Hello every one
can any one tell me how to draw contour with this data set
c zshock
1 0.45450237 0
2 0.02663337 0
3 -2.08444556 0
4 -0.12715275 0
5 0.67066360 0
6 -0.73540081 0
I want to draw contour for shock i.e my z matrix is shock
Great! Thank you ever so much Rui.
Santi
From: Rui Barradas ruipbarra...@sapo.pt
To: Santiago Guallar sgual...@yahoo.com
Cc: r-help@r-project.org
Sent: Wednesday, July 18, 2012 8:29 PM
Subject: Re: [R] Imposing more than one condition to if
Hello,
You're
On Wed, Jul 18, 2012 at 06:02:27PM -0700, bilelsan wrote:
Leave the Taylor expansion aside, how is it possible to compute with [R]:
f(e) = e1 + e2 #for r = 1
+ 1/2!*e1^2 + 1/2!*e2^2 + 1/2!*e1*e2 #for r = 2, excluding e2*e1
+ 1/3!*e1^3 + 1/3!*e1^2*e2 + 1/3!*e2^2*e1 + 1/3!*e2^3 #for r = 3,
Dear R helpers,
I have one trivial problem while writing an output file in csv format.
I have two dataframes say df1 and df2 which I am reading from two different csv
files.
df1 has column names as date, r1, r2, r3 while the dataframe df2 has column
names as date, 1w, 2w.
(the dates in
On Wed, Jul 18, 2012 at 06:34:49PM -0700, Noah Silverman wrote:
Hi,
I'm looking for an easy way to setup a decision tree.
This is *not* any kind of regression, but a very simple DAG with probability
on each edge and a value at each node. I just need a way to input the graph
and
thank you Mr. Carlson
you're right, the matrix that I used as an example does not make sense.
When I built the array I was not thinking of the formula
s[n]=1+s[n]+rnorm(1,mean=0,sd=1), I just wanted to give an example of the
structure of the matrix that I needed.
Anyway I think you have solved
Oh, sorry: X3D is an interchange format for 3D scenes (s [1])...
X3D's IndexedTriangleSet is a visual component to describe a complex figure
to be visualized by triangles (s. [2])
I would like to triangulate a pointset (2.5D DEM consisting of xy coords
and height) using deldir and convert the
Hello all,
I have a dataframe that looks like this:
head(df)
datey
1 2010-09-27 1356
2 2010-10-04 1968
3 2010-10-11 2602
4 2010-10-17 3116
5 2010-10-24 3496
6 2010-10-31 3958
I need a function that, given any date, returns the y value
corresponding to the given date or the last day
Hi Vincy,
have you checked
names(df2)
and
names(df_new)
because by default 'data.frame' checks the column names to ensure that
they are syntactically valid variable names and 1w and 2w aren't, so an
X is prepended (see ?data.frame and ?make.names).
compare
On 19/07/12 18:55, Vincy Pyne wrote:
Dear R helpers,
I have one trivial problem while writing an output file in csv format.
I have two dataframes say df1 and df2 which I am reading from two different csv
files.
df1 has column names as date, r1, r2, r3 while the dataframe df2 has column
Hello,
Try the following.
dat - read.table(text=
datey
1 2010-09-27 1356
2 2010-10-04 1968
3 2010-10-11 2602
4 2010-10-17 3116
5 2010-10-24 3496
6 2010-10-31 3958
, header=TRUE, stringsAsFactors=FALSE)
dat$date - as.Date(dat$date)
str(dat)
search - as.Date(2010-10-06)
Hi Robert,
how about this (assuming your data.frame is ordered by date):
tmp-read.table(textConnection(datey
1 2010-09-27 1356
2 2010-10-04 1968
3 2010-10-11 2602
4 2010-10-17 3116
5 2010-10-24 3496
6 2010-10-31 3958),header=T,colClasses=c(date=Date))
dat - structure(list(date = structure(c(14879, 14886, 14893, 14899,
14906, 14913), class = Date), y = c(1356L, 1968L, 2602L, 3116L,
3496L, 3958L)), .Names = c(date, y), row.names = c(1, 2,
3, 4, 5, 6), class = data.frame)
x - as.Date( 2010-10-06 )
getY - function( x ) { dat[ dat$date = x, 2][
I think Michael already gave you some really good hints. To add a bit more
details to this, first of all, I suggest to use:
q2 - rma(yi, vi, mods=cbind(grupo), data=qim)
forest(q2, transf=transf.ztor, digits=3, alim=c(0,1), refline=.5, addfit=FALSE)
so that the fitted values for that
Martin,
I've just submitted mgcv_1.7-19 to CRAN, which includes a major upgrade
of the p-value computation for random effect terms (and any other smooth
term which can be penalized to zero as part of estimation). The new
p-values are still conditional on the smoothing parameter/variance
On Wed, Jul 18, 2012 at 10:06 AM, Patrick Burns pbu...@pburns.seanet.comwrote:
It looks to me like the following
should do what you want:
f2 - function(dotot) array(FALSE, c(dotot, 1))
What am I missing?
