readLines() is for a text-mode connection; readChar() is for a
binary-mode connection. Given that you asked for possible re-encoding
by the 'encoding' argument, you cannot safely mix them (text-mode
access is re-encoded, binary-mode is not). However, we don't know if
re-encoding was active
Thx so much, especially to Dennis.
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Hi,
I am trying to run th mclust package on a variable Tuberculin indurations
recorded as mm. The file has only one variable. When I run the package I get
NULL value for mu and sigma. Can anybody say why?
This is the program:
library(mclust)
mc-Mclust(x.trab,G=1:9,warn=TRUE)
mc
mc$mu
Hi R-users
I would like to extract the random effects (1|SITE, 1+SPECIES|SITE
and BA|SITE) from this model formula:
Full_model - formula (VAR ~ (1|SITE) + (1+SPECIES|SITE) + (BA|SITE) +
HEIGHT + COND + NN_DIST)
I tried:
terms(Full_model)
labels(terms(Full_model))
but I could
Hi Dennis,
Thanks for your answer (for some reason, the only one).
I could do that for sure, but I think it looks better if the labels are
all horizontal.
Anyway, it looks like what I wanted to do is not really possible. I have
to play with the arguments I already know...
Regards,
Ivan
Le
Thank you everyone for your helpful feedback.
To clarify a few things that people have been asking:
Part of the problem comes from the fact that these are external
students, so our conversations are over email and I'm not there to try
things.
I admit my shortcomings when it comes to R script
Hi,
I agree with you that levels should not be automatically dropped after
subsetting.
However, I think there should/can be an extra argument to make it
possible (the default being no dropping). I have no example in mind, but
I guess it is possible that sometimes, one want to show only some
Dear all, I have following piece of codes:
xx - seq(-2,2, length.out=11)
mat1 - cbind(rep(43, 5), rnorm(5))
mat2 - cbind(rep(53, 5), rnorm(5))
outer(c(1,-1), xx, function(x,y) {
AA - sign(x)
BB - sign(y)
CC - abs(y)
Hi Jannis,
thanks a lot for your reply. Unfortunately the solution you proposed
does not work.
Maybe the reason is, that plot.Map only accepts hsv colours and I do not
know to convert the rgb colours to the right colour space.
Isn't there a possibility to specify minimal an maximal values
On Wed, Sep 22, 2010 at 9:56 AM, schaber kscha...@ipp.mpg.de wrote:
Hi Jannis,
thanks a lot for your reply. Unfortunately the solution you proposed does
not work.
Maybe the reason is, that plot.Map only accepts hsv colours and I do not
know to convert the rgb colours to the right colour
Hello everyone,
I need some help with lists inside lists (a good way to emulate a struct)\
Assume that there is a small list called fred:
fred - list(happy = 1:10, name = squash)
and a big list called bigfred that includes fred list 5 times
bigfred - rep(fred,5)
Is it possible somehow to
I'm confused by what you are looking for.
There's some slight possibility that you
are looking for double bracket subscripting
with a vector:
list(a=1:5, b=letters)[[c(2,4)]]
[1] d
On 22/09/2010 10:58, Alaios wrote:
Hello everyone,
I need some help with lists inside lists (a good way to
From: Vijayan Padmanabhan
Dear R Group
I had an observation that in some cases, when I use the
randomForest model
to create partialPlot in R using the package randomForest
the y-axis displays values that are more than -1!
It is a classification problem that i was trying to address.
Any
On 09/22/2010 08:52 AM, Simon Kiss wrote:
hello, can someone tell me how to generate the means for a data frame
that looks like this? My data frame has many more variables, but I won't
bother you with those; these are the one's that I'm interested in.
Needless to say, z is the variable in which
Also, it may be that you want bigfred to be a list of 5 lists each of 2
elements (happy and name) rather than a list of 10 elements.
Thus (also using double bracketing)
fred- list(happy = 1:10, name = squash)
bigfred - replicate(5, fred, FALSE)
bigfred[[2]][[2]]
hth
Keith J
Patrick Burns
% correct is an improper scoring rule and a discontinuous one to boot. So it
will not always agree with more proper scoring rules.
