hi
I need some help in dbWriteTable.
I'm not able to insert the rows in the table if the column order are not
same in the database and in the dataframe which i'm inserting. Also facing
issue if the table is already created externally and inserting it thru
dbWrite.
is there some way that we can
Hi All,
I get the message failed with message Dimension out of range when using
cuhre in package R2Cuba. Does anyone know what this mean? Or would I need
to email the package author?
The funny thing is it does give a result and comparing it to
"adaptIntegrate" in package cubature, the two numbe
On Dec 1, 2011, at 1:13 PM, lglew wrote:
Hi R-users,
I'm trying to produce decompositions of a multiple time-series,
grouped by a
factor (called "area"). I'm modifying the code in the STLperArea
function of
package ndvits, as this function only plots produces stl plots, it
does not
retu
On Dec 1, 2011, at 10:50 PM, Worik R wrote:
I really would like to be able to read about this in a document but I
cannot find my way around the documentation properly
Given the code...
M <- matrix(runif(5*20), nrow=20)
colnames(M) <- c('a', 'b', 'c', 'd', 'e')
ind <- c(1,2,3,4)
dep <- 5
I ca
On Dec 1, 2011, at 11:11 PM, Bert Gunter wrote:
strsplit is certainly an alternative, but your approach is
unnecessarily complicated and inefficient. Do this, instead:
sapply(strsplit(x,"/"),length)-1
Definitely more compact that the regex alternates I came up with, but
one of these still
Inefficient, maybe, but what you suggest does not work if a string
starts or ends with a slash.
On Thu, Dec 1, 2011 at 11:11 PM, Bert Gunter wrote:
> strsplit is certainly an alternative, but your approach is
> unnecessarily complicated and inefficient. Do this, instead:
>
> sapply(strsplit(x,"/"
Tena koe Lara
If I understand your question correctly, I use the colorspace package for that
sort of thing, but you could also use the built-in colour palettes such as
rainbow and topo.colors
HTH
Peter Alspach
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-he
strsplit is certainly an alternative, but your approach is
unnecessarily complicated and inefficient. Do this, instead:
sapply(strsplit(x,"/"),length)-1
Cheers,
Bert
On Thu, Dec 1, 2011 at 7:44 PM, Florent D. wrote:
> Resending my code, not sure why the linebreaks got eaten:
>
>> x <- data.fram
Similarly, this might work:
unlist(lapply(Version1_, `[`,"First"))
Michael
On Thu, Dec 1, 2011 at 9:41 PM, Sarah Goslee wrote:
> How about:
>
> lapply(Version1_, subset, subset=c(TRUE, FALSE))
> or sapply() depending on what you want the result to look like.
>
> Thanks for the reproducible exam
I have this. I have modified your input structure to be a matrix. I
think it is generally recommended to use matrices over data.frames
when your data allows it, i.e., when you only have one type of data,
here character(). Matrices are easier to work with.
x <- matrix( c('BMW', '', '', 'Mercedes
I really would like to be able to read about this in a document but I
cannot find my way around the documentation properly
Given the code...
M <- matrix(runif(5*20), nrow=20)
colnames(M) <- c('a', 'b', 'c', 'd', 'e')
ind <- c(1,2,3,4)
dep <- 5
I can then do...
l2 <- lm(M[,dep]~M[,ind]) ## Clearl
Resending my code, not sure why the linebreaks got eaten:
> x <- data.frame(Col1 = c("abc/def", "ghi/jkl/mno"), stringsAsFactors = FALSE)
> count.slashes <- function(string)sum(unlist(strsplit(string, NULL)) == "/")
> within(x, Col2 <- vapply(Col1, count.slashes, 1))
Col1 Col2
1 abc/d
I used within and vapply:
x <- data.frame(Col1 = c("abc/def", "ghi/jkl/mno"), stringsAsFactors = FALSE)
count.slashes <- function(string)sum(unlist(strsplit(string, NULL)) ==
"/")within(x, Col2 <- vapply(Col1, count.slashes, 1))
Col1 Col21 abc/def 12 ghi/jkl/mno 2
On Thu, Dec 1
How about:
lapply(Version1_, subset, subset=c(TRUE, FALSE))
or sapply() depending on what you want the result to look like.