ah, the output of this is purely a toy example. the point is that the
add*()
Dear people,
I am including an example of a dataframe:
mydataframe-data.frame(X=c(1:4),total_bill=c(16.99,10.34,21.01,23.68),tip=c(1.01,1.66,3.50,3.31),sex=c(Male,Male,Male,Female))
When I use the sapply function getting the information about the factors
works:
Hi,
Recently I have installed R in my Linux operating system , after
installation I was trying to install some packages and I was getting error
installed.packages(hdf5)
{ Package LibPath Version Priority Depends
Imports LinkingTo Suggests Enhances OS_type
License Archs Built}
this
Roger Koenker-3 wrote
There are obviously a large variety of non-smooth problems;
for CVAR problems, if by this you mean conditional value at
risk portfolio problems, you can use modern interior point
linear programming methods. Further details are here:
Hans W Borchers wrote
The most robust solver for non-smooth functions I know of in R is
Nelder-Mead
in the 'dfoptim' package (that also allows for box constraints).
First throw out the equality constraint by using c(w1, w1, 1-w1-w2) as
input.
This will enlarge the domain a bit, but
On 19/07/12 18:36, Erdal Karaca wrote:
Oh, sorry: X3D is an interchange format for 3D scenes (s [1])...
X3D's IndexedTriangleSet is a visual component to describe a complex
figure to be visualized by triangles (s. [2])
I would like to triangulate a pointset (2.5D DEM consisting of xy
Roger Koenker-3 wrote
There are obviously a large variety of non-smooth problems;
for CVAR problems, if by this you mean conditional value at
risk portfolio problems, you can use modern interior point
linear programming methods. Further details are here:
Oh, I dont want anyone to do anything for me. I am just asking for
hints/infos to be able to do it myself...
I was hoping that someone has already done this...
But anyways, I will post my solution once I have succeeded...
Thanks for the support!
2012/7/19 Rolf Turner rolf.tur...@xtra.co.nz
Dear all,
apologies for this (perhaps recurrent) question but I did not found a question
when searching mailing lists.
How to write a list of a simple kind, e.g.:
abc - list(one=(1:2), two=(1:5))
# to a file? I understand that write() co. cannot work but when I try
sink(aa.txt, append=T,
(Oh, and yes I've tried setting 'inches=F', but I then get a circle correct on
X axis, but too large on the Y axis.)
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Stuart Leask
Sent: 18 July 2012 11:03
To: r-help@r-project.org
what is that you want to do with this data after it is written? you can easily
write a function using 'cat' to create any format that you want.
Sent from my iPad
On Jul 19, 2012, at 5:09, Skála, Zdeněk (INCOMA GfK)zdenek.sk...@gfk.com
wrote:
Dear all,
apologies for this (perhaps recurrent)
Problem has gone away!
Last night, on this computer (64-bit i7 Windows 7), setting inches=F plotted
wrongly on Y.
Tried it again this morning - still wrong on Y the first time, BUT repeating it
- okay on Y!
Repeating previous syntax, then repeating this - still okay.
Odd. I will get back to
apply(mydataframe,2,function(x){ print(x);is.factor(x)})
[1] 1 2 3 4
[1] 16.99 10.34 21.01 23.68
[1] 1.01 1.66 3.50 3.31
[1] Male Male Male Female
X total_billtipsex
FALSE FALSE FALSE FALSE
sapply(mydataframe,function(x){
On Thursday, July 19, 2012, Stuart Leask wrote:
(Oh, and yes I've tried setting 'inches=F', but I then get a circle
correct on X axis, but too large on the Y axis.)
