On Sat, 2011-07-09 at 09:35 -0700, VG wrote:
Hi,
I was wondering if someone can tell me what is the difference between
strata argument (function adonis in vegan package) and
using random effects in PERMANOVA+ add-on package to PRIMER6 when doing
permutational MANOVA-s? Is the way
Prof Brian Ripley ripley at stats.ox.ac.uk writes:
On Mon, 11 Jul 2011, Tomaz wrote:
I upgraded R on windows xp from 2.12.2 to 2.13.1 and now I can not
process Rnw files with windows cp1250 encoding. Sweave complains:
Which is of course not an ISO Standard encoding. One way out is to
Any NA values or values outside the support region of your distribution?
UWe
On 11.07.2011 23:21, Peter Maclean wrote:
I am trying to estimate a gamma function using real data and I am getting the
following error messages.
When I set a lower limit; the error message is L-BFGS-B needs finite
On 12.07.2011 09:01, Tomaz wrote:
Prof Brian Ripleyripleyat stats.ox.ac.uk writes:
On Mon, 11 Jul 2011, Tomaz wrote:
I upgraded R on windows xp from 2.12.2 to 2.13.1 and now I can not
process Rnw files with windows cp1250 encoding. Sweave complains:
Which is of course not an ISO
flags - c(rep(1, length(patient_indices)), rep(0,
length(control_indices)))
# dataset is a data.frame and param the parameter to be analysed:
data1 - dataset[,param][c(patient_indices, control_indices)]
fit1 - glm(flags ~ data1, family = binomial)
new.data- seq(0, 300, 10)
new.p
On 12/07/11 09:04, Joseph Park wrote:
Greetings,
I would like to estimate a spectral coherence between
two timeseries. The stats : spectrum() returns a coh matrix
which estimates coherence (squared).
A basic test which from which i expect near-zero coherence:
x =
Hi,
Am 11.07.2011 22:57, schrieb Lyndon Estes:
ctch[ctch$threshold == 3.5, ]
# [1] threshold val tpfptnfntpr
fpr tnr fnr
#0 rows (or 0-length row.names)
this is the very effective FAQ 7.31 trap.
Dear all,
I would like to use the apply or a similar function belonging to this
family, but applying for each column (or row) but let say for each q
columns. For example I would like to apply a function FUN for the first q
columns of matrix X then for q+1:2*q and so on. If I do apply (X, 2, FUN)
Hi all,
I first create a matrix/data frame called d2 if another matrix
accomplishes some restrictions dacc2
da2-da1[colSums(dacc2)9,]
da2-da2[(da2[,13]=24),]
write.csv(da2, file =paste('hggi', i,'.csv',sep = ''))
The thing is if finally da2 cannot get/passs the filters, it cannot writte a
csv
Dear all,
I would like to obtain the Brier score prediction error at different times t
for an extended Cox model. Previously I have used the 'pec' function
(pec{pec}) to obtain prediction error curves for standard Cox PH models but
now I have data in the counting process format (I have a
Dear R user,
After I imported data (csv format) in R, I called it out. But it is in
non-numeric format.
Then using as.numeric function.
However, the output is really awful !
PE[1,90:99]
V90 V91 V92 V93 V94
V95 V96
Dear R community,
cloudnumbers.com provides researchers and companies with the resources
to perform high performance calculations in the cloud. As
cloudnumbers.com's community manager I may invite you to register and
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Hi,
You don't provide us with a reproducible example, so I can't provide you with
actual code. But two approaches come to mind:
1. Create da2 with one row and n columns, then change the appropriate elements,
if any, based on your conditions.
2. Do the conditional parts, then check to see whether
Jessica,
This would be easier to solve if you gave us more information, like str(PE).
However, my guess is that your data somewhere has a nonnumeric value in that
column, so the entire column is being imported as factor. It's not
really awful -
R is converting those factor values to their
On 07/12/2011 06:38 PM, Jessica Lam wrote:
Dear R user,
After I imported data (csv format) in R, I called it out. But it is in
non-numeric format.
Then using as.numeric function.
However, the output is really awful !
