Hi,
I am wondering if anyone knows of an easy way to fit a saturated model
using the sem package on raw data? Say the data were:
mtcars[, c("mpg", "hp", "wt")]
The model would estimate the three means (intercepts) of c("mpg",
"hp", "wt"). The variances of c("mpg", "hp", "wt"). The covariance
Hi,
Sure, you could do a qqplot for each variable between two datasets.
In a 2d graph, it will be hard to reasonably compare more than 2
datasets (you can put many such graphs on a single page, but it would
be pairwise sets of comparisons, I think. Perhaps you could plots
multiple qqplots on top
Something like this?
library(KernSmooth)
x <- rnorm(100)
KS <- bkde(x)
diff <- KS$x[2] - KS$x[1]
yc <- cumsum(KS$y*diff)
approx(yc, KS$x, runif(1000))
If you are going to use approx() repeatedly, you can create a
function with approxfun().
-
David L Carlson
A
Hi,
I am trying to call a sas program (test.sas) in R, the directory for
test.sas is testfolder.
It will work very well in the sas9.2 platform with the following command
system(paste('"C://Program Files/SAS/SASFoundation/9.2(32-bit)/sas.exe"',
'C:/testfolder/test.sas'), wait = TRUE)
However, w
hi all.
for example :
Table 1 variable:
BERTH TIME
A 9:52
D 7:43
C 8:33
A 10:13
I want to plot maybe only Berth 'A' timing. How can i do that?
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Hi Charlie,
Thanks for replying top my post, I really appreciate it.
I understand whats going on now.
Thank you once again
David
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Sent from the R help mail
Thanks Jessi,
your insights are extremely helpful.
If you would indulge me one more quick question on your script.
You have written...
newData<-data.frame(JVeg5=factor(Jdata[,"JVeg5"]),scale(Jdata[,c("Elevation","Lat_Y_pos","Coast_dist","Stream_dist")]))
I wish to expand this analysis for all ot
Hi,
dat1<-read.table(text="
weekly.returns
2010-1-4 -0.015933327
2010-1-11 -0.015042868
2010-1-18 0.005350297
2010-1-25 -0.049324703
2010-2-1 -0.052674121
",sep="",header=TRUE)
str(dat1)
'data.frame': 5 obs. of 1 variable:
$ weekly.returns: num -0.01
Dear R users,
I have created a Loess surface in R, in which x is relative longitude by
miles, y is relative latitude by miles, and z is population density at the
neighborhood level. The purpose is to identify some population centers in
the region. I'm wondering if there is a way to determine the c
You have to explicitly add the quotes:
Example=paste0("select * from LibDB where Date_Entered
='",as.character(StartDate), "'") # notice the single quotes within
the double quotes
> sqldf(Example,verbose=TRUE)
On Wed, Jul 11, 2012 at 11:16 AM, scstrein wrote:
> Hey guys,
>
> So I'm working with
Take a look at the 'plyr' package; it probably does what you want.
On Wed, Jul 11, 2012 at 7:53 PM, Charles Stangor
wrote:
> Say I want to perform transformations on row subsets of my dataframe.
>
> Do I have to break the dataframe into subsets, perform the analysis on each
> subset, and rbind()
Go back and reread the section about the scoping of variables and that
functions do not have side effects; they only return values. You are
changinga "local" copy of df1 within the function which is returning
the changed values to df3.
On Wed, Jul 11, 2012 at 7:36 PM, Charles Stangor
wrote:
> Wh
"R. Michael Weylandt" writes:
> On Wed, Jul 11, 2012 at 10:05 AM, Russell Bowdrey
> wrote:
>>
>> Dear all,
>>
>> This is what I'd like to do (I have an implementation using for
>> loops, which I designed before I realised just how slow R is at
>> executing them - this process currently takes day
Say I want to perform transformations on row subsets of my dataframe.
Do I have to break the dataframe into subsets, perform the analysis on each
subset, and rbind() them together again?
..or is there another way.
Thank you!
[[alternative HTML version deleted]]
Why does this sapply code change df3 but not df1?
