er to deal with any text file newline decoding
> choice/task after the file transfer is completed.
>
> On December 6, 2018 7:03:48 AM PST, Duncan Murdoch <
> murdoch.dun...@gmail.com> wrote:
> >On 06/12/2018 7:45 AM, Kate Stone wrote:
> >> Hello r-help,
> >&g
is is an R problem or something to do with my
OS specifically, or even with the file itself. Any ideas?? I've
attached a small script demonstrating the issue.
Many thanks,
Kate
--
Kate Stone
PhD candidate
Vasishth Lab | Department of Linguistics
Potsdam University, 14467 Potsdam, Germany
https:/
When reading in a tab delimited file using args I keep getting the error:
Error: unexpected symbol in Name index
Execution halted
The code is this:
a - read.table(args[1],sep=\t,header=T, stringsAsFactors=F)
When inputting the file directly, as follows, this produces no errors:
a -
.
-- Clifford Stoll
On Fri, Jun 26, 2015 at 10:58 AM, Kate Ignatius kate.ignat...@gmail.com
wrote:
reading in a tab delimited file using args
What I mean by that is that I'm using a bash script to call in an R
script and using the command: args - commandArgs(TRUE) in my R
script
]] - match(data[[j]], uniqStrings, nomatch = 0L)
}
data
}
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Fri, May 29, 2015 at 9:58 AM, Kate Ignatius kate.ignat...@gmail.com
wrote:
I have a pedigree file as so:
X0001 BYX859 0 0 2 1 BYX859
X0001 BYX894 0 0
I have a pedigree file as so:
X0001 BYX859 0 0 2 1 BYX859
X0001 BYX894 0 0 1 1 BYX894
X0001 BYX862 BYX894 BYX859 2 2 BYX862
X0001 BYX863 BYX894 BYX859 2 2 BYX863
X0001 BYX864 BYX894 BYX859 2 2 BYX864
X0001 BYX865 BYX894 BYX859 2 2 BYX865
And I was hoping to change
---
Sent from my phone. Please excuse my brevity.
On May 9, 2015 7:59:31 AM PDT, Kate Ignatius kate.ignat...@gmail.com wrote:
I have some data that I've trouble importing...
A B C D E
A 1232 0.565
B 2323 0.5656 0.5656 0.5656
C 2323 0.5656
D 2323 0.5656
E 2323 0.5656
F 2323 0.5656
G 2323
I have some data that I've trouble importing...
A B C D E
A 1232 0.565
B 2323 0.5656 0.5656 0.5656
C 2323 0.5656
D 2323 0.5656
E 2323 0.5656
F 2323 0.5656
G 2323 0.5656
G 2323 0.5656 0.5656 0.5656
When I input the data it seems to go like this:
SampleID ItemB ItemC ItemD ItemE
A 1232 0.565
B
MacQueen
Lawrence Livermore National Laboratory
7000 East Ave., L-627
Livermore, CA 94550
925-423-1062
On 5/9/15, 7:59 AM, Kate Ignatius kate.ignat...@gmail.com wrote:
I have some data that I've trouble importing...
A B C D E
A 1232 0.565
B 2323 0.5656 0.5656 0.5656
C 2323 0.5656
D 2323
), value=FIXED)]
aps - af[,grep(as.character(list(ap),colnames(af))]
and also aps - unique (grep(ap, colnames(af))
Is there another way I can do this - maybe without using grep?
Thanks!
Kate.
