On 24/09/14 17:31, Mohan Radhakrishnan wrote:
Hi,
I have streaming data(1 TB) that can't fit in memory. Is there a
way for me to find the median of these streaming integers assuming I can
fit only a small part in memory ? This is about the statistical approach to
find the median of a l
This list is not here to do your homework... Even if you plan to pay !
Olivier.
--
Olivier Crouzet
LLING - Laboratoire de Linguistique de Nantes - EA3827
Université de Nantes
-Original Message-
From: Richard Lerner
Sender: r-help-bounces@r-project.orgDate: Tue, 23 Sep 2014 22:20:07
To:
Hi,
I have streaming data(1 TB) that can't fit in memory. Is there a
way for me to find the median of these streaming integers assuming I can
fit only a small part in memory ? This is about the statistical approach to
find the median of a large number of values when I can inspect only a p
The "Value" section of the help page describes what the function returns... in
this case a vector of 1 named integer element that is zero (success), based on
the question you asked (is it readable).
Given this input I would suggest that you confirm that the software that
produced this file was
I have a project that I have to do this Friday-Sunday. I have 2 dirty excel
spreadsheets that I need brought into R, cleaned up, and then some
descriptive statistics run. I am very new to R and not having fun. I can
pay, but you will need to 1) be expert and 2) be willing to show me how the
work is
Hi,
Hope you are doing well.
My name is Tony I came across your company website and I could see
Optometrists and Ophthalmologists is your one of the targeted specialist, so
I thought of asking if you would be interested in acquiring a complete
data-set of Optometrists and Ophthalmologists?
I'm using a package and need to keep track of the version of the database used.
Initially, I was under the impression that the package was querying a remote
database live and therefore the data version would be the date of my query.
This turned out to be an incorrect assumption, and the database
Dear all,
I have a data frame (datos) of hourly wind speed and direction with 4columns
(1st date, 2nd hour, 3rd wind speed and 4rth wind direction). I have been able
to do the daily mean of the wind speed, but when I try to get the more frequent
wind direction of every day, it doesn't work. I ha
Hello,
Thank you so much for your guidance. This time I am providing more information.
R Script and R Console are appended below.
The list.files() below provides evidence the existence of this file in the
"temp" directory. Please note that the "heroin.rdata" file was created from
the SAS data
Insufficient information, and irrelevant information (the second error is a
direct consequence of the first).
We have no way of knowing based on this input that your file is there.
(?list.files). We also don't know if you have read access to that file
(?file.access).
Since you posted in HTML a
Hello,
I am running Rx64 3.03 under Windows 8 environment. I have been getting the
following error.
when running some of my old R applications. Below is a mock-up example.
Could someone please help me resolve the issue?
Thanks,
Pradip Muhuri
setwd ("D:/")
>
>
> #load Rdata f
On 23 Sep 2014, at 23:30 , Rolf Turner wrote:
> On 24/09/14 03:10, carol white wrote:
>> Hi, If I want to divide the column of a matrix by the sum of the
>> column, should I loop over the columns or can I use apply family?
>
> m1 <- apply(m,2,function(x){x/sum(x)})
>
> should do what you want
On 24/09/14 03:10, carol white wrote:
Hi, If I want to divide the column of a matrix by the sum of the
column, should I loop over the columns or can I use apply family?
m1 <- apply(m,2,function(x){x/sum(x)})
should do what you want IIUYC.
cheers,
Rolf Turner
--
Rolf Turner
Technical Editor
Hi David,
My initial reaction (not that the decision is mine to make), is that from a
technical perspective, obviously indexing by name is common.
There are two considerations, off the top of my head:
1. There would be a difference, of course, between:
> month.abb["1"]
NA
and
> month.ab
Marc;
Feature request:
Would it make sense to construct month.abb as a named vector so that the
operation that was attempted would have succeeded? Adding alphanumeric names
c("01", "02", "03", "04", "05", "06",
"07", "08", "09", "10", "11", "12") would allow character extraction from
substring
Two things:
1. You need to convert the result of the paste() to a Date related class.
2. R's standard Date classes require a full date, so you would have to add in
some default day of the month:
See ?as.Date
NewDate <- as.Date(paste(month.abb[as.numeric(ddf$month)], "01", ddf$Year,
sep="-"),
Many thanks for your quick answer which has created what I wished. May
I ask followup question on the same issue. I failed to convert the new
column into date format with this code. The class of MonthDay is still
character
df$MonthDay <- format(df$MonthDay, format=c("%b %Y"))
I would appreciate if
Try:
map(database= "worldHires","Canada", ylim=c(39,90),
xlim=c(-145,-25), col="grey95", fill=TRUE, projection="lambert",
param=c(50,65))
lines(mapproject(y=yValue, x=-xValue))
mapproject does care about the sign of the longitude, but if you
incompletely reset the projection it messes things u
> If I want to divide the column of a matrix by the sum of the column, should I
> loop over the columns or can I use apply family?
Looping's unnecessary.
See ?scale or ?sweep, with ?colSums for two non-looping answers; apply() also
works if you give it a suitable function argument.
