? cumsum
something like
library(chron) # Reporting packages you use is always considerate
mt <- times(c('00:05:00', '00:15:00', '00:30:00')) # Spaces are legible!
times("09:30:00") + cumsum(mt)
Michael
On Tue, Apr 24, 2012 at 11:35 AM, René Mayer
wrote:
> Dear List,
> given a vecor of times i
15 27 381
> [471] 1 6 6 47 341 3 19 6 2 62
> [481] 2761 196 19 49 100 7 10 40 22 5
> [491] 2040 8 68 41 17 4 44 3 2554 81
> 180 Levels: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 1
Take a look at nobs()
Michael
On Tue, Apr 24, 2012 at 10:05 AM, Eiko Fried wrote:
> I have a dataset with plenty of variables and lots of missing data. As far
> as I understand, R automatically removes subjects with missing values.
>
> I'm trying to fit a mixed effects model, adding covariate by
What is the output of your sessionInfo()? Many folks have ggplot2
running on Snow Leopard (myself included) -- most likely, you need to
update something or other...
Michael
On Tue, Apr 24, 2012 at 9:26 AM, ramonovelar wrote:
> Hello,
>
> I have a similar error, running R in Snow Leopard too
>
>>
The scatter plot is easy:
plot(pH1 ~ pH2, data = OBJ)
When you say a loess for each -- how do you break them up? Are there
repeat values for pH1? If so, this might be hard to do in base
graphics, but ggplot2 would make it easy:
library(ggplot2)
ggplot(OBJ, aes(x = pH1, y = pH2)) + geom_point() +
It depends how you want to ensure that condition -- if you just want
to censor at an upper bound of 6k, this is easy:
pmin(zmsample, 6000)
If you want to sample "as before" but it just happens to all be less
than 6000 -- that's not really a rigorous statement, but just go with
it -- intuitively,
It is indeed the fact you're plotting factors, but unless you say what
"as intended" is, it's hard to provide exactly what you're seeking.
Perhaps this will help though:
X <- factor(sample(letters[1:5], 15, TRUE))
Y <- rnorm(15)
dats <- data.frame(X, Y)
plot(Y ~ X, data = dats) # No good
plot(X
Lists of numbers of length ~1000 are no problem for the wilcox.test()
function (Mann-Whitney is a special case) if you leave the default
exact = NULL.
The choice of test is all yours.
Michael
On Mon, Apr 23, 2012 at 12:39 PM, aoife doherty wrote:
> Hello
> I have two lists of numbers, each list
There are, but it's generally considered better style to keep them all
in a single list and use lapply() if you want to do things to each
element.
Michael
On Mon, Apr 23, 2012 at 5:33 PM, cyclondude wrote:
> Yes. That is what I was looking for. Is there a simple way to (in this
> scenario)
>
>
That's not the ifelse() that's the for loop returning NULL
(everything's a function!). If you put the assignment inside you'll
get expected behavior.
x <- (for(i in 1:5) i) # Strange
for(i in 1:5) x<- i # Normal (but notice you only get the last value
because previous ones are overwritten)
Michae
Look at ?ifelse, a combination of logical subscripting and mean(), or even
better ?ave -- I can't say too much more; there's a no homework policy on this
list and I recognize that first solution as mine already... (I should have
noted that the first time)
Michael
On Apr 22, 2012, at 2:54 PM,
Yes dput() for a reproducible example with some minimal
reproducible code (and the packages "day.of.week" and wday() come
from...)
x <- xts(10, Sys.Date())
wday(x)
seems fine for me.
"precisely the wrong answers" -- interesting turn of phrase.
Michael
On Sun, Apr 22, 2012 at 12:53 PM, Rag
The C implementation of the algorithm is here:
http://svn.r-project.org/R/trunk/src/main/plot3d.c
(grep filledcontour) but I don't see a reference other than to Ross Ihaka.
