On Wed, 17 Oct 2012, swertie wrote:
Hello!
When I am analyzing proportion data, I usually apply logistic regression
using a glm model with binomial family. For example:
m - glm( cbind(not realized, realized) ~ v1 + v2 , family=binomial)
However, sometimes I don't have the number of cases
No. You need to test more carefully.
a - factor(c(1,3,5))
b - factor(c(5,7))
c(a,b)
[1] 1 2 3 1 2
lev - sort(unique(f - c(a,b)))
f - factor(f,levels=lev)
f
[1] 1 2 3 1 2
Levels: 1 2 3
## but
unlist(list(a,b),use.names=FALSE)
[1] 1 3 5 5 7
Levels: 1 3 5 7
However, Is level 5 in 'a' the
Thank you for clarifying, Dr. Gunter. My bad.
Regards,
Jorge.-
On Thu, Oct 18, 2012 at 5:21 PM, Bert Gunter gunter.ber...@gene.com wrote:
No. You need to test more carefully.
a - factor(c(1,3,5))
b - factor(c(5,7))
c(a,b)
[1] 1 2 3 1 2
lev - sort(unique(f - c(a,b)))
f -
I need to load ICS calendar event files in R. Is there a package with a
function that does this already?
Many thanks!
[[alternative HTML version deleted]]
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PLEASE
HI,
Hi,
Try this:
str(PriceList)
'data.frame': 161 obs. of 3 variables:
$ Price : num 0 8.18 8.27 10.42 10.5 ...
$ Size : int 664640 440 407 180 690 851 190 480 720 74 ...
$ bandNum:List of 161
PriceList1-within(PriceList,{bandNum-as.numeric(bandNum)})
str(PriceList1)
HI,
I tried the code with unsorted ac_names column and found to be
working. So, couldn't identify exactly the problem. If you can provide
a subset of your dataset using ?dput(), then it would be much helpful.
set.seed(1)
HI,
Try this:
seq1 = seq(0, 100, by = 5)
seq2 = seq(100, 1000, by = 100)
Bands = c(seq1, seq2)
DF1 - data.frame(matrix(ncol=2,nrow=200))
colnames(DF1)- c(Price, Size)
set.seed(1)
DF1$Price - sample(1:1000, 200, replace=F)
set.seed(300)
DF1$Size - sample(1:1000, 200, replace=F)
how R implement qnorm()
I wonder anyone knows the mathematical process that R calculated the
quantile?
The reason I asked is soly by curiosity. I know the probability of a normal
distribution is calculated through integrate the Gaussian function, which
can be implemented easily (see code),
Thanks for your response as well. It turned out that this was already a date
typed file but I'll file this away for future reference when I come up
against a string of characters
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Hi arun,
Thanks,
I used your code, however, there's error message Error: could not find
function count.
Thanks,
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Hi Duncan,
Thanks for taking the time to look at my problem. I apologize for sending
in my data in an improper format, I have included the dput output for my
data at the bottom of this message. I guess I was not clear on the exact
problem I was trying to solve. I want to keep each trellis
Dr. Varadhan:
Perfect. Package Rsolnp is exactly what I was looking for. Thank you very
much.
r/
Greg
-Original Message-
From: Ravi Varadhan ravi.varad...@jhu.edu
Sent: Oct 17, 2012 10:24 AM
To: 'boyla...@earthlink.net' boyla...@earthlink.net
Cc: r-help@r-project.org
I want to iterate the program until it converge? can anybody help me? I just
write it manual, but i need to generalize it
so that simulation can stop if the difference between two iteration is less
than 0.0001.
The following are the program :
rm(list=ls(all=T))
x=rnorm(50,2,0.7)
Hi there,
a very simple question: I would like to add nice labels to
SpatialLinesDataFrame which have been plotted using the 'plot' function.