When you have a more difficult task, e.g., discriminating more categories,
indexes such as the generalized c-index that utilize all the categories will
recognize
Simple problem - I want the ylab to automatically pick up x1 rather than
having to define x1 in the plot statement.
x1-c(1.2,2,3);x2-c(1,2.1,2.6)
y-x1
plot(1:3,y, ylab=x1)
There must be a way of accessing the name x1 somehow, but unfortunately I
don't know how to search for it. Any help
Hi:
I believe we had this discussion yesterday,
http://r.789695.n4.nabble.com/Object-oriented-programming-in-R-td2538541.html#a2538916
but since you chose to repeat that message, it clearly wasn't enough, so
start with
http://cran.r-project.org/doc/manuals/R-intro.html#Lists
Hi:
I presume you mean
x1-c(1.2,2,3)
y - 'x1'
plot(1:3, get(y), ylab = y)
??
HTH,
Dennis
On Wed, Sep 22, 2010 at 4:44 AM, Paul Chatfield p.s.chatfi...@reading.ac.uk
wrote:
Simple problem - I want the ylab to automatically pick up x1 rather than
having to define x1 in the plot statement.
Thats great thanks
I guess it is hard to not use % as a performance measure when that is what is
commonly used in everyday life.
So when i come to predicting the response of new data ( using the estimated
mean Y ) which i am more comfortable with i can say -
Species A - 2.12 - Therefore this
plot.ts has an argument yax.flip, plot.zoo does not.
Is there a way to make the yaxis flip in plot.zoo?
I tried using a custom panel function:
panel.yaxis-function(...) {
npnl-parent.frame$panel.number
if (npnl %% 2 == 0) {
axis(side=3)
} else {
axis(side=2)
}
}
do the same thing for female and then take the weighted average of the two
means.
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do the same thing for female and then take the weighted average of the two
means.
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On Wed, Sep 22, 2010 at 8:07 AM, Alex van der Spek do...@xs4all.nl wrote:
plot.ts has an argument yax.flip, plot.zoo does not.
Is there a way to make the yaxis flip in plot.zoo?
I tried using a custom panel function:
panel.yaxis-function(...) {
npnl-parent.frame$panel.number
if
Dear R-users
Idea:
Analysing tree height frequency with hist(), normal distribution (ks.test
shapiro.test) and skewness (package e1071 - thanks a lot for this useful
package)as an indication of possible self-thinning in an experimental tree
stand.
Problem:
Results from the ks.test and the
unique(unlist(list.array))
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Great one!
Thanks this will simplify a lot addresing.
Best REgards
From: Keith Jewell k.jew...@campden.co.uk
To: r-h...@stat.math.ethz.ch
Sent: Wed, September 22, 2010 1:25:27 PM
Subject: Re: [R] efficient list indexing
Also, it may be that you want bigfred to
Hello,everyone:
I facing a problem in Rimage package.I cannot read the images with size 364X364
in jpeg format.What happen to it?After I run it, it give me an error Can't Open
fie.I try many time already. But I can read an JPEG image than have smaller
size or bigger size than 364X364.What
Dear R-helpers,
could anybody explain me briefly what is the difference between
eigenvectors returned by 'eigen' and 'svd' functions and how they are
related?
Thanks in advance
Ondrej Mikula
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For some reason I did not receive your email.
Sorry for the inconvenience caused
From: Dennis Murphy djmu...@gmail.com
Cc: Rhelp r-help@r-project.org
Sent: Wed, September 22, 2010 1:46:28 PM
Subject: Re: [R] efficient list indexing
Hi:
I believe we had this
dear R experts: I am writing my own little newey-west standard error
function, with heteroskedasticity and arbitrary x period
autocorrelation corrections. including my function in this post here
may help others searching for something similar. it is working quite
well, except on occasion, it
In general Shapiro's normality test is more powerful than the KS. For this
specific case, I don't see the significantly different results from both
tests. The normality assumption in this example seems to be questionable.
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Hi,
I have a data set that I'd like to run logistic regressions on, using
ddply to speed up the computation of many models with different
combinations of variables. I would like to run regressions on every
unique two-variable combination in a portion of my data set, but I
can't quite
Hi,
I'm using the Cp and Radj method for the selection of variable, with the
library leaps.
I need to save the best model in a file.
If I use the adjr2 method it's work, but I've a problem with Cp method:
adjr - leaps(x,y,method=adjr2)
maxadjr(adjr)
12312
0.713 0.56
ok...
but whit Cp
Thats great thanks,
I suppose it is hard to move away from a more traditional measure of
performance such a percentage correct, at least for the relatively amateur
statisticians among us who have been graded on such a system.
The difficulty comes in reporting the effectiveness of the model to
Hi again,
Sorry, probably I was not clear enough.
