Re: [R] lm and anova
anova uses sequential sums of squares (type 1), summary adjusted sums of squares (type 3) Take for example the first line of each output. In summary this tests whether vole1 is needed ASSUMING volelag and year are already in the model (conclusion would then be: it isn't needed p=.89). Whereas in anova, it's testing do we need vole1 assuming nothing else is in the model (conclusion: vole1 is better than nothing. p=.0009). anova assumes all terms above it are in the model but terms below it are not so volelag assumes vole1 is in the model but not year. You can see how anova changes but summary doesn't by varying the order you put them in. so the final model I would fit here would probably end up being either year+volelag or just year, HTH, Paul -- View this message in context: http://r.789695.n4.nabble.com/lm-and-anova-tp3516748p3517356.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] All other variables in upper scope arg for stepAIC
Hi - I am trying to substitute for the_other_y in the code below. I want
y2 and y3 to be there when i is 1, y1 and y3 to be there when i is 2 and y1
and y2 to be there when i is 3. I'm sure it's to do with what format the
data should be in and I've tried alldata[,-i], but it fits all the columns
of alldata except i rather than each column one at a time. I've tried
changing the data i.e. as.matrix or as.list but that's still not the right
format. I'm sure it's a simple trick, but don't know how to search it out.
Any help on how to search this as well as the solution would be greatly
appreciated.
y1-rnorm(20);y2-rnorm(20);y3-rnorm(20);alldata-cbind(y1,y2,y3)
library(MASS)
for (i in 1:3)
{mod-lm(get(paste(y, i, sep=))~1)
stepAIC(mod, scope=list(lower=~1, upper=~the_other_y))}
Thanks
Paul
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[R] Referencing factor name
Simple problem - I want the ylab to automatically pick up x1 rather than having to define x1 in the plot statement. x1-c(1.2,2,3);x2-c(1,2.1,2.6) y-x1 plot(1:3,y, ylab=x1) There must be a way of accessing the name x1 somehow, but unfortunately I don't know how to search for it. Any help would be great, -- View this message in context: http://r.789695.n4.nabble.com/Referencing-factor-name-tp2550131p2550131.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] A question about conducting crossed random effects in R
There is a mixed effects e-mail list you might want to join for more in depth discussion of these topics - you can subscribe here https://stat.ethz.ch/mailman/listinfo/r-sig-mixed-models. In general the format for crossed effects would be lmer(y~f1+f2+(1|r1)+(1|r2)) where f1, f2 are fixed effects and r1, r2 are random. I believe that the random lines correspond to those in SAS as below random int/subject=r1; (1|r1) for random varying intercept with each r1 random f1/subject=r1; (f1|r1)for random varying slope with each r1 However, in your case, I suspect that the model might not run because crossed random effects generally take time and with 3 way interactions that's probably too much for R. Either way, you can check that by building your model up slowly and seeing if it runs for the simpler case. Let me know how far you get, Paul -- View this message in context: http://r.789695.n4.nabble.com/A-question-about-conducting-crossed-random-effects-in-R-tp2278443p2280578.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] information reduction-database management question
If you redefine your NAs as below to be detected as some arbitrary large
number, then the code should work through. Any 5's left in your dataset can
be replaced just as easily by NAs again. Not elegant, but effective.
site - c(s1, s1, s1, s2,s2, s2)
pref - c(1, 2, 3, 1, 2, 3)
R1 - c(NA, NA, 1, NA,NA,NA)
R2 - c(NA, 0, 1, 1, NA, 1)
R3 - c(NA, 1, 1, NA, 1, 1)
R4 - c(0, NA, 0, 1, NA, 0)
R5 - c(NA, 0, 1, NA, 1, 1)
datum - data.frame(site, pref, R1, R2, R3, R4, R5)
## For 1 column;
datum$R1[is.na(datum$R1)==T]-5
tapply(datum$R1, datum$site, min, na.rm=T)
## Can loop this over all columns;
new-matrix(0,5,2)
for (i in 3:7)
{datum[,i][is.na(datum[,i])==T]-5
new[i-2,]-tapply(datum[,i], datum$site, min, na.rm=T)}
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Re: [R] A question about conducting crossed random effects in R
Sounds distinctly like an assignment you've been set for which we wouldn't help. All I'll say is crossed random effects can be dealt with effectively in lmer. See that for more. -- View this message in context: http://r.789695.n4.nabble.com/A-question-about-conducting-crossed-random-effects-in-R-tp2278443p2279508.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] information reduction-database management question
I don't think the approach would change much with text. You would have
to write a function which picks the 'min' or whatever that means to you
with text and then it should work ok,
Paul
From: Brad Patrick Schneid [via R]
[mailto:ml-node+2279677-1095983982-120...@n4.nabble.com]
Sent: 06 July 2010 15:48
To: Paul Chatfield
Subject: Re: information reduction-database management question
Thanks Paul,
I appreciate your time and this is an interesting approach.