We're back to read the help. ?symbols explictly states that if
inches=FALSE and the axis scales are different, then the x axis
At 22:27 18/07/2012, Chet Seligman wrote:
This doesn't work, what should I do?
Try reposting on the r-sig-debian list.
It might help to tell them what version of Ubuntu you are using and
to paste in your sources.list
sudo apt-get install r-base
[sudo] password for cseligman:
Reading
Hi Marion,
as stated in the help file ?apply coerces a data.frame object into an
array. Since an array has only one type of data, this coercion turns all
your variables into strings (because this data type can hold all
information given without loss).
If it happens that your data.frame consists
Hello,
There is a mistake. Type
install.packages(hdf5)
instead of
installed.packages(hdf5)
in order to install a package.
Best Regards,
Passcal
Le 19/07/2012 17:02, uday a écrit :
Hi,
Recently I have installed R in my Linux operating system , after
installation I was trying to install
On 12-07-19 2:01 AM, Akhil dua wrote:
Hello every one
can any one tell me how to draw contour with this data set
c zshock
1 0.45450237 0
2 0.02663337 0
3 -2.08444556 0
4 -0.12715275 0
5 0.67066360 0
6 -0.73540081 0
I want to
As Duncan Murdoch pointed out, the example data frame that you provided
doesn't give very interesting results (all the shock values are zero), so
I created a different shock variable for illustration. I suggest using
the interp() function in the R package akima.
df - structure(list(c = 1:6, z
Hello,
I have the following design, counts were collected at different transects,
different depths and different sites at different times. Time is continuous
and assumed to be random, all the others are categorical fixed where
transect is nested within depth which is nested within site.
I would
This should definitely be posted on the r-sig-mixed-models list, not here.
-- Bert
On Thu, Jul 19, 2012 at 6:16 AM, Yolande Tra yolande@gmail.com wrote:
Hello,
I have the following design, counts were collected at different transects,
different depths and different sites at different
Code?
Sample data?
John Kane
Kingston ON Canada
-Original Message-
From: rshep...@appl-ecosys.com
Sent: Wed, 18 Jul 2012 14:30:24 -0700 (PDT)
To: r-help@r-project.org
Subject: [R] cenbox(): Changing Default x-axis Group Labels
I've looked at the lattice book and the 'R
As far as I can see those instructions are not all that helpful.
The answer below, from R-sig-Debian worked very nicely for me a little over a
month ago.
-Original Message-
From: marutter@gmail.com
Sent: Tue, 12 Jun 2012 16:56:16 -0400
To: jrkrid...@inbox.com
Subject: Re:
?read.table
Using my normal file path and th data as you supplied it.
x - read.csv(/home/john/rdata/ages.csv, sep = , header = TRUE)
Change the file path to whatever yours is.
Example might be
x - read.csv(C:/mydata/ages.csv, sep = , header = TRUE)
in Windows.
Note that you can do
When i make Boxplots with a lot of boxes, the names of them get only written
down every second column.
Since they aren't in any way ordered, you don't see anymore to what they belong.
example:
l-rep(list(1:5),20); boxplot(l,names=sample(20,1:20))
Is there a way to show them all, or do i have
Dear Zdenek
You could generate this file using a loop.
# the data
abc - list(one=(1:2), two=(1:5))
# create a connection
sink(aa.txt, append=T, split=T)
# for each element in the list, print
for (list_name in names(abc)) {
cat(list_name,
On 2012-07-19 06:58, Jessica Streicher wrote:
When i make Boxplots with a lot of boxes, the names of them get only written down every
second column.
Since they aren't in any way ordered, you don't see anymore to what they belong.
example:
l-rep(list(1:5),20); boxplot(l,names=sample(20,1:20))
Great, Thanks! This is really helpful!
On Thu, Jul 19, 2012 at 12:25 AM, ilai ke...@math.montana.edu wrote:
Maybe I'm missing something too but from your example seems like you are
looking for
xyplot(rnorm(12) ~ 1:12 , type=l,
scales=list(x=list(at=seq(2,12,2),labels=c(1, ' ', 3 , ' ' , 5 ,
You're right - easily tested by just re-sizing the graphics box - sort of
counter-intuitive until I remember the clue is in the name - this generates a
CIRCLE, come what may...