I was wondering if there is a way in R to identify (and extract) only
the random effects, which, because I am using the lmer function, are the
terms with the symbol | on the left side of the grouping variable
(SITE in my example).
hi,
can anyone tell me how to merge a vector and a matrix?
v=c(1,4,2)
names(v)=c(e,r,t)
m=matrix(c(r,t,r,s,e,5,6,7,8,9),nr=5)
colnames(m)=c(c1,c2)
I want to do like
merge(v, m, by.x=names,by.y=c1)
I got error
Error in fix.by(by.x, x) : 'by' must specify valid column(s)
thanks
jian
Hi,
I am using ggplot2 to create a boxplot that summarizes a continuous
variable. This code works fine for me on one PC however when I use it on
another it doesnt.
The structure of the dataset AHT_TopCD is SubReason=Categorical variable,
AHT=Continuous variable.
The code for the boxplot:
actually I need to extract the random effect from the formula, not the model
any idea?
From: Sacha Viquerat [mailto:sacha.v...@googlemail.com]
Sent: Wed 22/09/2010 5:55 PM
To: Lorenzo Cattarino
Subject: Re: [R] extracting random effects from model formula
Am
Hi, are you sure you have the same version R + packages version on both ?
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Hello,
I just figured out that there was some problem with my dataset. So, the
Regression is working fine now. Thanks a lot for all your help and
suggestions.
Uttara
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Sent from
Hello everybody,
I have a list.array which is a list containing arrays, lets say
list.array= list( c(1,2,3),c(5,6,7),c(1,4,6,7,8) );
I would like to apply the union function to those 3 vectors to get the union
of the three : [1,2,3,4,5,6,7,8]
I tried do.call(what=union,args=list.array) but this
svd() does not return eigeinvectors!
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I think you need to set up a cut-off of Cp and then get the good values of
Cp from adjr$Cp.
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X = U D V' ## D are the singular values of X
X'X = V D^2 V' ## D^2 are the eigenvalues of X'X
V is the same in both factorizations.
2010/9/22 OndÅej Mikula onmik...@gmail.com
Dear R-helpers,
could anybody explain me briefly what is the difference between
eigenvectors returned by 'eigen'
Try this:
merge(m, v, by.x = 'c1', by.y = 0, all = TRUE, sort = FALSE)
On Wed, Sep 22, 2010 at 4:57 AM, Yuan Jian jayuan2...@yahoo.com wrote:
hi,
can anyone tell me how to merge a vector and a matrix?
v=c(1,4,2)
names(v)=c(e,r,t)
m=matrix(c(r,t,r,s,e,5,6,7,8,9),nr=5)
v=data.frame(c1=c(e,r,t),v=c(1,4,2) )
m=matrix(c(r,t,r,s,e,5,6,7,8,9),nr=5)
colnames(m)=c(c1,c2)
m=as.data.frame(m)
merge(v, m, by =c1 )
c1 v c2
1 e 1 9
2 r 4 5
3 r 4 7
4 t 2 6
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On Sep 22, 2010, at 3:57 AM, Yuan Jian wrote:
hi,
can anyone tell me how to merge a vector and a matrix?
v=c(1,4,2)
names(v)=c(e,r,t)
m=matrix(c(r,t,r,s,e,5,6,7,8,9),nr=5)
colnames(m)=c(c1,c2)
I want to do like
merge(v, m, by.x=names,by.y=c1)
I got error
Error in fix.by(by.x, x) : 'by' must
On Sep 22, 2010, at 5:09 AM, Lorenzo Cattarino wrote:
actually I need to extract the random effect from the formula, not
the model
any idea?
From: Sacha Viquerat [mailto:sacha.v...@googlemail.com]
Sent: Wed 22/09/2010 5:55 PM
To: Lorenzo Cattarino
Subject:
The AICs do not seem right to me either. Unless I am missing something, it
appears that the formula:
AIC= -2x logLik -2k
is being applied, rather than:
AIC= -2x logLik +2k
Meaning models with fewer degrees of freedom are being penalised.
So in your example I make the degrees of freedom
9.61
It does for real-symmetric and complex Hermitian matrices, i.e. the $u and
$v from svd() are the same as $vectors from eigen() for Hermitian matrices.
There might be differences in signs, but that does not matter. Of course the
singular values and eigenvalues are identical too.
Ravi.
I think I've figured it out, the AIC column is the IMPROVEMENT in AIC
compared to the null model. So bigger values are better.