Thanks for the reproducible example.
Sarah
On Thu, Dec 1, 2011 at 5:17 PM, LCOG1 wrote:
> Hi everyone,
> I looked around the list for a while but couldn't find a soluti
Hi everyone,
I looked around the list for a while but couldn't find a solution to my
problem. I am storing some results to a simulation in a list and for each
element i have two separate vectors(is that what they are called, correct my
vocab if necessary). See below
Version1_<-list()
for(i in
Thanks Michael! Yeah, that dealt with one of the problems.
I still get the following error message:
Error in cat(list(...), file, sep, fill, labels, append) :
argument 1 (type 'list') cannot be handled by 'cat'
I know that this has something to do with writing the names of the output
file, an
Hi everyone,
I'm having problems with plotting my data. I have a set of positions with
different attributes and I'm wondering if I can plot it, as x,y plot, with
gradient colours according to a 3rd factor. Is it possible to show gradient
segments between postions and not one-coloured segment betwe
Sorry -- I meant to write 'row names,' but the question specifically is
where those unlikely numbers come from. So I guess it comes down to
why, when 'bar' is the first item, the row name is assigned '2' rather
than '1' .
On 12/1/11 6:26 PM, Sarah Goslee wrote:
Those are row *names*, not r
Sorry for that, and thanks Gabor,
I could have sworn that it wouldn't.
Gabor Grothendieck wrote
>
> On Thu, Dec 1, 2011 at 7:13 PM, B77S wrote:
>> # need zoo to use rollapply()
>>
>> # your data (I called df)
>> df <- structure(list(a = 1:2, b = 2:3, c = c(5L, 9L), d = c(9L, 6L),
>>
On Thu, Dec 1, 2011 at 7:13 PM, B77S wrote:
> # need zoo to use rollapply()
>
> # your data (I called df)
> df <- structure(list(a = 1:2, b = 2:3, c = c(5L, 9L), d = c(9L, 6L),
> e = c(1L, 5L), f = c(4, 7)), .Names = c("a", "b", "c", "d",
> "e", "f"), class = "data.frame", row.names = c(NA, -2L
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On 2011-12-01 08:54, emorway wrote:
Hello,
A bit of fairly simple code, yet I don't seem to be able to manipulate it
quite as much as I would like:
1) It would be nice if the objects appearing in the legend were aligned,
and by aligned I mean the boxes are centered over the lines. Do I need t
Thank you so much for your help.
The data I am using is the last file called l_yx.RData at this link (the
second file contains the plots from earlier):
http://scientia.crescat.net/static/ben/
Seems like the warning went away with pmin(x,1) but now the OR is over
15k. If I multiple my x's by 1000
# need zoo to use rollapply()
# your data (I called df)
df <- structure(list(a = 1:2, b = 2:3, c = c(5L, 9L), d = c(9L, 6L),
e = c(1L, 5L), f = c(4, 7)), .Names = c("a", "b", "c", "d",
"e", "f"), class = "data.frame", row.names = c(NA, -2L))
# transpose and make a zoo object
df2 <- zoo(t(df
On Dec 1, 2011, at 23:43 , Ben quant wrote:
> I'm not proposing this as a permanent solution, just investigating the
> warning. I zeroed out the three outliers and received no warning. Can someone
> tell me why I am getting no warning now?
It's easier to explain why you got the warning before.
Those are row *names*, not row *numbers*. It's just that if you don't specify,
numbers are assigned by default when creating a data frame.
Your rbind() statements are implicitly creating a data frame, so the likely
information is in ?data.frame:
check.names: logical. If ‘TRUE’ then the names of
Someone is bound to know a better way, but...
subset(unlist(Version1_), subset=names(unlist(Version1_))=="First")
LCOG1 wrote
>
> Hi everyone,
>I looked around the list for a while but couldn't find a solution to my
> problem. I am storing some results to a simulation in a list and fo
"Not that it really matters, but"
Can someone explain how the row numbers get assigned in the following
sequence? It looks like something funky happens when rbind() coerces
'bar' into a dataframe.