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Hi,
Supposed I had a data frame like so:
A B C D
0 1 0 7
0 2 0 7
0 3 0 7
0 4 0 7
0 1 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 1 5
0 5 1 5
0 4 1 5
0 8 4 7
0 0 3 0
0 0 3 4
0 0 3 4
0 0 0 5
0 2 0 6
0 0 4 0
0 0 4 0
0 0 4 0
For each row, I want to count how many max column values appear to
adventurely get the
Hi,
I've got a complicated grep problem (or not)... I currently have a
file with the headings as follows:
DAY
MONTH
YEAR
SA_TUES
SA_MON
SU_WED
CH_TUES
CH_WED
CH_MON
AR_TUES
AR_WED
AR_MON
SA_THUR
SU_FRI
CH_THUR
CH_FRI
AR_THUR
AR_FRI
I want to grep out all columns that have SA at the beginning
- lapply(days, function(x) grep(x,columns))
selected - sort(unique(unlist(all_ind)))
columns[selected]
[1] SA_TUES SA_MON CH_TUES CH_MON AR_TUES AR_MON
SA_THUR CH_THUR AR_THUR
On Wed, Feb 18, 2015 at 2:55 PM, David Winsemius dwinsem...@comcast.net
wrote:
On Feb 18, 2015, at 12:27 PM, Kate
I have genetic data as follows (simple example, actual data is much larger):
comb =
ID1 A A T G C T G C G T C G T A
ID2 G C T G C C T G C T G T T T
And I wish to get an output like this:
ID1 AA TG CT GC GT CG TA
ID2 GC TG CC TG CT GT TT
That is, paste every two columns together.
I have
550
15: RA AARR C 0 554
On Fri, Jan 2, 2015 at 12:29 PM, David Winsemius [via R]
ml-node+s789695n4701316...@n4.nabble.com wrote:
On Jan 2, 2015, at 12:07 AM, Kate Ignatius wrote:
Ah, crap. Yep you're right. This is not going too well
...@comcast.net wrote:
On Jan 1, 2015, at 5:07 PM, Kate Ignatius kate.ignat...@gmail.com wrote:
Apologies - mix up of syntax all over the place, a habit of mine. The
last line was in there because of code beforehand so it really doesn't
need to be there. Here is the proper code I hope
4 55
7: AA AAAA B 4 55
8: AA AARA C 3 30
9: AA AARA C 3 30
10: AA RRRA C 3 30
On Tue, 30 Dec 2014, Kate Ignatius wrote:
I'm trying to use both these packages
#. with
/Software/Embedded Controllers) .OO#. .OO#. rocks...1k
---
Sent from my phone. Please excuse my brevity.
On January 1, 2015 4:16:52 AM PST, Kate Ignatius kate.ignat...@gmail.com
wrote
7: AA AAAA B 4 55
8: AA AARA C 3 30
9: AA AARA C 3 30
10: AA RRRA C 3 30
On Tue, 30 Dec 2014, Kate Ignatius wrote:
I'm trying to use both these packages and wondering whether
correct code:
childseg-0
x:=sumchild -0
span-rle(x)$lengths[rle(x)$values==TRUE
childseg[x]-rep(seq_along(span), times = span)
childseg[childseg == 0]-''
On Thu, Jan 1, 2015 at 1:56 AM, Kate Ignatius kate.ignat...@gmail.com wrote:
Is it possible to add the following code or similar
I'm trying to use both these packages and wondering whether they are possible...
To make this simple, my ultimate goal is determine long stretches of
1s, but I want to do this within groups (hence using the data.table as
I use the set key option. However, I'm I'm not having much luck
making this
screen
On Mon, Dec 8, 2014 at 9:31 PM, Kate Ignatius kate.ignat...@gmail.com wrote:
Hi,
I have a simple question. I know there are plenty of packages out
there that can provide code to generate a table in latex. But I was
wondering whether there was one out there where I can generate a table
---
Sent from my phone. Please excuse my brevity.
On December 9, 2014 8:43:02 AM PST, Kate Ignatius kate.ignat...@gmail.com
wrote:
Thanks! I do get several errors though when running on Linux.
Running your code, I get this:
Error in system(cmd, intern = TRUE
function doesn't do pdflatex (by default it does regular
latex) unless you set the options
as I indicated.
