Re: [R] plotmath vector problem; full program enclosed
On Tue, Jul 6, 2010 at 12:41 PM, Duncan Murdoch murdoch.dun...@gmail.com wrote: On 06/07/2010 10:54 AM, Paul Johnson wrote: Here's another example of my plotmath whipping boy, the Normal distribution. You want as.expression(b1), not expression(b1). The latter means the expression consisting of the symbol b1. The former means take the object stored in b1, and convert it to an expression.. It's not perfect, because you'll end up with mu - -1.96sigma (i.e. two minus signs), but it's closer than what you had. Duncan Murdoch Hi, Duncan and David Thanks for looking. I suspect from the comment you did not run the code. The expression examples I give do work fine already. But I have to explicitly put in values like 1.96 to make them work. I'm trying to avid that with substitute, which does work for b2, b3, b4, b5, all but b1. Why just one? I'm uploading a picture of it so you can see for yourself: http://pj.freefaculty.org/R/plotmathwrong.pdf please look in the middle axis. Why does only b1 not work, but the rest do? -- Paul E. Johnson Professor, Political Science 1541 Lilac Lane, Room 504 University of Kansas __ 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] Gray level mosaic plot with shading_Friendly
On Tue, 6 Jul 2010, Michael Friendly wrote:
Michael Kubovy wrote:
Suppose we start with
data(Titanic)
mosaic(Titanic, shade = TRUE)
How do I combine the dashed box contours of shading_Friendly to indicate
negative residuals, with three levels of gray: dark for abs(Pearson Resid)
4, lighter for 4 abs(Pearson Resid) 2, and lightest for bs(Pearson
Resid) 2 ?
Do you mean [1] you want to plot positive residuals in color and negative in
gray scale?
Or [2] to fold + and - residuals by shading all according to abs(resid), and
distinguishing + from - by the dashed box outlines?
In fact, I designed this coding scheme so that mosaic plots in color (with my
blue - white - red scheme) would approximately do exactly what
you might want under [2], when rendered in B/W, since the fully saturated red
and blue are close in darkness in B/W.
And shading_hcl() has been written to do exactly what you want under [2].
While it is hard to come up with colors of different hues in HSV or HLS
space that have the same brightness (aka lightness/luminance) and the same
colorfulness (aka chroma), this is easy in HCL.
Try
mosaic(Titanic, gp=shading_Friendly)
save as a jpg/png and try converting to B/W with an image program and see if
this is good enough.
mosaic(Titanic, shade = TRUE)
is the same as
mosaic(Titanic, gp = shading_hcl)
which you can then modify to have different line types
mosaic(Titanic, gp = shading_hcl, gp_args = list(lty = 1:2))
If you print that on a grayscale printer you will see the same plot
without any chroma, i.e.,
mosaic(Titanic, gp = shading_hcl, gp_args = list(lty = 1:2, c = 0))
The shading_hcl() function is introduced in Zeileis et al. (2007, JCGS),
see ?shading_hcl, which provides more detailed references to HCL colors
etc.
Best,
Z
Alternatively, write your own, shading_Kubovy, modeled on
shading_Friendly -
function (observed = NULL, residuals = NULL, expected = NULL,
df = NULL, h = c(2/3, 0), lty = 1:2, interpolate = c(2, 4),
eps = 0.01, line_col = black, ...)
{
shading_hsv(observed = NULL, residuals = NULL, expected = NULL,
df = NULL, h = h, v = 1, lty = lty, interpolate = interpolate,
eps = eps, line_col = line_col, p.value = NA, ...)
}
environment: namespace:vcd
attr(,class)
[1] grapcon_generator
In the defaults, lty=1:2 is what distinguishes + and - for outline line type
hope this helps,
-Michael
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[R] R C# (Mono)
Hello did anyone ever use C# in connection with R ? i am looking into R extension but would like to use C# instead of C or C++ i wonder whether anyone has experience in particular with Mono for doing so many thanks in advance bernd __ 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] plotmath vector problem; full program enclosed
On 07/07/10 06:03, Paul Johnson wrote:
[...]