Unfortunately, I need it to work for all types of information, including
character data (i.e. text).
Again.. thanks for your suggestion!
Brad
Paul Chatfield wrote:
If you redefine your NAs as below to be detected as some arbitrary large
number, then the code should work through. Any 5's left in your dataset
can be replaced just as easily by NAs again. Not elegant, but
effective.
site - c(s1, s1, s1, s2,s2, s2)
pref - c(1, 2, 3, 1, 2, 3)
R1 - c(NA, NA, 1, NA,NA,NA)
R2 - c(NA, 0, 1, 1, NA, 1)
R3 - c(NA, 1, 1, NA, 1, 1)
R4 - c(0, NA, 0, 1, NA, 0)
R5 - c(NA, 0, 1, NA, 1, 1)
datum - data.frame(site, pref, R1, R2, R3, R4, R5)
## For 1 column;
datum$R1[is.na(datum$R1)==T]-5
tapply(datum$R1, datum$site, min, na.rm=T)
## Can loop this over all columns;
new-matrix(0,5,2)
for (i in 3:7)
{datum[,i][is.na(datum[,i])==T]-5
new[i-2,]-tapply(datum[,i], datum$site, min, na.rm=T)}
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[R] How to say if error
Hi - I am looping over a structural equation model for a variety of datasets. Occasionally, the model returns an error and the loop then breaks. I'd like to set a condition which says something like if error, then print NAs rather than the loop breaking, but I don't know how to say if error. I'm sure there's a simple answer; I've tried searching Nabble, but error as a hitword comes up rather often on Nabble lists :) Thanks Paul -- View this message in context: http://r.789695.n4.nabble.com/How-to-say-if-error-tp2266619p2266619.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to say if error
I've had a look at the conditions in base and I can't get the ones to work
I've looked at but it is all new to me.
For example, I can work the examples for tryCatch, but it won't print a
finally message for me when I apply it to my model. Even if I could get
this to work, I think it would still cause a break e.g.
for (j in 1:10)
{tryCatch(ifelse(j==5, stop(j), j), finally=print(oh dear))}
Thanks for the suggestion though - any others?
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Re: [R] How to say if error
Thanks Roman - you're right it can do more than I thought. We're close now to solving it I feel. Essentially I'm trying to get the code below to work. If the condition is satisfied, it prints 2, but it doesn't save it in z. I want it to save it even though there's an error. Perhaps you can easily see what I'm missing, x-rnorm(0);y-rnorm(0) z-ifelse(tryCatch(lm(y~x), finally=print(2))==1, NA,lm(rnorm(10)~rnorm(10))) -- View this message in context: http://r.789695.n4.nabble.com/How-to-say-if-error-tp2266619p2266782.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to say if error
That's great. That solves it. I can work on eloquence later :) I just to
sort out that model problem:
dof-numeric(10)
for (i in 1:10){
x-rnorm(i-1);y-rnorm(i-1)
cc - try(lm(y~x), silent=T)
if(is(cc,try-error)) {dof[i]-NA}
else {dof[i]-(summary(lm(rnorm(10)~rgamma(10,1,1)))$coef[1,1])}
}
Thank you!
Paul
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Re: [R] How to say if error
On a similar issue, how can you detect a warning in a loop - e.g. the
following gives a warning, so I'd like to set up code to recognise that and
then carry on in a loop
x-rnorm(2);y-c(1,0)
ff-glm(y/23~x, family=binomial)
so this would be incorporated into a loop that might be
x-rnorm(10);y-rep(c(1,0),5)
for (i in 1:10)
{ee-glm(y~x, family=binomial)
ff-glm(y/23~x, family=binomial)}
from which I would recognise the warning in ff and not those in ee, saving
results from ee and not from ff. The last bit would be easy adding a line
if(there_is_a_warning_message) {newvector-NA} else {use results} but how do
you detect the warning message?