Those lines are bathymetry and I would like to have labels for the different
depths. The function 'contour' does it nicely but does not directly read
The Quantitative Decision Strategies group at Janssen Research Development,
Johnson Johnson, is looking for a candidate to represent QDS in Beijing,
China in the subsidiary company of Xian-Janssen Pharmacetical Ltd. The basic
requirements for this candidate are 1) 3+ years experience in
On 18/10/2012 00:16, Sheng Liu wrote:
how R implement qnorm()
I wonder anyone knows the mathematical process that R calculated the
quantile?
It's on the help page!
'For qnorm, the code is a C translation of
Wichura, M. J. (1988) Algorithm AS 241: The Percentage Points of the
Normal
Dear Romita,
It is not quite clear to me what exactly you want the graph to look like. What
do you mean by plots of the unadjusted and adjusted models?
In general though, it sounds to me as if the graph you have in mind is rather
complex. It may be possible to accomplish this with the forest()
Hello,
Try the following.
dat - read.table(text=
ID
1001
1001
1001
1122
1122
1122
1421
1421
1789
1789
, header = TRUE)
r - rle(dat$ID)
dat$SID - rep(seq_along(r$lengths), r$lengths) + 1000
Hope this helps,
Rui Barradas
Em 18-10-2012 02:16, york8866 escreveu:
Hi all,
I have a dataset with
Thank you very much for replies and the nice explanation about variance
stabilization. I heard about the arcsin transformation, but some recent
papers were very critical about it (i.e., Warton Hui, 2011), so that I
would better try another way. I will have a look at beta regression.
Best,
V.
On 18-10-2012, at 09:55, namrata mohapatra wrote:
Hello Sir/Madam
I want to reverse the colour distribution . I want the lowest value of error
to be in blue and highest in red .
Please provide a reproducible example in the form of R code.
Berend
With Regards
Namrata
On 10/18/2012 06:55 PM, namrata mohapatra wrote:
Hello Sir/Madam
I want to reverse the colour distribution . I want the lowest value of error to
be in blue and highest in red .
With Regards
Namrata Mohapatra
Hi Namrata,
Let's see, where is my crystal ball?
Ah, right. I see that you are
On Oct 18, 2012, at 09:55 , Prof Brian Ripley wrote:
R is a bit confusing as it requires inverse error function (X =
- sqrt(2)* erf-1 (2*P)), while R doesn't have a build in one. The InvErf
function most people use is through qnorm( InvErf=function(x)
I think you are wrong about 'most
Hello,
Yes it is.
Hope this helps,
Rui Barradas
P.S. You must be much more specific in your question.
1. Provide a data example using ?dput.
dput( head(MyData, 50) ) # paste the output of this in a post.
2. What type of model are you talking about? What exactly does
prediction mean?
3. At
Hi,
I agree with Ista that you need the second option. So long as you don't want to
look at interactions between condition and item it shouldn't be a problem that
you don't have the same items in each condition. However, it may be difficult
to find a main effect of condition that couldn't also
As Rui said, this is a pretty vague question... Nonetheless, if you are using
lm() for example you should use the newdata argument in predict(). Have a look
at ?predict.lm for example.
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On
Hello,
I am trying to run the demoscript for the package siar. However, using
the example script and example data supplied I cannot get 1) the model to
run unless I specify iterations =500 and burnin = 500 and 2) if the model
does work without crashing R then the function siarhistograms does
Thanks for the reply!!
Inga
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Dear all!
I'm quite new to R and at this moment a little bit lost in trying to make a
function work with a for loop
that has two variables (i and j) and multiple if statements..
I hope that someone of you could help me. I would very appreciate it!
What do I want to do? I am analysing the data
I generated maps with the function symbols (graphics). These are basic
maps generated with :
symbols(x,y,circles=myvariable)
where x et y are spatial coordinates corresponding to replicates of
myvariable.
I would associate legend to this kind of maps, is it possible?
Regards,
Marion.
--
Hi List,
I want to show three matrices in a 3D plot, one horizontal, and the other
two as vertical slices.
Ideally this would use the persp or drape.plot functions.