In either sequence of rbind below, once you get past the first two rows,
the row numbers count nor
I'm not proposing this as a permanent solution, just investigating the
warning. I zeroed out the three outliers and received no warning. Can
someone tell me why I am getting no warning now?
I did this 3 times to get rid of the 3 outliers:
mx_dims = arrayInd(which.max(l_yx), dim(l_yx))
l_yx[mx_dims
Oops! Please ignore my last post. I mistakenly gave you different data I
was testing with. This is the correct data:
Here you go:
> attach(as.data.frame(l_yx))
> range(x[y==0])
[1] 0.0 14.66518
> range(x[y==1])
[1] 0.0 13.49791
How do I know what is acceptable?
Also, here are the scr
Here you go:
> attach(as.data.frame(l_yx))
> range(x[y==1])
[1] -22500.746.
> range(x[y==0])
[1] -10076.5303653.0228
How do I know what is acceptable?
Also, here are the screen shots of my data that I tried to send earlier
(two screen shots, two pages):
http://scientia.crescat.n
I ran the following code:
And I run into problems with the last line of code (when it says
hn<-..). I keep getting an error code: Error in distsamp(~hab ~ 1,
peldist, keyfun = "halfnorm", output = "density", :
object 'A' not found
I would appreciate any and all help.
rm(list=ls(all
Hi, I have a problem with mixed effect model (multilevel).
My model that is written in following formulation using the lme4
package and Zelig package:
mylogit<- lmer(OVERFLOW ~ ALTEZZA + INTENSITA + ( 1 | CODICE), family=
binomial(link = "probit"),data = dati)
OVERFLOW can be 0 or 1 and represen
Hello,
I have the following problem. After a PACS upgrade in our hospital, the dynamic
3D images acquired after contrast media injection with a 3T Philips MRI are all
saved in a single 4D file instead of a series of 2D images containing multiple
3D images over time. I used to convert this serie
Thanks Michael, I'll keep that in mind if I want to do anything more
complicated.
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Sent from the R help mailing list ar
On Thu, Dec 1, 2011 at 11:54 AM, emorway wrote:
> Hello,
>
> A bit of fairly simple code, yet I don't seem to be able to manipulate it
> quite as much as I would like:
>
> 1) It would be nice if the objects appearing in the legend were aligned,
> and by aligned I mean the boxes are centered over
A new version of RSQLite has been uploaded to CRAN. Details on the enhancements
and fixes are below. Please direct questions to the R-sig-db mailing list.
Version 0.11.0
- Enhance type detection in sqliteDataType (dbDataType). The storage
mode of a data.frame column is now used as part of the
With binomial/binary responses (0|1) running MCMCglmm with
family="multinomial" terminates with
Error in if (nJ < 1) { : missing value where TRUE/FALSE needed
with family="categorical" there are no errors
I have not looked in the code, do I need format the responses
TRUE/FALSE , or is this just
On Dec 1, 2011, at 21:32 , Ben quant wrote:
> Thank you for the feedback, but my data looks fine to me. Please tell me if
> I'm not understanding.
Hum, then maybe it really is a case of a transition region being short relative
to the range of your data. Notice that the warning is just that: a
NAME
ID
a
b
c
d
1
Control_1
probe~B01R01C01
381
213
345
653
2
Control_2
probe~B01R01C02
574
629
563
783
3
Control_1
probe~B01R09C01
673
511
521
967
4
Control_3
probe~B01R09C02
53
809
999
50
5
MM0289~RFU:11810.15
probe~B29R13C06
681
34
115
587
6
MM0289~RFU:9238.41
probe~B29R13C05
784
443
20
I think your attachment got scrubbed so I can't verify this, but i
think the difficulty is that the "inner" apply returns a whole set of
orders, instead of just one and that throws arima off: would this
work?
getOrder <- function(x) ar(na.omit(x), method = "mle")$order
all.equal(apply(TSX, 2, get
Dear all, I have a data frame (DF) in the following format:
NAME
ID
a
b
c
d
1
Control_1
probe~B01R01C01
381
213
345
653
2
Control_2
probe~B01R01C02
574
629
563
783
3
Control_1
probe~B01R09C01
673
511
521
967
4
Control_3
probe~B01R09C02
53
809
999
50
5
MM0289~RFU:11810.15
probe~B29R13
Thank you for the feedback, but my data looks fine to me. Please tell me if
I'm not understanding.
