{
> >
> > m <- mean(x, na.rm = T)
> >
> > if (!is.Nan (m)) {
> > m <- NA
> > }
> >
> > return (m)
> > }
> >
> > Would do what you suggested in your reply to Bert.
> >
> > On Mon, 16 Sep 2024, 19:48 Rui Barradas,
All' Na Is Na.
Il lun 16 set 2024, 16:29 Bert Gunter ha scritto:
> See the na.rm argument of ?mean
>
> But what happens if all values are NA?
>
> -- Bert
>
>
> On Mon, Sep 16, 2024 at 7:24 AM Francesca
> wrote:
> >
> > Sorry for posting a non u
,
replacing instead of the value of the single, the group mean.
But when NA appears, all the group gets NA.
Perhaps there is a different way to obtain the same result.
On Mon, 16 Sept 2024 at 11:35, Rui Barradas wrote:
> Às 08:28 de 16/09/2024, Francesca escreveu:
> > Dear Contributors,
the NA is
replaced.
Here the mean function has not the na.rm=T option associated, but it
appears that this solution cannot be implemented in this case. I am not
even sure that this would be enough to solve my problem.
Thanks for any help provid
~ifelse(. == NA)) & id!=1,1,0 )
but the problem with across is that it will implement the condition only on
cp_ columns. How do I tell R to use the column id with all the other
columns?
Thanks for any help provided.
Francesca
--
.
--
Francesca Pancotto
Associate Professor Political Economy
University of Modena, Largo Santa Eufemia, 19, Modena
Office Phone: +39 0522 523264
Web: *https://sites.google.com/view/francescapancotto/home
<https://sites.google.com/view/francescapancotto/h
I apologize, I solved the problem, sorry for that.
f.
Il giorno gio 13 giu 2024 alle ore 16:42 Francesca PANCOTTO <
francesca.panco...@unimore.it> ha scritto:
> Dear Contributors
> I am trying to create a numeric series with repeated numbers, not
> difficult task, but I do not
Dear Contributors
I am trying to create a numeric series with repeated numbers, not difficult
task, but I do not seem to find an efficient way.
This is my solution
blocB <- c(rep(x = 1, times = 84), rep(x = 2, times = 84), rep(x = 3, times
= 84), rep(x = 4, times = 84), rep(x = 5, times = 84), re
Hello,
The problem was that version 4.0.4 did not support the package so I tried with
several old versions until 3.6.2 installs both climtrend and Rcmdr with its
graphical interface !! solved and thanks again Davide !!Francesca
(from Italy)
[[alternative HTML version deleted
it, what should I do to use it?
thank you in advance for your kindness,
Regards
Francesca
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0,50),ylim=c(0,max(density11_1$y)),type="n",
> xlab="delta",ylab="density",main="participation == 1")
> plot_bg("lightgray")
> grid(col="white")
> polygon(density11_1,col="#ff773344")
> polygon(density2_1,col="#
Exactly. I was trying to remelt data in the right way, but I could not get
there yet. Can you suggest me this code?
Thanks a lot
F.
--
> Il giorno 5 dic 2019, alle ore 11:11, Jim Lemon ha
> scritto:
>
> Hi Francesca,
> Do you want something like
like to have in the same plot with
mixed participation.
I hope it is clear.
Nonetheless, the previous plot is useful to understand something I had not
thought about.
Thanks again for your time.
F.
--
> Il giorno 4 dic 2019, alle ore 15:27, Francesca
> ha s
Hi!
It is not exactly what I wanted but more than I suspected I could get.
Thanks a lot, this is awesome!
Francesca
On Wed, 4 Dec 2019 at 14:04, Rui Barradas wrote:
> Hello,
>
> Please keep R-Help in the thread.
>
> As for the question, the following divides by facets, particip
, 0, 0, -10, -20), .Dim = c(236L, 6L
), .Dimnames = list(NULL, c("participation1", "participation2",
"ParticipantsNOPUN", "ParticipantsPUN", "delta11_L", "delta2_L"
)))
Francesca Pancotto
--
Fran
Dear Contributors,
I would like to ask help on how to create a plot that is the overlapping
of two other plots.
It is a geom_bar structure, where I want to count the occurrences of two
variables, participation1 and participation2 that I recoded as factors as
ParticipationNOPUN and ParticipationPUN
Thanks for the answer.
Il gio 19 lug 2018, 01:04 Jim Lemon ha scritto:
> Hi Francesca,
> This looks like a fairly simple task. Try this:
>
> fpdf<-read.table(text="PASP SUBJC
> 0 0
> 4 1
> 0 0
> 8 0
> 4
provide.
Francesca
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and provide commented, minimal, self
I did not need to select the whole character sentence, otherwise I would know
how to do it.. from basic introduction to R as you suggest.
