Dear guRus
I have data frame as:
str(voldf)
'data.frame': 130 obs. of 8 variables:
$ Date: Date, format: "2014-08-01" "2014-08-05" "2014-08-08" ...
$ kc : num 0.453 0.424 0.468 0.481 0.485 ...
$ sb : num 0.1128 0.123 0.1272 0.1128 0.0949 ...
$ qc : num 0.0626 0.0661 0.0777 0.0765 0.0763
This works:
Change the destination directory to suit you.
MyURL1 = "http://www.ishares.com/us/239726/fund-download.dl";
download.file(MyURL1,paste("C:/Data/Rtest1",date1,"r.xls",sep=""),method="wget",quiet=TRUE,mode="wb",
extra="--header=\"User-Agent: Mozilla/5.0
(X11; Lin
Dear guRus
I have xts data as:
str(volsA)
An ‘xts’ object on 2014-05-13/2015-04-07 containing:
Data: num [1:221, 1:2] 18.8 18.5 18.4 22.2 22 ...
- attr(*, "dimnames")=List of 2
..$ : NULL
..$ : chr [1:2] "vol1" "vol2"
Indexed by objects of class: [Date] TZ: UTC
xts Attributes:
List of 1
Thanks Josh. I am not sure if I can attach a Jpeg. Please find attached.
On Tue, Apr 7, 2015 at 5:47 PM, Joshua Ulrich wrote:
> On Tue, Apr 7, 2015 at 11:30 AM, Raghuraman Ramachandran
> wrote:
>> Dear guRus
>>
>> I have xts data as:
>> str(volsA)
>> An ‘
, Apr 7, 2015 at 5:47 PM, Joshua Ulrich wrote:
> On Tue, Apr 7, 2015 at 11:30 AM, Raghuraman Ramachandran
> wrote:
>> Dear guRus
>>
>> I have xts data as:
>> str(volsA)
>> An ‘xts’ object on 2014-05-13/2015-04-07 containing:
>> Data: num [1:221,
Dear guRus
How can I round of time in R to the nearest 30th minute please?
For example suppose if
>Sys.time()
[1] "2014-11-21 22:49:05.59042 GMT"
then I would like a function that outputs 22:30:00.
if Sys.time is 12:13:22 then I would like to get 12:00:00 etc.
Any help would be appreciated.
Ma
Sorry I forgot to mention it clearly. I like to round it to the
nearest 30th minute that is past. So 12:28:59 will be again 12:00:00
and
12:59:59 will be 12:30:00 etc. Apologies for the lack of clarity in
the beginning.
Many thanks
Raghu
On Fri, Nov 21, 2014 at 10:52 PM, Raghuraman Ramachandran
21, 2014, at 2:55 PM, Raghuraman Ramachandran wrote:
>>
>>> Sorry I forgot to mention it clearly. I like to round it to the
>>> nearest 30th minute that is past. So 12:28:59 will be again 12:00:00
>>> and
>>> 12:59:59 will be 12:30:00 etc. Apologies for
Dear guRus
Can I use R inside Microsoft access Macros and programmes? I understand
that R-Excel is compatible but wish to know if I can use R inside Access
programmes please?
Many thanks
Raghu
[[alternative HTML version deleted]]
__
R-help@r-p
Try this:
mod = lm(methcol ~ exposure+X1+X2+X3+batch, data = meth_matrix)
res=coef(summary(mod))[2,]
It will good if you give a subset of data you are working on by dput.
Raghu
On Sat, Jul 12, 2014 at 3:11 PM, Jessica Timms
wrote:
> Hi,
>
>
>
> I'm quite new to R and currently trying to use
Try package CCA.
On Sat, Jul 12, 2014 at 11:13 PM, Monaly Mistry
wrote:
> Hi,
>
> I was wondering if it's possible in R to do a canonical correlation with
> only one dependent variable and several independent variables.
>
> I've tried using cc(X,Y) but I got an error message. In this case I had
gt; >
>> > Best,
>> > John
>> >
>> > --------
>> > John Fox, Professor
>> > McMaster University
>> > Hamilton, Ontario, Canada
>> > http://socserv.mcmaster.ca/jfox/
>>
Hi Fabio
Can you please reproduce your output? When I tried it curl are the
following issues I get:
>download.file("
http://www.nseindia.com/content/historical/DERIVATIVES/2014/JUL/fo09JUL2014bhav.csv.zip";,
temp, method="curl")
Error in download.file("
http://www.nseindia.com/content/historical/
stall curl(http://curl.haxx.se/) binaries and try the curl
> method again.
> #! Fábio
>
>
> On Thu, Jul 17, 2014 at 5:26 AM, Raghuraman Ramachandran
> wrote:
> > Hi Fabio
> >
> > Can you please reproduce your output? When I tried it curl are the
> following
> &
Try 'shell'? There must be a workaround to do this.