On Tue, Dec 9, 2014 at 3:11 PM, Kate Ignatius kate.ignat...@gmail.com wrote:
Ah yes, you're right.
The log has this error:
! LaTeX Error: Missing \begin{document}.
Though can't really
use the
system standard.
I don't know what the linux equivalent is, either the exact program or
the instruction to use the standard.
xdg-open (but like OS X it depends on having the right associations set).
On Tue, Dec 9, 2014 at 3:36 PM, Kate Ignatius kate.ignat...@gmail.com
wrote:
I set
Hi,
I have a simple question. I know there are plenty of packages out
there that can provide code to generate a table in latex. But I was
wondering whether there was one out there where I can generate a table
from my data (which ever way I please) then allow me to save it as a
pdf?
Thanks
K.
I have genetic information for several thousand individuals:
A/T
T/G
C/G etc
For some individuals there are some genotypes that are like this: A/,
C/, T/, G/ or even just / which represents missing and I want to
change these to the following:
A/ A/.
C/ C/.
G/ G/.
T/ T/.
/ ./.
/A ./A
/C ./C
/G
I'm having an issue with grep:
I have numerous columns that end with .at... when I use grep like so:
df[,grep(.at,colnames(df))]
it works fine. When I have one column that ends with .at, it does not
work. Why is that? As this is loop with varying number of columns
ending in .at I would like
it will have a problem grepping out this single column.
On Tue, Oct 14, 2014 at 10:38 AM, John McKown
john.archie.mck...@gmail.com wrote:
On Tue, Oct 14, 2014 at 9:23 AM, Kate Ignatius kate.ignat...@gmail.com
wrote:
I'm having an issue with grep:
I have numerous columns that end
---
Sent from my phone. Please excuse my brevity.
On October 14, 2014 7:23:55 AM PDT, Kate Ignatius kate.ignat...@gmail.com
wrote:
I'm having an issue with grep:
I have numerous columns that end with .at... when I use grep like so:
df[,grep(.at,colnames(df))]
it works fine
Hi all,
I need help with a function. I'm trying to write a function to apply
to varying number of columns in a lot of files - hence the function...
but I'm getting stuck. Here it is:
gt- function(x) {
alleles - sapply(x, function(.) strsplit(as.character(.), /))
gt - apply(x,
'closure' is not subsettable.
The vcf is my original file that I want to match with so not sure
whether this a problem.
On Mon, Oct 13, 2014 at 4:46 PM, Kate Ignatius kate.ignat...@gmail.com wrote:
Hi all,
I need help with a function. I'm trying to write a function to apply
to varying number
Is there an easy way to check whether a variable is within +/- 10%
range of another variable in R?
Say, if I have a variable 'A', whether its in +/- 10% range of
variable 'B' and if so, create another variable 'C' to say whether it
is or not?
Is there a function that is able to do that?
, 2014 at 6:54 PM, Peter Alspach
peter.alsp...@plantandfood.co.nz wrote:
Tena koe Kate
If kateDF is a data.frame with your data, then
apply(kateDF, 1, function(x) isTRUE(all.equal(x[2], x[1], check.attributes =
FALSE, tolerance=0.1)))
comes close to (what I think) you want (but not to what you
computation.
Regards
Petr
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of Kate Ignatius
Sent: Sunday, September 28, 2014 9:14 PM
To: r-help
Subject: [R] if else statement in loop
I have two data frames
For simplicity
its all about factors and data frames and
characters...
K.
On Sun, Sep 28, 2014 at 1:15 AM, Jim Lemon j...@bitwrit.com.au wrote:
On Sun, 28 Sep 2014 12:49:41 AM Kate Ignatius wrote:
Quick question:
I am running the following code on some variables that are factors:
dbpmn$IID1new - ifelse
not wisdom.