I followed your instructions and here is a sample of the first 500 values :
(info on 'd' is below that)
> d <- as.data.frame(l_yx)
> x = with(d, y[order(x)])
> x[1:500] # I have 1's and 0's dispers
On Thu, Dec 1, 2011 at 3:11 PM, " Majid " wrote:
> Hi,
> yes, It is homework,
Then ask your TA/instructor for help.
>
> These are 2 command:
> first for generating data:
> (1:10)
> that output is 1 2 310
> ok ?
> second is :
> a1<-c( 1:10)
> what is the output ?I didnot see any thing.
Exact
On Dec 1, 2011, at 1:00 PM, William Dunlap wrote:
Terry will correct me if I'm wrong, but I don't think the
answer to this question is specific to the coxph function.
It depends on our interpretation of the questioner's intent. My answer
was predicated on the assumption that the phrase "bas
Hi,
On Thu, Dec 1, 2011 at 2:43 PM, " Majid " wrote:
> Dear Sarah.
> Thanks so much,Really I am new in this software,I am wrking to learn the
> software.
First, you should always send your replies to the list. That way
information can help others,
and more people are available to provide advice.
saschaview wrote on 12/01/2011 12:30:18 PM:
> Hello
>
> I have a data frame, x, holding 5 persons answering the question which
> cars they have used:
>
> # the data frame
> x <- as.data.frame(
>matrix(
> c('BMW', '', '',
>'Mercedes', 'VW', '',
>'Skoda', 'VW', 'BMW',
>
On 01/12/2011 1:12 PM, jcano wrote:
Hi all
How can I change the limits (xlim or ylim) in a plot that has been already
created?
You can't, if you're using classic R graphics. They use an "ink on
paper" model of graphics. If you want to change what you've drawn, you
get a new piece of paper.
I think part of your problem is the loop:
you probably mean
for(i in unique(area))
Michael
On Thu, Dec 1, 2011 at 1:13 PM, lglew wrote:
> Hi R-users,
>
> I'm trying to produce decompositions of a multiple time-series, grouped by a
> factor (called "area"). I'm modifying the code in the STLperA
Thanks David and Rubén!
@David: indeed 15 betas I forgot the interaction terms, thanks for correcting!
@Rubén: the re-parameterize would be done within nls()? how to do
this practically with including the factor predictor?
and yes, we can solve within each group for Y=0 getting
0=b0+b1*X |
jcano wrote on 12/01/2011 12:12:03 PM:
> Hi all
>
> How can I change the limits (xlim or ylim) in a plot that has been
already
> created?
>
> For example, consider this naive example
> curve(dbeta(x,2,4))
> curve(dbeta(x,8,13),add=T,col=2)
>
> When adding the second curve, it goes off the ori
Hi,
With some help I learned how to use plsr(Y~x, 2, data=my) function in R (and
build "my" from vector "Y" and matrix "x").
But still I have question about results interpretation. In the end I want to
construct prediction function in form:
Y=a1x1+a2x2
But I do not understand how to do it. Docu
Hi,
I have a multilevel situation where subjects are nested within clinics, and
each subject has multiple measurements.
For simplicity, suppose there 4 clinics, 3 subjects per clinic, and each
subject has 3 repeated measures.
Outcome is continuous. I am trying to implement this model with lme
Hi R-users,
I'm trying to produce decompositions of a multiple time-series, grouped by a
factor (called "area"). I'm modifying the code in the STLperArea function of
package ndvits, as this function only plots produces stl plots, it does not
return the underlying data.
I want to extract the tren
Hi all
How can I change the limits (xlim or ylim) in a plot that has been already
created?
For example, consider this naive example
curve(dbeta(x,2,4))
curve(dbeta(x,8,13),add=T,col=2)
When adding the second curve, it goes off the original limits computed by R
for the first graph, which are rou
On Dec 1, 2011, at 18:54 , Ben quant wrote:
> Sorry if this is a duplicate: This is a re-post because the pdf's mentioned
> below did not go through.
