Displaying 20 results from an estimated 200 matches similar to: "lapply() not converting columns to factors (no error message)"
2009 Sep 18
3
Error: length(f1) == length(f2) is not TRUE
Dear R users,
I am trying to fit an lmer model with only random effects which is giving
me the following error:
Error : length(f1) == length(f2) is not TRUE
In addition: Warning messages:
1: In P1L55:family :
numerical expression has 390 elements: only the first used
2: In P1L55:family :
numerical expression has 390 elements: only the first used
I am trying to extract variance components
2009 Jul 15
2
Differing Variable Length Inconsistencies in Random Effects/Regression Models
Dear All,
I am quite new to R and am having a problem trying to run a linear model
with random effects/ a regression- with particular regard to my variable
lengths being different and the models refusing to compute any further.
The codes I have been using are as follows:
vc<-read.table("P:\\R\\Testvcomp10.txt",header=T)
>> attach(vc)
>
> family<-factor(family)
>
2009 Nov 19
1
Splitting massive output into multiple text files
Dear List,
I thought it would be much easier to put a second query into a second mail.
I need to print 426*10000 blocks of variance components data, where 426 is
the number of columns of data that have 10000 permutations of variance
generated for each of them.
I have tried printing out a smaller number of permutations for a smaller
number of markers and that has worked.
However, since a
2009 Sep 18
0
Error: length(f1) == length(f2) is not TRUE (fwd)
---------- Forwarded Message ----------
Date: 18 September 2009 19:24 +0100
From: A Singh <bzwas at bristol.ac.uk>
To: William Dunlap <wdunlap at tibco.com>
Subject: RE: [R] Error: length(f1) == length(f2) is not TRUE
Yup, they are all factors- and its still doesn't work.
Getting to the stage where I can use 'summary()' is the problem- the error
stalls the process before a
2009 Sep 11
0
How to block data across multiple columns?
Dear all,
Does anyone have any suggestions on how to block multiple columns of data
one at a time in the midst of an analysis, having specified the blocking
variable?
I am running a random effects model using lmer, and my data set has
multiple columns.
Individuals in the study are grouped into 60 families- which is the
blocking factor.
The random effects are markers (labeled Pxlyy below)
2009 Nov 19
0
Printing labeled summary to text file ?
Dear List,
I am trying to run a mixed model which, on the R console, prints output as
follows:
[1] "Marker"
[1] "perm no."
[1] NA
Linear mixed model fit by REML
Formula: peg.no.prm ~ 1 + (1 | family/f)
Data: modeldf
AIC BIC logLik deviance REMLdev
3119 3134 -1555 3112 3111
Random effects:
Groups Name Variance Std.Dev.
f:family (Intercept) 0.0
2009 Sep 21
4
Working around 256 byte variable names? + trouble opening large file
Dear R users,
I am trying to read in a file with 105 columns, and when trying to attach
it, get an error as follows:
> vc1<-read.table("P:\\R\\Everything-I.txt", header=T, sep=" ", dec=".",
na.strings=NA, strip.white=T)
> attach(vc1)
Error in attach(vc1) : variable names are limited to 256 bytes
Is there a way to get around this, and make R accept the
2009 Dec 08
1
Printing 'k' levels of factors 'n' times each, but 'n' is unequal for all levels ?
Dear List,
I need to print out each of 'k' levels of a factor 'n' times each, where
'n' is the number of elements belonging to each factor.
I know that this can normally be done using the gl() command,
but in my case, each level 'k' has an unequal number of elements.
Example with code is as below:
vc<-read.table("P:\\Transit\\CORRECT
2009 Jul 28
1
Multiple lmer runs using 2 'for' loops
Dear Mark and the R list,
I apologise greatly for not having thanked you earlier for your help with
my last mail, which involved trying to sort out variance components using
two methods: a direct looped lmer, and a two-step analysis involving
regressions of AFLP markers on lmer residuals for the same result (or so I
hoped).