I have found a means of assigning one matrix to the lower xy plane (like a
carpet), thus:
mat_2D -
Hello,
I am trying to use this fix for the convergence problem in polr, but I don't
seem to get the change of code right. I redefine the function polr by the
lines
tjb wrote
if(missing(start)) {
# try something that should always work -tjb
u - as.integer(table(y))
u -
Without commenting on the method itself, shouldn't you be using both x and y
twice within each loop instead of x three times and y only once?
Please provide a snippet of your data by including the output of
dput(head(data,30)) for example.
-Original Message-
From:
Dear all,
I am using the nlsLM function to fit a Lorentzian function to my experimental
data.
The LM algorithm should allow to specify limits, but the upper limit appears
not to work as expected in my code.
The parameter 'w', which is peak width at half maximuim always hits the upper
limit if
Hi,
May be this also works:
a - factor(c(1,3,5))
b - factor(c(5,7))
f1-as.numeric(c(as.character(a),as.character(b)))
lev-as.numeric(c(levels(a),setdiff(levels(b),levels(a
f2-factor(f1,levels=lev)
f2
#[1] 1 3 5 5 7
#Levels: 1 3 5 7
a1-factor(5:1,levels=1:9)
b1-factor(9:1,levels=1:9)
If you use ifelse() you don't need the loops at all since ifelse() is
vectorized.
But I don't have any idea why you need two loops: do you really need to
compare all pairs of observations?
Also, the logic in your if statements is not the same as the logic in your
problem description.
Sarah
On
Hi,
I downloaded a dataset from UCI repositories named Bag of Words:
http://archive.ics.uci.edu/ml/machine-learning-databases/bag-of-words/readme.txt
The dataset is in a text file with the following structure:
---
docID1 wordID1 count
docID1 wordID2 count
docID1 wordID3 count
docID1 wordID4
- Why are you making a matrix for lookzone if it is ever only one number?
- changing a variable used for iteration is bad practice (input in this case)
though you aren't changing the rows i guess.
- There are too many x and too few y in those ifs.
Plus this is shorter:
Better would be to use interval censored data. Create your data set so that
you have
(time1, time2) pairs, each of which describes the interval of time over which
the tag was
lost. So an animal first captured at time 10 sans tag would be (0,10); with
tag at 5 and
without at 20 would be
Hello,
It's much easier than you think, the first two columns of the input
matrix are the row and column numbers into the output matrix, therefore
those columns form an index matrix. Just see:
x - scan(text=
1 1 3
1 2 54
1 3 11
1 4 17
2 1 5
2 4 78
2 5 20
)
mat - matrix(x, ncol = 3,
Hi everybody,
I have a little problem about filling some gaps of NAs in my data.
These gaps are between nearly constant data (temperature under snow). Here's
a fake example to illustrate how it looks like approximately:
DF -
Hello,
The function below fails if any of the followind conditions is met.
1. The first value in the vector is NA
2. There are less than n = 5 (or 500) values before/after the
first/last, respectively, NA.
fun - function(x, n){
na - is.na(x)
rna - rle(na)
sna -
R 2.15.1
OS X
Colleagues,
I am reading a 1 GB file into R using read.table. The file consists of 100
tables, each of which is headed by two lines of characters.
The first of these lines is:
TABLE NO. 1
The second is a list of column headers.
For example:
TABLE NO. 1
COL1
Hello,
Try this, It'll maybe help you:
a - 1,2
b - strsplit(a,,) #split your data according to ,
b - unlist(b) # it creates a list, so we unlist the result to obtain a
vector like c(1,2)
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Still so perfect Rui! A bit much more complicated as what I thought,
nevertheless it's what I want!
Thank you Rui!
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On 10/18/2012 09:57 AM, Fisher Dennis wrote:
R 2.15.1
OS X
Colleagues,
I am reading a 1 GB file into R using read.table. The file consists of 100
tables, each of which is headed by two lines of characters.
The first of these lines is:
TABLE NO. 1
The second is a list of column
Jason
Are you suggesting grep in R or grep in the system? If the latter, this won't
work because I need to implement this same procedure in Windows (sorry about
not mentioning this), in which grep does not exist. If in R, the syntax is not
obvious -- could you provide an example?