Grep works perfectly.
f.
--
Francesca Pancotto, PhD
> Il giorno 19 ott 2017, alle ore 18:01, Jeff Newmiller
> ha s
Thanks a lot, so simple so efficient!
I will study more the grep command I did not know.
Thanks!
Francesca Pancotto
> Il giorno 19 ott 2017, alle ore 12:12, Enrico Schumann
> ha scritto:
>
> df[grep("strat", row.names(df)), ]
[[alternativ
simple, but they are not and involve also spaces.
I tried with select matches from dplyr but works for column names but I did not
find how to use it on row names, which are of course character values.
Thanks for any help you can provide.
--
Francesca P
(i in c(ra,rb,rc,rd,re,rf,rg,rh))
{
calcolati[[i]] <- i$coefficients[1]
}
Thanks for any help you can provide.
f.
--
Francesca Pancotto
Web: https://sites.google.com/site/francescapancotto/
<https://sites.google.com/site/francescapancotto/>
En
0, regions=nomi)
The problem is to fill the provinces level with the values of a variable that
is the variable of interest:
I found a series of examples based on US data extracted from very hard to get
databases.
Can anyone provide an easy example where to start from?
Thanks in advance
,
wald_fin12.2$result$chi2,wald_fin21.2$result$chi2)
My idea is that it is possible to create all this variable with a loop
across the objects names but it is a level of coding much higher than my
personal knowledge and ability.
I hope anyone can help
Thanks in advance
--
Fr
I only replicated the values of the day.spot variable.
Any help?
Thanks for your time and patience!
Francescaa
--
Francesca
------
Francesca Pancotto, PhD
Università di Modena e Reggio Emilia
Viale A. Allegri, 9
40121 Reggio Emilia
Office: +39 0522 523264
Web: https://
.42 1.42
I need to create a variable to attach to the data frame which is composed of
11500 rows that takes values
1.32 when p_f==1
1.34 when p_f==2
It seems so easy but I cannot find a way to do it in an efficient way.
Thanks in advance for any help.
Francesca
[[alternative H
if (p_int$p_made[i]==211) p_m[i]<-19
}
Can anyone help to find something more efficient?
Thanks in advance.
Francesca
--
Francesca
--
Francesca Pancotto
Associate Professor
University of Modena and Reggio Emilia
Viale A
requencies of the same allele found in the
1000 Genomes, I get concordance up to frequency= 0.5, then a shift in
direction happens and I get discordance up to 1 for the reference
frequency.
Thank you very much for any suggestions you may have,
Francesca
2014-05-27 2:07 GMT+01:00 David Duffy
).
Thanks for your help.
Francesca
2014-05-26 11:13 GMT+01:00 francesca casalino :
> But then how do you know which allele is the reference and which the risk
> allele (between A/T/C/G)?
>
>
> 2014-05-26 1:41 GMT+01:00 David Duffy :
>
> francesca casalino asked:
>>
But then how do you know which allele is the reference and which the risk
allele (between A/T/C/G)?
2014-05-26 1:41 GMT+01:00 David Duffy :
> francesca casalino asked:
>
>
>> Does anyone know how to find the reference allele used for genetic
>> associations ran in snpS
Hi,
Does anyone know how to find the reference allele used for genetic
associations ran in snpStats?
I have ran several associations using snp.rhs.tests, but I cannot tell which
allele was used as the "effect allele". Is it the one coded as "Al1" in the
SNP.support file? I can find the RAF (risk
Thanks a lot, it works perfectly!
f.
--
Francesca Pancotto
Università degli Studi di Modena e Reggio Emilia
Palazzo Dossetti - Viale Allegri, 9 - 42121 Reggio Emilia
Office: +39 0522 523264
Web: https://sites.google.com/site/francescapancotto
though some of them are NA?
Thanks for any help you can provide,
Best,
Francesca
row.names bank_name date px_last Q_Y p_made p_for
1 2 1 11/30/061.31 p406-q406406 406
1
2 47 1 02/26/09
Hi,
I am trying to produce a ggplot graph using specific characters in the
labels, but ggplots doesn't seem to support certain symbols.
For example, when I type:
print("\u25E9")
it shows a square which is half black, but when I try to use it in ggplot
it doesn't print.
I am using facet_wrap, bu
s
What am I doing wrong?
f.
--
Francesca Pancotto
Università degli Studi di Modena e Reggio Emilia
Palazzo Dossetti - Viale Allegri, 9 - 42121 Reggio Emilia
Office: +39 0522 523264
Web: https://sites.google.com/site/francescapancotto/
help you can provide.
Francesca
On 4 February 2014 12:42, Francesca wrote:
> Dear Contributors
> I am asking some advice on how to solve the following problem.