Hi, Duncan and David
Thanks for looking. I suspect from the comment you did not run the
code. The expression examples I give do work fine already. But I
have to explicitly put in values like 1.96 to make them work. I'm
trying to avid that with substitute, which does work for b2, b3, b4,
b5, all but b1. Why just one?
I'm uploading a picture of it so you can see for yourself:
http://pj.freefaculty.org/R/plotmathwrong.pdf
please look in the middle axis.
Why does only b1 not work, but the rest do?
Because you only had one (as.)expression in your original code,
perhaps? This version works for me:
### Filename: plotMathProblem.R
### Paul Johnson July 5, 2010
### email mepaulj...@ku.edu
### 2010-07-06 AE : Changes.
sigma- 10.0
mu- 4.0
myx- seq( mu - 3.5*sigma, mu+ 3.5*sigma, length.out=500)
myDensity- dnorm(myx,mean=mu,sd=sigma)
### xpd needed to allow writing outside strict box of graph
### Need big bottom margin to add several x axes
par(xpd=TRUE, ps=10, mar=c(18,2,2,2))
plot(myx, myDensity, type=l, xlab=, ylab=Probability Density ,
main=myTitle1, axes=FALSE)
axis(2, pos= mu - 3.6*sigma)
axis(1, pos=0)
lines(c(myx[1],myx[length(myx)]),c(0,0)) ### closes off axes
addInteriorLine- function(x, m, sd){
for (i in 1:(length(x))){
lines( c(x[i],x[i]), c(0, dnorm(x[i],m=m,sd=sd)), lty= 14, lwd=.2)
}
}
dividers- c(qnorm(0.025), -1, 0, 1, qnorm(0.975))
addInteriorLine(mu+sigma*dividers, mu,sigma)
# bquote creates an expression that text plotters can use
t1- bquote( mu== .(mu))
mtext(bquote( mu == .(mu)), 1, at=mu, line=-1)
addInteriorLabel- function(pos1, pos2, m, s){
area- abs(100*( pnorm(m+pos1*s,m,s)-pnorm(m+pos2*s, m,s)))
mid- m+0.5*(pos1+pos2)*s
text(mid, 0.5*dnorm(mid,m,s),label=paste(round(area,2),%))
}
addInteriorLabel(dividers[1],dividers[2], mu, sigma)
addInteriorLabel(dividers[2],dividers[3], mu, sigma)
addInteriorLabel(dividers[3],dividers[4], mu, sigma)
addInteriorLabel(dividers[4],dividers[5], mu, sigma)
### Following is problem point: axis will
### end up with correct labels, except for first point,
### where we end up with b1 instead of mu - 1.96*sigma.
b1- substitute( mu - d*sigma, list(d=*-round(dividers[1],2))* )
b2- substitute( mu - sigma )
b3- substitute( mu )
b4- substitute( mu + sigma )
b5- substitute( mu + d*sigma, list(d=round(dividers[5],2)) )
## plot(-20:50,-20:50,type=n,axes=F)
axis(1, line=4,at=mu+dividers*sigma,
labels=*as.expression(c(b1,b2,b3,b4,b5))*, padj=-1)
### This gets right result but have to hard code the dividers
b1- expression( mu - 1.96*sigma )
b2- expression( mu - sigma )
b3- expression( mu )
b4- expression( mu + sigma )
b5- expression( mu + 1.96*sigma )
axis(1, line=8,at=mu+dividers*sigma, labels=c(b1,b2,b3,b4,b5), padj=-1)
Allan
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Re: [R] grayscale wireframe??