Clifford Stoll
On Sun, Sep 28, 2014 at 6:38 AM, Kate Ignatius kate.ignat...@gmail.com
wrote:
Strange that,
I did put everything with as.character but all I got was the same...
class of dbpmn[,2]) = factor
class of dbpmn[,21] = factor
class of dbpmn[,20] = data.frame
This has
I have two data frames
For simplicity:
X=
V1 V2 V3 V4 V5 V6
samas4 samas5 samas6 samas4_father samas5_mother samas6_sibling
samas4 samas5 samas6 samas4_father samas5_mother samas6_sibling
samas4 samas5 samas6 samas4_father samas5_mother samas6_sibling
Y=
FID IID
FAM01 samas4
FAM01 samas5
Quick question:
I am running the following code on some variables that are factors:
dbpmn$IID1new - ifelse(as.character(dbpmn[,2]) ==
as.character(dbpmn[,(21)]), dbpmn[,20], '')
Instead of returning some value it gives me this:
c(1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1))
Playing around
Hi,
I hope I can explain my problem clearly
I have a plink output file that I want to graph a heat map of the
PI_HAT estimates. I have the following code that I has worked in the
past but this time I'm getting the error:
In `levels-`(`*tmp*`, value = if (nl == nL) as.character(labels) else
Hi,
I have a data.table question (as well as if else statement query).
I have a large list of families (file has 935 individuals that are
sorted by famiy of varying sizes). At the moment the file has the
columns:
SampleID FamilyID Relationship
To prevent from having to make a pedigree file by
Thanks!
I think I know what is being done here but not sure how to fix the
following error:
Error in l$PID[l$\Relationship == sibling] - l$Sample.ID[father] :
replacement has length zero
On Sat, Aug 16, 2014 at 6:48 PM, Jorge I Velez jorgeivanve...@gmail.com wrote:
Dear Kate,
Assuming
for this?
On Sat, Aug 16, 2014 at 8:02 PM, Kate Ignatius kate.ignat...@gmail.com wrote:
Thanks!
I think I know what is being done here but not sure how to fix the
following error:
Error in l$PID[l$\Relationship == sibling] - l$Sample.ID[father] :
replacement has length zero
On Sat, Aug 16, 2014
5sibling 879 880
3064 86 sibling 879 880
3064 87 sibling 879 880
On Sat, Aug 16, 2014 at 9:31 PM, Jorge I Velez jorgeivanve...@gmail.com wrote:
Dear Kate,
Try this:
res - do.call(rbind, lapply(xs, function(l){
l$PID - l$MID - 0
father - with(l, Relationship == 'father')
mother
sibling 4 3
#3064.7 306486 sibling 4 3
#3064.8 306487 sibling 4 3
HTH,
Jorge.-
On Sun, Aug 17, 2014 at 11:47 AM, Kate Ignatius kate.ignat...@gmail.com
wrote:
Yep - you're right - missing parents are indicated as zero in the M/PID
field
I have 4 columns, and about 300K plus rows with 0s and 1s.
I'm trying to count how many rows satisfy a certain criteria... for
instance, how many rows are there that have the first column == 1 as
well as the second column == 1.
I've tried using rowSums and colSums but it keeps giving me this
Thanks!
On Sat, Jun 21, 2014 at 11:05 AM, Jorge I Velez
jorgeivanve...@gmail.com wrote:
Hi Kate,
You could try
sum(X[, 1] == 1 X[, 2] == 1)
where X is your data set.
HTH,
Jorge.-
On Sun, Jun 22, 2014 at 12:57 AM, Kate Ignatius kate.ignat...@gmail.com
wrote:
I have 4 columns
I'm trying to have a layout of two graphs on a page... this has worked
before... but I changed up the way I do my venn diagrams so now
instead of the Venn Diagram being at the bottom of the page below the
bar/line graph it takes up the whole page and its overlays the
bar/line graph placed on the
Hi All,
I'm trying to merge two files together using:
combinedfiles - merge(comb1,comb2,by=c(Place,Stall,Menu))
comb1 is about 2 million + rows (158MB) and comb2 is about 600K+ rows (52MB).