Thanks
df1 <- read.table(text="
cola colb colc cold cole
1NA59 NA 17
2NA6 NA 14 NA
3 3NA 11 15 19
4 48 12 NA 20
", header=TRUE)
df2 <-df1*2
df1
df2
df3 <-sapply(names(df1),function(x) {df
I think I just learned this myself:
Don't put the $ extension in the bracket :
df1$cola[is.na(df1$cola)]<-
>
> df2$cola
>
Instead substitute using brackets within the brackets:
df1["cola"]is.na(df1["cola"])]<-
>
> df2["cola"]
> then the "cola" s can be substituted.
>
Maybe this will help
Hi,
It is a bit difficult to give suggestions when there are no example datasets.
Have you checked whether both belong to "Date" class?
You could use,
difftime(intakeDS$DOB, intakeDS$IntakeDate_d,units="days")
to get the difference between dates
A.K.
- Original Message -
From: Willia
Which Linux are you running?
For distros with package managers, that's usually all you need. Also Check up
on the R-SIG-Debian (including Ubuntu) or R-SIG-Fedora as appropriate if you
need more help.
Michael
On Jul 11, 2012, at 6:26 PM, Hui Du wrote:
> Thanks, all. I will upgrade my R in li
Thanks a lot, everyone for your helpful suggestions!
Eloi, this is very elegant - thank you!
I did not know 2 columns are allowed to have the same names! Always
good to learn something new.
Thanks again!
Dimitri
On Wed, Jul 11, 2012 at 6:58 PM, Mercier Eloi wrote:
> This should do the trick :
>
On 12-07-11 2:34 PM, Jonas Stein wrote:
Take a look at the predicted values at your starting fit: there's a
discontinuity at 0.4, which sure makes it look as though overflow is
occurring. I'd recommend expanding tanh() in terms of exponentials and
rewrite the prediction in a way that won't over
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
> On Behalf Of William Dunlap
> Sent: Wednesday, July 11, 2012 4:23 PM
> To: F86; r-help@r-project.org
> Subject: Re: [R] HELP me please with import of csv to R
>
> You or nabble left out my sug
Hello,
1. Post a data example,
dput(head(intakeDS, 20)) # post the output of this
2. You are subtracting IntakeDate from DOB, when it should be the other
way around. Like this you get negative dates and those are illegal.
If this point 2 doesn't solve it, follow step 1.
Hope this helps,
R
Thanks, all. I will upgrade my R in linux but I suspect the root cause is some
set up problem although I don't know how to trace it yet. I am not that
familiar with R in Linux. If somebody could give me a clue how to diagnose this
kind of problem, that will be great.
HXD
-Original Message
You or nabble left out my suggestion to use dec="," in your
call to read.table (or read.csv, etc.)
Did you use read.table(filename, ..., dec=",") when importing
the data (so "30,3" is read as the number 30 and 3 tenths instead
of as the character string "30,3")?
along with an example of wha
Hello William,
- I used str() and got this
'data.frame': 290 obs. of 2 variables:
$ Kommuner: Factor w/ 289 levels "Ale","Alingsas",..: 34 40 44 79 95 99 132
162 169 173 ...
$ Skatt : Factor w/ 177 levels "28,89","28,9",..: 86 7 47 67 74 25 24 23
85 74 ...
- and summary()
- got this
Hello,
About many columns like 'e1' and 'e2' I don't know but with the provided
example the following does NOT depend on them, only on 'a', 'b' and 'a2'
and 'a3'.
z <- lapply(c("a2", "a3"), function(cc) merge(x, y, by.x=c("a", "b"),
by.y=c(cc, "b")))
z <- lapply(seq_along(z), function(i)
> Take a look at the predicted values at your starting fit: there's a
> discontinuity at 0.4, which sure makes it look as though overflow is
> occurring. I'd recommend expanding tanh() in terms of exponentials and
> rewrite the prediction in a way that won't overflow.
>
> Duncan Murdoch
Hi Du
You're rather out of date with your version of R -- if you want to use
the CRAN binaries provided by install.packages(), you usually have to
be at most one minor rev behind.
Try install.packages("car", type = "source")
But even better -- update to a poppin-fresh R 2.15.1: you get
parallelization,
Dear Umesh Khatri,
The vif() function in the car package will compute (generalized) variance
inflation factors for GLMs.