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Hi Alison,
On Wed, Sep 22, 2010 at 11:05 AM, Alison Macalady a...@kmhome.org wrote:
Hi,
I have a data set that I'd like to run logistic regressions on, using ddply
to speed up the computation of many models with different combinations of
variables.
In my experience ddply is not
There has been a recent addition of parallel processing capabilities
to plyr (I believe v1.2 and later), along with a dataframe iterator
construct. Both have improved performance of ddply greatly for
multicore/cluster computing. So we now have the niceness of plyr's
grammar with pretty good
Hello everyone,
I want to estimate the reliability *of the level*-*1 random coefficients
(the one *reported in the HLM output) using the software R. Does anyone know
how to get this statistic from R?
I'm using the function lmer of the package lme4 to estimate multilevel
models.
I tried to use
Hi
It is difficult to give some help as you did not provide any clue what the
result shall be.
basically outer takes 2 vectors and evaluate function for each combination
of elements in both vectors. However the function has to be vectorised and
your function is not.
fff=function(x,y) {
+
Hi
what exactly do you want.
You say you can read your data
x - read.jpeg(system.file(data, cat.jpg, package=rimage))
plot(x)
What do you mean by pixel image? By reading a picture you get an object
imagematrix
str(x)
imagematrix [1:420, 1:418, 1:3] 0.255 0.251 0.247 0.247 0.255 ...
-
Dear fellow R users,
I am trying to conduct a regression analysis. I have thousands of
variables. The names are V1, V2,V2000
Is there an easy way to include these variables in the regression?
my model is something like that:
model- lm(y~V1+V2+.+V2000, data=data)
Thanks so much in
Ozlem,
Just read ?formula, where it says:
There are two special interpretations of ‘.’ in a formula. The
usual one is in the context of a ‘data’ argument of model fitting
functions and means ‘all columns not otherwise in the formula’:
see ‘terms.formula’. In the context of
Hi,
Try with .:
model- lm(y~., data=data)
From ?formula:
There are two special interpretations of |.| in a formula. The usual
one is in the context of a |data| argument of model fitting functions
and means 'all columns not otherwise in the formula': see |terms.formula
Hello R users,
I have 2 files (file1 and f2) and I am trying to sum columns 6:10 of a
specific row in f2 and append it in
file 1 if the state variable in file 1 equals the rowname in f2. Below is
an example of the code I wrote
using a for loop, but it not working (i.e it only works for the
That implies you need to update your version of plyr.
Hadley
On Wed, Sep 22, 2010 at 4:10 AM, RaoulD raoul.t.dso...@gmail.com wrote:
Hi,
I am using ggplot2 to create a boxplot that summarizes a continuous
variable. This code works fine for me on one PC however when I use it on
another it
I am helping a fellow worker get R up and running, and he has run into a
peculiar problem I've not encountered in previous install situations.
From the Rconsole menu choice, he can set CRAN mirror to USA CA2, but when he
selects load packages a very truncated list of packages appears (many
thanks. Duncan.
previously I try to install the 32 bit ucminf on a 64 bit R thus it cannot fit
in.
thanks for the link and I have downloaded and installed it. it works perfect.
Nan
- Original Message
From: Duncan Murdoch murdoch.dun...@gmail.com
To: Hey Sky
On Sep 22, 2010, at 11:42 AM, Pele wrote:
Hello R users,
I have 2 files (file1 and f2) and I am trying to sum columns 6:10 of a
specific row in f2 and append it in
file 1 if the state variable in file 1 equals the rowname in f2.
Below is
an example of the code I wrote
using a for loop,
Hello,
I'm currently trying to model the movement of a slope (v.obs) with a
regression model.
The data can be found following the given links:
either
http://www.sendspace.com/file/dnugwc
or
http://rapidshare.com/files/420569660/sel.day.txt
I want to use the Box-Cox transformation to
hey, Ravi
yes. I have tried the hessian() in the numDeriv package and it is the same
package ucminf() uses to calculate the hessian matrix while having the option
hessian=1.