Dennis
* Bert Gunter thagre.ore...@trar.pbz [2012-10-17 23:21:44 -0700]:
However, Is level 5 in 'a' the same as level 5 in 'b' ?
yes, of course.
would anyone want to _different_ factors with identical string representations?!
--
Sam Steingold (http://sds.podval.org/) on Ubuntu 12.04 (precise) X
mood - factor(c(blue, sunny))
skycolor - factor(c(azure,blue,teal)
If factors are not defined with levels specifications, automatic merging should
never be allowed. The fact that read.table automatically generates factors
using default levels is why I nearly always import using as.is=TRUE,
On Thursday, October 18, 2012, Sam Steingold wrote:
* Bert Gunter thagre.ore...@trar.pbz [2012-10-17 23:21:44 -0700]:
However, Is level 5 in 'a' the same as level 5 in 'b' ?
yes, of course.
would anyone want to _different_ factors with identical string
representations?!
Off the cuff,
hi Jorge,
* Jorge I Velez wbetrvinair...@tznvy.pbz [2012-10-18 16:43:58 +1100]:
a - factor(5:1,levels=1:9)
b - factor(9:1,levels=1:9)
lev - sort(unique(f - c(a, b)))
f - factor(f, levels = lev)
str(f)
Factor w/ 9 levels 1,2,3,4,..: 5 4 3 2 1 9 8 7 6 5 ...
is sort(unique()) really
Dear R community,
is there any efficient way to use aaply on different datacubes? I have 3
dimesniolan datacubes/arrays with dimensions lon x lat x time. Now I
would like to do caclulations on each individual time series (e.g. all
vectors along the third dimension) using a time series (or
Thank you very much. That appears to be what I wanted.
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Hi All,
Thanks in advance for your help. I'm trying to convert a string to an
integer vector. For instance, I will start with
a - 1,2
The result I want to end up with will be the equivalent of
c(1,2)
What's the best way to make the conversion? I've tried using as.integer(a),
but R
Hi,
I'm trying to obtain a table of coefficients and confidence intervals from a
logistic regression analysis in R. My code is as follows:
# read in csv file
datafile-read.csv(file.csv, row.names=1)
# read in the variable list
varlist-names(datafile)[66:180]
models-lapply(varlist, function(x)
* R. Michael Weylandt zvpunry.jrlyn...@tznvy.pbz [2012-10-18 16:01:37
+0100]:
On Thursday, October 18, 2012, Sam Steingold wrote:
* Bert Gunter thagre.ore...@trar.pbz [2012-10-17 23:21:44 -0700]:
However, Is level 5 in 'a' the same as level 5 in 'b' ?
yes, of course.
would anyone
Depending on what exactly you are trying to accomplish:
as.numeric(unlist(strsplit(a, ,)))
[1] 1 2
read.csv(textConnection(a), header=FALSE)
V1 V2
1 1 2
Sarah
On Thu, Oct 18, 2012 at 9:08 AM, BenM bmmos...@amath.washington.edu wrote:
Hi All,
Thanks in advance for your help. I'm
Dear all,
Thanks for the many replies!
Indeed, the function does not represent my explanation. It should have been:
if (x[i] 170 x[i] 1250 y[j] 150 y[j] 480)
Sorry, my mistake.
My data looks like this:
head(input[,3:4])
x y
1 701 209
2 685 209
3 566 248
4 562 234
5 601 225
6
Hi
I have one text file which containing 4 variables with 10 observations.
I would like to import with scan() function. Please give some
suggestion
Thanks...
Mydata set is.
id namesex age
111 HELEN f 22
112 DONNA f 22
113 ERICm
Hello fellow R users,
I am currently learning to use R, so please forgive me if there is an
obvious explanation for the following problem. My goal is to perform WGCNA
on a dataset of 19776 genes, so I opted to follow the block-wise network
construction (Section 2c) in the WGCNA R Tutorial by
Dear R users,
I'm currently trying to re-project a geotiff in another coordinate system.