Stuart
From: Sarah Goslee [mailto:sarah.gos...@gmail.com]
Sent: 19 July 2012 11:18
To: Stuart Leask
Cc:
Copied the wrong lines, sry
l-rep(list(1:5),20); boxplot(l,names=sample(1:20,20))
of course.
thanks for the answer
.
On 19.07.2012, at 16:17, Peter Ehlers wrote:
On 2012-07-19 06:58, Jessica Streicher wrote:
When i make Boxplots with a lot of boxes, the names of them get only written
down
Johan,
Your 'list' and 'array doubling' code can be written much more efficient.
The following function is faster than your g and easier to read:
g2 - function(dotot) {
v - list()
for (i in seq_len(dotot)) {
v[[i]] - FALSE
}
}
In the following line in you array doubling function
You will probably need to write a custom function, but it could
use the built-in write.dcf() or formatDL() functions. E.g.,
# options(width=50)
write.dcf(list(One=paste(1:50,collapse= ), Two=paste(state.abb[1:5],
collapse=, )))
One: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
17 18 19
You might alternatively find the horizontal = TRUE with las=2 to be useful; e.g.
dat - data.frame(val=rnorm(100),
grp=rep(apply(matrix(sample(letters,100,rep=TRUE),nr=5),2,paste,collapse=),5))
boxplot(val~grp,horizontal=TRUE,data=dat,las=2)
## Note that las=2 might also help with
If you need to do this for a lot of dates findInterval() will do
it faster. E.g.,
f - function(dates, data) {
+ i - findInterval(dates, data$date)
+ i[i==0] - NA # before first data$date
+ data[i, ]
+ }
xx - as.Date(c(2010-10-06, 2010-10-25, 2009-01-01, 2011-12-31))
hi
My inputs is min=(10,10,10,10,10) and max=(100,100,100,100,100)
total = 300
i have to generate 5 numbers between min and max and those numbers should
sum upto total
Can anyone help?
-
Thanks in Advance
Arun
--
View this message in context:
Dear professor Harrell,
I probably have the same problem as Haleh Ghaem Maralani.
I am using the rms package and the rcspline.plot function to assess the
relation of a continuous predictor to the log hazard function.
I would like to use the adj statement, for example using this test
dataset:
Thanks William, that works fantastically!
I had a quick play with my data and have realised a potential problem in
that if an individual ends the series at home it records an additional
trip-no when one wasnt made. I was wondering whether you could think of a
way to alter it slightly so that the
I have this problem as well
Could you please kindly let me know what s is please. Is s equals to the
value of lambda?
Many thanks
Lini
--
View this message in context:
Is this homework?
runif() is one way to generate random numbers, but there are others
depending on the distribution desired. And of course the fifth number
is deterministic.
Sarah
On Thu, Jul 19, 2012 at 7:12 AM, arunkumar akpbond...@gmail.com wrote:
hi
My inputs is min=(10,10,10,10,10)
My data:
head(cbind(x,y,z))
x y z
[1,] -78.1444 -60.4424 -10.09
[2,] -78.1444 -58.4424 -10.26
[3,] -78.1444 -56.4424 -10.45
[4,] -78.1444 -54.4424 -10.64
[5,] -76.1444 -60.4424 -10.19
[6,] -76.1444 -58.4424 -10.34
tris - deldir(x, y)
triangs - triang.list(tris)
head(tris$delsgs)
x1 y1 x2 y2
I am trying to triangulate a point set as follows:
head(cbind(x,y))
x y
[1,] -78.1444 -60.4424
[2,] -78.1444 -58.4424
[3,] -78.1444 -56.4424
[4,] -78.1444 -54.4424
[5,] -76.1444 -60.4424
[6,] -76.1444 -58.4424
length(x)
[1] 5000
tri - tri.mesh(x, y)
Fehler in tri.mesh(x, y) : error in trmesh
Hello,
I didn't give enough information when I sent an query before, so I'm trying
again with a more detailed explanation:
In this data set, each patient has a different number of measured variables
(they represent tumors, so some people had 2 tumors, some had 5, etc). The
problem I have is that
Sorry, I just found this to be a common problem of tri.mesh:
I had to jitter one of my first three coords in the point set:
x[2] - x[2] + 0.01
Though, that does not seem to sound clean. Is there a better way?