maybe I should avoid the word fail but instead using some others. anyway, what
I
mean in the former post is the
Hi all,
I'm looking at a large data set, and I'm interested in removing rows where
only one variable is duplicated. Here's an example:
presidents
Qtr1 Qtr2 Qtr3 Qtr4
1945 NA 87 82 75
1946 63 50 43 32
1947 35 60 54 55
1948 36 39 NA NA
1949 69 57 57
Hi,
Take a look at ?duplicated and ?unique
HTH,
Ivan
Le 9/22/2010 16:55, AndrewPage a écrit :
Hi all,
I'm looking at a large data set, and I'm interested in removing rows where
only one variable is duplicated. Here's an example:
presidents
Qtr1 Qtr2 Qtr3 Qtr4
1945 NA 87
On 9/21/2010 8:04 PM, Henrik Bengtsson wrote:
Each of the following calls crash (core dumps) R (R --vanilla) on
various versions and OSes:
regexpr(a{2-}, )
sub(a{2-}, )
gsub(a{2-}, )
EXAMPLES:
To add another (windows) example it also crashes the 2.12.0 alpha build:
sessionInfo()
R version
Hi David - thanks for your suggestion, but I am trying to avoid doing any
merging and sorting for this step because the real file I will be working
with has about 20 million records. If I can get this loop or something
similar to work will be good enough.
thanks again..
--
View this message
The problem R ver is 2.10.1 - I am using 2.11.1
Running windows xp sp3 - same system as what I use
The suggested solution is for the user to upgrade to 2.11.1 to resolve package
incompatibilities
Gregory A. Graves, Lead Scientist
Everglades REstoration COoordination and VERification (RECOVER)
The way that you called ks.test below your null hypothesis is that your data
comes from a normal distribution with mean 0 and standard deviation 1. Now I
am not familiar with your data, but I would be very surprised in general to
find a population of trees where half of them had negative
Hello,
I am trying to build a generalised linear mixed model. My dependent variable
is
ordinal. I have a random factor (7 individuals), and a repeated measure where
the dependent variable was measured three times for each of four attempts (so
the repeats are nested). I also have a few
Hi There,
Just a question regarding the function that is specified to boot (I have
read the help, the manual and online examples.). The description of
boot says that the second argument of statistic (non parametric bootstrap)
must be a vector of indices, frequencies or weights which
I understand how duplicated and unique work for a list where all parts of a
given row are duplicated, or how to find duplicated values if I'm just
looking at that first column, but in this case the rows for 1954 and 1955
are not completely the same; only quarter 1 is duplicated, so I'm not sure
On Sep 22, 2010, at 12:35 PM, AndrewPage wrote:
I understand how duplicated and unique work for a list where all
parts of a
given row are duplicated, or how to find duplicated values if I'm just
looking at that first column, but in this case the rows for 1954
and 1955
are not completely
Hi,
Thanks for the advice! My locale and encoding setting follow:
Sys.getlocale()
[1] en_US.UTF-8/en_US.UTF-8/C/C/en_US.UTF-8/en_US.UTF-8
getOption(encoding)
[1] native.enc
I was indeed able to solve my immediate problem by using readLines to
read the whole response and parse it later,
I just figured that out, but the real data I'm using is a data frame for
sure, so I'll find another example.
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I can not open PDf device. Acrobat is closed. I have checked archives but
could not find a solution. What should I do?
cont.cdfplot(myanalysis.pdf, myanalysis$CDF, ylbl.r=Stream Length (km))
Error in pdf(file = pdffile, width = width, height = height) :
unable to start device pdf
In
Hi,
there is a function to plot survival curves:
library(survival)
plot.KM - function(survival, x, x_cut.off, main='', label='')
{
plot(survfit(survival ~ I(x = x_cut.off)), main=main)
legend('bottomleft', c(expression(label = x_cut.off),expression(label
x_cut.off)))
}
Now, I need to determine
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of Seth W Bigelow
Sent: Tuesday, September 21, 2010 4:22 PM
To: bill.venab...@csiro.au
Cc: r-help@r-project.org
Subject: Re: [R] Doing operations by grouping variable
Aah, that
Hi Andrew,
You can use duplicated() to index the rows you wish to keep, like this:
test.dat - data.frame(a=c(1,1:5,5:10), b=1:12, c=letters[1:12]) #make up data
duplicated(test.dat$a) # see what duplicated() function does
!duplicated(test.dat$a) # see how we can invert using the ! function
so
How about this:
s = c(aa, bb, cc, , aa, dd, , aa)
n = c(2, 3, 5, 6, 7, 8, 9, 3)
b = c(TRUE, FALSE, TRUE, TRUE, FALSE, TRUE, TRUE, FALSE)
df = data.frame(n, s, b) # df is a data frame
I want to display df with no value in s occurring more than once. Also, I
want to delete the
On 09/21/2010 09:44 PM, Dennis Murphy wrote:
Hi:
Reshaping multiple variables is nontrivial. Try the following (untested):
reshape(rcw, idvar = 'ICU', varying = list(c(paste('Q6.RC', 1:4, sep = '.'),
c(paste('Q6.FT.RC', 1:4, 'years', sep =
'.'),
Thanks-- that works for what I'm trying to do. I was also wondering, in the
data frame example you gave, if I just wanted to get rid of rows where the
a value is 5, how would I do that?