Still not there. Sometimes it's because your mailer doesn't label them with the
appropriate mime-type (e.g. as application/octet-stream, which is
I repeat myself:
Any more automated solution will depend on whether your data has
rownames or not. [...] create a plain text representation of R data
using the dput() command.
Another way that might make more sense is to cbind() the data you need
later on before the split and then it will be carr
Hello
Hello
I have a data frame, x, holding 5 persons answering the question which
cars they have used:
# the data frame
x <- as.data.frame(
matrix(
c('BMW', '', '',
'Mercedes', 'VW', '',
'Skoda', 'VW', 'BMW',
'', '', '',
'VW', 'Skoda', ''
),
ncol=3,
Just a heads up -- I don't think your code will work with an actual
.xls(x) file, only .txt, .csv, etc (aka, plain text files). I may be
wrong about that, but if you actually need to work with Excel files
directly you will need an additional package.
Michael
On Thu, Dec 1, 2011 at 9:10 AM, AOLear
## It's not a data frame -- it's just a vector.
> x
[1] "abc/def" "ghi/jkl/mno"
> gsub("[^/]","",x)
[1] "/" "//"
> nchar(gsub("[^/]","",x))
[1] 1 2
>
?gsub
?nchar
-- Bert
On Thu, Dec 1, 2011 at 8:32 AM, Douglas Esneault
wrote:
> I am new to R but am experienced SAS user and I was hoping t
Terry will correct me if I'm wrong, but I don't think the
answer to this question is specific to the coxph function.
For all the [well-written] formula-based modelling functions
(essentially, those that call model.frame and model.matrix to interpret
the formula) the option "contrasts" controls how
Sorry if this is a duplicate: This is a re-post because the pdf's mentioned
below did not go through.
Hello,
I'm new'ish to R, and very new to glm. I've read a lot about my issue:
Warning message:
glm.fit: fitted probabilities numerically 0 or 1 occurred
...including:
http://tolstoy.newcastle.e
On Dec 1, 2011, at 12:00 PM, Andreas Schlicker wrote:
Hi all,
I'm trying to fit a Cox regression model with two factor variables
but have some problems with the interpretation of the results.
Considering the following model, where var1 and var2 can assume
value 0 and 1:
coxph(Surv(time
Thanks again for your help.
I've been able to add several packages, bigmemory seems to be the only one
to fail and it
fails on isinf.
Is there a way I can download the code and change it to include a ininf
function or definition?
I'm using the GNU compiler; should I have been using the SUN St
Dear R users,
I have been trying to obtain the MLE of the following model
state 0: y_t = 2 + 0.5 * y_{t-1} + e_t
state 1: y_t = 0.5 + 0.9 * y_{t-1} + e_t
where e_t ~ iidN(0,1)
transition probability between states is 0.2
I've generated some fake data and tried to estimate the parameters using
I am new to R but am experienced SAS user and I was hoping to get some help on
counting the occurrences of a character within a string at a row level.
My dataframe, x, is structured as below:
Col1
abc/def
ghi/jkl/mno
I found this code on the board but it counts all occurrences of "/" in the
d
The references are here: http://manning.com/kabacoff/excerpt_references.pdf
(they will be included on the next printing too, got omitted by mistake)
regards,
Pablo
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http
Hi all,
I'm trying to fit a Cox regression model with two factor variables but
have some problems with the interpretation of the results. Considering
the following model, where var1 and var2 can assume value 0 and 1:
coxph(Surv(time, cens) ~ factor(var1) * factor(var2), data=temp)
What is t
Hi,
On Thu, Dec 1, 2011 at 10:35 AM, " Majid " wrote:
> Hi.
> Can you please answer to my questions about R ?
> 1.how can I write command for vector ?
>
> for exaple in this sample :
> I have this :
> a1 <- c (1:10)
> now how can I put in the vector ?
I'm afraid I don't understand your question.
Dear List,
I was wondering if any of you worked on an R package which has many branches on
a repository i.e. SVN. Is it recommended to branch an R package source tree
based on a specific project? I know it depends on project but it would be great
to
hear opinions from R community.
