"SD = ", S, "Fluo = ", q, "\n")
if(q wrote:
>
> Hello Luigi,
>
> Great follow-up — Looks like you’re on the right track using nls() for
> nonlinear regression. You're fitting a logistic-like sigmoidal model (as in
> Rutledge’s paper), I think both
nitial parameter estimates
>mod2
Error in qr.qty(QR.rhs, .swts * ddot(attr(rhs, "gradient"), lin)) :
NA/NaN/Inf in foreign function call (arg 5)
```
How can I properly set this regression model?
Thank you
On Wed, Apr 16, 2025 at 7:08 AM Luigi Marongiu wrote:
>
> Thank you. This
Thank you. This topic is more complicated than anticipated. Best regards, Luigi
On Tue, Apr 15, 2025 at 11:09 PM Andrew Robinson wrote:
>
> A statistical (off-topic!) point to consider: when the GLM was fitted, you
> conditioned on x and let y be the random variable. Therefore
I have fitted a glm model to some data; how can I find the inverse
function of this model? Since I don't know the y=f(x) implemented by
glm (this works under the hood), I can't define a f⁻¹(y).
Is there an R function that can find the inverse of a glm model?
Thank you.
The working example is:
```
the formula isn't convenient, you can approximate the
> answer using the R function uniroot().
>
> Duncan Murdoch
>
> On 2025-04-11 11:53 a.m., Luigi Marongiu wrote:
> > I am trying to predict the intersection value between a curve and a line.
> > I can fit a logistic
I am trying to predict the intersection value between a curve and a line.
I can fit a logistic model to the data by converting the data to the
range 0-1. How can I determine the intersection with a line?
Also, is there a way to do the same without converting the data?
Here is an example:
```
val =
ig = 3)
```
However, this level is too low; the Cq would probably be way higher on
the y-axis, and there are instances where I would like to use a
pre-fixed threshold value.
How can I determine a Cq with R?
Thank you
--
Best regards,
Luigi
__
R-help@r-proje
Thank you, that is exactly was I was looking for.
On Thu, Mar 13, 2025 at 8:16 AM Rui Barradas wrote:
>
> Hello,
>
> Inline.
>
> Às 05:19 de 13/03/2025, Luigi Marongiu escreveu:
> > Thank you, but then I will have long labels also on the axis,
>
>
> I am not g
Thank you, but then I will have long labels also on the axis, making the
plot too crowded. I would like instead to force the long labels only on the
legend...
On Wed, 12 Mar 2025, 12:47 Rui Barradas, wrote:
> Às 10:35 de 12/03/2025, Luigi Marongiu escreveu:
> > I have a data f
I have a data frame with measurements in different conditions. I set
the conditions as a factor using a notation for ease of use. I now
want to plot the data and assign meaningful labels to the factors. I
am using ggplot2; for the x axis I would like to keep the factors but
in the legend I would li
Already did...
On Tue, Apr 2, 2024 at 10:45 AM Eric Berger wrote:
>
> According to https://cran.r-project.org/web/packages/genoPlotR/index.html
> the maintainer of genoPlotR is
>
> Lionel Guy
>
> Send your question also to him.
>
> On Tue, Apr 2, 2024 at 11:27
-12 annotations in gff3
and genebank formats, but how do I load them in the system so that I
plot it here?
- Is it possible to zoom in?
- Is it possible to change the color scheme?
Thank you
--
Best regards,
Luigi
__
R-help@r-project.org mailing list -
Thank you. I though the values would have been reversed automatically.
Case closed then
On Tue, Mar 5, 2024 at 2:43 PM Duncan Murdoch wrote:
>
> On 05/03/2024 7:16 a.m., Luigi Marongiu wrote:
> > A <- rep(c(0, 3, 6, 12, 24), 3)
> > B <- c(rep(0,5), rep(1,5), rep(2,5))
&
Hello,
I am drawing some data with ggplot2 and would like to reverse the
order of the y axis and set a custom range for it.
I can do the former but when I add the key `limits` to
`scale_y_reverse` I get an error as shows below and, worse, no data
shown in the plot.
How can I properly set a reverse
Thank you for your answer. I will implement it, but still I reckon
ggplot2 cannot do the whole thing on its own terms: I need to prep the
data beforehand.
