utkarsh Singhal
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)
print(object.size(b -
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 either x or y.
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
To: 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)
Philip
--
A Smile costs
in this?
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...{{dropped:13
,
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...{{dropped:13
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
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
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