A standalone example is always appreciated (cf. the posting guide) but try and see if help(gray.colors, package=grDevices) is the sort of thing you are looking for. Hope this helps Allan On 06/07/10 23:30, Marlin Keith Cox wrote: I need grayscale formatting for a wireframe. The only col.regions that I can find are color palettes are all colored: rainbow(n, s = 1, v = 1, start = 0, end = max(1,n - 1)/n, gamma = 1, alpha = 1) heat.colors(n, alpha = 1) terrain.colors(n, alpha = 1) topo.colors(n, alpha = 1) cm.colors(n, alpha = 1) The code follows: X11() library(lattice) par(family=serif, cex=1.2) wireframe(z ~ y*x, mat.df, drape = TRUE,col.regions=rainbow(100), zlab = list(Water mass error (%),rot=90), zlim=c(-50,180), xlab = list(Resistance error (%),rot=-9), ylab = list(Length error (%),rot=38), scales = list(arrows = FALSE), screen = list(z = -35, x = -77, y = 10)) __ 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] plotmath vector problem; full program enclosed
Ooops, I didn't convert this one to text right for the list. b1- substitute( mu - d*sigma, list(d=*-round(dividers[1],2))* ) should be b1- substitute( mu - d*sigma, list(d=-round(dividers[1],2)) ) and similarly for labels=*as.expression(c(b1,b2,b3,b4,b5))*, padj=-1) read labels=as.expression(c(b1,b2,b3,b4,b5)), padj=-1) Apologies Allan __ 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] ctree ordering nodes
On Tue, 6 Jul 2010, Paras Sharma wrote: Hello, When using the ctree function, from library (party) what is the syntax to order the Variables in the nodes in a specific way? For example, how would I specify to make a binary come first, then a continuous variable? Not sure what you mean here. There is only one splitting variable in each node... The order in which the variables are tested is simply the order in which they appear in the model formula. If you want to force ctree() to use a specific variable in the first split and a specific variable in the second split, this is currently not possible. ctree() always selects the locally optimal splitting variable. Also is there a way to force ctree to show variables which are not significant? You can inspect the fitted ctree in various ways. See vignette(party, package = party) for a few helpful examples, especially Section 5.3. If you want to construct your own trees using different algorithms, the package partykit might be of interest which is currently under development on R-Forge. hth, Z __ 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] Visualization of coefficients
Hello David,
Thanks to your posting I started looking at the function in the arm package.
It appears this function is quite mature, and offers (for example) the
ability to easily overlap coefficients from several models.
I updated the post I published on the subject, so at the end of it I give an
example of comparing the coef of several models:
http://www.r-statistics.com/2010/07/visualization-of-regression-coefficients-in-r/
Thanks again for the pointer.
Best,
Tal
Contact
Details:---
Contact me: tal.gal...@gmail.com | 972-52-7275845
Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) |
www.r-statistics.com (English)
--
On Wed, Jul 7, 2010 at 12:02 AM, David Atkins datk...@u.washington.eduwrote:
FYI, there is already a function coefplot in the arm package; for example,
compare:
library(arm)
Loading required package: MASS
Loading required package: Matrix
[snip]
Attaching package: 'arm'
The following object(s) are masked from 'package:coda':
traceplot
data(Mroz, package = car)
fm - glm(lfp ~ ., data = Mroz, family = binomial)
coefplot(fm)
with version below.
cheeres, Dave
detach(package:arm)
coefplot - function(object, df = NULL, level = 0.95, parm = NULL,
+labels = TRUE, xlab = Coefficient confidence intervals, ylab = ,
+xlim = NULL, ylim = NULL,
+las = 1, lwd = 1, lty = c(1, 2), pch = 19, col = 1,
+length = 0, angle = 30, code = 3, ...)
