On Sat, Jul 25, 2020 at 12:40 AM Martin Maechler
wrote:
> Good answers to this question will depend very much on how many
> 'Machine' and 'Region' levels there are.
I second that.
And unless I missed something, the OP hasn't answered this question, as such.
But "10k+" combinations, does imply
Dear Byron,
On 2020-07-29 18:04 -0400, Duncan Murdoch wrote:
> The arrow3d function is also a pure R
> function, but not a generic. You can
> see the source by typing "arrow3d".
... but if I type rgl::shade3d, I get
> rgl::shade3d
function (x, ...)
On 2020-07-29 22:56 +0200, Ulrik Stervbo wrote:
> So, 50% on topic :-)
I guess so haha :-) Still I hope this
is useful for H ,
others, or he solved it but still no
bottom line volatile situation open for
new ideas.
__
R-help@r-project.org mailing
To begin with, don't assume it is in C++... R supports multiple compiled
languages, and rgl appears to have both C++ and C in it.
I googled "r rgl github" and found an online copy of the src (source) files
right away. The official way is to find the CRAN package page and download the
tar.gz
On 29/07/2020 5:35 p.m., Byron Dom via R-help wrote:
How can I access the C++ source associated with the rgl function shade3d. More
specifically, I'm interested in the part of the code used by arrow3d to draw
arrow heads.
I'm not familiar with how GitHub is organized and I've put in a lot of
How can I access the C++ source associated with the rgl function shade3d. More
specifically, I'm interested in the part of the code used by arrow3d to draw
arrow heads.
I'm not familiar with how GitHub is organized and I've put in a lot of effort
doing searches there and on the web in general.
True, but the question was also how to control for formats and naming columns
while loading the file.
The only way I know how to do this (sans work on my part) is through the
functions in readr. So, 50% on topic :-)
Best,
Ulrik
On 29 Jul 2020, 17:59, at 17:59, Rasmus Liland wrote:
>Dear
On 29/07/2020 4:16 p.m., Sebastien Bihorel wrote:
Thanks Duncan,
(Sorry for the repeated email)
People working in my field are frequently (and rightly) accused of
butchering statistical terminology. So I guess I'm guilty as charged
I will look into the suggested path. One question though
Thanks Duncan,
(Sorry for the repeated email)
People working in my field are frequently (and rightly) accused of butchering
statistical terminology. So I guess I'm guilty as charged
I will look into the suggested path. One question though in your expression of
loglik, p is "a + b*x/(c+x)".
Thanks Duncan,
People working in my field are frequently (and rightly) accused of butchering
statistical terminology. So I guess I'm guilty as charged
I will look into the suggested path. One question though in your expression:
loglik <- sum( resp*log(p) + (1-resp)*log1p(-p) )
a + b*x/(c+x)
Just a quick note about jargon: you are using the word "likelihood" in
a way that I (and maybe some others) find confusing. (In fact, I think
you used it two different ways, but maybe I'm just confused.) I would
say that likelihood is the probability of observing the entire data set,
Hello Ritwik,
There is another possibility.
You can count (crosstab) the number of elements for each Region and Machine
(with table() function) and represent this table with geom_tile() function.
Wit this you will get an equivalent of a heatmap which will give you a good
sense of which
My earlier posting on this thread was misleading. I thought the OP was trying to
fit a sigmoid to data. The problem is about fitting 0,1 responses.
The reproducible example cleared this up. Another strong demonstration that
a "simple reproducible example" can bring clarity so much more quickly
Dear Ulrik,
On 2020-07-29 17:14 +0200, Ulrik Stervbo via R-help wrote:
> library(readr)
> read_csv(
This thread was about
sqldf::read.csv.sql ...
What is the purpose of bringing up
readr::read_csv? I am unfamilliar with
it, so it might be a good one.
Best,
Rasmus
signature.asc
You might achieve this using readr:
```
library(readr)
lines <- "Id, Date, Time, Quality, Lat, Long
STM05-1, 2005/02/28, 17:35, Good, -35.562, 177.158
STM05-1, 2005/02/28, 19:44, Good, -35.487, 177.129
STM05-1, 2005/02/28, 23:01, Unknown, -35.399, 177.064
STM05-1, 2005/03/01,
You might achieve this using readr:
```
library(readr)
lines <- "Id, Date, Time, Quality, Lat, Long
STM05-1, 2005/02/28, 17:35, Good, -35.562, 177.158
STM05-1, 2005/02/28, 19:44, Good, -35.487, 177.129
STM05-1, 2005/02/28, 23:01, Unknown, -35.399, 177.064
STM05-1, 2005/03/01,
Hi Pedro,
I see you use dplyr and ggplot2. Are you looking for something like
this:
```
library(ggplot2)
library(dplyr)
test_data <- data.frame(
year = c(rep("2018", 10), rep("2019", 8), rep("2020", 6)),
value = sample(c(1:100), 24)
)
test_data <- test_data %>%
group_by(year) %>%
Hello,
Try %in% instead of == in:
m2<-m1[m1$`Classification Description` == levels(m1$`Classification
Description`)[c(1,15,2,4],]
Hope this helps,
Rui Barradas
Às 14:57 de 29/07/2020, Engin Yılmaz escreveu:
Dear
I try to create a new subset from my dataframe.
My dataframe's name is m1.
Dear Engin,
On 2020-07-29 16:57 +0300, Engin Yılmaz wrote:
> Dear
>
> I try to create a new subset from my dataframe.
> My dataframe's name is m1.
> "Classification Description" column has 15 different factors.
> The following code is used creating a subset for 1 factor.
>
Richard
Per your requests:
1. Plain text: no spaces
2. read_docx: spaces
3. read_rtf: no spaces
4. Not requested by you: copying from the Word document, then pasting
into “vim”: no spaces
The Word document was created by hand but #1, #3, and #4 confirm that
Dear
I try to create a new subset from my dataframe.
My dataframe's name is m1.
"Classification Description" column has 15 different factors.
The following code is used creating a subset for 1 factor.
m2<-m1[m1$`Classification Description` == levels(m1$`Classification
Description`)[1],]
My aim
Thank your, Pr. Nash, for your perspective on the issue.
Here is an example of binary data/response (resp) that were simulated and
re-estimated assuming a non linear effect of the predictor (x) on the
likelihood of response. For re-estimation, I have used gnlm::bnlr for the
logistic
Then this should work:
```
library(ggplot2)
library(cowplot)
p1 <- ggplot(iris, aes(x = Sepal.Length, y = Sepal.Width)) +
geom_point()
p2 <- ggplot(iris, aes(x = Petal.Length, y = Petal.Width * 1000)) +
geom_point()
plot_grid(p1, p2, ncol = 1, align = "hv", rel_heights = c(2, 1), axis =
Hi Ritwik,
I haven't seen any further answers to your request, so I'll make a
suggestion. I don't think there is any sensible way to illustrate that
many data points on a single plot. I would try to segment the data by
machine type or similar and plot a number of plots.
Jim
On Fri, Jul 24, 2020
Hi!
How about this?
--- snip ---
> x <- c("STRING 01. Remainder of the string","STR ING 01. Remainder
of the string","STRIN G 01. Remainder of the string","STR IN G
01. Remainder of the string")
> x1<-unlist(strsplit(x,"\\."))
> for (i in seq(1,length(x1),2)) { x[(i+1) %/%
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