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Hi Andrew,
Perhaps you did not notice my previous email. The answer is still the
same (see below):
On Wed, Sep 22, 2010 at 1:48 PM, AndrewPage savejar...@yahoo.com wrote:
How about this:
s = c(aa, bb, cc, , aa, dd, , aa)
n = c(2, 3, 5, 6, 7, 8, 9, 3)
b = c(TRUE, FALSE, TRUE, TRUE, FALSE,
Dear R users,
I would like to great a frequency table from raw data and then access
the classes/bins and
their respective frequencies separately. Here the code to create the
frequency tables:
x1 - c(1,5,1,1,2,2,3,4,5,3,2,3,6,4,3,8)
t1 - table(x1)
print(t1[1])
Its easy to plot this, but how do
I already gave you three examples of how this works. Your last request
can be done in exactly the same way. Give it a try and see what
happens (use example data of course!). As a last resort you could read
the documentation:
?Comparison
?Extract
-Ista
On Wed, Sep 22, 2010 at 2:22 PM, AndrewPage
Hi Ralf
try hist()
obl-hist(x1, plot=FALSE)
it returns midpoints and their respective frequencies. You can specify the
breakpoints as well.
?hist
for details.
Mike
On Wed, Sep 22, 2010 at 1:44 PM, Ralf B ralf.bie...@gmail.com wrote:
Dear R users,
I would like to great a frequency table
Ralf -
Try
bins = as.numeric(names(t1))
freqs = as.vector(t1)
- Phil Spector
Statistical Computing Facility
Department of Statistics
Try this:
as.data.frame(table(x1))
On Wed, Sep 22, 2010 at 3:44 PM, Ralf B ralf.bie...@gmail.com wrote:
Dear R users,
I would like to great a frequency table from raw data and then access
the classes/bins and
their respective frequencies separately. Here the code to create the
frequency
On Sep 22, 2010, at 1:39 PM, threshold wrote:
Hi,
there is a function to plot survival curves:
library(survival)
plot.KM - function(survival, x, x_cut.off, main='', label='')
{
plot(survfit(survival ~ I(x = x_cut.off)), main=main)
legend('bottomleft', c(expression(label =
On Sep 22, 2010, at 2:58 PM, David Winsemius wrote:
On Sep 22, 2010, at 1:39 PM, threshold wrote:
Hi,
there is a function to plot survival curves:
library(survival)
plot.KM - function(survival, x, x_cut.off, main='', label='')
{
plot(survfit(survival ~ I(x = x_cut.off)), main=main)
On 09/22/2010 05:28 PM, anupam wrote:
I can not open PDf device. Acrobat is closed. I have checked archives but
could not find a solution. What should I do?
cont.cdfplot(myanalysis.pdf, myanalysis$CDF, ylbl.r=Stream Length (km))
Error in pdf(file = pdffile, width = width, height = height) :
Hi group,
I would like to draw multiple Lorenz curves in a single plot using
data already prepared. Here is a simple example:
require(lawstat)
lorenz.curve(c(1,2,3),c(4,5,4))
lorenz.curve(c(1,2,3),c(4,2,1))
This example draws two separate graphs. How can I combine them in a
distinguishable way?
On Sep 22, 2010, at 3:32 PM, Ralf B wrote:
Hi group,
I would like to draw multiple Lorenz curves in a single plot using
data already prepared. Here is a simple example:
require(lawstat)
lorenz.curve(c(1,2,3),c(4,5,4))
#You can get a half-assed solution by separating the two plot calls
On Sep 22, 2010, at 3:03 PM, David Winsemius wrote:
On Sep 22, 2010, at 2:58 PM, David Winsemius wrote:
On Sep 22, 2010, at 1:39 PM, threshold wrote:
Hi,
there is a function to plot survival curves:
library(survival)
plot.KM - function(survival, x, x_cut.off, main='', label='')
{
Why do you want to do this?
If there is just a small part of the logistic regression that you are
interested in, then there may be a way to compute or approximate that more
quickly than doing a full glm fit on every pair. It seems unlikely that you
would get much meaning out of that many full
Oops, yeah I didn't see that.
Thanks,
Andrew
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