+ {
+cf - coef(object)
+se - sqrt(diag(vcov(object)))
+if(is.null(parm)) parm - seq_along(cf)
+if(is.numeric(parm) | is.logical(parm)) parm - names(cf)[parm]
+if(is.character(parm)) parm - which(names(cf) %in% parm)
+cf - cf[parm]
+se - se[parm]
+k - length(cf)
+
+if(is.null(df)) {
+ df - if(identical(class(object), lm)) df.residual(object) else 0
+}
+
+critval - if(df 0 is.finite(df)) {
+ qt((1 - level)/2, df = df)
+} else {
+ qnorm((1 - level)/2)
+}
+ci1 - cf + critval * se
+ci2 - cf - critval * se
+
+lwd - rep(lwd, length.out = 2)
+lty - rep(lty, length.out = 2)
+pch - rep(pch, length.out = k)
+col - rep(col, length.out = k)
+
+if(is.null(xlim)) xlim - range(c(0, min(ci1), max(ci2)))
+if(is.null(ylim)) ylim - c(1 - 0.05 * k, 1.05 * k)
+
+if(isTRUE(labels)) labels - names(cf)
+if(identical(labels, FALSE)) labels -
+labels - rep(labels, length.out = k)
+
+plot(0, 0, xlim = xlim, ylim = ylim, xlab = xlab, ylab = ylab,
+ axes = FALSE, type = n, las = las, ...)
+arrows(ci1, 1:k, ci2, 1:k, lty = lty[1], lwd = lwd[1], col = col,
+ length = length, angle = angle, code = code)
+points(cf, 1:k, pch = pch, col = col)
+abline(v = 0, lty = lty[2], lwd = lwd[2])
+axis(1)
+axis(2, at = 1:k, labels = labels, las = las)
+box()
+ }
coefplot(fm, parm = -1)
Achim Zeileis wrote:
I've thought about adding a plot() method for the coeftest() function in
the lmtest package. Essentially, it relies on a coef() and a vcov()
method being available - and that a central limit theorem holds. For
releasing it as a general function in the package the code is still too
raw, but maybe it's useful for someone on the list. Hence, I've included
it below.
An example would be to visualize all coefficients except the intercept for
the Mroz data:
data(Mroz, package = car)
fm - glm(lfp ~ ., data = Mroz, family = binomial)
coefplot(fm, parm = -1)
hth,
Z
coefplot - function(object, df = NULL, level = 0.95, parm = NULL,
labels = TRUE, xlab = Coefficient confidence intervals, ylab = ,
xlim = NULL, ylim = NULL,
las = 1, lwd = 1, lty = c(1, 2), pch = 19, col = 1,
length = 0, angle = 30, code = 3, ...)
{
cf - coef(object)
se - sqrt(diag(vcov(object)))
if(is.null(parm)) parm - seq_along(cf)
if(is.numeric(parm) | is.logical(parm)) parm - names(cf)[parm]
if(is.character(parm)) parm - which(names(cf) %in% parm)
cf - cf[parm]
se - se[parm]
k - length(cf)
if(is.null(df)) {
df - if(identical(class(object), lm)) df.residual(object) else 0
}
critval - if(df 0 is.finite(df)) {
qt((1 - level)/2, df = df)
} else {
qnorm((1 - level)/2)
}
ci1 - cf + critval * se
ci2 - cf - critval * se
lwd - rep(lwd, length.out = 2)
lty - rep(lty, length.out = 2)
pch - rep(pch, length.out = k)
col - rep(col, length.out = k)
if(is.null(xlim)) xlim - range(c(0, min(ci1), max(ci2)))
if(is.null(ylim)) ylim - c(1 - 0.05 * k, 1.05 * k)
if(isTRUE(labels)) labels - names(cf)
if(identical(labels, FALSE)) labels -
labels - rep(labels, length.out = k)
plot(0, 0, xlim = xlim, ylim = ylim, xlab = xlab, ylab = ylab,
axes = FALSE, type = n, las = las, ...)
arrows(ci1, 1:k, ci2,
[R] Wavelet
Hi useRs, Is it possible to get MORLET wavelet in R Thanks nuncio -- Nuncio.M Research Scientist National Center for Antarctic and Ocean research Head land Sada Vasco da Gamma Goa-403804 [[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.
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.
[R] How can I calculate Q-correlations?