I was in India for ten days and no CRAN mirror in the vicinity made
2006 Jun 04
3
passing a value in a hidden field inside a form
should be trivial but I can''t figure this out from api
<%= hidden_field ''certifications'', ''personnel_id'', {value =
#{@personnel.id}} %>
I want to pass the value of ''id'' field in my form to the certifications
controller as personnel_id
Craig
2012 Jan 04
1
rmpi vs snow - which one is better from communication overhead point of view
Hi,
I need to understand when is it best to use /rmpi/ and when is it best to
use /snow/ for parallel programming in R? I understand snow can be used for
a group of non-clustered work stations also. But I wish to understand from
the point of view of using both on clusters for a problem which has few
chunks of straightforward data-parallelism interleaved with some
communication. Since both are
2006 May 23
3
Pagination problem
hi everybody,
i am new to rails..
plz clear my doubt...
this is my question
In my table(public_topics), i have 11 rows of data..
whenever i run the following view, it shows the entire rows in the
current page..
and i have set the '':per_page'' option to ''2''.so it must show two data
per page..but it shows all the rows in a single page..and this case
continue
2016 Mar 23
1
We really need more community involvement with GSoC
On 03/20/2016 12:20 PM, John Criswell via llvm-dev wrote:
> On 3/19/16 6:26 AM, Bruce Hoult via llvm-dev wrote:
>> I work for one of those "top companies", but LLVM is only one of the
>> back-ends I'm working with so I have only pretty narrow knowledge of
>> it (but ok knowledge of compilers in general, I hope).
>>
>> I'm trying to post easy to
2006 Jun 18
5
on change of drop down showing a new textfield
hi,
i have a situation where i have a list of options from which a user selects
his options if its not in the list he/she selects Others/Not Present and a
new textfield should pop up and the data entered there should be considered
instead of dropdown.
i have done this using javascript/dhtml but i was wondering how this can be
done using pure rails i.e using some of the RoR apis?.
i would
2010 Apr 23
3
substract start from the end of the vector
Dear group,
Here is my df :
df <-
structure(list(DESCRIPTION = c("PRM HGH GD ALUMINIUM USD 09/07/10 ",
"PRM HGH GD ALUMINIUM USD 09/07/10 ", "PRIMARY NICKEL USD 04/06/10 "
), CREATED.DATE = structure(c(18361, 18361, 18325), class = "Date"),
QUANITY = c(-1L, 1L, 1L), CLOSING.PRICE = c("2,415.90", "2,415.90",
2006 Jan 10
3
Issue with c++ .C call
I am still having some difficulties with connecting R to a C++ function. I
am able to call the function as expected after compiling the shared library
and such. However, the call to the function is via .C; parameters from the
.C call are not being passed correctly to the function. As an example, I
have attached a GDB run of the code. I set a breakpoint on entry to the
function I am calling
2013 Feb 19
2
Cramer von Mises test for a discrete distribution
Hi,
?
I'm trying to carry out Cramer von Mises tests between pairs of vectors belonging to a discrete distribution (concretely frequencies from 0 to 200). However, the program crashes in the attempt. The problem seems to be that these vectors only have positive integer numbers (+ zero). When I add a random very small positive decimal to the non-decimal part everything works fine (files prm1
2010 Apr 16
1
data frame manipulation
Dear group,
Here is my data.frame :
df <-
structure(list(DESCRIPTION = c("PRM HGH GD ALU", "PRM HGH GD ALU",
"PRIMARY NICKEL", "PRIMARY NICKEL", "PRIMARY NICKEL", "PRIMARY NICKEL",
"STANDARD LEAD ", "STANDARD LEAD ", "STANDARD LEAD ", "STANDARD LEAD ",
"STANDARD LEAD ",
2011 Jul 10
1
question about release prm for 6
Hi,
I am running centos x686 and upgraded from 4 to 5 when it came out now I
want to try the same with centos 6 where can I get the centos i686 release
rpm I used does not have the rpm I used any help wood be greatfull
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2010 Apr 15
1
sum rows in a data.frame...solution
Found this solution. It is maybe not the most elegant way, but it does the
job.
> a=as.data.frame(substr(lme$DESCRIPTION,1,14))
> colnames(a)=c("DESCRIPTION")
> lme=as.data.frame(c(a,lme[,2:3]))
> lme
DESCRIPTION CLOSING.PRICE POSITION
1 PRIMARY NICKEL 25,755.7100 0
2 PRIMARY NICKEL 25,760.8600 0
3 PRM HGH GD ALU 2,415.9000 0