S
**
> testLines <- mapproject(yValue, xValue, proj="lambert", param=c(50,65))
For starters, if you give the x,y values in reverse order of what the
mapproject function expects you need to label them: y=yValue,
x=xValue.
(Also, I would have expected longitudes in the Americas to be
negative, but mappr
Here is a simple method I saw mentioned on this list a few years ago:
toExcel <- function(x, tag=FALSE) {write.table(x, "clipboard-128", sep="\t",
row.names=tag)}
***
This message and any attachments are for the named person's use on
Hi all,
Based on Ray's suggestions, I have tried the following script:
library(mapproj)
library(maps)
resuPdfFileName="C:/linesTest.pdf"
pdf(resuPdfFileName)
# Create a map of Canada in Lambert projection
map(database= "worldHires","Canada", ylim=c(39,90), xlim=c(-145,-25),
col=alpha("grey90",0.
On Sep 23, 2014, at 10:41 AM, Kuma Raj wrote:
> Dear R users,
>
> I have a data with month and year columns which are both characters
> and wanted to create a new column like Jan-1999
> with the following code. The result is all NA for the month part. What
> is wrong with the and what is the ri
On Tue, Sep 23, 2014 at 10:50 AM, ce wrote:
>
> Hello All,
>
> Is there any IPC tools like in UNIX/Linux systems in R ?
> I know there is mmap package but I am looking something more like sockets .
> Any example appreciated .
>
> Thx .
>
Try looking at the documentation for socketConnection()
?so
Hello All,
Is there any IPC tools like in UNIX/Linux systems in R ?
I know there is mmap package but I am looking something more like sockets .
Any example appreciated .
Thx .
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinf
Dear R users,
I have a data with month and year columns which are both characters
and wanted to create a new column like Jan-1999
with the following code. The result is all NA for the month part. What
is wrong with the and what is the right way to combine the two?
ddf$MonthDay <- paste(month.abb
Neither. Unless I misjudge, you really really need to do your homework
and read some basic R documentation -- e.g. An Intro to R, which ships
with R (and also the Posting Guide) -- before posting here further.
-- Bert
Bert Gunter
Genentech Nonclinical Biostatistics
(650) 467-7374
"Data is not in
Hi,
If I want to divide the column of a matrix by the sum of the column, should I
loop over the columns or can I use apply family?
Regards,
Carol
[[alternative HTML version deleted]]
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If you looked at the documentation for R2HTML you might have noticed that there
is no function HTML.matrix. Perhaps your recommendation from an unnamed source
is out of date? Assuming you loaded the package with library(R2HTML) as Ivan
suggested, the command would be
HTML( summary(iris), file("
On 23/09/2014 9:04 AM, Angel Rodriguez wrote:
Dear Subscribers,
I've found this recommendation to paste an R table to Excel:
HTML.matrix( summary(iris), file("clipboard", "w"), append=F )
# paste into Excel
After installing R2HTML and writting that command, I get:
Error: could not find fu
library(R2HTML) ??
Le 23/09/14 15:04, Angel Rodriguez a écrit :
Dear Subscribers,
I've found this recommendation to paste an R table to Excel:
HTML.matrix( summary(iris), file("clipboard", "w"), append=F )
# paste into Excel
After installing R2HTML and writting that command, I get:
Error
Dear Subscribers,
I've found this recommendation to paste an R table to Excel:
HTML.matrix( summary(iris), file("clipboard", "w"), append=F )
# paste into Excel
After installing R2HTML and writting that command, I get:
Error: could not find function "HTML.matrix"
Any clue?
Thank you very m
Beware of the trap of listening to people with no knowledge of basic
numerical methods!
It really is basic that the results of floating-point computer
calculations depends on the order in which they are done (and the
compiler can change the order). Using == on such calculations is warned
abo
On 23/09/2014 3:27 a.m., Alain Dubreuil wrote:
Hi. I have a requirement to plot a series of points on a map of Canada along
with boundaries defining search and rescue (SAR) regions. I have been
successful in plotting the map of Canada (Lambert projection) and the points,
but I have been unab
Hi Stephane,
This is the well known result of limitted floating point precision (e.g.,
http://www.validlab.com/goldberg/addendum.html). Using a test of
approximate rather than exact equality shows R yields the correct answer:
nperm <- 1
Fperm <- replicate(n=nperm, anova(lm(sample(Y) ~ F, dat
Recently, I came across a strange and potentially troublesome behaviour of the
lm and aov functions that ask questions about calculation accuracy. Let us
consider the 2 following datasets dat1 & dat2 :
> (dat1 <- data.frame(Y=c(1:3, 10+1:3), F=c(rep("A",3), rep("B",3
Y F
1 1 A
2 2 A
3
Hi,
With ggplot2 I can use the following to create a rectangle
geom_rect(aes(ymin=as.Date("8-Apr-2014", format="%d-%b-%Y"),
ymax=as.Date("30-Apr-2014", format="%d-%b-%Y"),
xmin="node002",xmax="node098"),
where the x values are levels of a factor. This works if I
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