Before going any further, I have to check, what is:
MinQuadProg qp
Also, if I'm following the C++ code correctly, H, is an identity matrix.
This implies the input to the C++ solver, requires the QP in a
different form to the R solver.
In which case, the C++ inputs and the R inputs, should be diffe
Those seem like useful properties if jitter() is used in plotting (as it
was originally intended), but that use isn't even mentioned in the help
page. Martin wanted to "add a small amount of noise to a numeric
vector" "in order to break ties" (quoting from that help page).
For Martin's use, i
Folks: Please note:
There is *no* way to "jitter" the 3 values 1,2, and 1e5 so that:
a) the jittered values differ from the original ones by a fraction of their
original value;
b) the plotting symbols for the jittered values will be distinguishable on
a linear scale holding all 3 values.
Cheers,
FWIW, there is a similar function called “dither” in the quantreg package.
> On Sep 24, 2020, at 8:08 AM, Martin Keller-Ressel
> wrote:
>
> Dear Duncan, Dear Rui,
>
> thanks for the responses and for pointing out that it is the ‚fuzz‘ part that
> is causing the problem. I agree that this is n
Hi Ana,
The ifelse function works like this:
*ifelse(condition, if.true, if.false)*
it will check the condition, and if, and only if, condition is true, it
will execute whatever is in if.true,
and if condition is false (and only if the condition is false) it will
execute what's in if.false.
wh
Dear Duncan, Dear Rui,
thanks for the responses and for pointing out that it is the ‚fuzz‘ part that
is causing the problem. I agree that this is not a bug, but could be
undesirable/surprising behaviour, since it causes a large ‚discontinuity‘ in
the jitter functions output depending on the inp
Hi All,
I am wondering if anyone has an R implementation of the James (1996) and/or
Bang (2004) blinding assessment indices for randomized, controlled clinical
trials. Ideally, I am looking for both.
I have Googled, search using rseek.org, and checked the CRAN clinical trials
task view and cam
Thank you for giving me your time!
The problem is the quadratic optimization part. Something goes wrong along
the way. In C++ loops run from 0 and in R they run from 1, and I've tried
to take that into account. Still I'm having hard time figuring out the
mistake I make, cause I get a result from m
Hi
And above what Jim suggested, count is only a placeholder for results of
tasks a-d.
Cheers
Petr
> -Original Message-
> From: R-help On Behalf Of Dan Bosio
> Sent: Wednesday, September 23, 2020 11:00 PM
> To: r-help@r-project.org
> Subject: [R] Help with for loops and if statements
>
Hi Dan,
This list has a "no homework" policy, but as you really do seem to be
struggling, I'll suggest that you read the help page for "abs"
carefully and learn about "<-" (assign a value). Perhaps these will
give you a start.
Jim
On Thu, Sep 24, 2020 at 6:19 PM Dan Bosio wrote:
>
> Hi there
>
>
Oh, sorry, forgot about the colors. A list beginning with the color
for the overall summary, then colors for the first factor and so on.
See the help page for examples.
Jim
On Thu, Sep 24, 2020 at 6:32 PM Jim Lemon wrote:
>
> Hi Luigi,
> I thought a lot about that when I was writing the function
Hi Luigi,
I thought a lot about that when I was writing the function. The only
way I could think of to show the nesting was dots with horizontal
lines and it looked messy and was quite hard to visualize the nesting.
If you do have any great ideas I always welcome contributions to
plotrix.
Jim
On
Hi there
I am in an intro to R course and the professor has not been much help. One of
the questions on the latest homework has me stumped. The question is below,
along with my answers so far.
8. [15 points] Given the following code,
#
# x <- rnorm(10)
#
# Do the following.
#
# (1) create a cou
Hi Luigi,
To display a nested breakdown like this I would suggest barNest. This
is one way to display the nesting. Note that if you change the order
of the factors in the formula you will get a different plot, so think
about how you want the summaries nested. Error bars can only be
displayed on the
Hi
instead of complicated ifelse construction I would try perform the task in
several steps
# make a new column
test$new <- NA
# select CURRELIG and RTNPTHY and set new to 1
test$new[which(test$CURRELIG==1 & test$RTNPTHY==1)] <- 1
# select CURRELIG and PLASER and set new to 2
test$new[which(te
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