What (if anything) to do about collinearity isn't in my opinion an answerable
question without knowing the details of your research. In particular, ridge
regression is not a
Hi there,
I'm new to some of these more advanced regression techniques and also new to
R. This looks like a great forum.
I am trying to examine the association with membership in a group and some
different variables, most of which are (approximately) normally distributed.
Would just do an ANOVA
This should do the trick :
colnames(y)[1:2]=c("a","a")
y2=rbind(y[,-1], y[,-2]) #duplicating the "y" matrix so the identifiers
are only in 1 column
merged = merge(x,y2)
merged
a b d e1 e2
1 aa 1 10 100 101
2 aa 2 20 200 201
3 ab 1 30 100 101
4 ab 2 40 200 201
5 ba 1 50 300 301
6 ba 2 60
Hi,
I wanted to calculate the age of people in my dataset by subtracting the
individual's date of birth from their intake into a program.
After several hours, searches of help archives, and the downloading of
lubiridate, I have had no luck with this.
Below is the code I used.
> intakeDS$DOB <-
Hi,
Just try: dat4<-read.table(text="
ABC PQR XYZ MNO
3 6 7 15
2 12 24 15
20 5 1 2
25 50 15 35
",sep="",header=TRUE)
apply(dat4,2,quantile)
apply(dat4,2,var)
#For mode calculation, use package ?modalvalue {rattle}
A.K.
,
- Original Message -
From: Ranton
Hello,
Try this:
dat3<-read.table("dat3.txt",sep="",skip=3,header=TRUE,fill=TRUE)
dat4<-data.frame(variable1=paste(dat3[,1],dat3[,2],sep="
"),Variable2=dat3[,3],Variable3="",Variable4=dat3[,4])
dat4<-dat4[-1,]
row.names(dat4)<-1:nrow(dat4)
dat4
variable1 Variable2 Variable3 Variable4
1 11/
Jorge, thank you!
that seems to be working, but unfortunately in real life I have
thousands of variables (except for a, a2, a3 and b) so that manually
selecting columns (as in c(2:4, 8:9)) would be too difficult...
Dimitri
On Wed, Jul 11, 2012 at 6:36 PM, Jorge I Velez wrote:
> Hi Dimitri,
>
> Tr
Why did you use the 'lower.tri' syntax?
Does this work for you?
lme(Y~Random, data = DATA,
random = list(Random = pdSymm(CovM,~Random)))
Kevin
On Wed, Jul 11, 2012 at 9:27 AM, Marcio wrote:
> Dear Simon,
> Thanks for the quick reply.
> Unfortunately I don't have access to Pinheiro and Bates. I
Hui Du,
There is probably something wrong with either your setup or your choice of
CRAN
mirror. Both HH and car are present in CRAN for all platforms.
---Your session info just arrived. "R version 2.11.1 (2010-05-31)" is over
two years old.
The current version is "R version 2.15.1 (2012-06-22)".
In that case, I think that using a subscript of NA is the
best way to go. It works for both matrices and data.frames
(unlike an integer larger than nrow(data)) and its meaning
is pretty clear.
Also, you will probably get better results if the function
in your call to apply() returns the index (pe
Hi Dimitri,
Try creating a key for "x" and "y" and then merging the result by that
variable:
x$key <- with(x, paste(a, b, sep = "/"))
y$key <- with(y, paste(a2, b, sep = "/"))
merge(x, y, by = 'key')[, c(2:4, 8:9)]
HTH,
Jorge.-
On Wed, Jul 11, 2012 at 6:28 PM, Dimitri Liakhovitski <> wrote:
>
Thank you for your consideration.
> sessionInfo()
R version 2.11.1 (2010-05-31)
i686-pc-linux-gnu
locale:
[1] LC_CTYPE=en_US.UTF-8 LC_NUMERIC=C
[3] LC_TIME=en_US.UTF-8LC_COLLATE=en_US.UTF-8
[5] LC_MONETARY=C LC_MESSAGES=en_US.UTF-8
[7] LC_PAPER=en_US.UTF-8 LC_
Could you please include your sessionInfo() ?
Thank you,
Jorge.-
On Wed, Jul 11, 2012 at 6:27 PM, Hui Du wrote:
> Thanks. But in UNIX side, I got the same error
>
> ** **
>
> In getDependencies(pkgs, dependencies, available, lib) :
>
> package âcarâ is not available
>
> ** *
Dear R-ers,
I feel I am close, but can't get it quite right.