When I try to merge using the above syntax I get the error:
Error in merge.data.frame(comb1, comb2, by
I have a list of files that I have called like so:
main_dir - '/path/to/files/'
directories - list.files(main_dir, pattern = '[[:alnum:]]', full.names=T)
filenames - list.files(file.path(directories,/tmpdir/), pattern =
'[[:alnum:][:punct:]]_eat.txt+$', recursive = TRUE, full.names=T)
This
I'm trying to plot a GWAS (in you will) with lined segments
representing an overall p-value for each gene. Here is my code:
skatg - ggplot(comm, aes(x = position,y = p, colour = grey)) +
geom_point(size = 0.75) +
geom_segment(data=rare, aes(x = txStart,
Hi All,
I've successfully gotten out the colMeans for 60 columns using:
col - colMeans(x, na.rm = TRUE, dims = 1)
My next question is: is there a way of getting a mean of all the
column means (ie a mean of a mean)?
Thanks!
__
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reproducible example so who knows.
R set.seed(1)
R x - data.frame(matrix(runif(150), ncol=10))
R # col is a function, so not a good name
R col - colMeans(x)
R mean(col)
[1] 0.5119
It's polite to include the list on your reply.
Sarah
On Wed, May 21, 2014 at 2:50 PM, Kate Ignatius
I've used geom_point and geom_hline in ggplot2 and have gotten
satisfactory legends for both. However, I have one black line and one
blue line in the figure but in the legend they are both black - how
can I correct this in the legend to be the right colors?
mcgc - ggplot(sam, aes(x = m,y =
My code that I've used is:
mcgc - ggplot(sam, aes(x = person,y = m, colour = X)) +
geom_point(size = 0.75) +
scale_colour_gradient2(high=red, mid=green,
limits=c(0,1), guide = colourbar) +
geom_hline(aes(yintercept = mad, linetype = mad),
I'm not doing a Manhattan plot, but plotting AD (coloured by DP) along
the genome:
points - ggplot(sam,aes(x = midpoint,y = ad, colour = dp, size = 3)) +
geom_point() +
scale_y_continuous(breaks=c(0,20,30,40)) +
labs(x = chr,y = ad) +
scale_colour_gradient2(high=red, mid=green)
However,
I'm trying to work out the average of a certain value by chromosome.
I've done the following, but it doesn't seem to work:
Say, I want to find average AD for chromosome 1 only and paste the
value next to all the positions on chromosome 1:
sam$mmad[sam$chrom == '1'] -
'
*My question: is it possible to remove the all columns from above file
to *achieve* the desired result?*
* *
* **Thank you for help*
Kate Dresh
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, tangentresiduals = FALSE) : object 'out' not found
Any help is appreciated.
Thanks,
Kate
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to use merge command to deal with this?
Thanks,
Kate
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Thanks for all of your help. It works to me.
Kate
On Mon, Nov 15, 2010 at 10:06 AM, David Winsemius dwinsem...@comcast.netwrote:
On Nov 15, 2010, at 10:42 AM, Kate Hsu wrote:
Hi r users,
I have two data sets (X1, X2). For example,
time1-c( 0, 8, 15, 22, 43, 64, 85, 106, 127, 148
I want to replace NA by 0, when I tried the following command, I get som
error message.
data[is.na(data)]-0
Warning message:
In `[-.factor`(`*tmp*`, thisvar, value = 0) :
invalid factor level, NAs generated
Anyone knows how to deal with this?
Thanks,
Kate
[[alternative HTML
)(0.8)(0.1)
YYNY (0.6)(0.5)(0.2)(0.9)
..
..
..
Any efficient way to do this?
Thanks,
Kate
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the raw numbers for successes/failures since the ORs are all
adjusted for various factors.]
Thanks!
