-
AOR Dr. Gerrit Eichner Mathematical Institute, Room 212
gerrit.eich...@math.uni-giessen.de Justus-Liebig-University Giessen
Tel: +49-(0)641-99-32104 Arndtstr. 2, 35392 Giessen, Germany
Fax: +49-(0)641-99-32109http://www.uni-giessen.de/cms/eichner
than 2, 3, 4, and 5?
Regards -- Gerrit
-
AOR Dr. Gerrit Eichner Mathematical Institute, Room 212
gerrit.eich...@math.uni-giessen.de Justus-Liebig-University Giessen
Tel: +49-(0)641-99-32104 Arndtstr. 2
-- Gerrit
-
AOR Dr. Gerrit Eichner Mathematical Institute, Room 212
gerrit.eich...@math.uni-giessen.de Justus-Liebig-University Giessen
Tel: +49-(0)641-99-32104 Arndtstr. 2, 35392 Giessen, Germany
Fax: +49
Hello, Emily,
take a look at read.table() for importing (with or without header
depending on your file which holds the data). Maybe
X - read.table( yourfilename, header = FALSE, row.names = 1)
and then
pvalues - apply( X, 1,
function( x)
fisher.test(
in row i, column j, and layer k (if one
wants to think of A as a cuboidal grid).
Hth -- Gerrit
-
AOR Dr. Gerrit Eichner Mathematical Institute, Room 212
gerrit.eich...@math.uni-giessen.de Justus-Liebig
, minimal, self-contained, reproducible code.
-
AOR Dr. Gerrit Eichner Mathematical Institute, Room 212
gerrit.eich...@math.uni-giessen.de Justus-Liebig-University Giessen
Tel: +49-(0)641-99-32104 Arndtstr. 2
Hi, Doug,
maybe
columns - c( 0, 3, 0, 2, 0, 1)
lapply( columns[ columns 0],
function( o) array( -1, dim = c( 2, o)))
does what you want?
Regards -- Gerrit
-
AOR Dr. Gerrit Eichner Mathematical
utils datasets methods
[7] base
other attached packages:
Matrix lattice
0.999375-10.15-11
-
AR Dr. Gerrit Eichner Mathematical Institute
[EMAIL PROTECTED] Justus-Liebig-University
so far been unable to find
something that works.
Maybe a look at
?prop.table
and/or
?addmargins
could help you.
Regards -- Gerrit
-
AR Dr. Gerrit Eichner Mathematical Institute
[EMAIL PROTECTED
] rcompgen_0.1-17
-
AR Dr. Gerrit Eichner Mathematical Institute
[EMAIL PROTECTED] Justus-Liebig-University Giessen
Tel: +49-(0)641-99-32104 Arndtstr. 2, 35392 Giessen, Germany
Fax: +49-(0)641-99-32029
/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
See
?object.size
Regards -- Gerrit
-
AR Dr. Gerrit Eichner
Hi, Steven,
try
lapply( x, function( v) rownames(x)[ v == 1])
or
lapply( x, function( v, rn) rn[ v == 1], rn = rownames( x)))
which is faster.
Regards -- Gerrit
-
AOR Dr. Gerrit Eichner Mathematical
Maybe
expand.grid( rep( list( 0:1), 10))
does what you want.
Best regards -- Gerrit
-
AOR Dr. Gerrit Eichner Mathematical Institute, Room 305 E
gerrit.eich...@math.uni-giessen.de Justus-Liebig-University
guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
-
Dr. Gerrit Eichner Mathematical Institute, Room 212
gerrit.eich...@math.uni-giessen.de
-contained, reproducible code.
-
Dr. Gerrit Eichner Mathematical Institute, Room 212
gerrit.eich...@math.uni-giessen.de Justus-Liebig-University Giessen
Tel: +49-(0)641-99-32104 Arndtstr. 2, 35392 Giessen
What is the function to have the entire part of a number?
If entire refers to integer take a look at
?trunc
Hth -- Gerrit
-
Dr. Gerrit Eichner Mathematical Institute, Room 212
gerrit.eich...@math.uni
Hi, Georgina,
using quotation marks around the file name like in
read.delim( ken_data_try_anova)
might help.
Hth -- Gerrit
On Mon, 25 Apr 2011, Georgina Salazar wrote:
Hi!