PE[1,90:99]
V90 V91 V92 V93
Have you noted you sent your message to the R-help list only and forgot
to include the original poster? You also forgot to cite the original
question (and any other former parts of the thread as far as there was
any). Please do so when sending messages to a mailing list such as R-help.
gfcoeffs - function(s, n) {
t - seq(-n,n,1) ## assuming 2*n+1 taps
return ( exp(-(t^2/(2*s^2)))/sqrt(2*pi*s^2) )
}
2011/6/29 Martin Wilkes m.wil...@worc.ac.uk:
I want to filter my time series with a low-pass filter using a Gaussian
smoothing function defined as:
w(t) = (2πσ^2)^0.5
On Tue, Jul 12, 2011 at 7:15 AM, Markus Schmidberger schmi...@in.tum.de wrote:
This is only a selection of our top features. To get more information
check out our web-page (http://www.cloudnumbers.com/) or follow our blog
about cloud computing, HPC and HPC applications (with R):
Hello all,
Could someone help me with the time zones in understandable practical way?
I got completely stucked with this.
Have googled for a while and read the manuals, but without solutions...
---
When data imported from Excel
Hello all,
Could someone help me with the time zones in understandable practical way?
I got completely stucked with this.
Have googled for a while and read the manuals, but without solutions...
---
When data imported from
Uwe Ligges ligges at statistik.tu-dortmund.de writes:
On 12.07.2011 09:01, Tomaz wrote:
Prof Brian Ripleyripleyat stats.ox.ac.uk writes:
On Mon, 11 Jul 2011, Tomaz wrote:
I upgraded R on windows xp from 2.12.2 to 2.13.1 and now I can not
process Rnw files with windows cp1250
Thanks you! I should have realized that without explicitly engaging
some form of averaging (which raises a windowing question) that the
coh is always 1.
On 7/12/2011 4:48 AM, Rolf Turner wrote:
On 12/07/11 09:04, Joseph Park wrote:
Greetings,
I would like to
Július 7-től 14-ig irodán kívül vagyok, és az emailjeimet nem érem el.
Sürgős esetben kérem forduljon Kárpáti Edithez (karpati.e...@gyemszi.hu).
Üdvözlettel,
Mihalicza Péter
I will be out of the office from 7 July till 14 July with no access to my
emails.
In urgent cases please contact Ms.
Thanks both of you for help!
This is my conclusion based on Rolf's and David's suggestion: (x=data.frame)
By for-looping over a data.frame and deleting certain rows with
x=x[-i,]
its better to collect all rows which need to be deleted in a vector and do
one final delete step:
collecting:
Dear all,
I am new to programming in R.
I deal with microarray data,which is a data frame object type. I need to carry
out a few statistical procedures on this, one of them being the pearson
corelation. I need to do this between each row which is a gene. So the desired
result is a square
Dear R-Users,
I run a MC-Simulation using the the packages foreach and doMC on a
PowerMac with 24 cores. There are roughly a hundred parametersets and I
parallelized the program in a way, that each core computes one of these
parametersets completely.
The problem ist, that some parametersets take
Also note that the statistical method you are using does not seem in line
with decision theory, and you are assuming that the threshold actually
exists. It is seldom the case that the relationship of a predictor with the
response is flat on at least one side of the threshold. A smooth prediction
Hi Jim,
by dropping them down it gives 1 day less than it should do, on all timezone
notations CEST and CET.
start
[1] 2002-09-04 CEST 2000-07-27 CEST 2003-01-04 CET 2001-06-29 CEST
2005-01-12 CET 2000-05-28 CEST 2002-06-01 CEST 2000-06-02 CEST
2000-02-27 CET 2000-09-29 CEST 2003-10-22
On Tue, Jul 12, 2011 at 6:58 AM, B Laura gm.spam2...@gmail.com wrote:
Hello all,
Could someone help me with the time zones in understandable practical way?
I got completely stucked with this.
Have googled for a while and read the manuals, but without solutions...
It may be helpful to make sure that, in the dialog that pops up when saving a
spreadsheet to CSV, the option Save cell content as shown is checked - that
would leave numbers as numbers, not wrapping them in . That has helped me at
least in a similar situation!