Hope this helps,
Michael
On Sat, Apr 21, 2012 at 9:21 PM, Stoch astic wrote:
> First time user, so sorry if I don't under
On the R level, I believe you're limited by the type of numeric
representation being used: either 32-big integer or 64-bit double. See
the storage.mode() of your objects. External code can make use of
128-bit types if desired, but I don't believe those can be naturally
represented back at the R lev
Ahh, I understand -- unfortunately, I'm not aware of an easy way to do
this so you'll have to hack your own: this doesn't look too hard
however, if you call
getAnywhere(predict.Arima)
you can get the prediction scheme R uses. It seems that most of the
heavy lifting is already in C so you'd probab
whathaveyoutried.com
Michael
On Fri, Apr 20, 2012 at 5:47 AM, dcgf wrote:
> I am trying to integrate this function
>
> f_x(x) = (1/(2pi))*(1+sin(2x)) for o < x < 2
>
> I know the answer is (x/(2*pi))-(cos(2x)/(4*pi)) but I must be doing
> something wrong..
>
> --
> View this message in context:
You're thinking about it wrong. This is an arithmetic sequence:
seq(from = 1000, by = -30, length.out = 15)
Michael
On Fri, Apr 20, 2012 at 5:14 AM, uday wrote:
> I would like to calculate vector from existing value
> e.g
> v <- 1000
> s <- 30
> d1 <- v-s
> d
This really has little to do with sorting, but can be much more easily
done with logical subscripting:
x[x < 20]
which I read as "x such that x is less than 20".
Hope this helps,
Michael
On Fri, Apr 20, 2012 at 6:26 AM, Yellow wrote:
> Can anyone help me maybe with a question I can seem to fin
It looks like you might want to use read.fwf for fixed width files
Michael
On Thu, Apr 19, 2012 at 3:45 PM, John S wrote:
> Dear R-users,
>
> I have a large data file that I am trying to read in R where the file has a
> white space at the begging and at random places. The whitespace is actually
There's something in your data that makes the model computationally
singular when you take the various subsettings... Can you provide a
small reproducible example so we can help narrow it down? It looks
like you're using different data for each mlogit though so I'm not
sure how the comparison that
Use matrix subsetting like this:
x <- matrix(1:9,3)
rownames(x) <- letters[1:3]
colnames(x) <- LETTERS[1:3]
print(x)
usrs <- c("a","b","a")
vars <- c("C","C","A")
counts <- c(10,11,12)
x[cbind(usrs, vars)] <- counts
print(x)
Hope this helps,
Michael
On Thu, Apr 19, 2012 at 1:48 PM, Tim Stutt
Though if you do decide to use Levenstein, it's implemented here in R:
http://finzi.psych.upenn.edu/R/library/RecordLinkage/html/strcmp.html
I'm pretty sure this is in C code so it should be mighty fast.
Michael
On Thu, Apr 19, 2012 at 11:40 AM, Bert Gunter wrote:
> Wrong list.This is R, not st
I think the OP is looking for the construct length(unique(x)) but not really
sure what the rest of the question is.
Michael
On Apr 19, 2012, at 2:12 AM, Petr PIKAL wrote:
> Hi
>
> Your question is rather cryptic. Why the output shall be 3? What has
> unique count to do with match function?
No like I said, lapply gets columnwise results (if you stopped
there, it would tell you whether each country is in the EU separately)
and the Reduce() combines them. Re-read my explanation and play around
with it on your own data set.
To convert 1 -> True and 2 -> False, you could use Boolean
Your problem is that columns A & B contain something that can't be
ordered. (Likely "factor" (=categorical) data like Male / Female
rather than numeric like 10 and 5)
Use str(data_2) to see what classes your data are -- they sometimes
get converted in unexpected ways if you aren't careful in setti
Your problem is that length(x) != length(y)
approx uses linear interpolation but there's no way to make sense of
that if you can't match up the x and y coordinates -- and you can't
match up the x and y coordinates if there aren't the same number of
them.
Michael
On Wed, Apr 18, 2012 at 10:15 AM,
The preferred format for time series will be something like this:
Sensor1Sensor2 Sensor3
T1 Read1_1 Read2_1 Read3_1
T2 Read1_2 Read2_2 Read3_2
T3 Read1_3 Read2_3 Read3_3
if you can get that. CSV separations are nice but not as essential as
the columnar organization. (It's pos
I'll unpack it bit by bit:
stringsAsFactors = FALSE is a command to tell R how to construct the
data frame: this doesn't apply to you because you already have a data
frame, but I need it to set up my examples. By default, when you give
R data that looks like strings/labels, it wants to convert tho
In addition to whatever feedback you may get here, you might subscribe
to the SIG-Teaching list for another interested population.
