Displaying 20 results from an estimated 2000 matches similar to: "regex in maps package"
2023 Nov 18
1
Can someone please have a look at this query on stackoverflow?
And indeed again (I did not understand your previous question exactly
at first), the 'hard-coded' definition of a landscape a4 sheet would
therefore be:
paperwidth=29.7cm, paperheight=21cm
Olivier.
On Sat, 18 Nov 2023 13:20:49 +0530
Ashim Kapoor <ashimkapoor at gmail.com> wrote:
> Dear Olivier,
>
> Many thanks for your reply.
>
> Very cool.
>
> You know
2023 Nov 18
2
Can someone please have a look at this query on stackoverflow?
Dear Olivier,
Many thanks for your reply.
Very cool.
You know what I thought ? I thought you had modified the A4 sheet size
to compute the paperheight and paperwidth ?
I wonder if that's another way of proceeding.
Best,
Ashim
On Sat, Nov 18, 2023 at 12:51?PM Olivier Crouzet
<olivier.crouzet at univ-nantes.fr> wrote:
>
> Dear Ashim,
>
> these are documented in the LaTeX
2018 Jan 07
0
SpreadLevelPlot for more than one factor
Dear All,
we need to do :
library(car) for the spreadLevelPlot function
I forgot to say that.
Apologies,
Ashim
On Sun, Jan 7, 2018 at 10:37 AM, Ashim Kapoor <ashimkapoor at gmail.com> wrote:
> Dear All,
>
> I want a transformation which will make the spread of the response at all
> combinations
> of 2 factors the same.
>
> See for example :
>
>
2023 Nov 18
1
Can someone please have a look at this query on stackoverflow?
Dear Ashim,
these are documented in the LaTeX 'geometry' package (see for example
on CTAN: https://ctan.org/pkg/geometry). As I added in my response on
Stackoverflow, several parts in the RMarkdown header actually concern
information that are processed by LaTeX to actually generate the PDF,
among which the 'geometry' line. For someone who is used to working with
LaTeX, it is
2018 Jan 14
1
SpreadLevelPlot for more than one factor
Dear Ashim,
I?ll address your questions briefly but they?re really not appropriate for
this list, which is for questions about using R, not general statistical
questions.
(1) The relevant distribution is within cells of the wool x tension
cross-classification because it?s the deviations from the cell means that
are supposed to be normally distributed with equal variance. In the
warpbreaks data
2023 Jun 08
1
Cryptic error from stargazer
On Thu, 08 Jun 2023, Ashim Kapoor writes:
> Dear All,
>
> I had done an automatic upgrade of my Debian 10 system which had also
> upgraded R.
>
> I reinstalled the stargazer package and the error went away.
>
> Query : Do I need to reinstall all packages with each upgrade of R ?
>
> Best,
> Ashim
Your session info says "stargazer 5.2.2". What version
2023 Nov 18
1
Can someone please have a look at this query on stackoverflow?
Dear Olivier,
Many thanks for your reply.
This works well for me.
How did you come up with the pagewidth / pageheight numbers? I do
understand that their ratio = 16:9,
but how did you choose these numbers?
Best Regards,
Ashim
On Fri, Nov 17, 2023 at 9:25?PM Olivier Crouzet
<olivier.crouzet at univ-nantes.fr> wrote:
>
> Dear Ashim,
>
> I don't think the aspectratio is
2023 Jun 08
1
Cryptic error from stargazer
Dear All,
I had done an automatic upgrade of my Debian 10 system which had also
upgraded R.
I reinstalled the stargazer package and the error went away.
Query : Do I need to reinstall all packages with each upgrade of R ?
Best,
Ashim
On Thu, Jun 8, 2023 at 11:11?AM Ashim Kapoor <ashimkapoor at gmail.com> wrote:
>
> Dear All,
>
> Here is my reproducible example:
>
>
2018 Jan 07
2
SpreadLevelPlot for more than one factor
Dear All,
I want a transformation which will make the spread of the response at all
combinations
of 2 factors the same.
See for example :
boxplot(breaks ~ tension * wool, warpbreaks)
The closest I can do is :
spreadLevelPlot(breaks ~tension , warpbreaks)
spreadLevelPlot(breaks ~ wool , warpbreaks)
I want to do :
spreadLevelPlot(breaks ~tension * wool, warpbreaks)
But I get :
>
2018 Apr 18
0
Understanding which
Look at
which(x>100)
This is a zero-length vector. The negative of nothing is nothing, not a list of all possible index values.
Do you want
x[ !( x > 100 ) ]
?