Cheers
Luigi
On Thu, Nov 16, 2023 at 1:39 PM Rui Barradas wrote:
>
> Às 11:59 de 16/11/2023, Luigi Marongiu escreveu:
> > Hello
Hello,
I have triplicate (column A) readings (column D) of samples exposed to
different concentrations (column C) over time (column B).
Is it possible to draw a line plot of the mean values for each
concentration (C)? At the moment, I get a single line.
Also, is there a simple way to draw the 95% C
1 BACT2 88.3
> > 31 BACT3 18.0
> >
> > Or using a different form, that might be more straightforward to you:
> >
> >> aggregate(df[, c("OD", "ODnorm")], by = df[, c("Time", "Target", "
uot;, "Target", "Conc")],
> > data = df, FUN = "mean")
> Time Target Conc OD ODnorm
> 11 BACT1 765. 108.3
> 21 BACT2 745. 88.3
> 31 BACT3 675. 18.0
>
> Sarah
>
> On Tue, Oct 2
Hello,
I have a data frame with different groups (Time, Target, Conc) and
each entry has a triplicate value of the measurements OD and ODnorm.
How can I merge the triplicates into a single mean value?
I tried the following:
```
df = data.frame(Time=rep(1, 9), Well=paste("A", 1:9, sep=""),
Thank you that is exactly it.
I know these are basic questions but I can't find the solution on
either manuals or online...
Best regards
Luigi
On Tue, Sep 5, 2023 at 11:33 PM Bert Gunter wrote:
>
> Luigi:
>
> For base R graphics, you should always consult ?par for o
I would like to draw a graph where the y-lables are missing but the
marks still present.
In this example, I get marks from 2 to 140 000 with increments of
20 000. I could use `plot(... yaxt="n"...)` combined with `axis(2,
at..., label="")` but this needs to know exactly the sequence of marks
pr
Thank you, the problem was indeed about factors.
Case solved
On Fri, Jul 14, 2023 at 7:05 PM Rui Barradas wrote:
>
> Às 17:33 de 14/07/2023, Luigi Marongiu escreveu:
> > Hello,
> > I am measuring a certain variable at given time intervals and
> > different concentratio
Hello,
I am measuring a certain variable at given time intervals and
different concentrations of a reagent. I would like to make a scatter
plot of the values, joined by a line to highlight the temporal
measure.
I can plot this all right. Now, since I have more than one replicate,
I would like to ad
Thank you that is exactly it!
The problem was to connect each point of the series 'Conc' with a line.
Best regards
Luigi
On Sat, Jul 1, 2023 at 8:33 PM Chris Evans wrote:
>
>
> On 01/07/2023 19:20, Luigi Marongiu wrote:
> > Hello,
> > I have a dataframe with
observation. Do you need to
adjust the
group aesthetic?
```
and no lines are drawn.
Where am I going wrong?
Thank you
Luigi
```
df = data.frame(Conc = c(rep(1, 3), rep(2, 3), rep(5, 3)),
Time = rep(1:3, 3),
Value = c(0.91, 0.67, 0.71, 0.91, 0.65, 0.74, 0.95, 0.67, 0.67
as large or larger than the observed.
> This is your p-value.
>
> Tim
>
> -Original Message-
> From: R-help On Behalf Of Luigi Marongiu
> Sent: Wednesday, March 22, 2023 5:12 PM
> To: r-help
> Subject: [R] How to test the difference between paired correla
1 - xz)) : NaNs produced
```
What is the right way to run this test?
Shall I apply also yz?
Thank you
--
Best regards,
Luigi
__
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PLEASE do read
Thanks, I will.
On Tue, Dec 6, 2022 at 7:21 AM Jeff Newmiller wrote:
>
> I have never seen this package mentioned on this list. You should contact the
> package maintainer.
>
> On December 5, 2022 9:59:34 PM PST, Luigi Marongiu
> wrote:
> >Thanks for the tip! I think
lished as it gets.
>
> I would look for an example of what you are trying and see if you can get the
> example to work.
>
> LMH
>
>
> Luigi Marongiu wrote:
> > Hello,'
> > I have seen from this link
> > https://cran.r-project.org/web/packages/rcdk/vigne
Hello,'
I have seen from this link
https://cran.r-project.org/web/packages/rcdk/vignettes/using-rcdk.html
that there is a way to draw chemical structures using R via rcdk package.
