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
I would like to use your genoPlotR package
(doi:10.1093/bioinformatics/btq413) to compare the genomes of two
isolates of E. coli K-12 that I have. One is a K-12 that was in my
lab's fridge; the other is a derivative of K-12 bought some time ago,
HB101.
I tried to use genoPlotR, but I could not unde
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=""),
as ?axis describes
>
> Again, if this is not what you wanted, I hope you can use the above to modify
> stripchart() to your desired specification.
>
> Cheers,
> Bert
>
> On Tue, Sep 5, 2023 at 12:57 PM Luigi Marongiu
> wrote:
>>
>> I would like to dr
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
Hello,
I have a dataframe with measurements stratified by the concentration
of a certain substance. I would like to plot the points of the
measures and connect the points within each series of concentrations.
When I launch ggplot2 I get the error
```
geom_path: Each group consists of only one obser
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
Hello,
I have three numerical variables and I would like to test if their
correlation is significantly different.
I have seen that there is a package that "Test the difference between
two (paired or unpaired) correlations".
[https://www.personality-project.org/r/html/paired.r.html]
However, there i
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
Hello,
I have a data frame where some lines containing strings including digits.
How do I select those rows and change their values?
In essence, I have a data frame with different values assigned to the
column "val". I am formatting everything to either "POS" and "NEG",
but values entered as numbe
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
Hello,
I have a data frame with a string column. All data that are neither
"POS" nor "NEG" should've replaced by an NA. How can I implement that
(even with extra libraries)? My attempts actually wipe out POS and
NEG...
Thank you
```
df = data.frame(a = 1:5, b = c("", "31.35", "POS", "20.61"
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
Hello,
I am using the package meta to plot a forest plot. My data looks like this:
```
X
Entry Cases Ca_pos Controls Co_pos Method Tissue
Disease Virus Set
6 de Villiers, 200782 72 82 61PCR colon/rectum
cancer TTV 3
7 de Villiers, 2002 162
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
&
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
= element_text(face = "bold"))` but it gives an error (same in both
cases, even when using `strip.text`):
```
No summary function supplied, defaulting to `mean_
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
> > facet_grid(. ~ Species, scale
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
Hello,
I am following this post
https://www.datanovia.com/en/lessons/ggplot-error-bars/ to superimpose
a jitter plot and a barplot with error bars.
Te example works:
```
df2 <- ToothGrowth
df2$dose <- as.factor(df2$dose)
head(df2, 3)
df2.summary <- df2 %>%
group_by(dose) %>%
summarise(
sd
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
Hello,
I am trying to apply multiple correspondence analysis to a data frame
using the FactoMineR package following the post
http://sebastien.ledien.free.fr/unofficial_factominer/factosbest/dimensions-description.html
My data frame has this structure:
```
> str(df)
'data.frame': 27 obs. of 33 var
$int<- interaction(iris$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's very handy the next time you a
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
Sure! I am attaching a simplified version of it. I did not see the
quote text; it might be difficult to call all table's names since
there are hundreds of columns..
On Wed, Nov 10, 2021 at 4:15 PM Rasmus Liland wrote:
>
> Are you willing to send me a part of the
> the proof.odb so I can run the q
Thank you.
I have used ODB, it looks easier to use (and install since it needs
way fewer dependencies than odbc) but I can't run the SQL statement:
```
> library(ODB)
Loading required package: DBI
Loading required package: RJDBC
Loading required package: rJava
> db = odb.open("proof.odb", jarFile
how can I connect to a LibreOffice base odb file using the package odbc for R?
I tried with:
> library(odbc)
> library(DBI)
> con <- dbConnect(odbc::odbc(),
+ driver = "PostgreSQL Driver",
+ database = "proof.odb",
+ uid = "",
+ p
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
calculated
> in much the same way.
>
> Jim
>
> On Sun, Oct 24, 2021 at 7:16 AM Luigi Marongiu
> wrote:
> >
> > 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).
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)
Hello,
I have two peaks distributions (and some noise values in between), and
I used mixtools to characterize the two overlapping data. How can I
find the minimum between the peak distributions?
The sigma value of the second population is close to that value, but I
am not sure if this is really the
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."