Rgds,
Rainer
On Tuesday 12 July
peter_petersen henning.bumann at gmail.com writes:
I run a MC-Simulation using the the packages foreach and doMC on a
PowerMac with 24 cores. There are roughly a hundred parametersets and I
parallelized the program in a way, that each core computes one of these
parametersets completely.
On 11-07-12 6:42 AM, Tomaz wrote:
Uwe Liggesliggesat statistik.tu-dortmund.de writes:
On 12.07.2011 09:01, Tomaz wrote:
Prof Brian Ripleyripleyat stats.ox.ac.uk writes:
On Mon, 11 Jul 2011, Tomaz wrote:
I upgraded R on windows xp from 2.12.2 to 2.13.1 and now I can not
process
matplot(timestamp,xymatrix,type='l')
where timestamp is a vector filled with POSIXct objects and xymatrix is
a numeric 2x2 matrix plots but the horizontal axis labels are raw
unformatted timestamps.
I would like to format these in any of the available codes for strftime,
for instance
Dear Gabor
http://rwiki.sciviews.org/doku.php?id=tips:data-io:ms_windowss=excel
doesnt describe handling dates with daylight saving time issues.
R classes Date can remove time and timezone, however calculating days
difference between two manipulated variables same problem appear if handling
Hi,
I want to apply a function to a matrix, taking the columns 3 by 3. I could use
a for loop:
for(i in 1:3){ # here I assume my data matrix has 9 columns
j = i*3
set = my.data[,c(j-2,j-1,j)]
my.function(set)
}
which looks cumbersome and possibly slow. I was hoping there is some function
in
Hi all,
Is there any code to run fixed effects Tobit models in the style of Honore
(1992) in R?
(The original Honore article is here:
http://www.jstor.org/sici?sici=0012-9682%28199205%2960%3A3%3C533%3ATLALSE%3E2.0.CO%3B2-2)
Cheers
David
[[alternative HTML version deleted]]
Hello,
Are there any built in or user defined functions for printing the date created
or date updated for a given file? Ideally a function that works across
operating systems.
Thanks!
Scott Chamberlain
[[alternative HTML version deleted]]
On Jul 12, 2011, at 7:27 AM, Mitra, Sumona wrote:
Dear all,
I am new to programming in R.
You see to think there is a ++ operation in R. That is not so.
I deal with microarray data,which is a data frame object type. I
need to carry out a few statistical procedures on this, one of them
Hi,
file.info()
does that.
Cheers
Am 12.07.2011 15:29, schrieb Scott Chamberlain:
Hello,
Are there any built in or user defined functions for printing the date
created or date updated for a given file? Ideally a function that works
across operating systems.
Thanks!
Scott
On 12 July 2011 12:27, Mitra, Sumona sumona.mi...@kcl.ac.uk wrote:
Dear all,
I am new to programming in R.
You sure are ;-)
I deal with microarray data,which is a data frame object type. I need to
carry out a few statistical procedures on this, one of them being the
pearson corelation.
Hi Frederico. I would keep the data as it is, create two small vectors
referring to the ranges and use a mapply (as a sapply but with multiple
variables) for the function. Hope the example below is helpful, although as
usual someone out there will have a better solution for it.
dta - c()
for
Eik,
Thanks very much!
Scott
On Tuesday, July 12, 2011 at 8:34 AM, Eik Vettorazzi wrote:
Hi,
file.info (http://file.info)()
does that.
Cheers
Am 12.07.2011 15:29, schrieb Scott Chamberlain:
Hello,
Are there any built in or user defined functions for printing the date
Hi,
I am trying to do a lasso regression using the lars package with the following
data (see attached):
FastestTime
WinPercentage
PlacePercentage
ShowPercentage
BreakAverage
FinishAverage
Time7Average
Time3Average
Finish
116.90
0.14
0.14
0.29
4.43
3.29
117.56
117.77
5.00
I just realised that:
apply(matrix(1:dim(my.data)[2], nrow =3), 2,
function(x){my.function(my.data[,x])})
is the simplest possible method.