I tried to draw a simple structure but I got an error. What is it
missing? What is the correct syntax?
Thanks
```
libr
Thank you, those are all viable solutions.
Regards
Luigi
On Wed, Nov 30, 2022 at 8:59 PM Rolf Turner wrote:
>
>
> On Wed, 30 Nov 2022 13:40:50 +0100
> Luigi Marongiu wrote:
>
> > Hello,
> > I have a data frame where some lines containing strings including
> > d
Thank you,
I have been trying with [:digit:] but did not work. It worked with
`df$val[grepl('[0-9]', df$val)] = "NUM"`
On Wed, Nov 30, 2022 at 2:02 PM Ivan Krylov wrote:
>
> В Wed, 30 Nov 2022 13:40:50 +0100
> Luigi Marongiu пишет:
>
> > I am formatti
G",
but values entered as number should get the value "NUM".
How do I change such values?
--
Best regards,
Luigi
```
df = data.frame(id = runif(10, 1, 100),
val = c("", "POs", "Pos", "P", "Y",
&
Perfect, thank you!
On Fri, Oct 28, 2022 at 11:53 AM Rui Barradas wrote:
>
> Às 10:43 de 28/10/2022, Luigi Marongiu escreveu:
> > Hello,
> > I have a data frame with a string column. All data that are neither
> > "POS" nor "NEG" should've repl
uot;, "31.35", "POS", "20.61", "NEG"),
stringsAsFactors = F)
df$b[!(df$b == "POS") & (df$b == "NEG")] = NA
df$b[(df$b == "POS") | (df$b == "NEG")] = NA
```
--
Best regards,
Luigi
insemius
> Sent: Sunday, September 11, 2022 11:50 PM
> To: Ebert,Timothy Aaron ; Luigi Marongiu
>
> Cc: r-help
> Subject: Re: [R] Interpreting fa.diagram from package psych
>
> [External Email]
>
> On 9/11/22 07:17, Ebert,Timothy Aaron wrote:
> > It is a bad grap
they represent?
Is there a way to show an auto-legend?
Thank you
On Sat, Sep 10, 2022 at 11:33 PM David Winsemius wrote:
>
>
> On 9/10/22 14:08, Luigi Marongiu wrote:
> > Hello,
> > I have plotted data from exploratory factor analysis, and I got a
> > graph similar to
Hello,
I need to convert an R vector to a python array. Essentially, I got a
vector of strings where each element must be enclosed in single
quotes. The problem is that each closing single quote should contain a
comma. What is the regex trick to do that?
I tried with:
```
> (raws = c("field_1", "fi
Thank you
On Thu, May 26, 2022 at 12:38 PM Ivan Krylov wrote:
>
> В Thu, 26 May 2022 12:30:12 +0200
> Luigi Marongiu пишет:
>
> > > Error in forest.meta(m10, sortvar = TE, predict = TRUE, print.tau2
> > > = TRUE, :
> > argument 3 matches multiple formal ar
predict = TRUE, print.tau2 = TRUE, :
argument 3 matches multiple formal arguments
```
What am I getting wrong? How can I solve it?
--
Best regards,
Luigi
__
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xt = element_text(size = 10,face = 'bold'))
> ```
>
> Ron.
>
>
> On 22/02/2022 14:25, Luigi Marongiu wrote:
> > I am trying instead to increase the size with `theme(strip.text.x =
> > element_text(size = 10))` (evem from 1 onwards) or `theme(strip.text.x
&
Removed 24 rows containing missing values (geom_point).
On Tue, Feb 22, 2022 at 12:41 PM Ron Crump wrote:
>
> Hi Luigi,
>
> > # FROM HERE IT DOES NOT WORK
> > facet_grid(. ~ Species, scales="free_y") + # THE SCALE IS FIXED FOR ALL
> > PANELS
>
> I don
No wonder it was not working: I was using the wrong functions (so far
of manuals)
This is great, thank you! All tips worked fine, including the free axis scale.