>
Hello,
I would like to show a density plot of the Y axis. To do that, I would
like to split the plot into a panel 2/3 long and a density plot 1/3
long. The problem is that, since the density is on the Y axis, the
density plot should be rotated byb90 degrees. I tried with the package
gridGraphics bu
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,
>>
Hello,
I have drawn data from a dataframe as follows:
```
Substance = rep(c("A", "B", "C", "D"),4)
Concentration = rep(1:4,4)
Value = c(62.8067, 116.2633, 92.2600, 9.8733, 14.8233,
92.3733, 98.9567, 1.4833, 0.6467, 50.6600,
25.7533, 0.6900,
> "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
Hello,
I am trying to plot data using lattice. The basic plot works:
```
Substance = rep(c("A", "B", "C", "D"),4)
Concentration = rep(1:4,4),
Value = c(62.8067, 116.2633, 92.2600, 9.8733, 14.8233,
92.3733, 98.9567, 1.4833, 0.6467, 50.6600,
Hello,
I am drawing some data with lattice using:
```
library(lattice)
COLS = c("gold", "forestgreen", "darkslategray3", "purple")
xyplot(Value ~ Concentration,
group = Substance, data = inf_avg,
pch = 16, cex = 1.2, type = "b",
xlab=expression(bold(paste("Concentration (", mu,
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
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:
```
f2<-data.frame(age=c(10,20,30),weight=c(100,200,300))
f3<-data.frame(age=c(15,25))
f4<-data.frame(age=18)
mod<-lm(weight~age,data=f2)
> predict(mod,f3)
1
150
> predict(mod,f4)
Hello,
I would like to calculate the 95% success rate of a test. I have a
series of dilutions and the proportion of positive results out of 37
attempts for each of them. I would like to find the concentration that
gives 95% success and I used logit regression:
```
df <- data.frame(concentration = c
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
Hello
I have R version 4.1.1 (2021-08-10) -- "Kick Things", on an Ubuntu 21
machine. I am trying to install the package meta but I get the
following error:
```
...
ERROR: dependency ‘RcppEigen’ is not available for package ‘lme4’
* removing ‘/home/gigiux/R/x86_64-pc-linux-gnu-library/4.1/lme4’
Warn
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
Hello,
this is a very simple question but...
what is the vector alternative to `subset(dataframe, select = - column)`?
I tried with:
```
> x = new[-ID]
Error in `[.data.frame`(new, -ID) : object 'ID' not found
> x = new[-"ID"]
Error in -"ID" : invalid argument to unary operator
> x = new[[-"ID"]]
E
-liner modifying df, not df1 is
>
>
> df[sapply(df, is.factor)] <- lapply(df[sapply(df, is.factor)], as.character)
>
>
> Hope this helps,
>
> Rui Barradas
>
> Às 11:03 de 19/09/21, Luigi Marongiu escreveu:
> > Thank you Jim, but I obtain:
> > ```
> >&
)
> else return(x)
> df1<-lapply(df,factor2character)
>
> Jim
>
> On Sun, Sep 19, 2021 at 8:03 PM Luigi Marongiu
> wrote:
> >
> > Thank you Jim, but I obtain:
> > ```
> > > str(df)
> > 'data.frame': 5 obs. of 3 variables:
> >
..: 1 2 3 4 5
$ sales : num 13 16 22 27 34
$ country: char "a","b","c","d",..: 1 2 3 4 5
```
On Sun, Sep 19, 2021 at 11:37 AM Jim Lemon wrote:
>
> Hi Luigi,
> It's easy:
>
> df1<-df[,!unlist(lapply(df,is.factor))]
>
> _except
Hello,
I woul dlike to remove factors from all the columns of a dataframe.
I can do it n a column at the time with
```
df <- data.frame(region=factor(c('A', 'B', 'C', 'D', 'E')),
sales = c(13, 16, 22, 27, 34), country=factor(c('a',
'b', 'c', 'd', 'e')))
new_df$region <- droplevel
Thanks
On Wed, Sep 15, 2021 at 5:12 PM Jeff Newmiller wrote:
>
> Really?
>
> str( df[ , grep(".*_a*.", names(df)) ] )
>
>
> On September 15, 2021 7:53:17 AM PDT, Luigi Marongiu
> wrote:
> >Hello,
> >I have a dataframe and I would like to browse
Hello,
I have a dataframe and I would like to browse the information of its
structure only for a subset of columns (since there are hundreds of
them).
For instance, I tried with grepping some columns as in:
```
df <- data.frame(var_a1 = c(letters[1:3], letters[1:4]),
var2 = c(LETTE
NA NA NA
> 6 NA NA NA
> 7 NA NA 0.7987658
> 8 NA NA -0.5225229
> 9 NA NA 0.7673103
> 10 NA NA -0.5263897
>
> So my "nice compact command"
> dat[sapply(dat, is.nan)] <- NA
>
> works as expected, but summary gives as mean NAN.
>
> Che
Thanks, that is perfect!
On Thu, Sep 2, 2021 at 7:02 PM Deepayan Sarkar
wrote:
>
> On Thu, Sep 2, 2021 at 9:26 PM Enrico Schumann
> wrote:
> >
> > On Thu, 02 Sep 2021, Luigi Marongiu writes:
> >
> > > Hello, is it possible to show only the header (that i
as.matrix(x)
> x[is.nan(x)] <- NA_real_
>
> I'd also suggest this same solution for the other question you posted,
>
> x[x == 0] <- NA
>
> On Thu, Sep 2, 2021 at 10:01 AM Luigi Marongiu
> wrote:
>>
>> Sorry,
>> still I don't get it:
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