For instance, I have a tif image in UTM 19 zone which I would like to
reproject into UTM 18. I was wondering if it was possible in R.
Furthermore, I looked into 'rgdal' package, but I can't really find out
* Jeff Newmiller wqarj...@qpa.qnivf.pn.hf [2012-10-18 07:53:24 -0700]:
If you HAVE defined your factors using explicit levels definitions, you
should have no trouble combining them.
http://article.gmane.org/gmane.comp.lang.r.general:277719
--
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Hello,
I am totally new in the field of time series analysis and forecasting and R.
I read that R is a powerful tool for time series. Could anyone give me
navigation what models of time series are availiable in R etc?
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Dear all,
I am trying to project my LongLat-maps to a plane.
The ultimate purpose is to do a search of points in vicinity of other points
using overlay-commands (sp) with radius in km.
I am applying spTransform (package rgdal) and it gives my some curious results.
An example.
Let's take a
Hi Arun,
Thanks, it works now.
Hi Rui,
Your code also works.
Thanks,
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I analyzed the kalman filter based approaches like mean reverting, random
coefficient and random walk.
At this point Automatic package is inadequate and need some constraints. I
also found Kalman Filter code
in Shumway$Stoffer book, but it did not provide the correct optimization.
Can you
Hello,
Try the following, readaing your file into 'x', using readLines.
tc - textConnection(
TABLE NO. 1
COL1COL2COL3COL4COL5 COL6
COL7COL8COL9COL10 COL11 COL12
1.0010E+05 0.E+00 1.E+00 1.E+03 -1.E+00
Hello!
I am trying to model data on species abundance (count data) with a poisson
error distribution. I have a fixed and a random variables and thus needs a
mixed model. I strongly doubt that my model is overdispersed but I don't
know how to get the overdispersion parameter in a mixed model.
c() has an unfortunate history. Originally, c(x) stripped the attributes,
except names but including dim, dimnames, and class, from x.
Also, c(x,y) stripped the attributes from both x and y and concatenated
them. Also, c(nameA=1,nameB=2) constructed a vector with a names attribute.
Then c()
Why do you need to use scan()? read.table() would be much easier.
Sarah
On Thu, Oct 18, 2012 at 9:14 AM, killerkarthick karthick@gmail.com wrote:
Hi
I have one text file which containing 4 variables with 10 observations.
I would like to import with scan() function. Please give
Sure:
http://cran.r-project.org/web/views/TimeSeries.html
Next time please try google first.
Best,
Ista
On Thu, Oct 18, 2012 at 11:29 AM, vas mpapoutsog...@gmail.com wrote:
Hello,
I am totally new in the field of time series analysis and forecasting and R.
I read that R is a powerful tool
There are a bunch of things wrong with your code, from not
understanding subsetting to only returning a single value rather than
a vector of values. I highly recommend that you read the Introduction
to R document that came with your R installation and is readily
available online.
In the meantime:
Dear all,
I want to calculate mean values for multiple rows:
structure(list(Name = structure(c(1L, 1L, 1L, 1L, 1L, 2L, 2L,
2L, 2L), .Label = c(AKT, CKT), class = factor), val1 = c(2,
3, 2, 2, 2, 5, 3, 8, 2), val2. = c(4, 5, 4, 8, 4, 8, 4, 7, 4),
val3 = c(5, 6, 5, 9, 5, 9, 5, 9, 5)), .Names =
? .onAttach
? .onLoad
at slightly different points in the load process.
Cheers,
Michael
On Wed, Oct 17, 2012 at 10:12 PM, Marc Girondot marc_...@yahoo.fr wrote:
Dear List Member
Is there a mechanism to automatically run a piece of R code when a package
is loaded using library(xxx) ?
* William Dunlap jqha...@gvopb.pbz [2012-10-18 15:33:38 +]:
c() has an unfortunate history.
