Dear Sir,
Thank you very much for your suggestions.
Due to certain inconveniences, I was unable to work on your suggestions.
Today I worked on both suggestions and got the result that I really wanted
that monthly averages for each country.
Here, I am asking one more query (just for learning pur
Hello,
Richard's idea is good but shouldn't it be `[.myclass` instead?
`[.myclass` <- function(x, i, j, drop = if (missing(i)) TRUE else
length(cols) == 1){
SaveAt <- lapply(X, attributes)
X <- NextMethod()
lX <- lapply(names(X),function(nm, x, Sat){
attributes(x[[nm]]) <- Sat[[nm]]
Dear listers,
I am trying to pass an expression inlcuding a variable and a
superpscript to a legend. What I want to obtain is e.g. with densren = 1.25
1.25 ind./km^2
I have tried many variants of the following:
legend=bquote(.(densren) (ind./km)^2)
but if not errors, do obtain
1.25 (ind./km^
I did a Google search on
R plot superscript in legend
and the first search result was
https://stackoverflow.com/questions/20453408/superscript-r-squared-for-legend
which looks like it might address your question.
On Sun, Oct 20, 2019 at 3:30 PM Patrick Giraudoux <
patrick.giraud...@univ-fcomte
Thanks Eric. I got it too already (and already tried some variations
based on it), but to my understanding it does not include a variable
whose contents is used in the expression as in the case submitted...
Le 20/10/2019 à 14:56, Eric Berger a écrit :
> I did a Google search on
>
> R plot super
Assuming you are using base graphics, your syntax for adding the legend
appears to be wrong.
legend() is a separate function, not a parameter of plot.default afaics.
The following works for me:
> densren <- 1.25
> plot(1:10)
> legend (x="center", legend =bquote(.(densren) (ind./km)^2))
See ?lege
Thanks Bert and Peter,
Yes Bert, I was aware of the legend() function syntax, and just quoting
the legend argument within the function.
However, Bert and Peter, I do not understand why it works with your
absolutely reproducible examples and not in the slightly (not so
slightly apparently) diff
Does this do what you want:
> library(tidyverse)
> input <- read_delim("PERMNO DATE Spread
+ 111 19940103 0.025464308
+ 111 19940104 0.064424296
+ 111 19940105 0.018579337
+ 111 19940106 0.018872211
..." ... [TRUNCATED]
> # drop last two digits to get the month
> monthly <- input %>%
+ group_
The legend must be "an expression vector."
c("Sans renard",bquote(.(densren) (ind./km)^2)) is not because the first
element is a character string.
This works:
plot(1:100,1:100,type="n")
legend(list(x=0,y=100),legend=c(expression("Sans
renard"),bquote(.(densren)
(ind./km)^2)),lty=c(1,2),co
Hello,
Here are two other ways using aggregate.
The difference is in the way to create a MONTH grouping column.
The second way is base R only.
df1$MONTH <- zoo::as.yearmon(as.Date(as.character(df1$DATE), '%Y%m%d'))
aggregate(Spread ~ PERMNO + MONTH, df1, mean)
df1$MONTH <- df1$DATE %/% 100
agg
Great ! You have helped to solve a problem on which I was sweating
(sporadically, however) since months...
Thanks,
Best,
Le 20/10/2019 à 18:29, Bert Gunter a écrit :
> The legend must be "an expression vector."
> c("Sans renard",bquote(.(densren) (ind./km)^2)) is not because the
> first el
It's tricky, but I think what you want is
legend(list(x=0,y=100),
legend=as.expression(list(
"Sans renard",
bquote(.(densren) * " ind."/"km"^2)
)),
lty=c(1,2),col=c("black","red"),bty="n")
Generally, if you want a vector of unevaluated expressions, you need an object
of mode "
Now, we have two solutions working. This is great since I did not find
any example of the kind searching r-help archives and google...
Thanks !
Le 20/10/2019 à 19:31, Peter Dalgaard a écrit :
It's tricky, but I think what you want is
legend(list(x=0,y=100),
legend=as.expression(list(
Would be nice to put those two way examples in the documentation of the
function 'expression' and 'bquote' in the next R version (we are in the
base) for other users ;-) I am sure many would enjoy.
Le 20/10/2019 à 19:15, Patrick Giraudoux a écrit :
> Great ! You have helped to solve a prob
To continue down this rabbit hole ...
Actually, both solutions are the same; Peter's is just more general than
mine, as it works more conveniently for more lines in the legend.
However, note that:
> class(c("Sans renard", bquote(.(densren) (ind./km)^2)))
[1] "list" # by coercion
so it does not
You're right. I was worried that c() would create a character vector and
deparse the unevaluated call in the process, but apparently it is an implicit
as.character _inside_ legend that is doing us in. (I can't offhand see where it
is happening, but there might be scope for improvement if legend(
On 21/10/19 1:15 AM, Rui Barradas wrote:
Hello,
Richard's idea is good but shouldn't it be `[.myclass` instead?
Yes, I kind of thought that, and cobbled together something on that
basis that seemed to work. However my code was rather a hodge-podge. I
kept having to work around errors th
Hello,
Sorry, you're right, in the method it's x, X is the test dataframe.
Repost:
`[.myclass` <- function(x, i, j, drop = if (missing(i)) TRUE else
length(cols) == 1){
SaveAt <- lapply(x, attributes)
x <- NextMethod()
lX <- lapply(names(x),function(nm, x, Sat){
attributes(x[[nm]]) <
Well, the direct answer is "no", but then again I did not know the answer to
the other question until I Googled it either.
When I do the same for grid.draw, it appears to be a generic function for
drawing graphical objects... data frames are not grobs, so you must be doing
something to convert
On 21/10/19 11:07 AM, Rui Barradas wrote:
Hello,
Sorry, you're right, in the method it's x, X is the test dataframe.
Repost:
`[.myclass` <- function(x, i, j, drop = if (missing(i)) TRUE else
length(cols) == 1){
SaveAt <- lapply(x, attributes)
x <- NextMethod()
lX <- lapply(names(x
Hello everyone again,
I much appreciated the explanations.
On Wed, Sep 25, 2019 at 11:02:42AM +0200, Francesco Ariis wrote:
> Maybe the Introduction should link to it (or similar page) with text
> "In case you are interest in the difference between static and lexical
> scope, check this explan
Hi Subhamitra,
This is not the only way to do this, but if you only want the monthly
averages, it is simple:
# I had to change the "soft" tabs in your email to commas
# in order to read the data in
spdf<-read.table(text="PERMNO,DATE,Spread
111,19940103,0.025464308
111,19940104,0.064424296
111,1994
Hai Rui,
It seems doesnt work for me, the "" still there.
So I used this one (Bert suggestion),
test<-lapply(test,function(x){x$RR[x$RR==] <- NA; x})
Best,
Ani
On Sat, Oct 19, 2019 at 6:55 PM Rui Barradas wrote:
> Hello,
>
> Why not use read.xlsx argument 'na.strings', an argumen
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