On Tue, Feb 22, 2022 at 12:41 PM Ron Crump wrote:
>
> Hi Luigi,
>
> > # FROM HERE IT DOES NOT WORK
> > f
Hello,
I have a dataframe with 3 columns: the actual measurement
(Concentration), and two groups (Vitamin and Group). I would like to
plot a barplot with superimposed a jitterplot, with error bars. I am
using facet_grid. All that I can do, but the customization does not
work. The customization is a
(data = df.summary, fill = NA, color = "black") +
+ geom_errorbar(aes(ymin = len, ymax = len+sd),
+ data = df.summary, width = 0.2) +
+ geom_jitter(position = position_jitter(0.2), color = "black&q
Mon, Jan 17, 2022 at 4:22 AM Rolf Turner wrote:
>
>
> On Thu, 13 Jan 2022 20:38:04 +0100
> Luigi Marongiu wrote:
>
> > Hello,
> > I have a numerical variable (x) and a series of categories. I would
> > like to make a box plot of x against each of the categories.
Hello,
I have a numerical variable (x) and a series of categories. I would
like to make a box plot of x against each of the categories. How can I
arrange the data so that I can accomplish it with lattice?
At the moment I got this:
```
df = data.frame(x = c(rep(1,5), rep(2,5), rep(3,5)),
y = rnorm(1
df, quanti.sup=2, quali.sup=c(11:33))
Error in dimnames(res) <- list(attributes(tab)$row.names, listModa) :
length of 'dimnames' [2] not equal to array extent
> nrow(df[10])
[1] 27
> nrow(df[11:33])
[1] 27
```
What would be the problem?
Thank you
--
Best regards,
Luigi
__
$int<- interaction(iris$bin, iris$Species)
> boxM(iris[,1:4], iris[,7])
>
> Cheers
> Petr
>
> > -Original Message-
> > From: R-help On Behalf Of Luigi Marongiu
> > Sent: Tuesday, January 4, 2022 11:56 AM
> > To: r-help
> > Subject: [R] how to r
I have a data frame containing a half dozen continuous measurements
and over a dozen ordinal variables (such as, death, fever, symptoms
etc).
I would like to run a box matrix test and I am using biotools' boxM,
but it allows to run only one ordinal group at the time. For instance:
```
>data(iris)
>
Got it, thanks!
On Tue, Dec 14, 2021 at 11:59 AM Ivan Krylov wrote:
>
> On Tue, 14 Dec 2021 11:54:08 +0100
> Luigi Marongiu wrote:
>
> > "9/29/2021"
>
> > format = "%d/%m/%y"
>
> > Why the conversion did not work?
>
> The according
Hello,
I have these kind of dates:
```
> ori[[n[3]]]
[1] "9/20/2021" "9/20/2021" "9/20/2021" "9/20/2021" "9/20/2021" "9/20/2021"
[7] "9/20/2021" "9/20/2021" "9/20/2021" "9/20/2021" "9/21/2021" "9/21/2021"
[13] "9/21/2021" "9/21/2021" "9/21/2021" "9/21/2021" "9/21/2021" "9/21/2021"
[19] "9/21/2021"
Awesome, thanks!
On Wed, Nov 17, 2021 at 7:50 PM Bill Dunlap wrote:
>
> Try using at=c(1.8, 2.8) to specify the contour levels you want (and omit the
> cuts= argument).
>
> -Bill
>
> On Wed, Nov 17, 2021 at 5:41 AM Luigi Marongiu
> wrote:
>>
>> I have a d
icking things into it."
> -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
>
>
> On Wed, Nov 17, 2021 at 1:42 AM Luigi Marongiu
> wrote:
>>
>> sorry, it was easier than expected: just add `lwd` to the main cal.
>> sorry I coul
I have a dataframe of three variables: x, y, z. The value of z are:
```
> unique(df$z)
[1] 1.0 1.2 1.4 1.6 1.8 2.0 2.2 2.6 3.0 2.4 2.8
```
I would like to plot the contour where the data get integer values
(1.0, 2.0, 3.0) but also highlight where the 1.8 and 2.8 values
occurred. Thus, I am plotting
Hello,
I have a dataframe with 3 variables. I want to loop through it to get
the mean value of the variable `z`, as follows:
```
df = data.frame(x = c(rep(1,5), rep(2,5), rep(3,5)),
y = rep(letters[1:5],3),
z = rnorm(15),
stringsAsFactors = FALSE)
m = vector()
for (i in unique(df$y)) {
s = df[df$y
sorry, it was easier than expected: just add `lwd` to the main cal.
sorry I could not stop the message before checking...