:-)
ISTR reading in the R manual ~15(?) years ago that the language was in a
flux and one could not expect code written for the current release to
work in the next release. I was considering R as
Hi,
Are you looking for something like this?
#dat1
aggregate(dat1[,2:4],by=list(dat1[,1]),FUN=mean)
# Group.1 val1 val2. val3
#1 AKT 2.2 5.00 6
#2 CKT 4.5 5.75 7
A.K.
- Original Message -
From: Nico Met nicome...@gmail.com
To: R help r-help@r-project.org
Cc:
Sent:
Replace the y[1] with x[2] in the apply function - there is no 'y' defined in
the function passed to apply. In the apply function, x is an 2-element vector
representing a single row of the data (since the 1 in the second argument of
apply says that it is applied row-wise). So the second
And there's a wealth of textbooks on time series based in R. Two to start with:
Paul S.P. Cowpertwait and Andrew V. Metcalfe (2009). Introductory Time Series
with R. Use R! Series. Springer.
Robert H. Shumway and David S. Stoffer (2006). Time Series Analysis and Its
Applications With R
On Thu, Oct 18, 2012 at 4:20 PM, Sam Steingold s...@gnu.org wrote:
* R. Michael Weylandt zvpunry.jrlyn...@tznvy.pbz [2012-10-18 16:01:37
+0100]:
On Thursday, October 18, 2012, Sam Steingold wrote:
* Bert Gunter thagre.ore...@trar.pbz [2012-10-17 23:21:44 -0700]:
However, Is level 5 in
Package dlm does it, as well as other contributed packages (KFAS, sspir,
dse,...)
Best,
Giovanni
From: r-help-boun...@r-project.org [r-help-boun...@r-project.org] on behalf of
nserdar [snes1...@hotmail.com]
Sent: Thursday, October 18, 2012 8:40 AM
To:
Hi,
You can also try this:
dat1-read.table(text=
1 1 3
1 2 54
1 3 11
1 4 17
2 1 5
2 4 78
2 5 20
,sep=,header=FALSE)
library(reshape2)
dat2-cast(dat1,V1~V2)
dat2-dat2[,-1]
dat2[is.na(dat2)]-0
dat3-as.matrix(dat2)
dat3
# [,1] [,2] [,3] [,4] [,5]
#[1,] 3 54 11 17 0
#[2,] 5 0
On 18-10-2012, at 14:16, Martin Hehn wrote:
Dear all,
I am using the nlsLM function to fit a Lorentzian function to my experimental
data.
The LM algorithm should allow to specify limits, but the upper limit appears
not to work as expected in my code.
The parameter 'w', which is peak
What is the correct format for the shebang line and what options are
allowed or necessary along with this?
I find plenty of blogs and opinions, but few authoritative answers.
I want an R script to run and update packages periodically, with a
cron job that launches it. What is necessary to put in
On 18/10/2012 12:56 PM, Paul Johnson wrote:
What is the correct format for the shebang line and what options are
allowed or necessary along with this?
I find plenty of blogs and opinions, but few authoritative answers.
The authoritative source for the options is the R help page ?Rscript.
The
Possible? Yes. (see fortune(Yoda))
Automated using the legend function? No
Automated using another function? possbly somewhere in the 4,000+
packages on CRAN, but I don't know which.
It is doable with the basic tools. You could either find a part of
your graph with open area to put the legend
Hi Derek,
the simple answer is that the block-specific dendrograms cannot be
meaningfully combined into a single dendrogram. You have to plot them
separately. You can create a multi-panel figure that shows all block
dendrograms in one big figure, although with 10 blocks I would not
necessarily
You have a list of models, the coef and confint functions only work on
a single model, so you need to use lapply or sapply to get the
information from each model. Possibly something like (untested):
tableOfOddsRatios - sapply( models, function(x) exp(c( coef(x)[2],
confint(x)[2,]) )
I included
Another option would be to read the data using read.table or similar
to get the data into a data frame then use the xtabs function,
something like:
result - xtabs( count ~ docID + wordID, data=mydf)
On Thu, Oct 18, 2012 at 6:44 AM, Rui Esteves ruimax...@gmail.com wrote:
Hi,
I downloaded a
Hi Rainer,
Thanks for notifying me. You are right.