Kate Snedeker
--
Kate Snedeker, PhD
Postdoctoral Fellow
Centre for Public Health and Zoonoses Department of Population Medicine
MacNabb House
Ontario Veterinary College
University of Guelph
movement (meters),xlab=x
coordinate movement (meters))
abline(h=0, v=0, col = grey, lty=2)
#legend??
Any help would be appreciated!
Kate
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PLEASE do read the posting guide
)
{
log(prod(dcauchy(x,mu,s)))
}
maxNR(loglik,start=median(x))$estimate
Does anyone know how to solve this problem?
Thanks,
Kate
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would be greatly appreciated!
Kate
. You need to check the @hole
slot of the @polygons[[1]] object if you care whether it's a hole
or an island!
Barry
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PLEASE do read
efforts (and
much reading!), I'm at a lost.
Many thanks,
Kate
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and provide commented, minimal
individually for each tree ensemble? Is it possible to calculate these
predictors for the new random forest object after calling the combine
function? Any help would be greatly apprecaited. Thanks, Kate
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separate new datasets if I have to, but
there HAS to be a more elegant way.
Thank you for ANY feedback!
Kate
Kate Rohrbaugh
Independent Project Analysis, Inc.
44426 Atwater Drive
Ashburn, VA 20147
office - 703.726.5465
fax - 703.729.8301
email - [EMAIL PROTECTED]
website
separate new datasets if I have to, but
there HAS to be a more elegant way.
Thank you for ANY feedback!
Kate
Kate Rohrbaugh
Independent Project Analysis, Inc.
44426 Atwater Drive
Ashburn, VA 20147
office - 703.726.5465
fax - 703.729.8301
email - [EMAIL PROTECTED]
website - www.ipaglobal.com
they generated regressions of data
I don't care about (i.e., when model1==0).
Regards -- Kate
Kate Rohrbaugh
Independent Project Analysis, Inc.
44426 Atwater Drive
Ashburn, VA 20147
office - 703.726.5465
fax - 703.729.8301
email - [EMAIL PROTECTED]
website - www.ipaglobal.com ( http
In lm command, we can use vcov option to get variance-covariance matrix.
Does anyone know how to get variance-covariance matrix in nlrq?
Thanks,
Kate
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Has anybody used fuzzy time series to forecast enrollments? I have some code
that does not work so well. Thanks.
Kate
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PLEASE
',type='l')
The shapes are what I want, but I want x-axis consistent to the correspond to
n. Does anyone know the solution?
Thanks,
Kate
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!!
Kate
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and provide commented, minimal, self
class mle object: invalid object for slot fullcoef in class
mle: got class list, should be or extend class numeric
When I typed warnings(), I get
In dt(x, ncp = ncp, df = df, log = TRUE) :
full precision was not achieved in 'pnt'
Does anyone know how to solve it?
Thanks,
Kate
- Original
? Thanks.
Kate
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and provide commented, minimal, self
version.
On Tue, 6 May 2008, Kate wrote:
Hi there,
I've tried to install the A2R package using the files from
http://addictedtor.free.fr/packages/A2R/lastVersion/
This is the error I get when trying to load the library:
library(A2R)
Error in library(A2R) :
'A2R
need to use the observed y
and x run regression first and then assign the value to the missing y later.
Thanks,
Kate
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PLEASE do
: TSE[i,j] corresponds to V[i] and sigmaV[j].
Thanks,
Kate
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=0.7. I do not want to have the line at
0.7, and would like a solid dot at payoff=100 and a hollow dot at payoff=40 at
the jump point 0.7. How would I do this?
Thanks,
Kate
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Hi,
My example is as the following,
W-matrix(c(2,4,2,8,1,3),ncol=3)
which(W==min(W))
I would like to find the index for min of the matrix W, i.e. (1, 3) instead of
5 as the output. What command could I use?
Thanks,
Kate
[[alternative HTML version deleted
could I do
it in R?
Thanks,
Kate
[[alternative HTML version deleted]]
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