Best,
M
On Thu, Dec 1, 2011 at 10:02 AM, Berry, David I. wrote:
> Hi List
>
> Apologies if this isn't the correct place for this query (I've tried a search
> of the mail archives but not had much joy).
>
> I'm running R (2.14.0) on a Mac (OSX v 10.5.8, 2.66GHz, 4GB memory) and am
> having a few perform
On 12/01/2011 05:25 PM, Dustin Fife wrote:
What is your goal? I have used and like mice pretty well, but using
mice + sem to try to address missingness seems like more work than
using FIML in OpenMx or lavaan to try to address it. Is there a
reason you want to use the sem package or a reason y
Hello,
A bit of fairly simple code, yet I don't seem to be able to manipulate it
quite as much as I would like:
1) It would be nice if the objects appearing in the legend were aligned,
and by aligned I mean the boxes are centered over the lines. Do I need to
adjust the use of "NA" in the code b
On Dec 1, 2011, at 10:49 AM, lincoln wrote:
Thanks.
So, suppose for one specific year (first year over 10) the
percentage of
successes deriving from 100 trials with 38 successes (and 62
failures), its
value would be 38/100=0.38.
I could calculate its confidence intervals this way:
succes
> Hi, I want to do a 2 sample hotelling's test but i can't
> figure out how. When i type T2.test it says there is no such
> test and when i tried library(rrcov) it says there is no such
> program.
Have you installed rrcov using install.packages?
S Ellison**
On Dec 1, 2011, at 8:24 AM, René Mayer wrote:
Dear All,
I have a binomial response with one continuous predictor (d) and one
factor (g) (8 levels dummy-coded).
glm(resp~d*g, data, family=binomial)
Y=b0+b1*X1+b2*X2 ... b7*X7
Dear Dr Mayer;
I think it might be a bit more complex than tha
Thanks for all the response,
I manage to write a small function to complete what I want based on the
wonderful helps.
iround <- function(x, interval){
## Round numbers to desired interval
##
## Args:
## x:Numeric vector to be rounded
## interval: The interval th
You might be interested in package "bit".
---
Jeff NewmillerThe . . Go Live...
DCN:Basics: ##.#. ##.#. Live Go...
Live: OO#.. Dea
The first version of the package was created by re-writing the main program in
the original Fortran as C, and calls other Fortran subroutines that were mostly
untouched, so dynamic memory allocation can be done. Later versions have most
of the Fortran code translated/re-written in C. Currently
On 01/12/2011 10:55 AM, Michael Kao wrote:
Dear R users/helpers,
I am wondering is there an existing function in which you can round
numbers to a set of values. I know you can use 5 * round(x/5) for
rounding to the nearest 5 or so, but what if the interval size is not
constant.
For example:
##
Got it.
Thank you so much for the help!
Min Yi MD, PhD
Department of Surgical Oncology - Unit 1484
1400 Pressler Street, #FCT17.6061
University of Texas M.D. Anderson Cancer Center
P.O. Box 301402
Houston, TX 77230-1402
Phone: 713-563-1874
Fax: 713-792-4689
From: Frank Harrell [via R] [mailto:
Hi.
Can you please answer to my questions about R ?
1.how can I write command for vector ?
for exaple in this sample :
I have this :
a1 <- c (1:10)
now how can I put in the vector ?
bye for now,
Thanks a lot.
Majid.
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_
>
> What is your goal? I have used and like mice pretty well, but using
> mice + sem to try to address missingness seems like more work than
> using FIML in OpenMx or lavaan to try to address it. Is there a
> reason you want to use the sem package or a reason you do not want to
> use the others?
On 12/01/2011 02:46 PM, jim holtman wrote:
If we assume that you are just convert 8 bits, here is one way with a
table lookup. If more than 8 bits, just partition the data and
repeat. This sets up a mapping table one time for the lookup. Does
1M in 0.3 seconds on my computer; is this fast enou
Thanks.
So, suppose for one specific year (first year over 10) the percentage of
successes deriving from 100 trials with 38 successes (and 62 failures), its
value would be 38/100=0.38.
