Displaying 20 results from an estimated 2000 matches similar to: "how to replace <NA> values"
2012 Feb 26
1
Matrix problem to extract animal associations
Dear List,
I have been trying to extract associations from a matrix whereby individual locations are within a certain distance threshold from one another.
I have been able to extract those individuals where there is 'no interaction' (i.e. where these individuals are not within a specified distance threshold from another individual) and give these individuals a unique Group ID containing
2006 Mar 05
2
Wishlist: merge and subset to keep attributes (PR#8658)
Full_Name: Ulrike Gr?mping
Version: 2.2.1
OS: Windows
Submission from: (NULL) (84.190.139.94)
When importing data from SPSS, it is a nice feature of the package foreign that
it allows (option use.value.labels=F) to work with the original SPSS codes while
keeping the value labels as information in an attribute. Unfortunately, after
merging or subsetting, these attributes disappear.
The code
2013 Jul 02
2
Recoding variables based on reference values in data frame
I'm new to R (previously used SAS primarily) and I have a genetics data
frame consisting of genotypes for each of 300+ subjects (ID1, ID2, ID3,
...) at 3000+ genetic locations (SNP1, SNP2, SNP3...). A small subset of
the data is shown below:
SNP_ID SNP1 SNP2 SNP3 SNP4 Maj_Allele C G C A Min_Allele T A T G ID1
CC GG CT AA ID2 CC GG CC AA ID3 CC GG
nc
AA
2016 Dec 27
0
colnames for data.frame could be greatly improved
Hi there,
Any update on this?
Should I create bugzilla ticket and submit patch?
Regards
Jan Gorecki
On 20 December 2016 at 01:27, Jan Gorecki <J.Gorecki at wit.edu.pl> wrote:
> Hello,
>
> colnames seems to be not optimized well for data.frame. It escapes
> processing for data.frame in
>
> if (is.data.frame(x) && do.NULL)
> return(names(x))
>
> but
2011 Apr 20
1
How to check if a value of a variable is in a list
Hi all,
I am working with some social network analysis in R and ran into a problem I
just cannot solve.
Each observation in my data consists of a respondent, some characteristics
and up to five friends. The problem is that all of these five friends might
no show up later as a respondent (observation). Therefore I might not have
characteristics on all the friends listed in the data and I want to
2016 Dec 20
2
colnames for data.frame could be greatly improved
Hello,
colnames seems to be not optimized well for data.frame. It escapes
processing for data.frame in
if (is.data.frame(x) && do.NULL)
return(names(x))
but only when do.NULL true. This makes huge difference when do.NULL
false. Minimal edit to `colnames`:
if (is.data.frame(x)) {
nm <- names(x)
if (do.NULL || !is.null(nm))
return(nm)
else
2006 Mar 05
0
Wishlist: xtabs and table to optionally use attribute value labels (PR#8659)
Full_Name: Ulrike Gr?mping
Version: 2.2.1
OS: Windows
Submission from: (NULL) (84.190.139.94)
A wish somehow related to my wish 8658: Package foreign allows to import
categorical data from SPSS (and possibly other software) using the original
codes, which are often useful for data manipulation, since one can use already
available lists of codes from others who don't use R etc. The original
2016 Apr 04
0
multiple bar plot annotation text labelling
Use only plain text emails. Don't attach file types that will be stripped. See the footer at the bottom of your email for more information.
Do give us the data using dput():
> dput(testbarplot)
structure(list(V1 = c(1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L), V2 = c(3L,
8L, 1L, 9L, 5L, 4L, 2L, 0L), V3 = structure(c(1L, 2L, 3L, 4L,
3L, 5L, 6L, 7L), .Label = c("A", "B",
2007 Apr 20
2
Fastest way to repeatedly subset a data frame?
Hi -
I have a data frame with a large number of observations (62,000 rows,
but only 2 columns - a character ID and a result list).