On Wed, Nov 17, 2021 at 10:31 AM Luigi Marongiu
wrote:
>
> Hello,
> I have generated a contourplot with lattice. How do I set the line
> width? I tried with:
>
Hello,
I have generated a contourplot with lattice. How do I set the line
width? I tried with:
```
library(lattice)
contourplot(Z ~ X*Y, data = df, cuts = 3,
panel=function(x,y,...){
panel.contourplot(lty=1, lwd = 3)
})
```
but did not work...
Thank you
Hello,
I have a dataframe with 3 values and that I would like to plot with contour:
```
> head(df)
Y X Z
1 0.0008094667 50 1
2 0.0012360955 50 1
3 0.0016627243 50 1
4 0.0020893531 50 1
5 0.0025159819 50 1
6 0.0029426108 50 1
> contour(df$X
Hello,
I have a large database with a column containing a factor:
```
> str(df)
'data.frame': 500 obs. of 4 variables:
$ MR : num 0.000809 0.001236 0.001663 0.002089 0.002516 ...
$ FCN : num 2 2 2 2 2 2 2 2 2 2 ...
$ Class: Factor w/ 3 levels "negative","positive",..: 1 1 1 1 1 1 1 1 1 1
Awesome, thanks!
On Wed, Nov 10, 2021 at 10:33 PM Rasmus Liland wrote:
>
> Dear Luigi,
>
> Yes, from the named list of dataframes,
> you access the demographic table like
> that.
>
> You can remove dput() now, I used it
> only to print the output; read ?dput,
> it&
nt: chr "2020-11-01" "2022-01-01"
$ EnrolTime : chr "13:24:00" "14:25:00"
```
Still, I thought it might have been easier to run `odb.read`.
But it works, thank you again!
On Wed, Nov 10, 2021 at 7:03 PM Rasmus Liland wrote:
>
> On 2021-11-10 18
I can run the query? R
--
Best regards,
Luigi
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and pro
able not found in statement [SELECT * FROM demographic]
Statement: SELECT * FROM demographic"
```
the database contains a table `demographic` but I can't select it.
Would you know what is the correct syntax?
On Wed, Nov 10, 2021 at 3:43 PM Rasmus Liland wrote:
>
> Dear Luigi,
>
>
/lib/psqlodbcw.so' : file not found
> con <- dbConnect(odbc::odbc(),
+ driver = "/usr/local/lib/libsqlite3odbc.dylib",
+ database = "proof.odb",
+ uid = "",
+ pwd = "",
+ host = "localhost",
+ port = 5432)
Erro
Hello,
I am connecting R to a libreoffice-generated database (.odb) using the
ODB package.
I would like to link a dataframe to a specific table, but I cannot
manage to extract a given table:
```
> db = odb.open(".../LOdatabase.odb", jarFile = NULL)
> odb.tables(db)
$demographic
field.name field.ty
Thanks for the tip! I'll check it out.
On Sun, Oct 24, 2021 at 8:07 AM Jim Lemon wrote:
>
> Hi Luigi,
> In that case you will want a binomial confidence interval.
>
> Jim
>
> On Sun, Oct 24, 2021 at 4:39 PM Luigi Marongiu
> wrote:
> >
> > Thank you. So
nto it."
> -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
>
>
> On Sat, Oct 23, 2021 at 1:19 PM Luigi Marongiu
> wrote:
>>
>> Hello,
>> I am using the ksvm function from the library kernlab to generate an
>> SVM classification
ot numbers.
On Sun, Oct 24, 2021 at 12:08 AM Jim Lemon wrote:
>
> Hi Luigi,
> I may be missing the point, but:
>
> matrix((z1+z2+z3)/3,ncol=10)
>
> gives you the mean rating for each item, and depending upon what
> distribution you choose, the confidence intervals could be
Hello,
I am using the ksvm function from the library kernlab to generate an
SVM classification. I am running the model with k-mean
cross-validation, thus obtaining different accuracy.
Is it possible to merge the different models obtained with the
separate data set to generate a kind of median model
Hello,
I have a series of classifications of the same data. I saved this
classification in a single dataframe (but it could be a list). X and Y
are the variable and Z is the classification by three raters. `I` is
the individual identifier of each entry:
```
z1 = c(0,0,0,0,0,1,0,0,0,2,
0,1,1,1,0,0,0
Hello,
I have run some support vector machine analysis. If I draw a grid of
10*10 points in a space, the model I built will assign the points to a
given group. Lets' say:
```
results1 = data.frame(row_1 = c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0),
row_2 = c(0, 0, 0, 1, 0, 0
hings into it."