Thanks a lot for your help!
Dimitri
# I have 2 data frames:
x<-data.frame(a=c("aa","aa","ab","ab","ba","ba","bb","bb"),b=c(1:2,1:2,1:2,1:2),d=c(10,20,30,40,50,60,70,80))
y<-data.frame(a2=c("aa","aa","ba","ba"),a3=c("ab","ab","bb","bb"
Thanks. But in UNIX side, I got the same error
In getDependencies(pkgs, dependencies, available, lib) :
package ââ¬Ëcarââ¬â¢ is not available
HXD
From: Jorge I Velez [mailto:jorgeivanve...@gmail.com]
Sent: Wednesday, July 11, 2012 3:19 PM
To: Hui Du
Cc: R-help
Subject: Re: [R] Variance I
See the examples at
# install.pacages('car')
require(car)
?vif
HTH,
Jorge.-
On Wed, Jul 11, 2012 at 6:10 PM, Hui Du <> wrote:
> Hi All,
>
>
> I need to calculate VIF (variance inflation factor) for my linear
> regression model. I found there was a function named vif in 'HH' package.
> I have
Hello,
That seems easy.
dat$variable1 <- with(dat, paste(variable1, variable2))
dat$variable2 <- dat$variable3
dat$variable3 <- ""
Then convert variable1 to date/time using as.POSIXct or strptime
See ?strptime.
Hope this helps,
Rui Barradas
Em 11-07-2012 13:30, vioravis escreveu:
Thanks a
Hi All,
I need to calculate VIF (variance inflation factor) for my linear regression
model. I found there was a function named vif in 'HH' package. I have two
questions:
1) I was able to install that package in my R under windows. But while
trying to install that package in UNIX, I got
On Wed, Jul 11, 2012 at 10:05 AM, Russell Bowdrey
wrote:
>
> Dear all,
>
> This is what I'd like to do (I have an implementation using for loops, which
> I designed before I realised just how slow R is at executing them - this
> process currently takes days to run).
>
> I have a large dataframe
On Wed, Jul 11, 2012 at 11:16 AM, scstrein wrote:
> Hey guys,
>
> So I'm working with a project where I manage a database within R, and I'm
> developing a script/function that will automatically run my queries in R
> depending on the date parameters passed in.
>
> The problem is that when I create
Hello,
A one-liner could be
df1 <- read.table(text="
cola colb colc cold cole
1NA59 NA 17
2NA6 NA 14 NA
3 3NA 11 15 19
4 48 12 NA 20
", header=TRUE)
df2 <- read.table(text="
cola colb colc cold cole
1 1.4 0.8 0.02 1.6 0.6
", header=
It should. Reproducible example of it not?
Michael
On Jul 11, 2012, at 3:06 PM, Charles Stangor wrote:
> I can't seem to determine how to get the name of a list member to
> substitute as a variable name:
>
> ll <- list("a1" = "a","a2" = "b")
>
> t1[t==ll[1], "v"] <- 99
>
> why doesn't this s
Hello,
Please don't post datasets like this, it's unusable by us. Use dput().
?dput
dput(head(myData, 20)) # post the output of this.
Paste the output of that command in a post. It starts with 'structure'.
Don't worry if it looks awkward, it is, on the contrary, very usefull.
All we need to
Hi, Prof. Ripley, Bert:
Thanks for the comments.
Might there be a function similar to "c" but retains more
attributes than just names?
Before I write such, I felt a need to ask if anyone knows where
it may already have been done -- and if people have suggestions for how
On Wed, Jul 11, 2012 at 9:56 PM, William Dunlap wrote:
> Why does one want to replace a zero-row data.frame
> with a one-row data.frame of NA's? Unless this is for
> an external program that cannot handle zero-row inputs,
> this suggests that there is an unnecessary limitation (i.e.,
> a bug) in
thanks!
2012/7/11 Yasir
> Your error comes from the source file makebin
> it's because your sequence identifiers are not in ascending order:
>
> // check if the sequence and event/time
> // identifiers are in ascending order and
> // determine the number of sequences.