Michael
On Tue, Apr 17, 2012 at 10:46 PM, Christopher W Ryan
wrote:
> I participate peripherally on a listserve for middle- and high-school
> science teachers. Somet
Ok, so you simply need to use the methods David or I provided -- it
looks like you don't even have to change the variable names...
Is there something that isn't clear now?
Also, reproducible examples are much better if you use dput() so they
are actually reproducible
Michael
On Tue, Apr 17,
What exactly do you mean by "apply" it to a different data set?
Unlike regular regressions, time series models don't (generally) use
new data to make forecasts ...
By the way, this is a good guide to the time series functionality
available in R: http://cran.r-project.org/web/views/TimeSeries.html
Reproducible example not provided... try this -- it should generalize
to multiple columns:
EU <- c("UK","FR","DE") # Yes, I know there are more
countries <- data.frame( country1 = c("US","CH","UK","AU"), country2 =
c( "CA", "MX", "FR", "DE"), stringsAsFactors = FALSE)
cbind(countries, EU.Memb
It's a rather risky idea to call your function aggregate.zoo -- that's
actually a pre-existing function and it could (will!) confuse the
method dispatch system.
In short, S3 methods use the name convention generic.class (e.g.,
aggregate is the generic and zoo is the class) so when you just use
ag
Do you have lists or something like a data.frame? Your printout
suggests you don't actually have a list. (well, data.frames really are
lists but let's just not talk about that)
Perhaps %in% is what you are looking for...
EU <- c("UK","FR","DE") # Yes, I know there are more
countries <- c("US
A slightly quieter response:
Please use dput() to create a reproducible example -- for this case,
if x and y aren't too long, it seems that dput(x) and dput(y) would
comprise one. str() helps (and thank you for that -- it gave me a
place to start), but it doesn't provide quite enough to reproduce
I'm not sure that's francy's problem. This seems to work for me:
# Some fake data
nms <- letters[1:5]
lapply(nms, function(x) assign(x, rnorm(10), .GlobalEnv))
# Make some .Rdata files
lapply(nms, function(x) save(list =x, file = paste0(x, ".Rdata")))
# Check they are there
list.files()
# Bring
It looks to me like it's written in the penultimate paragraph of the
details section for ?strptime
Michael
On Mon, Apr 16, 2012 at 12:50 PM, Gene Leynes wrote:
> I would like to suggest that there is some documentation missing from
> strptime.
>
> There appears to be a way to show second decimal
He's saying it looks like you need to transpose your matrix with the
t() function -- documentation can be attained by typing ?t at the
console -- and that you'll need to subset to get the rows you want.
It's not quite clear to us which order your rows are actually in
because you sent HTML email whi
Certainly in theory, though the CRAN installers might over-write the
old version automatically. [I'm pretty sure they do for mac, don't
know about Windows] The way to ensure both versions remain will depend
on the OP's unstated OS.
Michael
On Mon, Apr 16, 2012 at 6:57 AM, ya wrote:
> Hi Partha,
as.Date() attemps to coerce a character string to a date where you
specify the input format -- if you want to specify an output format,
you need ?strftime [str + f + time == string format time]
E.g.,
titleDate <- as.Date("2011-05-03", format = "%Y-%m-%d")
plot(1:10, main = strftime(titleDate, "%
Strange, that isn't the error I get:
> mouter(wl, k1, k2, k3, FUN = function(w, k1, k2, k3) k1 *exp(k2 / (w + k3)))
Error in FUN(X, Y, ...) : argument "k2" is missing, with no default
Still, it's a problem with my mouter() function which was only tested
on binary operators (and then only really
I'm not quite sure what you are asking -- when you say an XY plot I
presume you mean a scatter plot of X against Y, but how do the values
underneath the headers play in? Do you just want different symbols?
With a little bit of data wrangling, I think this actually seems quite
well suited to the gg
You can also get the official getting started tutorial by typing
help.start() at the command prompt.
Michael
On Thu, Apr 12, 2012 at 2:52 PM, steven mosher wrote:
> Welcome to R and the list.
>
> Others may suggest books ( Nutshell was my first ) but first there are
> some things that will h
It seems that your first problem is syntax:
2x
will thrown an error, while
2*x
won't.
