On Sep 2, 2012, at 03:38 , David Winsemius wrote:
>
> Why should predict not complain when it is offered a newdata argument that
> does no contain a vector of values for "x"? The whole point of the terms
> method of prediction is to offer estimates for specific values of items on
> the RHS of
On Sep 1, 2012, at 4:33 AM, Rui Barradas wrote:
Hello,
John is right, there seems to be an error in predict.lm. It can be
made to work but if the model is fitted with lm(...,data) then it
messes things up. Apparently predict.lm disregards the data argument
and uses whatever it finds in t
On 02/09/12 10:52, CHEL HEE LEE wrote:
I have some trouble to deal the value of 'NaN'. For example,
exp(1e3)
[1] Inf
exp(1e3)*0
[1] NaN
The correct answer should be 0 rather than NaN. I will very appreciate
if anyone can share some technique to get a correct answer.
There is no techniqu
Hi,
Try this:
dat1 <- read.table(text="
PT_ID IDX_DT OBS_DATE DAYS_DIFF OBS_VALUE CATEGORY
13 4549 2002-08-21 2002-08-20 -1 183 2
14 4549 2002-08-21 2002-11-14 85 91 1
15 4549 2002-08-21 2003-02-18 181 89 1
16 4549 2002-08-21 2003
Following the beginning tutorial, I typed help.start() and was given a
choice of browsers. I picked chrome but got back that chrome is not found.
I cannot seem to change or get rid of it now.
I looked at /etc/R/Renviron but don't see anywhere to set browser
(R_BROWSER = ${R_BROWSER ...etc) - nothin
Em 02-09-2012 00:10, Jeff Newmiller escreveu:
I disagree that this answer is "wrong". If you want a mathematically correct
answer you are going to have to obtain it by applying intelligence to the algorithm in
which this calculation occurred.
Logarithms are the product of intelligence.
And t
I disagree that this answer is "wrong". If you want a mathematically correct
answer you are going to have to obtain it by applying intelligence to the
algorithm in which this calculation occurred. This is not a mailing list about
numerical methods in general, so it probably isn't appropriate to
I have some trouble to deal the value of 'NaN'. For example,
> exp(1e3)
[1] Inf
> exp(1e3)*0
[1] NaN
The correct answer should be 0 rather than NaN. I will very appreciate
if anyone can share some technique to get a correct answer.
__
R-help@r-proje
Hello,
Try the following.
dat <- read.table(text="
PT_ID IDX_DT OBS_DATE DAYS_DIFF OBS_VALUE CATEGORY
13 4549 2002-08-21 2002-08-20-1 1832
14 4549 2002-08-21 2002-11-1485911
15 4549 2002-08-21 2003-02-18 181891
16
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
> On Behalf Of arun
> Sent: Friday, August 31, 2012 8:47 AM
> To: Andras Farkas
> Cc: R help
> Subject: Re: [R] Help on numerical object and ifelse function
>
> Hi,
> Try this:
> z1<-c(z,z,z,z,z
On Sep 1, 2012, at 19:20 , Petr Savicky wrote:
> On Sat, Sep 01, 2012 at 02:29:40AM -0700, Andras Farkas wrote:
>> Dear All,
>> ?
>> is there a way to set low and high limits to a simulation with rlnorm()?
>> ?
>> as an example:
>> ?a <-rlnorm(500,0.7,1)
>> ?
>> ?
>> I get the summary of
>> ?
>>
On Sat, Sep 01, 2012 at 02:29:40AM -0700, Andras Farkas wrote:
> Dear All,
> ?
> is there a way to set low and high limits to a simulation with rlnorm()?
> ?
> as an example:
> ?a <-rlnorm(500,0.7,1)
> ?
> ?
> I get the summary of
> ?
> Min. 1st Qu. MedianMean 3rd Qu.Max.
0.1175 1.0590
Hi Rui:
Good points. However, as my offering is, I think, little more than a
disguised loop -- as compared to yours' and Ted's truly vectorized
solutions --I think it is far too inefficient to useful. I offered it
merely for its "aesthetic" interest, which is what you noted in your
comments below.
Yes, Rui's suggestion is what I would do, too.
However, for fun, here is a different, way less efficient, but perhaps
instructive (or amusing?) alternative using Vectorize() and all.equal()
directly:
> foo <- function(target,current,...)isTRUE(all.equal(target,current,...))
> equals <- Vectorize(
Hi,
May be this might help you.
set.seed(1)
a <-rlnorm(500,0.7,1)
a1<-a[a>30|a<0.5]
set.seed(2)
a2 <-rlnorm(500,0.7,1)
a3<-a2[a2>30|a2<0.5]
a4<-a2[!a2%in%a3]
summary(a4)
# Min. 1st Qu. Median Mean 3rd Qu. Max.
# 0.5105 1.2500 2.2990 3.6850 4.7140 25.3600
a[a%in%a1]<-sample(a4,leng
Hi,
I have encountered an issue about finding a date closest to another date
So this is how the data frame looks like:
PT_ID IDX_DT OBS_DATE DAYS_DIFF OBS_VALUE CATEGORY
13 4549 2002-08-21 2002-08-20-1 1832
14 4549 2002-08-21 2002-11-148591
Hello,
The two approaches are conceptually different. With yours we get
equality given a certain number of (significant) digits, which might be
what you are looking for if the measurements are known to be made with
those digits, a frequent case.
And there's also Bert's Vectorize way, with th
Thanks, Rui, but that was not my (apparently implicit) question.
Apologies if my wording gave the wrong impression. I was really
asking if such a function was already available somewhere in R.
Yes, your definition of equals() is one possibility; or, with the
work-round I was using,
equals <- fu
Hello,
Like this?
equals <- function(x, y, tol = .Machine$double.eps^0.5) abs(x - y) < tol
x <- rnorm(5)
y <- x[2]
equals(x, y)
[1] FALSE TRUE FALSE FALSE FALSE
y <- rnorm(3)
y[2] <- x[2]
equals(x, y)
[1] FALSE TRUE FALSE FALSE FALSE
Warning message:
In x - y : longer object length is not a
Greetings All.
Once again, I am probably missing something fairly accessible,
but since I can't find it I'd welcome advice!
I have a dataframe derived from a text file of data in tabular
format. For one of the variables, say X, I want to select the
subsets which in which X equals a particular val
On 09/01/2012 08:16 PM, Roopashree Shrivastava wrote:
Dear all,
I am coming across a funny problem in taylor diagram. When I plot the
diagram for given values of a variable some points are going outside the
diagram. i have never seen such a diagram and hence cannot understand the
reason for it.
Dear All,
is there a way to set low and high limits to a simulation with rlnorm()?
as an example:
a <-rlnorm(500,0.7,1)
I get the summary of
Min. 1st Qu. MedianMean 3rd Qu.Max.
0.1175 1.0590 2.1270 3.4870 4.0260 45.3800
I would like to set limits so that the simulated
Hello,
John is right, there seems to be an error in predict.lm. It can be made
to work but if the model is fitted with lm(...,data) then it messes
things up. Apparently predict.lm disregards the data argument and uses
whatever it finds in the global environment. In the examples below,
after r
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