> ns = l = h =
Hello,
If I understanding it, what you want are the values below.
half.hours <- function(x){
s <- strsplit(as.character(x), ":")
H <- as.integer(sapply(s, `[`, 1))
M <- as.integer(sapply(s, `[`, 2))
h <- (H*60 + M)/60
floor(h/0.5)*0.5
}
half.hours(dat$Sun
I can't seem to determine how to get the name of a list member to
substitute as a variable name:
ll <- list("a1" = "a","a2" = "b")
t1[t==ll[1], "v"] <- 99
why doesn't this substitute to:
t1[t=="a", "v"] <- 99
Thank you!
--
Charles Stangor
Professor
--
Charles Stangor
Professor
Thanks Peter.
We did manage to solve the problem using the approach bellow.
Hab <- cbind(seq(1,21),habitat)
hab <- matrix(NA,nrow=3,ncol=7)
for (i in 1:3) {hab[i,] <- Hab[habitat==i,1]}
The problem was that the hab matrix was not a multiple of habitat.
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ht
Hey guys,
So I'm working with a project where I manage a database within R, and I'm
developing a script/function that will automatically run my queries in R
depending on the date parameters passed in.
The problem is that when I create variables for the dates, and use those
variables in my sqldf s
Dear all,
This is what I'd like to do (I have an implementation using for loops, which I
designed before I realised just how slow R is at executing them - this process
currently takes days to run).
I have a large dataframe containing corporate bond data, columns are:
BondID
Date (goes back 5ye
Din't you try sapply function?
I tried it for you.
Just convert your matrix into a data frame using as.data.frame and then
*> rantony*
ABC PQR XYZ MNO
[1,] 3 6 7 15
[2,] 2 12 24 15
[3,] 20 5 1 2
[4,] 25 50 15 35
*> rantony=as.data.frame(rantony)*
*> sapply(rantony,va
Did you use read.table(filename, ..., dec=",") when importing
the data (so "30,3" is read as the number 30 and 3 tenths instead
of as the character string "30,3")?
Whenever importing data follow up by using str() or summary()
on its output, before doing any further analysis. E.g.,
> bad <- read
Hi,
I've installed the igraph package and have been otherwise using it
successfully, but when I try to use graph.bfs I get the error:
could not find function "graph.bfs"
Moreover, I don't seem to have the documentation installed either. (per
?graph.bfs and ??graph.bfs).
I'm using RStudio
Hello,
I wanted to know if there is a simple way of getting the inverse cdf for a
KDE estimate of a density (using the ks or KernSmooth packages) in R ?
The method I'm using now is to perform a numerical integration of the pdf
to get the cdf and then doing a search for the desired probablity valu
Thanks. That worked.
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__
R-help@r-project.org mailing list
https:
yes just use %in% instead of == AllMax.. but also use table for count
-
Yasir Kaheil
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__
Dear Simon,
Thanks for the quick reply.
Unfortunately I don't have access to Pinheiro and Bates. I tried googling
the pdSymm and lme but I still cannot get the syntax right.
In my model, I only have 1 random factor with repetitions (groups) (e.g. 2
records per each level)
I am pasting bellow a ver
I want the output to be 1,1,1,1,7,7,7,7
The multiple values of AllMax are 1 and 7.
I think I've figured it out though, I added a loop at the end:
test <- as.matrix(c(1,1,1,1,3,3,7,7,7,7))
Count <- tapply(test[,1], test[,1], length) # count for each value
spp <- unique(test[,1])
Count1 <- as.dat
Dear everyone,
I'm student of Masters in Statistics (Actuarial) from Central
University of Rajasthan, India. I am doing a major project work as a
part of the degree. My major project deals with fitting a glm model
for the data of car insurance. I'm facing the problem of
multicollinearity for this
Hi,
I have two dataframes:
The first, df1, contains some missing data:
cola colb colc cold cole
1NA59 NA 17
2NA6 NA 14 NA
3 3NA 11 15 19
4 48 12 NA 20
The second, df2, contains the following:
cola colb colc cold cole
1 1.4 0.8 0.