bw
F
On 12 Jul 2011, at 14:44, Filipe Leme Botelho wrote:
Hi Frederico. I would keep the data as it is, create two small vectors
referring to the
Hi,
I am trying to do a lasso regression using the lars package with the following
data:
FastestTime
WinPercentage
PlacePercentage
ShowPercentage
BreakAverage
FinishAverage
Time7Average
Time3Average
Finish
116.90
0.14
0.14
0.29
4.43
3.29
117.56
117.77
5.00
116.23
0.29
0.43
* David Winsemius qjvafrz...@pbzpnfg.arg [2011-07-11 18:16:25 -0400]:
What is the point of offering this code?
To illustrate what I was talking about (code is its own specification).
I hoped that there was already a package doing that (and more in that
direction).
It seems to be doing what
On Jul 12, 2011, at 9:53 AM, Heiman, Thomas J. wrote:
Hi,
I am trying to do a lasso regression using the lars package with the
following data (see attached):
Nothing attached. (And now you have also sent an exact duplicate.)
snipped failed attempt to include data inline that was
Hi,
Hopefully I got the formatting down.. I am trying to do a lasso regression
using the lars package with the following data (the data files is in .csv
format):
V1 V2 V3 V4
V5 V6 V7
I have two data frames:
str(ysmd)
'data.frame': 8325 obs. of 6 variables:
$ X.stock : Factor w/ 8325 levels A,AA,AA-,..: 2702
6547 4118 7664 7587 6350 3341 5640 5107 7589 ...
$ market.cap : num -1.00 2.97e+10 3.54e+08 3.46e+08 -1.00
...
$
How can perform logarithmic binning in the scatterplot? I could only take the
log of the variables and plot them, but I am sure that is not the way. I
have a very huge data, and would want to plot those high density
scatterplots and code then with different colors for the bins/density.
--
View
If you switch directly to the multicore package you can use the
mclapply() function. There, check for the parameter mc.preschedule=T /
F. You can use this parameter to improve the load balancing.
I do not know a parameter to tune foreach with this parameter.
Best
Markus
Am Dienstag, den
Many Thanks¡¡¡
I will try this night, I have read this I think could help me.
I´m conscient the question was badly formulated now, I will try to explain
better next time¡¡¡
On a side note: apply always accesses the function you use at least once. If
the input is a dataframe without any rows but
Hi all,
I have this information on a file ht.txt, imagine it is a data frame without
labels:
1 1 1 8 1 1 6 4 1 3 1 3 3
And on other table called pru.txt I have sequences similar this
4 1 1 8 1 1 6 4 1 3 1 3 3
1 6 1 8 1 1 6 4 1 3 1 3 3
1 1 1 8 1 1 6 4 1 3 1 3 3
6 6 6 8 1 1 6 4 1 3 1 3 3
I want
Hi,
I am trying to do a lasso regression using the lars package with the following
data (see attached):
FastestTime
WinPercentage
PlacePercentage
ShowPercentage
BreakAverage
FinishAverage
Time7Average
Time3Average
Finish
116.90
0.14
0.14
0.29
4.43
3.29
117.56
117.77
5.00
Hi,
I have a data frame of about 700 000 rows which look something like this:
DateTemperature Category
2007102 16 A
2007102 17 B
2007102 18 C
but need it to be:
Date TemperatureA TemperatureB TemperatureC
Dear All,
I have a collections of spatial data. I have to analyze pairs of these
point patterns to test their spatial interaction. I was moving towards
the cross K Ripley's function. The problem, however, are the following:
1) What is the best way to get a single real value that represents the
On Jul 12, 2011, at 10:12 AM, Sam Steingold wrote:
I have two data frames:
str(ysmd)
'data.frame': 8325 obs. of 6 variables:
$ X.stock : Factor w/ 8325 levels
A,AA,AA-,..: 2702 6547 4118 7664 7587 6350 3341 5640 5107
7589 ...
$ market.cap : num
On Tue, 2011-07-12 at 10:12 -0400, Heiman, Thomas J. wrote:
Hi,
Hopefully I got the formatting down.. I am trying to do a lasso regression
using the lars package with the following data (the data files is in .csv
format):
V1 V2 V3 V4
Hi Trying,
It would be helpful if you provided reproducible examples. It would
also be polite to sign a name so that we have something by which to
address you.