I need to calculate Q-correlations, in order to quantify differences between the profiles of my tested persons. Has anybody any experience doing that? Which command/package targets Q-correlations? Timo __ 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] Visualization of coefficients
Thanks Tal. Nice summary on the web page. I think the last example would be even better if it was a stand-alone piece of code (i.e.: with data) that we could run. For example library(arm) data(Mroz, package = car) M1- glm(lfp ~ ., data = Mroz, family = binomial) M2- bayesglm(lfp ~ ., data = Mroz, family = binomial) M3- glm(lfp ~ ., data = Mroz, family = binomial(probit)) coefplot(M2, xlim=c(-2, 6),intercept=TRUE) coefplot(M1, add=TRUE, col.pts=red, intercept=TRUE) coefplot(M3, add=TRUE, col.pts=blue, intercept=TRUE, offset=0.2) Allan On 07/07/10 09:16, Tal Galili wrote: Hello David, Thanks to your posting I started looking at the function in the arm package. It appears this function is quite mature, and offers (for example) the ability to easily overlap coefficients from several models. I updated the post I published on the subject, so at the end of it I give an example of comparing the coef of several models: http://www.r-statistics.com/2010/07/visualization-of-regression-coefficients-in-r/ Thanks again for the pointer. Best, Tal [...] __ 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] Wavelet
Try RSiteSearch(MORLET) before you post. Allan On 07/07/10 09:38, nuncio m wrote: Hi useRs, Is it possible to get MORLET wavelet in R Thanks nuncio __ 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] Visualization of coefficients
On Wed, 7 Jul 2010, Tal Galili wrote:
Hello David,
Thanks to your posting I started looking at the function in the arm package.
It appears this function is quite mature, and offers (for example) the
ability to easily overlap coefficients from several models.
Re: more mature. arm's coefplot() is more flexible in certain respects,
mine is more convenient in others. The overlay functionality is something
arm's coefplot() is better in and it also as some further options
(vertical vs. horizontal etc.). My coefplot() has the advantage that it
does not need any modification as long as coef() and vcov() methods are
available. Furthermore, level can specify the significance level
(instead of always using one and two standard errors, respectively).
But it shouldn't be too hard to create a superset of all options.
I updated the post I published on the subject, so at the end of it I give an
example of comparing the coef of several models:
http://www.r-statistics.com/2010/07/visualization-of-regression-coefficient
s-in-r/
As Allan pointed out in his reply, something fully reproducible would be
nice. Also, if you keep the example with quasi-complete separation, it
would be worth pointing this out. (Because the maximum likelihood
estimator is Infinity in this case.)
Finally, the Poisson model in comparison with the binomial models does not
make much sense, I guess.
Best,
Z
Thanks again for the pointer.
Best,
Tal
Contact
Details:---
Contact me: tal.gal...@gmail.com | 972-52-7275845
Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) |
www.r-statistics.com (English)
---
---
On Wed, Jul 7, 2010 at 12:02 AM, David Atkins datk...@u.washington.edu
wrote:
FYI, there is already a function coefplot in the arm package;
for example, compare:
library(arm)
Loading required package: MASS
Loading required package: Matrix
[snip]
Attaching package: 'arm'
The following object(s) are masked from 'package:coda':
traceplot
data(Mroz, package = car)
fm - glm(lfp ~ ., data = Mroz, family = binomial)
coefplot(fm)
with version below.
cheeres, Dave
detach(package:arm)
coefplot - function(object, df = NULL, level = 0.95, parm = NULL,
+ labels = TRUE, xlab = Coefficient confidence intervals, ylab =
,
+ xlim = NULL, ylim = NULL,
+ las = 1, lwd = 1, lty = c(1, 2), pch = 19, col = 1,
+ length = 0, angle = 30, code = 3, ...)