Google around for a good intro tutorial (there's the main one you can
access by typing help.start() and it's quite good) and these sorts of
things will be explained.
You might also want to use the curve func
Perhaps ?outer -- well, not outer directly, but a multivariate outer
-- I keep this one around for personal use:
`mouter` <- function(..., FUN = "*"){
dotArgs <- list(...)
FUN <- match.fun(FUN)
if(length(dotArgs) == 1L)
return(unlist(dotArgs))
if (length(dotArgs) == 2L)
Slightly on topic, just yesterday, the CFPB announced they'll be using
R in their work: http://blog.revolutionanalytics.com/applications/
Depending on the scale of your organization, it may not be the
open-source nature that's quite the problem of the "IT Gang" but
rather the support (or lack ther
Just a joke -- since you post through Nabble instead of being a list
subscriber, the vast majority of us didn't see your original post and
instead only saw your "cry for help"
Michael
On Thu, Apr 12, 2012 at 10:08 AM, helin_susam wrote:
> Sorry, I did not understand. What does it mean " call 999
Yes, this is as documented. See ? lag.xts under details for the
justification and how to change the default if desired.
Michael
On Wed, Apr 11, 2012 at 11:13 PM, jpm miao wrote:
> BTW, zoo is like ts in the application of lag.
> In other words, zoo and xts are opposite in this issue.
>
> 2012/4/
k you so much..
>
> is it possible to make arrows show up at the end of axis.. and make
> numberings show up in the middle axis??
>
> From: R. Michael Weylandt
> To: John Kim ; r-help
> Sent: Wednesday, April 11, 2012 4:06 PM
> Subject: Re: [R] r graphing
>
> Alternativel
Alternatively, you can use
curve(x^3, from = -5, to = 5); abline(h = 0, v = 0, lty = 2)
which will work even if the axes aren't in the middle of the image.
Michael
On Wed, Apr 11, 2012 at 6:54 PM, R. Michael Weylandt
wrote:
> Please reply all to the list and don't send HTML.
can realize that??
>
> john
>
>
> From: R. Michael Weylandt
> To: John Kim
> Cc: r-help@r-project.org; provicon2...@yahoo.com
> Sent: Wednesday, April 11, 2012 2:37 PM
> Subject: Re: [R] r graphing
>
> The easiest way is to just use ?curve (type ?curve at the p
The easiest way is to just use ?curve (type ?curve at the prompt to
get documentation for curve): e.g., curve(x^3, from = -5, to = 5)
You could also build the plot yourself like:
x <- seq(-5, 5, length.out = 200)
y <- x^3
plot(x,y)
Michael
On Wed, Apr 11, 2012 at 4:41 PM, John Kim wrote:
> ca
le to find one.
>
> Is there an mclapply included in 2.15? Is there a parallel package I'm
> missing? Or am I completely misunderstanding your response?
>
> Thanks!
>
> Wyatt
>
> -Original Message-
> From: R. Michael Weylandt [mailto:michael.weyla...
Can you give the line of code that gives the error? That'd make it
much easier to see what's there that should/shouldn't be.
Michael
On Wed, Apr 11, 2012 at 11:28 AM, krtek wrote:
> Thank you! Updating to R-patched really helped me.
>
> My problem were not been into my code. I've tried to run s
e answers:
http://stats.stackexchange.com/questions/26277/how-to-bootstrap-prediction-intervals-for-customized-regression-models-in-r
Michael Weylandt
On Wed, Apr 11, 2012 at 10:29 AM, Michael wrote:
> Hi all,
>
> Are there functions in R that could help me do the following?
>
> We have a special type
Simply pass all = FALSE to merge_all
merge_all(list_of_files, by = "Name", all = FALSE)
Michael
On Wed, Apr 11, 2012 at 1:09 AM, Chintanu wrote:
> Thanks to David and Michael
>
> Michael:
>
> That works, however with a glitch. Each of my 24 files has got two columns:
> "Name", and "Rank/score".
Well, obviously the interface to .Internal(identical) changed between
2.13.x and 2.15 -- as, ?.Internal says: Only true R wizards should
even consider using this function...
Anyways, find where you use it in your script and then we can help --
there's not much we can do without seeing the relevant
this:
> xts<-as.ts(x)
> y1<-ts(NA, start=start(y), end=end(y),frequency=frequency(y))
> y<-as.xts(y1)
>
> Is there any easier way to do it?