> -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
>
>
> On Thu, Oct 21, 2021 at 2:00 PM Luigi Marongiu
> wrote:
>>
>> Hello,
>> I would like to draw 5 figures in the same plot. The layout is:
>> first row: 1
Hello,
I would like to draw 5 figures in the same plot. The layout is:
first row: 1 column
second row: 2 columns
third row: 2 columns
I have used split.screen:
```
> split.screen(c(3, 1)) # split display into 3 screens
[1] 1 2 3
> split.screen(c(1, 2), screen = 2) # split second screen
), to=max(x1,x2), length=399)
> p <- a*dnorm(x, x1, s1) + (1-a)*dnorm(x, x2, s2)
> m <- min(p)
> x[x == m]
>
> On Wed, 13 Oct 2021 at 22:12, Luigi Marongiu wrote:
> >
> > Hello,
> > I have two peaks distributions (and some noise values in between), and
> &g
Got it, thanks!
On Thu, Oct 14, 2021 at 9:59 AM Eric Berger wrote:
>
> df[(df$Y>0.2) & (df$X<10),][2,]
>
> On Thu, Oct 14, 2021 at 10:52 AM Luigi Marongiu
> wrote:
>>
>> Hello,
>> I have selected a subset of a dataframe with the vector syntax (if
>
Hello,
I have selected a subset of a dataframe with the vector syntax (if
this is the name):
```
> df[(df$Y>0.2) & (df$X<10),]
YX
10 0.2200642 1.591589
13 0.2941828 1.485951
```
How can I select only the second row? I only managed to get the whole
of whole columns:
```
> df[(df
Thank you!
On Wed, Oct 13, 2021 at 12:00 PM Deepayan Sarkar
wrote:
>
> On Wed, Oct 13, 2021 at 12:13 PM Luigi Marongiu
> wrote:
> >
> > I have seen that the only package that easily rotate the plot is
> > ggplot, so I ran:
> > ```
> > library(ggplot2)
;]~df$X[df$GROUP=="positive"],
pch=16, cex=1.5)
points(df$Y[df$GROUP=="uncertain"]~df$X[df$GROUP=="uncertain"],
pch=16, cex=1.5, col="grey")
abline(h=Cut_off, col="blue", lwd=2)
abline(h=Cut_offE, col="blue", lwd=2, lty=2)
```
--
Best r
SE, ## this is the print.trellis method
> panel.width = list(1,"npc"),
> panel.height = list(1, "npc")
> )
>
> Bert Gunter
>
> "The trouble with having an open mind is that people keep coming along and
> sticking things into it."
>
I tried to assign the second
plot to an object p and then call grid.draw(p), or to assign g to the
second plot, but they did not work...
Thanks
--
Best regards,
Luigi
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.e
trouble with having an open mind is that people keep coming along and
> sticking things into it."
> -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
>
>
> On Tue, Oct 12, 2021 at 6:49 AM Luigi Marongiu
> wrote:
>>
>> Hello,
>>
(lattice)
xyplot(Value ~ Concentration,
group = Substance, data = df,
pch = 16, cex = 1.2, type = "b",
xlab=expression(bold(paste("Concentration (", mu, "M)"))),
ylab=expression(bold("Infection rate")),
col=COLS,
scales = list(x = list(log = 10, at=c(unique(df$Concentra
> "The trouble with having an open mind is that people keep coming along and
> sticking things into it."
> -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
>
>
> On Mon, Oct 11, 2021 at 12:24 PM Luigi Marongiu
> wrote:
>>
>&
function (x,y) {
panel.segments(x0 = log10(df$Concentration),
x1 = log10(df$Concentration),
y0 = df$Value - dfsd$Value,
y1 = df$Value + dfsd$Value,
col = COLS)
}
)
```
I will check xYplot out, I think it is the tool for the job.
On Mon, Oct 11, 2021 at 3:56 PM Deepayan Sarkar
wrote:
>
> O
Outlook for iOS
>
> From: R-help on behalf of Luigi Marongiu
>
> Sent: Monday, October 11, 2021 7:46:36 AM
> To: r-help
> Subject: [External] [R] Missing text in lattice key legend
>
> Hello,
> I am drawing some data with lattice using:
> ```
> library(lattice)
&
Awesome, thanks!