I have the data in a tab delimited text file titled ken_data_try_anova. I
tried to import it into R entering
as data frame)
Caveat: untested!
Hth -- Gerrit
-
Dr. Gerrit Eichner Mathematical Institute, Room 212
gerrit.eich...@math.uni-giessen.de Justus-Liebig-University Giessen
Tel: +49-(0)641-99-32104
) ) )
)
Still another possibility is to use xtabs() (with its summary-method)
which has a formula argument.
Hoping that you know what to do with the results -- Gerrit
-
Dr. Gerrit Eichner
Hi, Thomas,
if I'm not completely mistaken
Dat2 - match( t( Dat), ltable)
should do what you want.
Hth -- Gerrit
On Fri, 7 Oct 2011, thomas.chesney wrote:
Thank you Michael and Patrick for your responses. Michael - your code ran in
under 5 minutes, which I find stunning, and Patrick I
Marion,
try
rbind( Mymatrix, do.call( rbind, Z))
Hth -- Gerrit
On Tue, 11 Oct 2011, Marion Wenty wrote:
dear r-users,
i have got a problem which i am trying to solve:
i have got the following commands:
Mymatrix - matrix(1:9,ncol=3)
Z -
list(V1=c(a,,),V2=c(b,,),V3=c(c,,),V4=c(d,,))
Hello, uday,
e.g.,
(data - 1) %/% 3 + 1
would do the job in your very specific situation, but take a look at
?findInterval
for possibly more interesting (because more general) solutions.
Hth -- Gerrit
On Tue, 13 Mar 2012, uday wrote:
I want to replace some values in my Array
e.g
data-
()) to compare model 1 with model 2 using the MSE
of model 3 in a sequence of three nested (linear) models? (A short
RSiteSearch() and a google search didn't lead me far ...)
Thx in advance!
Best regards -- Gerrit
-
Dr. Gerrit Eichner
Hi, HJ,
see
?plotmath
Hth -- Gerrit
-
Dr. Gerrit Eichner Mathematical Institute, Room 212
gerrit.eich...@math.uni-giessen.de Justus-Liebig-University Giessen
Tel: +49-(0)641-99-32104 Arndtstr
Hello, uday,
there's presumably a typo in your code because you use path1t in
data2 - read.table(paste(path1t, file_wasaux2[i],header=TRUE))
and not path which you defined above.
Hth -- Gerrit
-
Dr. Gerrit Eichner
Hello, Alexander,
does
utest - unlist(test)
utest[ names( utest) == a]
come close to what you need?
Hth,
Gerrit
On Tue, 7 Dec 2010, Alexander Senger wrote:
Hello,
my data is contained in nested lists (which seems not necessarily to be
the best approach). What I need is a fast way to
- as.factor(as.numeric(cut(r,c(0,2,4,6,8,10,100
levels(col_no) - c(2%,2-4%,4-6%,6-8%,8-10%,10%)
col_no
[1] 2% 2% 2-4% 2% 2% 2%
Levels: 2% 2-4% 4-6% 6-8% 8-10% 10%Thanks,
Tianchan
Best regards -- Gerrit
Best regards -- Gerrit Eichner
Viele Grüße -- Gerrit
Viele Grüße -- Gerrit Eichner
On Tue, 14 Dec 2010, Matthew Rosett wrote:
How do I determine if my data deviate from the normal distribution?
The sample size is 1000 (weights of people).
See
?qqnorm
and/or
?shapiro.test
and/or
a text book on applied statistics
and/or
google for testing normality.
Hth,
Gerrit
PS:
Hello, Vassilis,
maybe
with( test.df, ave( vector, factor, FUN = function( x) x - mean( x)))
does what you want.
-- Gerrit
On Wed, 15 Dec 2010, Vassilis wrote:
I would like to de-mean the 'vector' column of the following dataframe by
factor:
set.seed(5444)
vector - rnorm(1:10)
Hello, Amy,
take a look at
?reshape
In your case, I think,
reshape( yourdatafame, varying = c( value1, value2, value3),
v.name = amount, times = c( value1, value2, value3),
timevar = name, direction = long)
should work.
Hth -- Gerrit
On Fri, 17 Dec 2010, Amy Milano
Hi, Luca,
if V is you data frame, maybe
with( V, tapply( v3, list( v1, v2), sum))
does what you want.