>
> Thanks
>
> miao
>
>
> 2012/4/11 R. Michael Weylandt
>>
>> Two ways around this:
Your problem is that you can't merge the file names, but you need to
load them into R and merge the resulting objects.
This should be straightforward enough to do:
file_list <- list.files()
list_of_files <- lapply(file_list, read.csv) # Read in each file
merge_all(list_of_files, by = "Name")
Mic
ruinf(1) {
> return(num1)
> }else{
> return(num2)
> }
> }
>
> And then, run 20 iterations of the function.
>
> On Tue, Apr 10, 2012 at 11:26 PM, R. Michael Weylandt
> wrote:
>>
>> The error message should make it pretty clear -- you are
You were told before this isn't a Mac question so please don't cc R-SIG-Mac.
I'm not sure what this bit of your reply means "My question is to find
any command to plot the data I got from the field;" but your reply
later suggests that your problem is that you are overriding previous
plots on a giv
Two ways around this:
I = Easy) Just use zoo/xts objects. ts objects a real pain in the
proverbial donkey because of things like this.
Something like:
library(xts)
PI1.yq <- as.xts(PI1) # Specialty class for quarterly data (or regular
zoo works)
lag(PI1.yq)
II = Hard) lag on a ts actually chang
This is the same malformatted message you posted on R-SIG-Mac even
after David specifically asked for clarification not to reward bad
behavior, but perhaps this will enlighten:
# Minimal reproducible data!
x <- runif(15, 0, 5)
y <- 3*x - 2 + runif(15)
dat <- data.frame(x = x, y = y)
rm(list =
The error message should make it pretty clear -- you aren't inside a
function so you can't return() a value which is, by definition, what
functions (and only functions) do.** Not sure there's a great
reference for this though beyond the error message
Incidentally, what are you trying to do her
So it's a machine/OS issue: if you really want to trace it down, take
a look here: http://svn.r-project.org/R/trunk/src/main/seq.c
seq.int() in R goes to do_seq() in C, but at this point it's probably
best to identify it as floating-point gremlins and to work around.
Michael
On Tue, Apr 10, 2012
I don't know the multicore package, but if possible, it might be
easier to upgrade to 2.15 and use the new built-in parallel package
that was introduced in R 2.14.
Then your syntax would be something like
mclapply(files, illumqc)
Michael
On Tue, Apr 10, 2012 at 11:33 AM, Wyatt McMahon wrote:
>
b
> 3 3 c
>
>> df$chaVec
> [1] "a" "b" "c"
>
> documentation of data.frame says the option is true by default.
>
>
> Am 10.04.2012 um 17:38 schrieb R. Michael Weylandt:
>
>> Don't use cbind() -- it forces everythi
What difference is it you are worried about:?
identical(seq.int(0,1,length.out = 11), seq.int(0,1, by = 0.1)) # TRUE
Though that may be OS dependent.
M
On Tue, Apr 10, 2012 at 10:51 AM, Alexander wrote:
>
> Berend Hasselman wrote
>>
>> On 10-04-2012, at 15:54, Alexander wrote:
>>
>>> I am work
You probably have more objects in your workspace than you did
previously. Clean them out (or just use a new R session) and things
should go back to normal.
You might also want to follow up on the help(memory.size) hint though
-- doesn't Windows impose a memory limit unless you ask it for more?
Mi
Don't use cbind() -- it forces everything into a single type, here
string, which in turn becomes factor.
Simply,
data.frame(a, b, c)
Like David mentioned a few days ago, I have no idea who is promoting
this data.frame(cbind(...)) idiom, but it's a terrible idea (albeit
one that seems to be very
You might want to use a while loop instead, something like:
while(TRUE){
# Do things
# Test: if your condition has occured
if(conditionHappened) break # break will end loop.
}
Michael
On Tue, Apr 10, 2012 at 10:48 AM, Steve Lavrenz
wrote:
> Everyone,
>
> I'm very new to R, especially when
You might try the Defaults package.