Sorry, I was working with library(reshape) instead of reshape2. So, I guess
the cast() will not work if we load only reshape2.
A.K.
- Original Message -
From: Rainer Schuermann rainer.schuerm...@gmx.net
To: arun
Hi all-
So sorry to bother you all with something pretty basic.
I am trying to add the lines method output from svysmooth to a svyplot with
style=grayhex. However, the line either appears in the wrong place or if I
am running in R Studio it causes the system to crash.
I know this is
Hi,
You can also try this:
a - 1,2
as.numeric(c(gsub((.*)\\,(.*),\\1,a), gsub((.*)\\,(.*),\\2,a)))
#[1] 1 2
- Original Message -
From: BenM bmmos...@amath.washington.edu
To: r-help@r-project.org
Cc:
Sent: Thursday, October 18, 2012 10:17 AM
Subject: Re: [R] converting a string to
Hi,
I agree with Sarah that read.table() will much easier in this case.
If you want scan(), then you can try this:
dat1-scan(what=list(numeric,character,character,numeric),text=
id name sex age
111 HELEN f 22
112 DONNA f 22
113 ERIC m 21
114 LINDA f
try adding legend = 0 to your svyplot()
On Thu, Oct 18, 2012 at 1:49 PM, Durant, James T. (ATSDR/DCHI/SSB)
h...@cdc.gov wrote:
Hi all-
So sorry to bother you all with something pretty basic.
I am trying to add the lines method output from svysmooth to a svyplot
with style=grayhex.
Good suggestion - I tried it - it does not work (line appears too high with the
data I am working with).
VR
James
James T. Durant, MSPH CIH
Environmental Health Scientist
US Agency for Toxic Substances and Disease Registry
Atlanta, GA 30341
770-488-0668
From: Anthony Damico
Hello,
Time down by a factor of 4. It still takes some minutes, 2 mins for a
file of 380Mb/3.6M lines. So maybe system commands (maybe awk?) can do
the job better.
fun - function(infile, outfile, lines = 1L){
remove - function(x){
i1 - grep(TABLE, x)
i2 - grep(COL, x)
Hello,
I am trying to set up a loop that can run the survdiff function with the
ultimate goal to generate a csv file with the p-values reported. However,
whenever I try a loop I get an error such as invalid type (list) for
variable 'survival_data_variables[i].
This is a subset of my data:
I have generated plot using ggplot, and i would like to add mean value to the
regression line as a marker. how to do it?
p - ggplot(data, aes(x = x, y = y,
color = a, shape = factor(sex),
linetype = factor(sex)))
p0 - p +
I know these package but I plan to analyse financial multi factorial data
set, and also estimate diffuse initial values for these
models.
I generated my own code, but I had problem with optim() package problem. I
need some constraints and I do not apply it
in my code.
Do you have any
Hi Peter,
Thank you for your time and input on this matter. I will take your advice
and reconstruct the blocks using a different maximum cutoff and plot the
figures next to one another. I also wanted to thank you for creating the
WGCNA package and including the section on exporting to other
So I do not find example what I expect.
I plan to estimate the multi-factor model for Kalman Filter Mean Reverting,
Random Walk and Random Coefficient.
For example:
R(it)= Alpha(it)+ Beta(it)R(mt)+Gamma(it)(R(mt)^2)+delta(it)(R(mt)^3)+ V(it)
KF Random walk
Alpha(it)= Alpha(it-1)+W(i1t)
Hi
On 19/10/2012 6:49 a.m., Durant, James T. (ATSDR/DCHI/SSB) wrote:
Hi all-
So sorry to bother you all with something pretty basic.
I am trying to add the lines method output from svysmooth to a
svyplot with style=grayhex. However, the line either appears in
the wrong place or if I am
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