Hth -- Gerrit
On Mon, 20 Dec 2010, Luca Meyer wrote:
Hi,
This must be an easy one but so far I haven't find a way out...
I have a data frame such as:
$ v1: Factor w/ 5 levels
$
Hello, Robert,
see hints below.
On Tue, 21 Dec 2010, Robert Ruser wrote:
Hello,
I'm wondering how to set a value of mar ( par( mar=c(...)) ) in
order to allow labels to be visible in barplot. Is there any relation
between the number of characters in a label and the second value of
mar?
On Wed, 22 Dec 2010, Marius Hofert wrote:
Dear expeRts,
I somehow don't see why the following does not work:
integrand - function(x, vec, mat, val) 1 # dummy return value
A - matrix(runif(16), ncol = 4)
u - c(0.4, 0.1, 0.2, 0.3)
integrand(0.3, u, A, 4)
integrate(integrand, lower = 0, upper =
Try
plot (1, 1, ylab = expression (Z[list(i,i)]))
Hth -- Gerrit
Original Message
Subject: puzzled with plotmath
Date: Thu, 20 Jan 2011 12:48:18 +0100
From: Claudia Beleites cbelei...@units.it
To: R Help r-help@r-project.org
Dear all,
I'm puzzled with matrix indices in
Claudia,
Mittelhessen! ;-)
thanks viele Grüße nach Oberhessen :-)
plot (1, 1, ylab = expression (Z[list(i,i)]))
though that cannot be evaluated, either (due to [ not knowing what to do
with an index list)
Something is missing; this last sentence of yours appears to be not
complete,
No bug. See ?options or ?print.default and note *significant* digits.
Hth -- Gerrit
On Fri, 18 Nov 2011, Ravi Kulkarni wrote:
This is in version 2.10.1
sqrt(3)
[1] 1.732051
6/7
[1] 0.8571429
options(digits=3)
sqrt(3)
[1] 1.73
6/7
[1] 0.857
sqrt(7)
[1] 2.65
7/9
[1] 0.778
Ravi
.
-
Dr. Gerrit Eichner Mathematical Institute, Room 212
gerrit.eich...@math.uni-giessen.de Justus-Liebig-University Giessen
Tel: +49-(0)641-99-32104 Arndtstr. 2, 35392 Giessen, Germany
Fax: +49-(0)641-99-32109http
Hello, iliketurtles (?),
for whatever strange reasons you want to regress all y-columns on all
x-columns, maybe
reg - apply( x, 2, function( xx) lm( y ~ xx))
do.call( cbind, lapply( reg, coef))
does what you want. (To understand what the code above does, check the
documentation for lm(): If
Hello, Nerak,
maybe
rbind( NA, head( results, -1))
does what you want (for all columns at once)?
Hth,
Gerrit
On Wed, 18 Jan 2012, Nerak wrote:
Dear all,
I have a question concerning manipulating data of several columns of a
dataframe at the same time. I manage to do it for one column
Hi, Johannes,
maybe
X - unlist( strsplit( as.character( x$ART), split = ;, fixed = TRUE))
X - strsplit( X, split = -, fixed = TRUE)
X - sapply( X, function( x)
if( length(x) == 2)
rep( x[1], as.numeric( x[2])) else x[1]
)
table(X, useNA = always)
Hi, Philip,
counter-questions:
1. Which/where is the grouping variable for the test of differences in
survival?
2. Assume the grouping variable is Gend in B27.vec. Then, why aren't you
using
survdiff( Surv( AgeOn, UV) ~ Gend, rho = 0, data = B27.vec)
?
Hth -- Gerrit
On Thu, 26 Jan
Hello, Dan,
you could probably use a combination of nchar(), substr() (or substring())
and paste(). Take a look at their online help pages.