On Tue, Jul 12, 2011 at 8:00 AM, Trying To learn again
tryingtolearnag...@gmail.com wrote:
Hi all,
I have this information on a file
On Jul 12, 2011, at 10:12 AM, Heiman, Thomas J. wrote:
Hi,
Hopefully I got the formatting down.. I am trying to do a lasso
regression using the lars package with the following data (the data
files is in .csv format):
V1 V2 V3 V4
On Jul 12, 2011, at 8:42 AM, anglor wrote:
Hi,
I have a data frame of about 700 000 rows which look something like
this:
DateTemperature Category
2007102 16 A
2007102 17 B
2007102 18 C
but need it to be:
Date
Hi,
(i) As David suggested, please use `dput` to provide examples of data!
(ii) The nut of your problem is that you are giving lars an object
that it is not expecting. It wants a *matrix* for its `x` variable, as
you'll see in the help for ?lars.
So, as long as this expression:
R is.numeric(x)
Hi:
Try the cast() function in the reshape package. Using d as the name of
your data frame,
library(reshape)
cast(d, Date ~ Category, value = 'Temperature')
Date A B C
1 2007102 16 17 18
HTH,
Dennis
On Tue, Jul 12, 2011 at 5:42 AM, anglor angelica.ekens...@dpes.gu.se wrote:
Hi,
I
On Tue, Jul 12, 2011 at 8:57 AM, B Laura gm.spam2...@gmail.com wrote:
Dear Gabor
http://rwiki.sciviews.org/doku.php?id=tips:data-io:ms_windowss=excel
doesnt describe handling dates with daylight saving time issues.
Two references were given and its discussed in the R News article. It
was
On Jul 11, 2011, at 9:16 PM, Steve Parker wrote:
Hi there,
I am using the RODBC library to connect to an Empress database.
I have installed the ODBC data source with the server DNs number and port,
and named the source Trawl.
It is the odbcDriverConnect that seems to have the problem, and I
If you don't need POSIXt types, as Gabor says don't use them. However, there
are good reasons to use them sometimes, and the most workable solution I have
found is to set your default timezone in R to a non-DST timezone before you
convert from character to POSIXct. This is dependent on your OS
Hello,
In my lab we use a four parameter logistic fit model for our ELISA data
(absorbance values). We are currently testing the use of different solvents
and need to find a way to add a correlation value (such as an R squared or
something similar) so we can test different solvents in making
Thanks Peter, Ted!
Best, Anirban
On Tue, Jul 12, 2011 at 4:54 AM, Ted Harding ted.hard...@wlandres.net wrote:
On 11-Jul-11 07:55:44, Anirban Mukherjee wrote:
Hi all,
I wanted to mark the estimation sample: mark what rows (observations)
are deleted by lm due to missingness. For eg, from the
I've written out codes for one particular file, and now I want to generate
the same kind of graphs and files for the rest of similar data files.
When I plugged in these codes, R produced only one plot for the file
eight, and it states my error(see below) I have edited and checked my
codes so many
On 2011-07-12 07:03, Sam Steingold wrote:
[snip]
the totally unnecessary semi-colons)
then why are they accepted?
optional syntax elements suck...
They're accepted because they *can* be useful (multiple
statements on one line).
Is there *any* language that can *not* be abused?
Peter
Hi Susie,
At a guess, there are no non-missing arguments to min or max.
But no, we can't help you. You haven't provided a minimal reproducible
example, and without knowing anything about your data it is impossible
for the list to offer any constructive suggestions.
The posting guide offers
On 07/12/2011 09:53 AM, Heiman, Thomas J. wrote:
## define x and y
x= x-crs[,9]#predictor variables
y= y-crs[1:8,] #response variable
This cannot be correct. The response variable is a vector, while the
predictor variables form a matrix. You have the response variable
consisting
This is just posed out of curiosity, (not as a criticism per se). But what is
the functional role of the argument na.rm inside the mean() function? If there
are missing values, mean() will always return an NA as in the example below.
