Displaying 16 results from an estimated 16 matches for "pancotto".
2017 Oct 19
2
Select part of character row name in a data frame
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 <es at enricoschumann.net> ha scritto:
>
> df[grep("strat", row.names(df)), ]
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2017 Oct 19
0
Select part of character row name in a data frame
...o strat:crt.dummy ",
"Common to strat, and crt.dummy ",
"Common to strat, and gender ",
"Common to strat, and age ") ,]
--
Sent from my phone. Please excuse my brevity.
On October 19, 2017 3:14:53 AM PDT, Francesca PANCOTTO <f.pancotto at unimore.it> wrote:
>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
><es at enricoschumann.net> h...
2012 May 25
2
Collecting results of a test with array
...bject 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 Pancotto, PhD
Università di Modena e Reggio Emilia
Viale A. Allegri, 9
40121 Reggio Emilia
Office: +39 0522 523264
Web: http://www2.dse.unibo.it/francesca.pancotto/
----------------------------------
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2024 Jun 13
1
Create a numeric series in an efficient way
...)
blocC <- arrange(.data = data.frame(blocC), blocC)
The second line sorts, but that may not be needed depending on application. The object class is also different in the sorted solution.
Tim
-----Original Message-----
From: R-help <r-help-bounces at r-project.org> On Behalf Of Francesca PANCOTTO via R-help
Sent: Thursday, June 13, 2024 2:22 PM
To: r-help at r-project.org
Subject: Re: [R] Create a numeric series in an efficient way
[External Email]
I apologize, I solved the problem, sorry for that.
f.
Il giorno gio 13 giu 2024 alle ore 16:42 Francesca PANCOTTO < francesca.pancotto a...
2024 Jun 13
1
Create a numeric series in an efficient way
I apologize, I solved the problem, sorry for that.
f.
Il giorno gio 13 giu 2024 alle ore 16:42 Francesca PANCOTTO <
francesca.pancotto at unimore.it> ha scritto:
> 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 =...
2017 Oct 19
2
Select part of character row name in a data frame
...ese names were 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 Pancotto, PhD
2017 Oct 19
0
Select part of character row name in a data frame
Quoting Francesca PANCOTTO <f.pancotto at unimore.it>:
> Dear R contributors,
>
> I have a problem in selecting in an efficient way, rows of a data
> frame according to a condition,
> which is a part of a row name of the table.
>
> The data frame is made of 64 rows and 2 columns, but the row nam...
2012 May 18
3
How to fix indeces in a loop
Dear Contributors,
I have an easy question for you which is puzzling me instead.
I am running loops similar to the following:
for (i in c(100,1000,10000)){
print((mean(i)))
#var<-var(rnorm(i,0,1))
}
This is what I obtain:
[1] 100
[1] 1000
[1] 10000
In this case I ask the software to print out the result, but I would
like to store it in an object.
I have tried a second loop, because if I
2024 Jun 13
2
Create a numeric series in an efficient way
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), rep(x = 6, times =
84), rep(x = 7, times = 84), rep(x = 8, times = 84), rep(x = 9, times =
84),
2011 Jul 27
3
Reorganize(stack data) a dataframe inducing names
...zation 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
--
Francesca
----------------------------------
Francesca Pancotto, PhD
Dipartimento di Economia
Università di Bologna
Piazza Scaravilli, 2
40126 Bologna
Office: +39 051 2098135
Cell: +39 393 6019138
Web: http://www2.dse.unibo.it/francesca.pancotto/
----------------------------------
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2013 Jan 27
2
Loops
...for (i in 1:3){
fa1b[i]<-(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 Modena e Reggio Emilia
Viale A. Allegri, 9
40121 Reggio Emilia
Office: +39 0522 523264
Web: https://sites.google.com/site/francescapancotto/
----------------------------------
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2013 Jan 29
0
On the calulation of crossed differences
...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 require
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/
----------------------------------
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2011 Nov 12
1
Simulation over data repeatedly for four loops
Dear Contributors,
I am trying to perform a simulation over sample data,
but I need to reproduce the same simulation over 4 groups of data. My
ability with for loop is null, in particular related
to dimensions as I always get, no matter what I try,
"number of items to replace is not a multiple of replacement length"
This is what I intend to do: replicate this operation for
four
2011 Sep 28
1
Wilcox test and data collection
Dear Contributors
I have a problem with the collection of data from the results of a test.
I need to perform a comparative test over groups of data , recall the value
of the pvalue and create a table.
My problem is in the way to replicate the analysis over and over again over
subsets of data according to a condition.
I have this database, called y:
gg t1 t2 d
40 1 1
2024 Sep 16
2
(no subject)
Dear Contributors,
I hope someone has found a similar issue.
I have this data set,
cp1
cp2
role
groupid
1
10
13
4
5
2
5
10
3
1
3
7
7
4
6
4
10
4
2
7
5
5
8
3
2
6
8
7
4
4
7
8
8
4
7
8
10
15
3
3
9
15
10
2
2
10
5
5
2
4
11
20
20
2
5
12
9
11
3
6
13
10
13
4
3
14
12
6
4
2
15
7
4
4
1
16
10
0
3
7
17
20
15
3
8
18
10
7
3
4
19
8
13
3
5
20
10
9
2
6
I need to to average of groups, using the values of column
2011 Nov 11
8
Help
Dear Contributors
I would like to perform this operation using a loop, instead of repeating
the same operation many times.
The numbers from 1 to 4 related to different groups that are in the
database and for which I have the same data.
x<-c(1,3,7)
datiP1 <- datiP[datiP$city ==1,x];
datiP2 <- datiP[datiP$city ==2,x];
datiP3 <- datiP[datiP$city ==3,x]
datiP4 <-