Hth -- Gerrit
Hello everyone,
I have a character vector of 24 hour time values in the format hm
without the delimiting :. How can I insert the :
-
Dr. Gerrit Eichner Mathematical Institute, Room 212
gerrit.eich...@math.uni-giessen.de Justus-Liebig-University Giessen
Tel: +49-(0)641-99-32104 Arndtstr. 2, 35392 Giessen, Germany
Fax: +49-(0)641-99-32109http://www.uni-giessen.de/cms/eichner
Hello, Dominic,
untested:
data - lapply( data, function( x) x[ x == ] - NA
Hth -- Gerrit
On Tue, 8 Jan 2013, Dominic Roye wrote:
Hello R-Helpers,
I have a slight problem with the expresion data[data==] - NA which works
well for a data.frame. But now i must use the same for a list of
Hi, Antti,
you should look at
?levels
(and particular its Examples section) to find out how to use
levels( X) - c( new1, ..., newk)
to achieve what you want.
Regards -- Gerrit
On Thu, 17 Jan 2013, Antti Simola wrote:
Hi,
This is quite simple data manipulation task and I need
Hi, Tammy,
maybe you find something interesting looking at
?interaction
and/or try (with df being your data frame)
df$Group - as.integer( with( df, interaction( Feature, OS)[, drop =
TRUE]))
HtH -- Gerrit
On Tue, 22 Jan 2013, Tammy Ma wrote:
HI,
I met this problem:
I have the
Hi, Patrick,
I think (with reshape from the stats package)
reshape( aa, idvar = ID, v.names = Eaten, timevar = Target,
direction = wide)
does the trick (followed by renaming the columns of the resulting data
frame).
Hth -- Gerrit
On Tue, 22 Jan 2013, Patrick Connolly wrote:
Hi, Daisy,
try
dat2 - reshape( dat, varying = c( species1, species2, species3),
v.name = presence, timevar = species,
times = c( species1, species2, species3),
direction=long)
dat2[ rev( order( dat2$region)), ]
Hth -- Gerrit
On Thu, 24
Hello, Marco,
I am not quite sure if understand correctly what you want, but maybe
DF - data.frame( Artist, Begin, End, Istitution)
AtSameInst - outer( DF$Istitution, DF$Istitution, ==)
Simultaneously - with( DF, outer( Begin, End, =) |
outer( End, Begin, =))
Hi, Benjamin,
have you tried for your list with NA-components to use is.na() as follows
(where x is assumed to be your list)?
x[ !is.na(x)]
Hth -- Gerrit
On Tue, 29 Jan 2013, Benjamin Ward (ENV) wrote:
Hi, This is probably a small query but one I'm struggling with: I have a
list in
Hello, John,
as a start take a look at
?merge
And to (maybe) get a bit overwhelmed at first sight use
RSiteSearch( merge)
Hth -- Gerrit
On Thu, 7 Feb 2013, John Smith wrote:
I know that the basic approach to append or merge data frames is using the
rbind and merge commands.
However,
Hello, sanshine,
maybe you should check the value of n.keep and make sure that it is
non-negative.
Hth -- Gerrit
On Tue, 12 Jun 2012, sanshine wrote:
Hello everyone,
I`m trying to normalize and analize an illumina SNP array.
But when i`m trying to segmentate i`m getting an error:
Error
in model.frame.default(formula = Sepal.Length ~ Sepal.Width, data = iris,
:
variable lengths differ (found for '(xlab)')
And are you -- the one with the work-around -- willing to share it? :)
Best regards -- Gerrit
-
Dr. Gerrit Eichner
-
Dr. Gerrit Eichner Mathematical Institute, Room 212
gerrit.eich...@math.uni-giessen.de Justus-Liebig-University Giessen
Tel: +49-(0)641-99-32104 Arndtstr. 2, 35392 Giessen, Germany
Fax: +49-(0)641-99-32109http://www.uni-giessen.de/cms/eichner
Hi, Arnold,
looking at the example section of
?stripchart
may help you.
Hth -- Gerrit
On Thu, 12 Jul 2012, darnold wrote:
Hi,
I'm looking for some ideas on how to reproduce the attached image in R.
There are three samples, each of size n = 10. The first is drawn from a
normal
Hello, Ivan,
you'll find argument frame (actually frame.plot) explained in
?plot.default
Regards -- Gerrit
Hi Peter,
I had never heard of this 'frame' argument and it's a breakthrough for me to
be finally able to get rid of this frame!
But where is this argument explained? I couldn't
Hello, Krunal,
try
summ.list[[2]]$coefficients[2]
Note the double square brackets (as summ.list is a list)!
Hth,
Gerrit
On Fri, 28 Sep 2012, Krunal Nanavati wrote:
Hi Rui,
Excellent!! This is what I was looking for. Thanks for the help.