> I have a list composed of 78 elements, each of which is a matrix of
> factors and numbers, similar to the following
&
--
Francesca
--
Francesca Pancotto, PhD
Università di Modena e Reggio Emilia
Viale A. Allegri, 9
40121 Reggio Emilia
Office: +39 0522 523264
Web: https://sites.google.com/site/francescapancotto/
--
[[alternative HTML
oes give the desired display.
>
> Best,
> Ista
>
> On Mon, Feb 18, 2013 at 6:04 AM, francesca casalino
> wrote:
>> Dear R experts,
>>
>> I am trying to arrange multiple plots, creating one graph for each
>> size1 factor variable in my data frame, and
Dear R experts,
I am trying to arrange multiple plots, creating one graph for each
size1 factor variable in my data frame, and each plot has the median
price on the y-axis and the size2 on the x-axis grouped by clarity:
library(ggplot2)
df <- data.frame(price=matrix(sample(1:1000, 100, replace =
I am sorry I have confused you, the logs are all base e:
ln(a) = 1347
ln(b) = 1351
And I am trying to solve this expression:
exp( ln(a) ) - exp( ln(0.1) + ln(b) )
Thank you.
2013/2/4 francesca casalino :
> Dear R experts,
>
> I have the logarithms of 2 values:
>
> log(a)
Dear R experts,
I have the logarithms of 2 values:
log(a) = 1347
log(b) = 1351
And I am trying to solve this expression:
exp( ln(a) ) - exp( ln(0.1) + ln(b) )
But of course every time I try to exponentiate the log(a) or log(b)
values I get Inf. Are there any tricks I can use to get a real resu
then
repeat this for each of the four row of the same matrix. The resulting
matrix should be composed of these distances.
I need to repeat this for each of the subsamples. I realize that there
arecalculations that are repeated but I did not find a strategy that does
not requir
Thanks to you all,
they are very useful and I am learning a lot.
Best,
Francesca
On 27 January 2013 19:20, arun wrote:
>
>
> Hi,
>
> You could use library(plyr) as well
> library(plyr)
> pnew<-colSums(aaply(laply(split(as.data.frame(p),((1:nrow(as.data.frame(p))-1)%/%
&
;-(100-(100*abs(fa1[i]/sum(fa1[i])-(1/3
}
fa2b<-c()
for (i in 1:3){
fa2b[i]<-(100-(100*abs(fa2[i]/sum(fa2[i])-(1/3
}
and so on.
Is there a more efficient way to do this?
Thanks for your time!
Francesca
--
Francesca Pancotto, PhD
Università di Mod
ional Hazard Model's one based on Laplace Tansforms'
ratio).
I can't find any documentation about AFT models with Gamma Frailty developed in
"survival package". Any reference paper to suggest?
Thank a million
Francesca
[[al
Dear R experts,
I have an R script that creates multiple scripts and submits these
simultaneously to a computer cluster, and after all of the multiple
scripts have completed and the output has been written in the
respective folders, I would like to automatically launch another R
script that works
ject like an array
of dimension (2,3,12) which contains each matrix @cval
produced by ca.jo for the 12 subjects that i tested.
Can anyone help me with that?
I hope my explanation of the problem is clear.
Thanks in advance for any help.
--
Francesca
--
Francesca Pan
Thanks a lot!
Francesca
On 18 May 2012 18:21, arun wrote:
> Hi Francesca,
>
>> for(i in 1:length(x1<-c(100,1000,1))){
> j<-x1[i]
> x1[i]<-mean(j)
> }
>
>> x1
> [1] 100 1000 1
>
>
>
> A.K.
>
>
>
> - Origina
dex in an efficient manner.
Thanks for any help you can provide.
Francesca
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,
Francesca
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Hi, I am trying to match the order of the rownames of a dataframe with
the rownames of another dataframe (I can't simply sort both sets
because I would have to change the order of many other connected
datasets if I did that): Also, the second dataset (snp.matrix$fam) is
a snp matrix slot:
so for e
in the command line) gives different pca scores
than running the same pca with the correlation matrix (using scores=FALSE in
the command line and therefore calculating the scores in the way you
suggested to Wolfgang). Is that normal?
Thanks
francesca
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View this message in context:
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Dear all,
we would like to point out the approaching XIII GRASS and
GFOSS italian Meeting which will take place at the University of Trieste from
Wednesday, February 15 until Friday, February 17, 2012.
Abstracts can be sent to gr...@units.it till January 20 while subscriptions
are open up to Fe
More infomations:
http://sites.google.com/site/grassts/
gr...@units.it
Kind regards. From meeting organizators
Dr. PhD Francesca Bader
Università degli Studi di Trieste
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ace = TRUE)
estr2<-sample(tmp2, 1000, replace = TRUE)
m1[i]<-mean(estr1,na.rm=TRUE)
m2[i]<-mean(estr2,na.rm=TRUE)
}
db<-data.frame(cbind(m1,m2))
Thanks for any help you can provide.