Hi,
I have a list called ds which has the following attributes:
attributes(ds)
$names
[1] "adj.r.squared" "fstatistic""intercept" "slope"
[5] "std.error" "tstatistic"
I want to replace all the NaN is ds with 0, and after searching past posts I
found I can hardcode it like t
I can't seem to determine how to get the name of a list member to
substitute:
ll <- list("a1" = "a","a2" = "b")
t1[t==ll[0], "v"] <- 99
why doesn't this substitute to:
t1[t=="a", "v"] <- 99
Thank you!
[[alternative HTML version deleted]]
__
Hi Everyone,
I could use help developing a for loop. I have a dataset with tallies for
a number of species within 7 different size classes. I need to uncollate the
data into the rawest form (ie: each row a different individual) & retrain
the metadata associated with each row (Date, Recorder, Sit
Hello,
I've been using: tmp.df = read.xls(filename, stringsAsFactors = FALSE) to
read in my files. Even though I get the "There were 50 or more warnings" thing,
for the most part most of the data is read in correctly. However, there are a
few select rows where there are values but they are
Hello everyone,
I'm relatively new to the RCurl package, and I'm having some issues with
reading in my data the way I want to. I have binary data that I've only been
able to read in correctly using getBinaryURL(), but I also want to read the
data in chunks (it's a huge timeseries dataset). Accord
I have a data assimilation problem that might be amenable to the use of GAMS,
but I am not sure how feasible it is to implement. I was told the R mailing
list was a great resource.
My observations are spatiotemporal salinity in the San Francisco Bay at a
number of instruments over a few days. T
Thank you! It is fixed now.
However, now when I'm trying with hist(skatter) i get this message: Error
in hist.default(skatter) : 'x' must be numeric
I don't know what I'm doing wrong but it worked perfectly on windows some
weeks ago.
My data skatter looks like this:
> skatter
I can't seem to determine how to get the name of a list member to
substitute:
ll <- list("a1" = "a","a2" = "b")
t1[t==ll[0], "v"] <- 99
why doesn't this substitute to:
t1[t=="a", "v"] <- 99
Thank you!
--
Charles Stangor
Professor
[[alternative HTML version deleted]]
Thank you! It is fixed now.
However, now when I'm trying with hist(skatter) i get this message: Error
in hist.default(skatter) : 'x' must be numeric
I don't know what I'm doing wrong but it worked perfectly on windows some
weeks ago.
My data skater looks like this:
> skatter
Ko
-- Forwarded message --
From: umesh khatri
Date: Wed, 11 Jul 2012 16:15:11 +0530
Subject: Poisson ridge regression
To: r-h...@rproject.org
I'm student of Masters in Statistics (Actuarial) from Central
University of Rajasthan, India. I am doing a major project work as a
part of the
Hi Dude!!
It seems that your column names are not sequentially arranged, I guess,
If so u just reorder them and do it your way...it will work..
for example,
colnames(data)[2]<–"weekly.returns" ## sometimes u can use like this as
well, if u want to assign at last most column
Hi everyone!
I already posted
http://r.789695.n4.nabble.com/Declaring-a-density-function-with-for-loop-td4635699.html
a question on finding density values of a new Binomial like distribution
which has the following pmf:
http://r.789695.n4.nabble.com/file/n4636134/kb.png
Thank fully
http://r.789
Hi Bill
Many thanks for your help.
Cheers
R
-Original Message-
From: William Dunlap [mailto:wdun...@tibco.com]
Sent: 10 July 2012 17:22
To: Raghuraman Ramachandran; r-help@r-project.org
Subject: RE: Help with vectors and rollapply
It looks like you already have the zoo package loaded so
On Wed, 11 Jul 2012, MacQueen, Don wrote:
An my "easy" but not very useful answer is that this particular subset
probably violates some assumption of the cenros() model. I myself would
start with simple inspections of the data, such as
with( subset(chem, param=='Ag'), table(ceneq1) )
with( s
Hi. Maybe this:
ct <- table(test)
as.numeric(names(ct[ct==max(ct)]))
test[test[,1]%in%as.numeric(names(ct[ct==max(ct)])),,drop=FALSE]
?