But, is there ever a purpose in computing a mean only to
In SQL, the default is to ignore NULL (equivalent to NA in R).
However, it can be dangerous to fail to verify how much data was actually used
in an aggregation, so the logic behind the default na.rm setting may be one of
encouraging the user to take responsibility for missing data.
On 12/07/2011 12:26 PM, Doran, Harold wrote:
This is just posed out of curiosity, (not as a criticism per se). But what is
the functional role of the argument na.rm inside the mean() function? If there
are missing values, mean() will always return an NA as in the example below.
But, is there
Hi Harold,
Many (most?) of the statistics function have a similar argument. I
suspect it is sort of to warn the user---you have to be explicit about
it rather than the program just silently removing or ignoring values
that would not work in the function called. I can think of one
example where
Dear all,
I have a problem and it is very difficult for me to get a code.
I am reading a file(attached with this mail) using the code-
df=read.table(summary.txt,fill=T,sep=,colClasses = character,header=T)
and dataframe df is like this-
V1V2 CaseA CaseC CaseG CaseT new
10
Hi Vikas,
Here is one way:
df - read.table(summary.txt, header = TRUE)
str(df)
df[, total] - rowSums(df[, 3:6])
df[, 3:6] - apply(df[, 3:6], 2, function(x) x / df[, total] * df[,
new] * 2)
head(df)
V1V2 CaseA CaseC CaseG CaseT new total
1 10 135344109 0 024 0 12
## Hello.. I have asked a similar question, but this is not fixed as before.
## I am running the following using Ubuntu OS:
R version 2.13.1 (2011-07-08)
Copyright (C) 2011 The R Foundation for Statistical Computing
ISBN 3-900051-07-0
Platform: x86_64-pc-linux-gnu (64-bit)
## when I do this:
On 7/7/2011 3:23 PM, elephann wrote:
Hi everyone!
I have a data frame with 1112 time series and I am going to randomly
sampling r samples for z times to compose different portfolio size(r
securities portfolio). As for r=2 and z=1,that's:
z=1
A=seq(1:1112)
x1=sample(A,z,replace =TRUE)
I've written out codes for one particular file, and now I want to generate
the same kind of graphs and files for the rest of similar data files.
For example, a file 8.csv would look like such:
enc_callee inout o_duration type
A out 342 de
B in 234 de
C
Probability - function(N, f, m, b, x, t) {
#N is the number of lymph nodes
#f is the fraction of Dendritic cells (in the correct node) that have
the
antigen
#m is the number of time steps
#b is the starting position (somewhere in the node or somewhere in the
gap
Hi,
I'm currently trying to calculate local Getis-Ord Gi* statistics for a
169x315 cell matrix of temperature values, below is the code I currently
have (diffc is the data vector I am removing NaN values from, and I am
moving said values to diffD; -999 represents NaN values; id contains ID
values
It works well. Thanks so much.
--
View this message in context:
http://r.789695.n4.nabble.com/as-numeric-tp3661739p3662671.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
Hello, I'm new to this list. Sorry if my question or parts of it already came
up before.
For my research in geostatistics, I am working with large sets of data in R
(basically large matrices containing discrete x and y coordinates and a
value for a certain parameter). These sets are obtained by
Dear Susie,
See inline for some suggestions, but generally, I think you would
benefit from breaking this down into smaller pieces. The error you
are getting indicates the problem has to do with the plotting, but
that will be trickier to isolate while also dealing with reading in
data, looping,
Dear list, I'm wondering if anyone can help me calculate the deviance
of either a zeroinfl or hurdle model from package pscl?
Even if someone could point me to the correct formula for calculating
the deviance, I could do the rest on my own.
I am trying to calculate a pseudo-R-squared measure
On 12-Jul-11 17:18:26, mousy0815 wrote:
Probability - function(N, f, m, b, x, t) {
#N is the number of lymph nodes
#f is the fraction of Dendritic cells (in the correct node) that
have
the
antigen
#m is the number of time steps
#b is the starting position (somewhere
Merging two posts (data and questions); see inline below.