On Fri, Dec 20, 2013 at 9:58 AM, R_Antony wrote:
> Hi,
>
> How to execute ms-Excel Macro(*.xlsm) using R function ? I tried but not
> get. There are method to call from R function from Excel macro, but i need
> Excel macro to execute from R. I
Dear R-users
I am looking for a very experienced R developer for a derivatives project
in Bangalore. The developer should preferably be well versed in GUI etc.
The contract is for 4 to 6 months.
Please contact me in this email.
Thanks
Raghu
[[alternative HTML version deleted]]
Hi
I wish to read a file from my local directory from inside a function. I am
passing the filename as the argument but this does not work.
Say for example
function(dat)
{
dat1=read.csv("D:\\dat.csv",header=TRUE)
}
If I call funtion(dat) I get the following error. 'Intuitively' i understand
this is
guRus
I have say a dataframe, d and I wish to do the following:
1) For each row, I want to take one particular value of the row and multiply
it by 2. How do I do it. Say the data frame is as below:
OPEN HIGH LOW CLOSE 1931.2 1931.2 1931.2 1931.2 0 0 0 999.05 0 0 0 1052.5
0 0 0 987.8 0 0 0 925.
?xts
xts will do what you want. as.xts or xts() are straightforward ways.
On Fri, Aug 6, 2010 at 5:11 PM, Serdar Akin wrote:
> Hi
>
> I want to create a TimeSeries object with already defined dates (in the
> first vector) so that all the data are coerced as a time series object with
> the dates
Dear guRus
Say I have a price vector, P which is P = c(20,50,40,50) for four
consecutive days.
My initial equity is say $100 and I re-invest the profits made in each
transaction so my equity curve increases.
If I go Long on day 1, short and short on day 2, long and long on day 3 and
finally shor
guRus!
I have a function f = exp(x^2-y+(1/z))
Also, x can take values from 1 to 37, y from 2 to 20 and Z from -13 to 51.
How can I find the maximum of f using any of the optimization functions
please?
Is there a way to store the possible values of x, y and Z in a single
variable like in a List
Hello
I have a vector a =(-2,0,0,0,1,0,0,3,0,0,-4)
I want to replace all zeros into previous non-zero state. So for instance the
above vector should be converted into:
a= (-2,-2,-2,-2,1,1,1,3,3,3,-4)
I tried many things and finally concluded that probably(?) rollapply may be the
best way?
I
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
GuRus
How do I use the write function (or write.table or write.csv) to achieve
the following please?
age=c(32,37,39)
names=c("john","peter","jake")
I would like create in a directory 3 files each named as john.csv,peter.csv
and jake.csv and each file have data from the age vector. That is jon.cs
7;cat'
Thanks once again.
Raghu
On Sat, Jul 14, 2012 at 3:22 PM, Raghuraman Ramachandran <
optionsra...@gmail.com> wrote:
> GuRus
>
> How do I use the write function (or write.table or write.csv) to achieve
> the following please?
>
> age=c(32,37,39)
> names=c("
rs) .OO#. .OO#. rocks...1k
> ---
> Sent from my phone. Please excuse my brevity.
>
> Raghuraman Ramachandran wrote:
>
> >GuRus
> >
> >How do I use the write function (or write.tab
Hello
I have a data frame like this:
dput(states)
structure(list(Date = c("24/07/2012", "25/07/2012", "26/07/2012",
"27/07/2012", "28/07/2012", "24/07/2012", "25/07/2012", "26/07/2012",
"27/07/2012", "28/07/2012"), State = c(1L, 1L, 1L, 1L, 1L, -1L,
-1L, -1L, 1L, -1L)), .Names = c("Date", "State")
lps,
>
> Rui Barradas
>
> Em 24-07-2012 07:40, Raghuraman Ramachandran escreveu:
>
>> Hello
>>
>> I have a data frame like this:
>> dput(states)
>> structure(list(Date = c("24/07/2012", "25/07/2012", "26/07/2012",
>> &quo
+ -10L))
> > # mark the state changes
> > states$diff <- c(TRUE, diff(states$State) != 0)
> > states[states$diff,]
> Date State diff
> 1 24/07/2012 1 TRUE
> 6 24/07/2012-1 TRUE
> 9 27/07/2012 1 TRUE
> 10 28/07/2012-1 TRUE
>
&g
Dear R-Gurus
I have a data frame (from CSV file) which has its first column called Date.
The Date is in the format mm/dd/. I was trying to get the weekday for
these dates and I tried using wday() and day.of.week() functions and both
of them gave me precisely the wrong answers. I think the issu
t;13/07/2010", "12/07/2010", "09/07/2010", "08/07/2010",
"07/07/2010", "06/07/2010", "05/07/2010", "02/07/2010", "01/07/2010",
"30/06/2010", "29/06/2010", "28/06/2010", "25/06/2010
ller wrote:
> On Sun, 22 Apr 2012, Hasan Diwan wrote:
>
> Raghu,
>>
>> On 22 April 2012 09:53, Raghuraman Ramachandran > >wrote:
>>
>> I have a data frame (from CSV file) which has its first column called
>>> Date.
>>> The Date is in the fo
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