Andrija
On Wed, Jul 11, 2012 at 8:33 PM, Amanduh320 wrote:
> I'm stuck on a seemingly simple problem. I'm trying to subset the data by
> several numbers and
On Wed, Jul 11, 2012 at 3:08 PM, R. Michael Weylandt
wrote:
> On Wed, Jul 11, 2012 at 3:07 PM, R. Michael Weylandt
> wrote:
>> On Wed, Jul 11, 2012 at 1:49 PM, Cren wrote:
>>> # One more question, Joshua: let instead of merging tickers
>>> # I would like to put prices from an OHLC object
>>> # i
On Wed, Jul 11, 2012 at 3:07 PM, R. Michael Weylandt
wrote:
> On Wed, Jul 11, 2012 at 1:49 PM, Cren wrote:
>> # One more question, Joshua: let instead of merging tickers
>> # I would like to put prices from an OHLC object
>> # in weekly format, then selecting just the close prices.
>> # What woul
On Wed, Jul 11, 2012 at 1:49 PM, Cren wrote:
> # One more question, Joshua: let instead of merging tickers
> # I would like to put prices from an OHLC object
> # in weekly format, then selecting just the close prices.
> # What would be a code to do it?
> # I guess:
>
> data = new.env()
> ticker.li
Why does one want to replace a zero-row data.frame
with a one-row data.frame of NA's? Unless this is for
an external program that cannot handle zero-row inputs,
this suggests that there is an unnecessary limitation (i.e.,
a bug) in the R code that uses this data.frame.
Bill Dunlap
Spotfire, TIBCO
An my "easy" but not very useful answer is that this particular subset
probably violates some assumption of the cenros() model. I myself would
start with simple inspections of the data, such as
with( subset(chem, param=='Ag'), table(ceneq1) )
with( subset(chem, param=='Ag'), qqnorm(quant) )
Hello useRs,
I'm having trouble trying to produce axis labels running parallel to the
axes in 3d space. Is their a way to do this when producing 3d graphs using
the rgl package?
Thanks for your help,
Patrick
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I see that this is your input: c(1,1,1,1,3,3,7,7,7,7)
what do you want the output to be?
what could the multiple values of AllMax be?
-
Yasir Kaheil
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Sent from the R help mai
I'm stuck on a seemingly simple problem. I'm trying to subset the data by
several numbers and it cuts out half of the rows. Here is the sample code:
test <- as.matrix(c(1,1,1,1,3,3,7,7,7,7))
Count <- tapply(test[,1], test[,1], length) # count for each value
spp <- unique(test[,1])
Count1 <- as.d
Thank you chamilka.
From: chamilka [via R] [mailto:ml-node+s789695n4636142...@n4.nabble.com]
Sent: Wednesday, July 11, 2012 8:13 PM
To: Akkara, Antony (GE Energy, Non-GE)
Subject: Re: MODE , VARIANCE , NTH PERCENTAILE
Din't you try sapply function?
I tried it for you.
Just convert your
You can try this:
for (i in 1:length(ds)) {
dummy<-ds[[i]];
dummy[is.nan(dummy)]<-0
ds[[i]]<-dummy
}
if is.nan doesn't work, replace with is.na
-
Yasir Kaheil
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Hi,
Here i have an matrix like this,
ABCPQRXYZ MNO
-- --- --
367 15
2 122415
20 5 1 2
25 50 15 35
i need to get the
"MO
Hi,
Try this:
.xa<-iris[1,][rep(NA,length(iris),1),]
.xa
# Sepal.Length Sepal.Width Petal.Length Petal.Width Species
#NA NA NA NA NA
#or
.xb<-iris[1,][rep(NA,ncol(iris),1),]
.xb
# Sepal.Length Sepal.Width Petal.Length Petal.Width Species
#NA
Hmm, I guess that's not exactly what I needed. It's my fault; you gave me
what I asked for, I was just asking for the wrong thing.
What I'm hoping to do is something similar to:
aggregate(dat$SunScore, by=list(h), median)
except I'd like to do it by half hour instead of hour. I guess what I ne
Your error comes from the source file makebin
it's because your sequence identifiers are not in ascending order:
// check if the sequence and event/time
// identifiers are in ascending order and
// determine the number of sequences.
ns = l = h = 0;
for (i = 0; i < LENGTH(sx); i++
Thanks a lot for the guidance. I have another text file with a time stamp and
an empty column as given below:
First line: Skip this line
Second line: skip this line
Third line: skip this line
variable1
read the part on power fitting:
http://www.itc.nl/~rossiter/teach/R/R_CurveFit.pdf
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