2012/7/19 Erdal Karaca erdal.karaca...@gmail.com
I am trying to triangulate a point
Hello,
I have a double matrix that I want to represent in a line chart. Although I
have seen some examples I still don't manage to get it. My data is this (a
double matrix called mymatrix) :
Blogs Wikis Redes Etiq. SPC LMS
Menor de 30 57.14 28.57 14.29 28.57 57.14 28.57
de 31 a
HI,
Possibly check.names=FALSE issue.
Try this:
dat1-read.table(text=
2.5a 3.6b 7.1c 7.9d
100 3 4 2 3
200 3.1 4 3 3
300 2.2 3.3 2 4
,sep=,header=TRUE)
dat1
#You can get rid of those X by either using check.names=FALSE while reading the
data
#with
Dear Chris,
many thanks! This is just what I had in mind! (namely the 'sapply' solution).
Thank you and best regards!
Zdenek
-Original Message-
From: Chris Campbell [mailto:ccampb...@mango-solutions.com]
Sent: Thursday, July 19, 2012 4:11 PM
To: Skála, Zdeněk (INCOMA GfK)
Cc:
Hi
Thanks for reply
after usinginstall.packages(hdf5)
I get error
{Installing package(s) into ‘/home/uday/R/x86_64-pc-linux-gnu-library/2.14’
(as ‘lib’ is unspecified)
trying URL
'http://cran.revolutionanalytics.com/src/contrib/hdf5_1.6.9.tar.gz'
Content type 'application/x-gzip' length 50870
On Thu, Jul 19, 2012 at 8:02 AM, Jan van der Laan rh...@eoos.dds.nl wrote:
Johan,
Your 'list' and 'array doubling' code can be written much more efficient.
The following function is faster than your g and easier to read:
g2 - function(dotot) {
v - list()
for (i in seq_len(dotot)) {
No you are not correct. The kde function estimates the density of 1 to 6
dimensions. To visualize, this try plotting the density of just the data
(using density() instead of kde):
plot(density(elevation$data))
rug(elevation$data)
The elevations are plotted along the x-axis and their density is
Dear Jessica,
You might try par(las=2) to rotate the tick labels to be perpendicular to the
axes.
I hope this helps,
John
John Fox
Sen. William McMaster Prof. of Social Statistics
Department of Sociology
McMaster University
Hamilton, Ontario,
The error message is here: configure: error: Can't find HDF5
The hdf5 package is an interface to the HDF5 library, so you need to
install hdf5 and hdf5-devel through your package manager (yum or
apt-get or whatever you use to install things on your linux distro).
Or you can get it straight from
Hadley et. al:
Indeed. And using a loop is a poor way to do it anyway.
v - as.list(rep(FALSE,dotot))
is way faster.
-- Bert
On Thu, Jul 19, 2012 at 8:50 AM, Hadley Wickham had...@rice.edu wrote:
On Thu, Jul 19, 2012 at 8:02 AM, Jan van der Laan rh...@eoos.dds.nl wrote:
Johan,
Your 'list'
When i make Boxplots with a lot of boxes, the names of them
get only written down every second column.
Since they aren't in any way ordered, you don't see anymore
to what they belong.
Jessica,
Another possibility if the names are long is to use abbreviated factor levels.
The labels appear
also rep.int()
system.time(for (i in 1:1000) x - rep.int(FALSE, 10))
user system elapsed
0.290.020.29
system.time(for (i in 1:1000) x - rep(FALSE, 10))
user system elapsed
1.960.082.05
On Thu, Jul 19, 2012 at 9:11 AM, Bert Gunter gunter.ber...@gene.com
Preallocation of lists does speed things up. The following shows
time quadratic in size when there is no preallocation and linear
growth when there is, for size in the c. 10^4 to 10^6 region:
f - function(n, preallocate) { v - if(preallocate)vector(list,n) else
list() ; for(i in seq_len(n))
Those restrictions you have given do not define a unique distribution!
so you need to think better
about what you need. For instance, if you want a uniform distribution
between min and max with n=5
independent observations from that, but conditional upon sum=total.
For that, you could use
Thanks, Bill, but:
system.time(z1 -f(n=1e5,pre=TRUE))
user system elapsed
0.320.000.32
system.time(z2 - as.list(seq_len(1e5)))
user system elapsed
0 0 0
identical(z1,z2)
[1] TRUE
So the point is not to use an R level loop at all.