On April 18, 2018 6:13:30 AM CDT, Ashim Kapoor <ashimkapoor at gmail.com> wrote:
>Dear All,
>
>Here is a reprex:
>
>> x<- 1:100
>> x[-which(x>100)]
>integer(0)
>
>In words, I am
2023 Nov 17
1
Can someone please have a look at this query on stackoverflow?
Dear Ashim,
I don't think the aspectratio is appropriate in this context because
it would imply that the beamer (LaTeX) class is used but you're actually
using the article (LaTeX) class.
You may use specifications of the geometry package rather than
specifying options to the class:
e.g. replace your current header:
---
title: "Testing landscape and aspect ratio"
output:
2017 Oct 02
0
Default value of the option initial in the ses function in the forecast package.
The first one, i.e. "optimal"; check help for match.arg() for the idiom.
-pd
> On 2 Oct 2017, at 11:48 , Ashim Kapoor <ashimkapoor at gmail.com> wrote:
>
> Dear All,
>
> I am trying to use the function ses from the forecast package.
>
> From its help I have :
>
> Usage:
>
> ses(y, h = 10, level = c(80, 95), fan = FALSE, initial =
2018 Apr 18
3
Understanding which
Dear All,
Here is a reprex:
> x<- 1:100
> x[-which(x>100)]
integer(0)
In words, I am finding out which indices correspond to values in x which
are greater than 100 ( there are no such items ) . Then I remove those
indices. I should get back the x that I started with since there are no
items in x which are bigger than 100 . Instead, it is returning an empty
vector.
Why is this ?
2017 Jun 08
0
regular expression help
Zitat von Ashim Kapoor <ashimkapoor at gmail.com>:
> Dear All,
>
> My query is:
>
> Do we always need to use perl = TRUE option when doing ignore.case=TRUE?
>
> A small example :
>
> my_text =
> "RECOVERY OFFICER-II\nDEBTS RECOVERY TRIBUNAL-III\n RC No. 162/2015\nSBI
> VS RAMESH GUPTA.\n Dated: 01.03.2016 Item no.01\n
> Present:
2018 Jan 09
0
SpreadLevelPlot for more than one factor
Dear Sir,
Many thanks for your reply.
I have a query.
I have a whole set of distributions which should be made normal /
homoscedastic. Take for instance the warpbreaks data set.
We have the following boxplots for the warpbreaks dataset:
a. boxplot(breaks ~ wool)
b. boxplot(breaks ~ tension)
c. boxplot(breaks ~ interaction(wool,tension))
d. boxplot(breaks ~ wool @ each level of tension)
e.
2018 Jan 07
2
SpreadLevelPlot for more than one factor
Dear Ashim,
Try spreadLevelPlot(breaks ~ interaction(tension, wool), data=warpbreaks) .
I hope this helps,
John
-----------------------------
John Fox, Professor Emeritus
McMaster University
Hamilton, Ontario, Canada
Web: socialsciences.mcmaster.ca/jfox/
> -----Original Message-----
> From: R-help [mailto:r-help-bounces at r-project.org] On Behalf Of Ashim
> Kapoor
> Sent:
2011 Sep 27
1
Does replacing some values of a zoo object by NA reduce it's size ?
Dear R-helpers,
Please have a look at the following. f1 is the same as f2 except that it has
some values replaced by NA. But it's corresponding file is slightly bigger
than the file containing f2. Could someone please tell me if this is an
anomaly ?
> load("file1")
> ls()
[1] "f1"
> load("file2")
> ls()
[1] "f1" "f2"
>
>
2023 Nov 14
2
Can someone please have a look at this query on stackoverflow?
Dear all,
I have posted a query which has received a response but that is not
working on my computer.
Here is the query:
https://stackoverflow.com/questions/77387434/pdf-from-rmarkdown-landscape-and-aspectratio-169
Can someone please help me ?
Best Regards,
Ashim
2017 Oct 02
2
Default value of the option initial in the ses function in the forecast package.
Dear All,
I am trying to use the function ses from the forecast package.
>From its help I have :
Usage:
ses(y, h = 10, level = c(80, 95), fan = FALSE, initial = c("optimal",
"simple"), alpha = NULL, lambda = NULL, biasadj = FALSE, x = y, ...)
My query is that if I do not mention the initial value will its default
value be "optimal".
A MWE would be
2011 Sep 27
2
Coercing a character zoo to a numeric
Dear R-helpers,
It seems to me that a character zoo cannot be coerced to a numeric zoo.
Below is a minimal example. Can someone tell me what I have done wrong?
> z<-zoo(1:4,order.by=1:4)
> coredata(z)<-as.character(coredata(z))
> str(z)
‘zoo’ series from 1 to 4
Data: chr [1:4] "1" "2" "3" "4"
Index: int [1:4] 1 2 3 4
>