Sample:
> my.df <- data.frame(id=c("ID1", "ID2", "ID3"), result=1:3)
> my.df
id result
1 ID1 1
2 ID2 2
3 ID3 3
I have a list of ID vectors. This list will have anywhere from 100 to
1000 members, and
2008 Jan 10
1
data.frame manipulation: Unbinding strings in a row
Hi all,
I have a data.frame I received with data that look like this (comma
separated strings in last row):
ID Shop Items
ID1 A1 item1, item2, item3
ID2 A2 item4, item5
ID3 A1 item1, item3, item4
But I would like to unbind the strings in col(2) items so that it will look
like this:
ID Shop Items
ID1 A1 item1
ID1 A1 item2
ID1 A1 item3
ID2 A2 item4
ID2 A2 item5
ID3 A1 item1
ID3 A1 item3
ID3 A1
2005 Nov 09
3
dataframe without repetition
Hello,
with a data.frame like this :
> toto <-
data.frame(id=c("id1","id1","id2","id3","id3","id3"),dpt=c("13","13","34","30","30","30"))
> toto
id dpt
1 id1 13
2 id1 13
3 id2 34
4 id3 30
5 id3 30
6 id3 30
what is the most efficient ways to obtain :
id
2011 Apr 25
2
Problem with ddply in the plyr-package: surprising output of a date-column
Hi Together,
I have a problem with the plyr package - more precisely with the ddply
function - and would be very grateful for any help. I hope the example
here is precise enough for someone to identify the problem. Basically,
in this step I want to identify observations that are identical in
terms of certain identifiers (ID1, ID2, ID3) and just want to save
those observations (in this step,
2016 Apr 04
2
multiple bar plot annotation text labelling
Readers,
The attempt is to create a bar plot with text labels adjacent to each
datum value.
Data file:
1,3,A
1,8,B
1,1,C
1,9,D
2,5,C
2,4,E
2,2,F
2,0,G
testbarplot<-read.csv('data1.csv', header=FALSE)
barplot(axes=FALSE, ann=FALSE, horiz=TRUE, testbarplot[,2], ylab=
'group', xlab= '(x values)', space=c(1,0,0,0, 1,0,0,0))
text(testbarplot[,2], testbarplot[,1],
2006 Feb 09
1
List Conversion
Hello,
I have a list (mode and class are list) in R that is many elements long and of the form:
>length(list)
[1] 5778
>list[1:4]
$ID1
[1] "num1"
$ID2
[1] "num2" "num3"
$ID3
[1] "num4"
$ID4
[1] NA
I'd like to convert the $ID2 value to be in one element rather than in two.?? It shows up as c(\"num2\", \"num3\") if I try to use
2010 Sep 07
1
average columns of data frame corresponding to replicates
Hi Group,
I have a data frame below. Within this data frame there are samples
(columns) that are measured more than once. Samples are indicated by
"idx". So "id1" is present in columns 1, 3, and 5. Not every id is
repeated. I would like to create a new data frame so that the repeated
ids are averaged. For example, in the new data frame, columns 1, 3,
and 5 of the original
2012 Apr 25
0
[LLVMdev] LLVM Backend for Z80. ADD -> replaced -> OR
Hi Peter,
I think the problem is that you did not explicitly define stack alignment
in Z80TargetMachine.cpp
DataLayout("e-p:16:8:8-i8:8:8-i16:8:8-i32:8:8-n8")
Try to add S16 to the string if your stack is 2-byte aligned. Refer to
http://llvm.org/docs/LangRef.html#datalayout .
If it does not work, try to specify the layout in the input module using
target layout directive.
David
On
2006 Jan 25
0
lazy evaluation (was RE: Number of replications of a term)
From: Thomas Lumley
>
> On Wed, 25 Jan 2006, Ray Brownrigg wrote:
>
> > There's an even faster one, which nobody seems to have
> mentioned yet:
> >
> > rep(l <- rle(ids)$lengths, l)
>
> I considered this but it wasn't clear to me from the initial
> post that
> each ID occupied a contiguous section of the vector.
>
> Also, lazy
2012 Apr 25
2
[LLVMdev] LLVM Backend for Z80. ADD -> replaced -> OR
Hello.
I am playing with LLVM and trying to make Z80 (Zilog Z80) backend.
The source code is attached.
I have succesfully made some simple test. But now I have problem with ADD
instruction.
The source C code is:
typedef struct
{
unsigned char id1;
unsigned char id2;
unsigned char id3;
} testS;
void simple()
{
testS test;
test.id1 = 0x40;
test.id2 = 0x80;
test.id3 = 0xc0;
}
It
2012 Apr 25
1
[LLVMdev] LLVM Backend for Z80. ADD -> replaced -> OR
Hello.
I have played with DataLayout and found a solution with is uknown to me.
I added S16 and also s0:16:16, but it had not worked.
Then I found that in Z80FrameLowering.h I am calling TargetFrameLowering
with stack aligment set to 8. So I changed it to 2 bytes. But this also
didn't help.
Then I changed llc to show TargetDataLayout and found that a option is set
to a0:0:64.
So I changed
2014 Jan 16
1
Doubt in simple merge
Dear R community
I have a two data set called "Elder" and "Younger".
This is my code for simple merge.
Elder <- data.frame(
ID=c("ID1","ID2","ID3"),
age=c(38,35,31))
Younger <- data.frame(
ID=c("ID4","ID5","ID3"),
age=c(29,21,31))
mer <- merge(Elder,Younger,by="ID", all=T)
Output I am