So, now I have stored the result of the 10
Hello, James,
see my comments inline.
... Main issue/question: In R the nparLD ANOVA-type Test showed a
significant p-value for diel period, no effect of season, and no
interaction between diel period and season. But a post-hoc Wilcoxon
Signed-Rank Test did NOT find a significant difference
Hi, Xochitl,
wrapping the call to histogram() inside your loop in a call to print()
should solve your problem:
print( histogram( .))
Regards -- Gerrit
On Wed, 5 Jun 2013, Xochitl CORMON wrote:
Hi all,
I'm encountering a problem I do not understand on my data:
library (lattice)
Dear Eliza,
the more unspecific a question is formulated, the more is the poster in an
urgent need for a statistical consultant nearby and -- at the same time --
the less likely is is to get a useful answer on this list ...
I suggest you to read the posting guide, look at CRAN's Task Views
Hi, Witold,
take a look at
?findIntervals
It might give want you need.
Hth -- Gerrit
On Thu, 27 Jun 2013, Witold E Wolski wrote:
Is there a build in function to create an index for tapply or by given a
a numeric vector x an a vector of breaks?
What I want to do is:
x - 1:100
breaks -
attached packages:
[1] fortunes_1.5-2
loaded via a namespace (and not attached):
[1] grid_3.0.2 lattice_0.20-27 tools_3.0.2
-
Dr. Gerrit Eichner Mathematical Institute, Room 212
gerrit.eich...@math.uni
/2014 7:32 AM, Gerrit Eichner wrote:
Hello, everyone,
I am struggling with an Sweave-problem that didn't occur sofar (and I have
no clue what I might have changed in my system; see below). The following
example *.Rnw file's only task is (for simplicity) to output text with a
little bit of TeX-code
for the respective code chunks (which also have
results=tex, and from which I want only linebreaks in the TeX-output).
Actually, embarrassingly simple.
Thanks again -- Gerrit
On Wed, 5 Mar 2014, Duncan Murdoch wrote:
On 05/03/2014 9:29 AM, Gerrit Eichner wrote:
Thanks, Duncan,
but, sorry
datasets methods base
other attached packages:
[1] fortunes_1.5-2
loaded via a namespace (and not attached):
[1] tools_3.0.2
-
Dr. Gerrit Eichner Mathematical Institute, Room 212
gerrit.eich...@math.uni
Hello, Catalin,
assume your data frame is as simple as
A - data.frame( year = years, month01 = values01, ,
+ month12 = values12)
then, e.g.,
reshape( A, varying = c( month01, , month12),
+ v.names = Values, timevar = Month, direction = long)
should do
2013 14:35, Gerrit Eichner
gerrit.eich...@math.uni-giessen.dewrote:
Hello, Catalin,
assume your data frame is as simple as
A - data.frame( year = years, month01 = values01, ,
+ month12 = values12)
then, e.g.,
reshape( A, varying = c( month01, , month12
-
Dr. Gerrit Eichner Mathematical Institute, Room 212
gerrit.eich...@math.uni-giessen.de Justus-Liebig-University Giessen
Tel: +49-(0)641-99-32104 Arndtstr. 2, 35392 Giessen, Germany
Fax: +49-(0)641-99-32109http://www.uni
package
just for convenience.
Regards -- Gerrit
Thank you in advance
Edouard Hardy
On Tue, Sep 3, 2013 at 11:49 AM, Gerrit Eichner
gerrit.eich...@math.uni-giessen.de wrote:
Hello, Edouard,
taking logs of A's elements (so that * turns into +, so to say), using a
left-multiplication
Hello, Francesco,
these could be considered as two of the central questions in statistics in
general ... but they do not necessarily have anything to do with R.
Regards -- Gerrit
On Fri, 6 Sep 2013, Francesco Miranda wrote:
how can i calculate the probability of occurrence of an event and
Hello, Paulito,
first, I think you haven't received an answer yet because you did not
provide commented, minimal, self-contained, reproducible code as the
posting guide does request it from you.
Second, see inline below.