Thanks all for your feedback so far,
Paul
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Re: [R] How to say if error
Thanks again Joris - you've been very helpful J
From: Joris FA Meys [via R]
[mailto:ml-node+2267176-1824205151-120...@n4.nabble.com]
Sent: 24 June 2010 16:40
To: Paul Chatfield
Subject: Re: How to say if error
You could do that using the options, eg :
set.seed(1)
x - rnorm(1:10)
y - letters[1:10]
z - rnorm(1:10)
warn -getOption(warn)
options(warn=2)
for (i in list(x,y,z)){
cc - try(mean(i), silent=T)
if(is(cc,try-error)) {next}
print(cc)
}
options(warn=warn)
see ?options under warn
Cheers
Joris
On Thu, Jun 24, 2010 at 5:12 PM, Paul Chatfield
[hidden email] wrote:
On a similar issue, how can you detect a warning in a loop - e.g. the
following gives a warning, so I'd like to set up code to recognise
that and
then carry on in a loop
x-rnorm(2);y-c(1,0)
ff-glm(y/23~x, family=binomial)
so this would be incorporated into a loop that might be
x-rnorm(10);y-rep(c(1,0),5)
for (i in 1:10)
{ee-glm(y~x, family=binomial)
ff-glm(y/23~x, family=binomial)}
from which I would recognise the warning in ff and not those in ee,
saving
results from ee and not from ff. The last bit would be easy adding a
line
if(there_is_a_warning_message) {newvector-NA} else {use results} but
how do
you detect the warning message?
Thanks all for your feedback so far,
Paul
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[R] Read in dataset without saving it
A simple question - I have a small dataset to read in and want to copy and paste part from Excel and paste it into an R script file without creating more files saving it as a .txt/.csv and then reading that in. I want to read in 3 columns e.g. 1 2.5 3.4 1 2.3 3.1 1 2.6 3.9 2 2.9 2.8 2 2.6 2.9 2 2.7 2.9 3 2.3 3.3 3 2.4 3.0 3 2.7 3.2 I thought I could use scan() but don't know how to extend it to multiple columns? I thought about using colwise (plyr) but think I am making this more complicated than it probably should be! Is there an easy way to do this? Any ideas gratefully received, Paul -- View this message in context: http://r.789695.n4.nabble.com/Read-in-dataset-without-saving-it-tp2248495p2248495.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] NAs are not allowed in subscripted assignments
I'm trying to assign NAs to values that satisfy certain conditions (more
complex than shown below) and it gives the right result, but breaks the loop
having done the first one viz:
new-c(rep(5,4),6)
for (i in 1:6)
{new[new[i]5.5][i]-NA}
gives the correct result, though an error message appears which causes a
break if it's in a loop. If I can get rid of the error message and get the
loop to continue, this should work fine. I'm sure I'm missing a simple
solution, but can't seem to see it,
Any help, as always, greatly appreciated,
Paul
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Re: [R] NAs are not allowed in subscripted assignments
Thank you â thatâs sorted it. Trying to make things too complicated! J
From: Alain Guillet-2 [via R]
[mailto:ml-node+1819104-1154170184-120...@n4.nabble.com]
Sent: 09 April 2010 10:34
To: Paul Chatfield
Subject: Re: NAs are not allowed in subscripted assignments
Maybe you can withdraw the [i] in your code...