Michael
On Tue, Apr 10, 2012 at 10:54 AM, Liviu Andronic wrote:
> Dear all
> Is it possible to set globally the option .progress = "text" to all
> the apply functions in 'plyr'. For example, current default is
> daply(..., .progress = "none"). I would like to
A slightly easier formulation of the second proposal from Jessica:
plot(c(0,0) , xlim = range(x1, x2, x3), ylim = range(y), type = "n")
will set the canvas correctly.
On Tue, Apr 10, 2012 at 10:10 AM, Jessica Streicher
wrote:
> Hello Arunkamar!
>
> Basically:
>
> plot(x1,y)
> lines(x2,y)
> line
ifferent R image file?
>
>
>
> On 2012-4-10 14:54, R. Michael Weylandt wrote:
>>
>> You'll need to save them manually to avoid name conflicts -- save.image()
>> is the function to do so but you need to give a file name.
>>
>> Michael
>>
>&g
You'll need to save them manually to avoid name conflicts -- save.image() is
the function to do so but you need to give a file name.
Michael
On Apr 10, 2012, at 7:41 AM, ya wrote:
> Hi guys,
>
> I have a question. I am running 3 R sessions simultaneously for different
> analysis. I found ou
ks for this. I tried to figure out how to download that
> version, but found the documentation on SVN's quite confusing. Is there
> anyway that you could make that version available?
>
> Much appreciated.
> --John Sparks
>
>
>
> On Fri, March 23, 2012 5:55 pm, R. M
You should look up what a t-test is.
Michael
On Mon, Apr 9, 2012 at 2:58 AM, ali_protocol
wrote:
> I am interested in the difference of 2 data:
>
> mat1= c(2.2, 2.3, 2.2,2.5)
> mat2= c(2.6, 2.8, 2.7,2.4)
>
> mat= mat2-mat1
>
> I perform an action on both mat1 and mat2, and I get
> mat1prime and
Just a heads up: I'm pretty sure Josh and the other QS developers have
access to intra-day data, but they can't provide an example using it
in the package because the EULAs of most data providers won't allow
direct redistribution of their data. The examples included all go
online and download data
This is, more often that not, statistically speaking a bad idea:
you're likely to get a false positive (https://xkcd.com/882/)
But it sounds like something along these lines should work for you to
get the models :
x <- data.frame(rnorm(50), rnorm(50), rnorm(50), rnorm(50))
m <- vector("list", nco
One last try -- though if you cannot figure out how to take the name
of your object and put it in dput( NameOfObjectGoesHere ) after I gave
you a worked example of how to do so, it seems unlikely that this will
help:
# Attempt to recreate your data
library(zoo)
x <- read.zoo(textConnection("monoMn
Are you sure you are executing that line of code?
Michael
On Sat, Apr 7, 2012 at 9:18 AM, Hurr wrote:
> readline("press return to continue")
> worked as a pause until I used it in 4 places in a different program where
> none of them wait.
> What problem do I look for?
>
>
> --
> View this me
So you want us to deliver an entire backtesting architecture for you?
That's a pretty hefty request...
But being R, it's already been done. This has all basically been made
available in the quantstrat project (google it) -- the documentation
is online and you can see lots of worked examples in the
This still isn't reproducible -- when I said use dput() I meant it.
x <- data.frame(x = 1:5, y = letters[1:5], z = factor(sample(1:3,5,
TRUE))) # complicated
dput(x) # Easy to copy and paste.
Also, what package is the linearizeTime function from? I'm having
trouble finding it on CRAN.
If you can
Try this:
x <- xts(as.character(1:10), Sys.Date() + 0:9)
storage.mode(x) <- "double"
Michael
On Fri, Apr 6, 2012 at 1:13 PM, Noah Silverman wrote:
> Hi,
>
> I have a rather large data frame (500 x 5000) that I want to convert to a
> proper xts object.
>
> I am able to properly generate an xts
Usually you can just use cor() and it will do all the possibilities directly:
x <- matrix(rnorm(100), ncol = 10)
cor(x)
But that works on the columns, so you'll need to transpose things if
you want all possible row combinations: cor(t(x))
Hope this helps,
Michael
On Fri, Apr 6, 2012 at 9:57 AM,
On Fri, Apr 6, 2012 at 10:27 AM, MSousa wrote:
> Good Afternoon,
>
> I have the following code, but it seems that something must be doing
> wrong, because it is giving the results I want.