On Wed, 11 Sep 2013, Paulito Palmes wrote:
Hi,
I have a data.frame
Hi, Jannis,
maybe
plot( 1, 1, main = bquote( paste( .(a), [, .(b), ])))
comes close to what you want, but I think you may even have to use the
following to get a varying exponent really printed elevated:
a - speed
b - m * s
cc - -2
plot( 1, 1, main = bquote( paste( .(a), [,
Hello, Adam,
I'm rather uncertain about your goal (and consequently even more so about
how to reach it), but anyway, maybe the function match.call() with its
argument expand.dots is of some help for you. From its help page:
match.call is most commonly used in two circumstances:
To
stats4_3.0.2tools_3.0.2
-
Dr. Gerrit Eichner Mathematical Institute, Room 212
gerrit.eich...@math.uni-giessen.de Justus-Liebig-University Giessen
Tel: +49-(0)641-99-32104 Arndtstr. 2, 35392 Giessen
Hello, Catalin,
check out
?coef
Regards -- Gerrit
On Tue, 15 Oct 2013, catalin roibu wrote:
Hello all!
I have a problem with R. I want to extract regression coefficients from
summary and use it for compute the theoretical values.
How can I do that in R?
thank you!
best regards,
--
---
to give you any hint. So, PLEASE do read the
posting guide http://www.R-project.org/posting-guide.html and provide
commented, minimal, self-contained, reproducible code.
Regards -- Gerrit
On 15 October 2013 14:59, Gerrit Eichner gerrit.eich...@math.uni-giessen.de
wrote:
Hello, Catalin
just forgot to take that possibility into consideration. :(
Regards -- Gerrit
On Tue, 15 Oct 2013, Duncan Murdoch wrote:
On 15/10/2013 7:53 AM, Gerrit Eichner wrote:
Dear list subscribers,
here is a small artificial example to demonstrate the problem that I
encountered when looking
Hi, Alaios,
check out
?which
and in particular its Examples and See Also section.
Hth -- Gerrit
On Tue, 22 Oct 2013, Alaios wrote:
Hi I have a vector like that
readCsvFile$V1
[1] 30 31 32 33 34 35 36 37 38 39 310 311 312 313 314 315 316 317 318
[20] 319 320 321 20 21 22 23
Hello, Vincent,
you may want to take a look at Nonparametric methods in factorial
designs by Edgar Brunner and Madan L. Puri in Statistical Papers 42, 1-52
(2001).
There is the R-package nparcomp for one-way layouts, but the paper goes
further (and mentions another software) and is maybe a
Hello, Alaois,
if x is your vector maybe
n - length( x)
positions - trunc( quantile( seq( n), prob = 0:5/5))
x[ positions]
comes close to what you want.
Hth -- Gerrit
Hi all, I have in my code some vectors that are not of equal size. I
would like to be able for each of these vectors
Hi, Jack,
well, I disagree: What do you expect to grab out of a bucket (= data
frame) if you do not at all grab into it (indexing with an _empty_ index,
i.e. with nothing)? And changing the sign of nothing is still nothing ...
Hth -- Gerrit
On Wed, 30 Oct 2013, Jack Tanner wrote:
I'm
Alex,
it is a bit unclear what you mean by remap etc., but maybe
y0 - 880e6; y1 - 1020e6
x0 - 1; x1 - ncol(x)
y0 + (x0-1):x1 * (y1 - y0)/(x1 - (x0-1))
gives what you want.
Hth -- Gerrit
On Thu, 31 Oct 2013, Alaios wrote:
Hi everyone,
I am plotting some legend and I am using
read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
-
Dr. Gerrit Eichner Mathematical Institute, Room 212
gerrit.eich...@math.uni
read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
-
Dr. Gerrit Eichner Mathematical Institute, Room 212
gerrit.eich...@math.uni
Alen, look at
?scan
and there particularly at the argument 'what'.
Hth -- Gerrit
On Mon, 12 Nov 2012, alen wrote:
hi,
I want to know just enter a string value
entering a numeric value is make with the function scan () or scan (nmax =
..)
but it does not work for string
my goal is
Hello, e-letter,
check
?read.delim
and look for the argument check.names and from there you'll should end up
reading
?make.names
Hth -- Gerrit
On Fri, 23 Nov 2012, e-letter wrote:
Readers,
The function 'read.delim' was used to import data into R:
columnnamea columnnameb
Hello, Mono,
your description of your problem is a bit overwhelming. You might get an
answer if you provide (simplified) reproducible code (maybe with example
data).