I could calculate its confidence intervals this way:
> success<-38
> total<-100
> prop.test(success,total,p=0.5,a
Hi, I want to do a 2 sample hotelling's test but i can't figure out how. When
i type T2.test it says there is no such test and when i tried library(rrcov)
it says there is no such program. Cheers.
--
View this message in context:
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Hi List
Apologies if this isn't the correct place for this query (I've tried a search
of the mail archives but not had much joy).
I'm running R (2.14.0) on a Mac (OSX v 10.5.8, 2.66GHz, 4GB memory) and am
having a few performance issues with reading data in from a Postres database
(using RPos
Thank you very much, that does exactly what I want it to! :)
Aodhán
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Sent from the R help mailing list archive at Nabb
?findInterval
.. would get you the endpoints and then you could determine which is nearer.
Of course in your "example", everything would get "rounded" to 1.
-- Bert
On Thu, Dec 1, 2011 at 7:55 AM, Michael Kao wrote:
> Dear R users/helpers,
>
> I am wondering is there an existing function in whi
Hazard() is not implemented except for parametric survival models.
You are not calling nomogram() correctly; in rms the plotting step is
separated from the nomogram computations.
To plot cumulative event rates do something like:
mort10 <- function(lp) 1 - surv(10,lp)
and tell nomogram about mor
Dear R users/helpers,
I am wondering is there an existing function in which you can round
numbers to a set of values. I know you can use 5 * round(x/5) for
rounding to the nearest 5 or so, but what if the interval size is not
constant.
For example:
## Not run
test <- rnorm(100)
round(test, c
I'd also suggest you read circle 2 of the "R inferno" (just google it)
which has some helpful tips on how to deal with these sorts of
problems.
Also, did you know that matrices can have column names and that
rbind() preserves them? E.g.,
m <- matrix(1:6, 3); colnames(m) <- letters[1:2]
print(m)
This really isn't the appropriate forum for most of your questions: I'd suggest
you work through the Wikipedia article on quantiles regression and direct
follow up to stats.stackexchange.com.
As to the R question: use the predict() function.
Michael
On Dec 1, 2011, at 8:12 AM, narendarreddy
Hi everybody,
Thanks for checking my code, Hans, it help to see where my initial mistake was.
I am sorry i assumed that there was a minimization problem.
In short i had 2 "wavy" lines (left and right) that didn't intersect and lots
of straight parallel lines that intersect the first 2 lines.
Assuming a logistic model, for each group solve for d at Y=0, or solve for d at
p=0.5, where d is your continuous predictor, Y is the logit score, and p is the
probability of success in the binomial model. After that you can get the
standard error of the inflection point by Taylor series (aka de
Good idea. I'll give it a try. Thanks!
On Thu, Dec 1, 2011 at 6:18 AM, John Fox wrote:
> Dear Dustin,
>
> > -Original Message-
> > From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> > project.org] On Behalf Of Dustin Fife
> > Sent: November-30-11 9:51 PM
> > To: r-help@r-proj
See this earlier post for SVG logos:
http://tolstoy.newcastle.edu.au/R/e12/devel/10/10/0112.html
Using Image Magick, do something like
convert logo.svg logo.eps
On Thu, 2011-12-01 at 10:56 +0700, Ben Madin wrote:
> G'day all,
>
> Sorry if this message has been posted before, but searching fo
Hi,
Alternatively you could have a look at the function "terrain" in the
raster package, it can calculate the slope for you using different
algorithms, not sure if the one below is included though.
For typical spatial requests like this, you could also use the
mailing-list r-sig-geo.
Cheer
On Dec 1, 2011, at 2:06 AM, yuying shi wrote:
Dear all,
I am doing an ordinal data simulation. I have a question
regarding the cut off values between categories. Suppose I have
three categories, if I do regression, there should be two cut off
values. I find some simulation code for the
First, dataframes can be much slower than matrices, for example, if
you are changing/accessing values a lot. I would suggest that you use
a matrix since is seems that all your values are numeric. Allocate a
large empty matrix to start (hopefully as large as you need). If you
exceed this, you hav
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