-- Bert
On Thu, Jul 19,
On Thu, Jul 19, 2012 at 04:12:07AM -0700, arunkumar wrote:
hi
My inputs is min=(10,10,10,10,10) and max=(100,100,100,100,100)
total = 300
i have to generate 5 numbers between min and max and those numbers should
sum upto total
Hi.
If we subtract the minimum from each number, then we
You should include your data using dput(mymatrix) to make things easier
structure(list(Label = structure(c(4L, 1L, 2L, 3L), .Label = c(de 31 a 40,
de 41 a 50, Mayor de 51, Menor de 30), class = factor),
Blogs = c(57.14, 63.83, 72.64, 62.07), Wikis = c(28.57, 61.7,
70.75, 58.62), Redes
Hello,
Try the following.
d - read.csv(text=
Patient, Cycle, Variable1, Variable2
A, 1, 4, 5
A, 2, 3, 3
A, 3, 4, NA
B, 1, 6, 6
B, 2, NA, 6
C, 1, 6, 5
C, 3, 2, 2
, header=TRUE)
d
compl - lapply(split(d, d$Patient), function(x) if(all(diff(x$Cycle) ==
1)) x)
holes - lapply(split(d, d$Patient),
On Thu, Jul 19, 2012 at 9:21 AM, William Dunlap wdun...@tibco.com wrote:
Preallocation of lists does speed things up. The following shows
time quadratic in size when there is no preallocation and linear
growth when there is, for size in the c. 10^4 to 10^6 region:
Interesting, thanks! I wish
On Thu, Jul 19, 2012 at 11:11 AM, Bert Gunter gunter.ber...@gene.com wrote:
Hadley et. al:
Indeed. And using a loop is a poor way to do it anyway.
v - as.list(rep(FALSE,dotot))
is way faster.
-- Bert
Its not entirely clear to me what we are supposed to conclude about this.
I can
Hi fellow R users,
I am desperately hoping there is an easy way to do this in R.
Say I have three functions:
f(x) = x^2
f(y) = 2y^2
f(z) = 3z^2
constrained such that x+y+z=c (let c=1 for simplicity).
I want to find the values of x,y,z that will minimize f(x) + f(y) + f(z).
I know I can use
Hi Linh,
Here is an approach:
f - function(v) {
v - v/sum(v)
(v[1]^2) + (2 * v[2]^2) + (3*v[3]^2)
}
(res - optim(c(.6, .3, .1), f))
res$par/sum(res$par)
This is a downright lazy way to implement the constraint. The main
idea is to combine all three functions into one function that takes
On 07/19/2012 05:50 PM, Hadley Wickham wrote:
On Thu, Jul 19, 2012 at 8:02 AM, Jan van der Laan rh...@eoos.dds.nl wrote:
The following function is faster than your g and easier to read:
g2 - function(dotot) {
v - list()
for (i in seq_len(dotot)) {
v[[i]] - FALSE
}
}
Except that
On 2012-07-19 05:56, penguins wrote:
Thanks William, that works fantastically!
I had a quick play with my data and have realised a potential problem in
that if an individual ends the series at home it records an additional
trip-no when one wasnt made. I was wondering whether you could think of
Can someone please give me an example of how to enter starting values for a
GRM (IRT) model using the ltm package?
The instructions from the ltm manual are below, but when I create either a
list of the values or a matrix I get the error, start.val not of proper
type. I can find no
On Thu, Jul 19, 2012 at 10:24:17AM -0700, Linh Tran wrote:
Hi fellow R users,
I am desperately hoping there is an easy way to do this in R.
Say I have three functions:
f(x) = x^2
f(y) = 2y^2
f(z) = 3z^2
constrained such that x+y+z=c (let c=1 for simplicity).
I want to find the
So I have the following data frame and I want to know how I can remove all
NA values from each string, and also
remove all | values from the START of the string. So they should
something like auto|insurance or auto|insurance|quote
one = data.frame(keyword=c(|auto, NA|auto|insurance|quote,
On 2012-07-19 07:10, Bart Ferket wrote:
Dear professor Harrell,
I probably have the same problem as Haleh Ghaem Maralani.
I am using the rms package and the rcspline.plot function to assess the
relation of a continuous predictor to the log hazard function.
