I am possibly in the market for a new laptop. Predominantly a Windows
user, I owned a macbook pro 10 years ago and am considering going that
route again. Does the standard advice still hold: Get the most
powerful processor (i7), most ram (16GB), and largest internal storage
(512GB), if affordable?
t; R. Mark Sharp, Ph.D.
>> msh...@txbiomed.org
>>
>>
>>
>>
>>
>> > On Feb 25, 2015, at 1:50 PM, Dan Murphy wrote:
>> >
>> > I am possibly in the market for a new laptop. Predominantly a Windows
>> > user, I owned a macbook p
To Users of Excel:
Following advice from Brian and Markus, I created an RMarkdown "vignette"
that shows an example of how the pasteFromExcel function in the excelRio
package on github could be used by an actuary to transfer a triangle from
Excel to R. See today's post at http://trinostics.blogspot
20, 2014 at 10:04 AM, David Winsemius
wrote:
>
> On Oct 19, 2014, at 11:18 PM, Dan Murphy wrote:
>
>> To Users of Excel:
>>
>> Following advice from Brian and Markus, I created an RMarkdown "vignette"
>> that shows an example of how the pasteFromExcel fu
3 4 5 7 # correct answer
*Now* one may remove unwanted characters from the well-formed strings.
And deal with the "negatives" of course .. and NAs. See how that's
done in excelRio.r in the excelRio package on github:
https://github.com/trinostics/excelRio
Thanks for your interest.
I suppose my biggest holdup is knowing how Excel formats currencies in
other denominations. Maybe there's a way for me to test euro,
sterling, etc. in my location, but I haven't discovered it yet. :(
Again, thanks for your help.
-Dan
On Mon, Oct 20, 2014 at 3:39 PM, David Winsemius wro
I have a vector of strings that contain mathematical expressions. E.g.,
x <- c("5 <= 7", "z = 1+2")
and I would like to decompose each expression into its left- and
right-hand-side components etc., output something like
"5" "<=" "7"
"z" "=" "1" "+" "2"
Is there something built into the R language
The lubridate and seq solutions work because they rely on your
specification that the returned date be the first day of the desired
subsequent months. Without that requirement they would not always
work. For example, if you wanted to add Vec months to the last day of
March rather than the first day
<- 2
> f <- function() x^2
> do.call(f, list(), envir = e)
Error in (function () : object 'x' not found
Thanks in advance for clarifying my misunderstanding.
Dan Murphy
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rdoch
wrote:
> On 25/06/2013 9:32 AM, Dan Murphy wrote:
>>
>> I am having difficulty understanding the envir argument of do.call.
>> The help page says
>>
>> envir an environment within which to evaluate the call.
>>
>> so I thought that in the follo
2013 11:56 AM, Dan Murphy wrote:
>>
>> So the trick is to put the function f into e and define its environment to
>> be e:
>
>
> Putting f into e, and defining the environment of f to be e solve different
> problems. Your toy example has both problems so it's a rea
I just noticed this annoyance, but I'm not the first one, apparently
-- see
http://lists.r-forge.r-project.org/pipermail/datatable-help/2012-May/001176.html
The thread never answered the OP's question "Is this a bug?" so I
assume the answer, unfortunately, is No.
If not a bug, do users of within
hin's assumed
backwards column order
Is there a way to capture the names of new objects created within?
On Fri, Apr 4, 2014 at 10:55 AM, Duncan Murdoch
wrote:
> On 04/04/2014 1:32 PM, Dan Murphy wrote:
>>
>> I just noticed this annoyance, but I'm not the first one, appare
e1, e2) e1 /
> as.numeric(e2))
[1] "/"
> x / y
[1] 2 2 2 2
Is it always necessary to define "array" methods for arithmetic on
user defined classes? Maybe the answer has to do with the error
message I did not understand.
Thanks,
Dan Murphy
> sessionInfo()
R version
and "years", the mondate package on CRAN
defines 'mondate' objects that store dates in units of the fraction of
the month that the close of business on that date represents relative
to the month it is in. Its primary users are financial analysts. There
are methods to convert between
Is there a way to determine which, if any, CRAN packages depend on my CRAN
package, mondate?
Thanks,
Dan Murphy
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PLEASE do
In the lubridate vignette
(http://cran.r-project.org/web/packages/lubridate/vignettes/lubridate.html)
setdiff(auckland, jsm)
results in
## [1] 2011-06-04 12:00:00 NZST--2011-07-20 NZST
But if Hadley's mentor had been called away earlier, say, July 31st,
wouldn't the set difference have resulted
I can do this in multiple steps with summarise, joins, etc., but can't
help thinking it can be accomplished in one plyr call.
Here's a small example:
> require(plyr)
> require(lubridate)
> data <- data.frame(
+ date = rep(as.Date(ymd(20140101 + (0:3) * 100)), 2),
+ state = rep(c("A", "B"), ea
13, 2014 at 8:36 AM, arun wrote:
> Hi,
> Try ?which.max() # unique values for the combination.
>
> ddply(data,.(state),summarize,max_date=value[which.max(date)])[,2]
> #or
> ddply(data,.(state),summarize,max_date=value[date == max(date)])[,2]
>
>
> A.K.
>
>
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