Didn't someone else have that problem just a few weeks ago? :-P
Michael
> The idea is to create segments
Probably to pull down the source of one and study it directly: if you
already know LaTeX and R, Sweave isn't much more to master: zoo does
vignettes nicely, but any package with vignettes should be pretty
good.
Michael
On Thu, Apr 5, 2012 at 4:33 PM, Erin Hodgess wrote:
> Hi R People:
>
> What i
http://www.rseek.org/ perhaps. [Take a look at the tabs on the RHS
after you do a search]
Michael
On Thu, Apr 5, 2012 at 11:36 AM, Jonathan Greenberg wrote:
> R-helpers:
>
> It looks like http://finzi.psych.upenn.edu/search.html has stopped
> spidering the mailing lists -- this used to be my go-
Authenticated access is sometimes a little harder than a simple
download -- the RCurl package provides tools that will be helpful, but
you'll need to tailor things to the sort of authentication used on the
other end. getURL iprovides one example of password authentication in
the examples. If you ar
Give a reproducible example (use dput()) and someone will be able to
help you. Otherwise, we're just guessing at what your data looks like.
Also, ?plot or ?matplot might help. Particularly, see the second
example for matplot. It might be what you are looking for.
A lesson of all this is though, p
Can't you just combine your matrices into a single matrix: rbind() or
cbind() should do the job.
Michael
On Wed, Apr 4, 2012 at 12:24 PM, uday wrote:
> Hi Liviu ,
> now I can see that function but the problem is that its only applicable for
> single data frame. as I wrote in my first post that I
I'm not sure what your definition of easier would be, but there are some style
things you might want to be aware of:
I) the name is likely to hit up against the S3 generic plot() when applied to a
glm object. This might lead to strange bugs at some point.
II) you can test !is.null once and use
No problem -- best of luck with it: the zoo package is one of the best
documentation-wise and I'd advise you to look at the available
vignettes when you have time.
Vignettes are extended documentation included in some packages that
give a more systematic presentation than can be given in the help
> [4] lattice_0.20-0
>
> loaded via a namespace (and not attached):
> [1] grid_2.14.2 tools_2.14.2
>>
> -Original Message-
> From: R. Michael Weylandt [mailto:michael.weyla...@gmail.com]
> Sent: Wednesday, 4 April 2012 2:35 PM
> To: Sue Xu
> Cc:
It works fine for me: can you give sessionInfo()?
Michael
On Tue, Apr 3, 2012 at 9:27 PM, Sue Xu wrote:
> Hi,
>
> I am trying to use the CSPADE function as part of the ArulesSequences
> package. When running with my own data I get a system invocation error, and
> also get the same when running
The zipfR package might be of help to you
Michael
On Sat, Mar 31, 2012 at 2:03 AM, ali_protocol
wrote:
> Hi everyone.
>
> Newbie to statistics.
>
> I have 40 matrices of ~400 values. how may I determine whether the
> distribution follows zips law?
>
> response <-sample (1:20,400*4, replace= TRUE
Setting a class is quite easy if you are in the S3 (read: easier) system:
x <- 1:15
class(x) <- "YourClass"
summary.YourClass <- function(x, ...) cat("The mean of your object is
", mean(x), "\n")
summary(x)
Michael
On Tue, Apr 3, 2012 at 12:54 PM, casperyc wrote:
> Hi all,
>
> I have a self w
Please take a look at my first reply to you:
ave(y, findInterval(y, quantile(y, c(0.33, 0.66
Then read ?ave for an explanation of the syntax. ave takes two
vectors, the first being the data to be averaged, the second being an
index to split by. You don't want to use split() here.
Michael
On
You need to load it once per session with library(np)
Michael
On Tue, Apr 3, 2012 at 2:46 PM, dnewbold wrote:
> Hi, Im trying to run a non parametric regression and I wish to use function
> npreg(), I installed the np package, but I am being told that npreg doesnt
> exist. Any advice on how I co
Like I said in my followup, please pass the maxgap argument: i.e.,
na.approx(x, maxgap = 4)
x <- zoo(1:20, Sys.Date() + 1:20)
x[2:4] <- NA # Short run of NA's
x[10:16] <- NA # Long run of NA's
na.approx(x) # All filled in
na.approx(x, maxgap = 4) # Only the short one filled in
Michael
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