I would like to use the adj
On 07/19/2012 06:11 PM, Bert Gunter wrote:
Hadley et. al:
Indeed. And using a loop is a poor way to do it anyway.
v - as.list(rep(FALSE,dotot))
is way faster.
-- Bert
I agree that not using a loop is much faster, but I assume that the
original question is about the situation where the
There are a couple of ambiguities in your request, but this should get
you started:
one$keyword - gsub(NA\\|, , one$keyword)
one$keyword - gsub(^\\|, , one$keyword)
one
keyword
1 auto
2 auto|insurance|quote
3 auto|insurance
4insurance
5
I couldn't find anything in the chron or timeDate packages, and a good search
yielded rounding to the nearest half hour, which I don't want.
The data:
structure(list(Date = structure(c(1209625080, 1209641460, 1209652500,
1209676800, 1209682860, 1209692100, 1209706980, 1209722580, 1209726300,
Hello,
Try this:
#Generate 4 random numbers.
set.seed(1)
rnorm(4,60)
#[1] 59.37355 60.18364 59.16437 61.59528
sum(rnorm(4,60))
#[1] 240.7348
fifthnumber-300-240.7348
fifthnumber
#[1] 59.2652
A.K.
- Original Message -
From: arunkumar akpbond...@gmail.com
To:
Hi,
Recently I have installed R version 2.14.1, after installation I am trying
to install some packages and I get error message. even I tried
install.packages(Rcmdr) but still I am unable to fix this problem.
I would be very grateful if somebody can help me to fix this problem.
When the length of the end result is not known, doubling the length of
the list is also much faster than increasing the size of the list with
single items.
f - function(n, preallocate) {
v - if(preallocate) vector(list,n) else list() ;
for(i in seq_len(n)) {
v[[i]] - i
}
Jan:
Point taken.
However, if possible, as Bill Dunlap indicated, it still may make
sense to create an oversized list first and then populate what you
need of it with your loop.
Note that a lot of this can be finessed with lapplyand friends
anyway, letting R worry about the details of creating
Thank you very much, Michael!
That was exactly the hint I needed. I ended up using try as below, in case
anyone happens to read this later on...
Works perfectly fine.
So yes, it _is_ amazing, what one can do with R ;-)
Regards,
Berry
internet - try(read.table(http://www...;), silent=T )
I would like to use a A priori mean comparison/orthogonal contrast test on my
data.
Im new to using R and I would like to know if its possible to perform this
test in R and how can be done.
/Anna
--
View this message in context:
Hi all,
I'm attempting to gap-fill a dataset, replacing the missing values with
each month's day or night median value.
The problem is that my code results in some, but not all the NA's being
replaced and I cannot figure out how this is possible. When I look at
the individual line's where
Hi, everyone.
I have some questions about quantile regression in R.
I am running an additive quantile regression first for a complete matrix and
then with some selected rows.
I am doing the following:
datos -read.table(Regresion multiple.txt,header=T)
Fit-rqss(datos$campings
Hi,
Try this:
one = data.frame(keyword=c(|auto, NA|auto|insurance|quote,
NA|auto|insurance,
NA|insurance, NA|auto|insurance, NA))
onenew-data.frame(keyword=gsub((NA){0,1}\\|,,one$keyword))
onenew1-data.frame(keyword=gsub((NA){0,1},,onenew$keyword))
onenew1
- Forwarded Message -
From: arun smartpink...@yahoo.com
To: Abraham Mathew abmathe...@gmail.com
Cc: R help r-help@r-project.org
Sent: Thursday, July 19, 2012 3:58 PM
Subject: Re: [R] Removing values from a string
Hi,
Try this:
one = data.frame(keyword=c(|auto,
On Thu, Jul 19, 2012 at 04:12:07AM -0700, arunkumar wrote:
hi
My inputs is min=(10,10,10,10,10) and max=(100,100,100,100,100)
total = 300
i have to generate 5 numbers between min and max and those numbers should
sum upto total
Hi.
Try the following.
while (1) {
x - 10 +
Dear list,
I am using the np package. With the npindex function I estimate a
semiparametric single index model using the method of Klein-Spady.
P(Z=1|X) = G(X’b)
I don’t have any problems to calculated the fitted values and standard
errors X’b:
bw = npindexbw(xdat=x, ydat=y_bi,
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