I can only guess that -- instead of factor( x[[1]]) -- you maybe should
use
factor( x[[1]][1])
in some place.
Hth --
Hi, Felipe,
two typos? See below!
On Sun, 2 Dec 2012, Felipe Carrillo wrote:
Hi,
Consider the small dataset below, I want to subset by two variables in
one line but it wont work...it works though if I subset separately. I have
to be missing something obvious that I did not realize before
Hi, T. Bal,
homework? Take a look at
?tapply
Regards -- Gerrit
On Tue, 4 Dec 2012, T Bal wrote:
Hi,
I have the following data:
0 12
1 10
1 4
1 6
1 7
1 13
2 21
2 23
2 20
3 18
3 17
3 16
3 27
3 33
4 11
4 8
4 19
4 16
4 9
In this data file I would
Hello, Simon,
see below!
On Tue, 4 Dec 2012, Simon wrote:
Hello all,
I have what feels like a simple problem, but I can't find an simple
answer. Consider this data frame:
x - data.frame(sample1=c(35,176,182,193,124),
sample2=c(198,176,190,23,15), sample3=c(12,154,21,191,156),
Hi, Jeremy!
... I have a matrix, measuring m X n. An example matrix will be this:
... [snip] ...
What I want to do is actually statistical manipulation of this matrix.
To be more precise, I want to calculate the probability of each value,
based on the assumption of a normal distribution,
producing far too many logical indices for your task (so
to say).
I assume you should be using
na.omit( df2)
instead.
Hth -- Gerrit
-
Dr. Gerrit Eichner Mathematical Institute, Room 212
gerrit.eich
Hi, Charles,
see
?which
and learn about the argument arr.ind.
Happy new year -- Gerrit
On Wed, 2 Jan 2013, Charles Novaes de Santana wrote:
Dear all, Happy New Year for all of you! I hope we have an year of
essential freedom for everyone!
I am trying to manipulate a matrix in order to
Hi, Aseem,
you may want to look at
?stack
or
?reshape
(and transform your matrix into a data frame).
Hth -- Gerrit
On Wed, 2 Jan 2013, Aseem Sharma wrote:
Hi,
I have a data matrix with 13 columns and 55 rows. First coulmn is year and
other are monthly values for 55 years.
Now i want
Hello, Maria,
take a look at
?errbar
in the package Hmisc.
Hth -- Gerrit
Dear all
I have a dataset of how metal concentrations change through time. I have
made a plot of date versus metal concentration. However I want to add
error bars in the plot.
Could you help me?
Thanks
Maria
Hello, Gregory,
for your first data set see
?read.table
and for you second
?read.fwf
may help solving your problem.
Hth -- Gerrit
On Thu, 3 Mar 2011, Gregory Ryslik wrote:
Hi,
I seem to be having somewhat of an unusual data input problem with some
of the data sets I'm working with
Hello, Taby,
for your (first) problem take a look at
?sample
and its argument replace.
Regards -- Gerrit
On Thu, 10 Mar 2011, taby gathoni wrote:
Please note is with replacement
From: taby gathoni tab...@yahoo.com
To: R help r-help@r-project.org
Sent: Thursday, March 10, 2011 11:53 AM
.
-
Dr. Gerrit Eichner Mathematical Institute, Room 212
gerrit.eich...@math.uni-giessen.de Justus-Liebig-University Giessen
Tel: +49-(0)641-99-32104 Arndtstr. 2, 35392 Giessen, Germany
Fax: +49-(0)641-99-32109http
Hi, Berry,
I think
for(i in -8:-3) axis(1, i, substitute(10^j, list( j = i)))
achieves what you want.
Regards -- Gerrit
On Wed, 13 Mar 2013, Berry Boessenkool wrote:
Hi all,
I want to label an axis with exponents, but can't get it done with expression.
Any hints would be very
Hi, Eric,
as a quick hack, does
spots - sapply( alist,
function( listcomp)
which( listcomp == value.to.look.for) )
spots[ sapply( spots, length) == 0] - NULL
do what you want?
Regards -- Gerrit
On Thu, 28 Mar 2013, Eric Rupley wrote:
Dear All,
This
1 - 100 of 192 matches
Mail list logo