On Mon, Oct 11, 2021 at 2:19 PM Deepayan Sarkar
wrote:
>
> On Mon, Oct 11, 2021 at 5:17 PM Luigi Marongiu
> wrote:
> >
> > Hello,
> > I am drawing some data with lattice using:
> > ```
> > library(lattice)
> > COLS = c(&qu
),
panel = function (x,y,) {
panel.segments(x0 = df$Concentration, x1 = df$Concentration,
y0 = df$Value - dfsd$Value,
y1 = df$Value + dfsd$Value,
col = COLS)
}
)
```
but the bars are plotted outside the
)
)
),
panel = function(x,y) {
panel.xyplot(x,y)
errbar()
}
)
```
It all works but the legend only shows the colored dots, there is no
text. Is it something missing from the syntax?
Thank
on, data = newdata, type="l")
text(lod+0.2, 0.87, labels = round(lod, 2), cex = 0.9)
```
Case closed
On Sat, Oct 2, 2021 at 4:33 PM Luigi Marongiu wrote:
>
> I tried with:
> ```
> library(chemCal)
> inverse.predict(model, 0.95)
> > inverse.predict(model, 0.95)
>
Hello,
I also tried with
```
library(MASS)
> dose.p(model,p=.95)
Dose SE
p = 0.95: 1.70912 96.26511
```
which is closer to the expected 1.72 but with a very large error (I
expected 1.10-2.34). Is this regression correct?
On Sat, Oct 2, 2021 at 10:14 AM Luigi Marongiu wr
out 1.7. Could
it be that model is based on glm whereas inverse.predict uses linear
regression?
On Sat, Oct 2, 2021 at 10:14 AM Luigi Marongiu wrote:
>
> Hello,
> I have set a glm model using probit. I would like to use it to predict
> X given Y. I have followed this example:
> ```
&g
for f4 but not e4? How do I get X given Y?
Do I need to find the inverse function of logit (which one?) and apply
this to the regression or is there a simpler method?
Also, is it possible to plot the model to get a smooter line than
`plot(positivity ~ concentration,
ce: 4.6727e-01 on 4 degrees of freedom
Residual deviance: 1.9445e-07 on 3 degrees of freedom
AIC: 4.3016
Number of Fisher Scoring iterations: 25
```
How can I now find the concentration that gives 95% positivity?
Thanks
--
Best regards,
Luigi
__
R
all(z == "")
>> [1] TRUE
>> > z <- c("a","")
>> > all(z == "")
>> [1] FALSE
>>
>> If this isn't it, just ignore without reply.
>>
>> Bert Gunter
>>
>> "The trouble with h
ly want an AND operator, & or &&, which will only
> return TRUE if all elements are TRUE,
>
> More on logical operators:
> https://stat.ethz.ch/R-manual/R-devel/library/base/html/Logic.html
>
> Sarah
>
> On Thu, Sep 30, 2021 at 9:07 AM Luigi Marongiu
>
Hello,
I have two data frames, each with three rows:
```
df_a <- data.frame(a = letters[1:3], b = LETTERS[1:3], q = c("", "", ""),
stringsAsFactors = FALSE)
df_b <- data.frame(a = letters[4:6], b = LETTERS[4:6], q = c("", "", "1.5"),
stringsAsFactors = FALSE)
```
I need to test whether the datafram
do not know the package names.
> >
> >
> >> On Sep 29, 2021, at 9:46 AM, Luigi Marongiu
> >> wrote:
> >>
> >> Hello
> >> I have R version 4.1.1 (2021-08-10) -- "Kick Things", on an Ubuntu 21
> >> machine. I am trying t
Eigen’ had non-zero exit status
```
and lme4 requires RcppEigen.
I launched
`$ sudo apt-get install r-cran-rcppeigen`
installation successful but I got the same error in installing RcppEigen.
What could be the error?
--
Best regards,
Luigi
__
R-help@r-p
ep 24, 2021 at 7:33 PM Rui Barradas wrote:
>
> Hello,
>
> Like this?
>
>
> mtcars[names(mtcars) != "mpg"]
>
>
> Hope this helps,
>
> Rui Barradas
>
> Às 15:09 de 24/09/21, Luigi Marongiu escreveu:
> > Hello,
> > this is a very
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