for (i in 1:6)
+ {new[new[i]5.5]-NA}
new
[1] 5 5 5 5 NA
Alain
On 09-Apr-10 11:23, Paul Chatfield wrote:
I'm trying to assign NAs to values that satisfy certain conditions (more
complex than shown below) and it gives the right result, but breaks the loop
having done the first one viz:
new-c(rep(5,4),6)
for (i in 1:6)
{new[new[i]5.5][i]-NA}
gives the correct result, though an error message appears which causes a
break if it's in a loop. If I can get rid of the error message and get the
loop to continue, this should work fine. I'm sure I'm missing a simple
solution, but can't seem to see it,
Any help, as always, greatly appreciated,
Paul
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Statistician and Computer Scientist
SMCS - IMMAQ - Université catholique de Louvain
Bureau c.316
Voie du Roman Pays, 20
B-1348 Louvain-la-Neuve
Belgium
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[R] Paste expression in graph title
This was my initial attempt at creating a title on a graph of the R squared value: x-rnorm(10) y-rnorm(10) plot(x,y, main=paste(expression(R^2), = ,round(summary(lm(y~ x))$r.squared, digits=3), sep=)) I've read various other posts that say expression needs to be taken outside the paste, but I can't seem to get it work as the following fails plot(x,y, main=expression(paste(R^2, = ,round(summary(lm(y~ x))$r.squared, digits=3), sep=))) I tried it with title() and didn't get much further either. Hmmm, any ideas what am I missing? Thanks Paul -- View this message in context: http://n4.nabble.com/Paste-expression-in-graph-title-tp1289272p1289272.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Paste expression in graph title
Thanks â thatâs the most helpful solution â itâs important for code to be neat too and that is fairly neat if unintuitive! I had tried bquote a bit, but hadnât figured it out properly. Cheers! Paul From: Henrique Dallazuanna [via R] [mailto:ml-node+1289286-317759...@n4.nabble.com] Sent: 25 January 2010 12:21 To: Paul Chatfield Subject: Re: [R] Paste expression in graph title Try this: plot(x, y, main = bquote(R^2 == .(round(summary(lm(y ~ x))$r.squared, 3 On Mon, Jan 25, 2010 at 10:06 AM, Paul Chatfield [hidden email] http://n4.nabble.com/user/SendEmail.jtp?type=nodenode=1289286i=0 wrote: This was my initial attempt at creating a title on a graph of the R squared value: x-rnorm(10) y-rnorm(10) plot(x,y, main=paste(expression(R^2), = ,round(summary(lm(y~ x))$r.squared, digits=3), sep=)) I've read various other posts that say expression needs to be taken outside the paste, but I can't seem to get it work as the following fails plot(x,y, main=expression(paste(R^2, = ,round(summary(lm(y~ x))$r.squared, digits=3), sep=))) I tried it with title() and didn't get much further either. Hmmm, any ideas what am I missing? Thanks Paul -- View this message in context: http://n4.nabble.com/Paste-expression-in-graph-title-tp1289272p1289272.html Sent from the R help mailing list archive at Nabble.com. __ [hidden email] http://n4.nabble.com/user/SendEmail.jtp?type=nodenode=1289286i=1 mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O __ [hidden email] http://n4.nabble.com/user/SendEmail.jtp?type=nodenode=1289286i=2 mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. View message @ http://n4.nabble.com/Paste-expression-in-graph-title-tp1289272p1289286.html To unsubscribe from Paste expression in graph title, click here (link removed) = . -- View this message in context: http://n4.nabble.com/Paste-expression-in-graph-title-tp1289272p1289291.html Sent from the R help mailing list archive at Nabble.com. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Patterned shading in ggplot
Am trying to produce a graph which prints out well in black and white using ggplot2. I have the following example set up nicely, but want to shade the red bars in one pattern and the blue in another so they print out clearly. I tried changing colours to 1 light, 1 dark, but then the overlapping colour looks virtually identical to the darker one. I noted the option density in barplot, but couldn't get this to work in ggplot. I could just replot this as density plots using kernel smoothing, but quite like this format for the data I have. Any help much appreciated. library(ggplot2) xy-data.frame(x=c(rnorm(1000), rnorm(1000,2,1)), grp=as.factor(rep(1:2, each=1000))) ggplot(xy, aes(x=x, fill=grp, group=grp)) + geom_histogram(binwidth=0.5,colour=black,position = identity, alpha = 0.5) + scale_fill_manual(values = c(red, blue)) -- View this message in context: http://old.nabble.com/Patterned-shading-in-ggplot-tp26193795p26193795.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Plotting 1 covariate, 3 factors
Cheers guys that's helpful. Doug, you're right, my code for ff should have
been
for (i in 1:length(y))
{if (f1[i]==after f3[i]==1) ff[i]-1, after
else if(f1[i]==after f3[i]==2) ff[i]-2, after
else if(f1[i]==before f3[i]==1) ff[i]-1, before
else if(f1[i]==before f3[i]==2) ff[i]-2, before}
As I have factors with only 2,2 and 3 levels respectively, your approach
suits the problem perfectly. Just to round this off, trying to reorient it
back to having the y on axis 2 seems to mean the line now does dot-to-dot
instead of fitting the average. Am I being dim in missing a key option in
my statement below which would correct this as your code did, Doug, when
oriented the other way, or does it require some kind of panel statement?