Best Regards
--
Francesca
datiP2 <- datiP[datiP$city ==2,x];
datiP3 <- datiP[datiP$city ==3,x]
datiP4 <- datiP[datiP$city ==4,x];
--
Thank you for any help you can provide.
Francesca
[[alternative HTML version deleted]]
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Dear experts,
I am trying to create a data frame from the residuals I get after
having applied a linear regression to each column of a data frame, but
I don't know how to create this data frame from the resulting list
since the list has differing numbers of rows.
So for example:
age<- c(5,6,10,14
without giving
any error…Does anybody have any idea what is going on?
Thank you very much for all the great help!
-f
2011/10/11 Weidong Gu :
> In your case, it may not be sensible to simply fill missing values by
> mean or mode as multiple imputation becomes the norm this day. For
> your sp
Dear R experts,
I have a large database made up of mixed data types (numeric,
character, factor, ordinal factor) with missing values, and I am
looking for a package that would help me impute the missing values
using either the mean if numerical or the mode if character/factor.
I maybe could use
Dear Peter and Tim,
Thank you very much for taking the time to explain this to me! It is much
more clear now.
And sorry for using the space here maybe inappropriately, I really hope this
is OK and gets posted, I think it is really important that non-statisticians
like myself get a good idea of the
Thank you very much to both Ken and Peter for the very helpful
explanations.
Just to understand this better (sorry for repeating but I am also new in
statistics
so please correct me where I am wrong):
Ken' method:
Random sampling of the mean, and then using these means to construct a
distribution
what I have to
explain better:
df1 is:
Name Position location
francesca A 75
maria A 75
cristina B 36
And df2 is:
location Country
75 UK
75 Italy
56 France
56 Austria
So I thought I had to first eliminate the duplicates like this:
df1_unique<-subset(df1, !duplicated(location))
df2_
Yes, your code did exactly what I needed.
Thank you!!
-f
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Ok I added quoting and it did work...Not sure why, but thank you for both
your replies!
-f
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PLEASE do read the posting guide ht
SE,
conf.int=TRUE,alternative = c("g"))
each of these tests will create an object, say x and then I extract the
value statistics using
x$statistics.
How to automatize this?
Thank you for any help you can provide.
Francesca
[[alternative HTML version deleted]]
1.48
> 10 13775 31215 DATA.Q211.WF.Index 04/14/11 1.44
> 12 13789 31865 DATA.Q211.WPC.Index 04/01/11 1.40
> 15 13804 32515 DATA.Q211.XTB.Index 04/29/11 1.50
> >
> >
>
>
>
> On Mon, Aug 1, 2011 at 11:13 AM, Francesca
> wrote:
> > Dear Contributors
>
).
I know how to select rows according to a substring condition on the
index column, but I cannot use it here because the substring changes
and moreover the number of occurrences per substring is variable.
Thank you for any help you can provide.
Francesca
[[alternative HTML version deleted
, but it
seems not to be reproducing the correct organization of the data, as
it takes date as the id values.
PS: the n1 index names are not ordered in the original database, so
I cannot fill in the NA with the names using a recursive formula.
Thank you for any help you can provide.
Francesca
s = FALSE) :
Reached total allocation of 1535Mb: see help(memory.size)
4: In unlist(vlist, recursive =
FALSE, use.names = FALSE) :
Reached total allocation of 1535Mb: see help(memory.size)
> Many thanks. Kind regards.
Dr. Francesca Bader
University of Trieste
Italy
cores PLSR) from "pls" package.
I wonder if K-OPLS performs the same discriminant analysis of OPLS-DA?
Is there any other available package for applying OPLS-DA?
Thanks in advance for any advice.
Best regards,
Francesca Chignola
--
---
Francesca Chignola, P
elp me a lot. Could anybody
make me understand how "ss" and "ncluster" are related, and how to build
the predictive densities of the random effect starting from "mub"s
elements? It would be a great gain in my work.
Looking forward to hear news from you
thanks
for this question, but I am just at the beginning of
my long path to handle and know statistics and R.
regards
Francesca
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?
Your kindly help is appreciated in advance
Francesca
--
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11A Sharon Gardens
E97RX
London
UK contact: +44 750 5485255
Italian contact: +39 349 7313294
Skype: jordanfrancesca
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?
Your kindly help is appreciated in advance
Francesca
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like the software to perform the ols regression
but to just ignore these data and consider the rest.
I tried to use the function na.action=na.omit in
lm( y~x, na.action=na.omit)
but it seems to exert no effect on the function.
Thank you for any available help.
Francesca
--
Post - doc Finance
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