On 7/11/2011 7:55 PM, Sigrid wrote:
Thank you, Dennis.
This is my regenerated dput codes. They should be correct as I closed off R
and re-ran them based on the dput output.
NB, this is the test dataset used later
Hello,
I have a sample file:
chr22 100 150 125 21 0.145 +
chr22 200 300 212 13 0.05+
chr22 345 365 351 12 0.09+
chr22 500 750 510 15 0.10+
chr22 500 750 642 9 0.02+
chr22 800
Hello:
R has an extensive Help system. Please learn to use it.
?histogram
?help
Also see the online manual tutorial An Introduction to R
-- Bert
On Tue, Jul 12, 2011 at 12:41 PM, a217 aj...@case.edu wrote:
Hello,
I have a sample file:
chr22 100 150 125 21 0.145 +
On Jul 12, 2011, at 3:41 PM, a217 wrote:
Hello,
I have a sample file:
chr22 100 150 125 21 0.145 +
chr22 200 300 212 13 0.05+
chr22 345 365 351 12 0.09+
chr22 500 750 510 15 0.10+
chr22 500 750
Carson Farmer carson.farmer at gmail.com writes:
Dear list, I'm wondering if anyone can help me calculate the deviance
of either a zeroinfl or hurdle model from package pscl?
Even if someone could point me to the correct formula for calculating
the deviance, I could do the rest on my own.
when do I need to use which()?
a - c(1,2,3,4,5,6)
a
[1] 1 2 3 4 5 6
a[a==4]
[1] 4
a[which(a==4)]
[1] 4
which(a==4)
[1] 4
a[which(a2)]
[1] 3 4 5 6
a[a2]
[1] 3 4 5 6
seems unnecessary...
--
Sam Steingold (http://sds.podval.org/) on CentOS release 5.6 (Final) X
11.0.60900031
To answer your questions:
Yes, yes, and probably no. You will have to pick up any introductory manual
of R where questions 1 and 2 will be discussed.
For 1: you index x as in x[452,682]. For 2: there are ways to write (and
avoid) loops in R (e.g. for or while loops). Often avoidance is
Well ...
which(a==4)^2
??
-- Bert
On Tue, Jul 12, 2011 at 1:17 PM, Sam Steingold s...@gnu.org wrote:
when do I need to use which()?
a - c(1,2,3,4,5,6)
a
[1] 1 2 3 4 5 6
a[a==4]
[1] 4
a[which(a==4)]
[1] 4
which(a==4)
[1] 4
a[which(a2)]
[1] 3 4 5 6
a[a2]
[1] 3 4 5 6
seems
On Jul 12, 2011, at 4:17 PM, Sam Steingold wrote:
when do I need to use which()?
a - c(1,2,3,4,5,6)
a
[1] 1 2 3 4 5 6
a[a==4]
[1] 4
a[which(a==4)]
[1] 4
which(a==4)
[1] 4
a[which(a2)]
[1] 3 4 5 6
a[a2]
[1] 3 4 5 6
seems unnecessary...
It is unnecessary when `a` is a toy case
On Tue, Jul 12, 2011 at 1:17 PM, Sam Steingold s...@gnu.org wrote:
when do I need to use which()?
See ?which
For examples, try:
example(which)
a - c(1,2,3,4,5,6)
a
[1] 1 2 3 4 5 6
a[a==4]
[1] 4
a[which(a==4)]
[1] 4
which(a==4)
[1] 4
a[which(a2)]
[1] 3 4 5 6
a[a2]
[1] 3 4 5 6
Here is a worked example. Can you point out to me where in temp rmean is
stored? Thanks.
Tom
library(survival)
library(ISwR)
dat.s - Surv(melanom$days,melanom$status==1)
fit - survfit(dat.s~1)
plot(fit)
summary(fit)
Call: survfit(formula = dat.s ~ 1)
time n.risk n.event survival
Hello all,
I am using AddHealth data to fit a cure, aka split population model using nltm.
I am not sure how to account for the complex survey design - does anyone have
any suggestions? Any help would be greatly appreciated!
Sincerely,
Sam
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