dotplot(y~f2|f1, groups=f3, layout=c(2,1), strip=T, type=c(a,p), pch=19)
Thanks
Paul
Douglas Bates-2 wrote:
I'm not sure if this is exactly what you are looking for but I would
generally create an interaction plot using the lattice 'dotplot' with
type = c(p,a) so I get both the original data and the lines
joining the averages for the different factor levels. I also prefer
the horizontal orientation to the vertical orientation. Combining all
these variations produces something like
dotplot(f2 ~ y | f1, groups = f3, aspect = 0.2, layout = c(1,2), type
= c(p,a), pch = 21, strip = FALSE, strip.left = TRUE, auto.key =
list(columns = 2, lines = TRUE))
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Re: [R] Plotting 1 covariate, 3 factors
That's solved it. Superb! All you probably need is to make f2 a factor (e.g., y ~ factor(f2) | f1). Otherwise dotplot() doesn't know which one to treat as categorical. -Deepayan -- View this message in context: http://www.nabble.com/Plotting-1-covariate%2C-3-factors-tp25789442p25800910.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Plotting 1 covariate, 3 factors
I'm interested in plotting a y with an x factor as the combination of 2
factors and colour with respect to a third, which the code below does with
interaction.plot(). However, this is because I redefine the x to be 1
factor. Is there a way of getting it to plot without redefining it, and
ideally to not join up the lines BETWEEN levels a and b, but just join those
between after and before for one level of f3. I figure this could be done
by manually drawing over blank lines using ?lines but am not sure what the
coordinates would be and figured there is probably an easier way where
someone has dealt with this before. Any thoughts greatly appreciated,
Paul
#
y-rnorm(36)
f1-rep(c(after,before), 18)
f2-rep(1:3,12)
f3-rep(1:2, each=18)
## Define new factor to be f1 and f3 for x axis - clumsy code, but gets its
done;
ff-numeric(length(y))
for (i in 1:length(y))
{if (f1[i]==a f3[i]==1) ff[i]-1, a
else if(f1[i]==a f3[i]==2) ff[i]-2, a
else if(f1[i]==b f3[i]==1) ff[i]-1, b
else ff[i]-2, b}
## Plot of interest;
interaction.plot(ff,f2,y)
--
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Re: [R] Statistician Needed
Hi Noah - I work for the statistical services centre and could help. Email p.s.chatfi...@rdg.ac.uk, Paul Noah Silverman-3 wrote: Hello, I've come up with some challenges with my process that are a bit too complicated for the mailing list. Is there anyone out there, preferably a real statistician, who is willing to consult with me via phone/email for a few hours. I'm happy to pay you for your time. Thanks, -Noah __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- View this message in context: http://www.nabble.com/Statistician-Needed-tp24909288p24916331.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Combine two matricies
The following code should do it. This assumes matrices a and x are of the same dimension which is why you can index a as below x[is.na(x)==TRUE]-a[is.na(x)==TRUE] -- View this message in context: http://www.nabble.com/Combine-two-matricies-tp24458609p24458797.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] random sampling or random replacement
If you want to average 20% missing values then you could try it in 1 step, viz: sample(c(1:2, rep(NA, 2000)),100) Otherwise, 2 steps is preferable. Use code as below: sample(1:2,100)-kk kk[sample(1:100,20)]-NA Paul -- View this message in context: http://www.nabble.com/random-sampling-or-random-replacement-tp24199695p24200736.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Snap axes to origin rather than around it
I'm trying to plot a graph where the axes go through 0,0, rather than around it combined with a box round the graph, so x-0:10;y-0:10 plot(x,y) gives me a box but doesn't go through the point 0,0, but stays at a distance. In trying to circumvent this problem, I wrote plot(x,y) axis(1, pos=c(0,0));axis(2,pos=c(0,0)) which gives me axes that go through the origin, but then I can't make a box round a plot (box() goes again at this fixed distance from the origin rather than through it) Any thoughts, ideas gratefully accepted. Perhaps I've been looking at it so long I'm now missing the obvious! :) Paul Any ideas? -- View this message in context: http://www.nabble.com/Snap-axes-to-origin-rather-than-around-it-tp23997802p23997802.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Curved arrows
I'm trying to draw an arrow with a curved shaft on the graph as a straight one looks messy on a detailed graph. I've looked in arrows but it doesn't seem to give an option. larrows doesn't look much more promising. I had a look in the archive and couldn't find anything. Any thoughts? Thanks Paul -- View this message in context: http://www.nabble.com/Curved-arrows-tp23312316p23312316.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