+ {
+ cf - coef(object)
+ se - sqrt(diag(vcov(object)))
+ if(is.null(parm)) parm - seq_along(cf)
+ if(is.numeric(parm) | is.logical(parm)) parm - names(cf)[parm]
+ if(is.character(parm)) parm - which(names(cf) %in% parm)
+ cf - cf[parm]
+ se - se[parm]
+ k - length(cf)
+
+ if(is.null(df)) {
+ df - if(identical(class(object), lm)) df.residual(object)
else 0
+ }
+
+ critval - if(df 0 is.finite(df)) {
+ qt((1 - level)/2, df = df)
+ } else {
+ qnorm((1 - level)/2)
+ }
+ ci1 - cf + critval * se
+ ci2 - cf - critval * se
+
+ lwd - rep(lwd, length.out = 2)
+ lty - rep(lty, length.out = 2)
+ pch - rep(pch, length.out = k)
+ col - rep(col, length.out = k)
+
+ if(is.null(xlim)) xlim - range(c(0, min(ci1), max(ci2)))
+ if(is.null(ylim)) ylim - c(1 - 0.05 * k, 1.05 * k)
+
+ if(isTRUE(labels)) labels - names(cf)
+ if(identical(labels, FALSE)) labels -
+ labels - rep(labels, length.out = k)
+
+ plot(0, 0, xlim = xlim, ylim = ylim, xlab = xlab, ylab = ylab,
+ axes = FALSE, type = n, las = las, ...)
+ arrows(ci1, 1:k, ci2, 1:k, lty = lty[1], lwd = lwd[1], col = col,
+ length = length, angle = angle, code = code)
+ points(cf, 1:k, pch = pch, col = col)
+ abline(v = 0, lty = lty[2], lwd = lwd[2])
+ axis(1)
+ axis(2, at = 1:k, labels = labels, las = las)
+ box()
+ }
coefplot(fm, parm = -1)
Achim Zeileis wrote:
I've thought about adding a plot() method for the coeftest() function
in
the lmtest package. Essentially, it relies on a coef() and a vcov()
method being available - and that a central limit theorem holds. For
releasing it as a general function in the package the code is still
too
raw, but maybe it's useful for someone on the list. Hence, I've
included
it below.
An example would be to visualize all coefficients except the intercept
for
the Mroz data:
data(Mroz, package = car)
fm - glm(lfp ~ ., data = Mroz, family = binomial)
coefplot(fm, parm = -1)
hth,
Z
coefplot - function(object, df = NULL, level = 0.95, parm = NULL,
labels = TRUE, xlab = Coefficient confidence intervals, ylab = ,
xlim = NULL, ylim = NULL,
las = 1, lwd = 1, lty = c(1, 2), pch = 19, col = 1,
length = 0, angle = 30, code = 3, ...)
{
cf - coef(object)
se - sqrt(diag(vcov(object)))
if(is.null(parm)) parm - seq_along(cf)
Re: [R] Wavelet
I would even look at the packages page... There are plenty of them. Please, at least, do minimal research before clogging up this list. Also, please read the posting guide that is appended to every mail to this list. On Wed, Jul 7, 2010 at 4:03 AM, Allan Engelhardt all...@cybaea.com wrote: Try RSiteSearch(MORLET) before you post. Allan On 07/07/10 09:38, nuncio m wrote: Hi useRs, Is it possible to get MORLET wavelet in R Thanks nuncio __ 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. -- Stephen Sefick | Auburn University | | Department of Biological Sciences | | 331 Funchess Hall | | Auburn, Alabama | | 36849| |___| | sas0...@auburn.edu | | http://www.auburn.edu/~sas0025 | |___| Let's not spend our time and resources thinking about things that are so little or so large that all they really do for us is puff us up and make us feel like gods. We are mammals, and have not exhausted the annoying little problems of being mammals. -K. Mullis __ 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] Visualization of coefficients
Hi Achim and Allan,
I updated the post with Allan's example (thanks Allan).
Achim, you wrote:
Finally, the Poisson model in comparison with the binomial models does not
make much sense, I guess.
I agree. I wanted something to showcase the function on 3 models (with the
same predictors), and that's the easiest I could think of. If you'd think
of a smarter example I'd be happy to incorporate it.
Best,
Tal
Contact
Details:---
Contact me: tal.gal...@gmail.com | 972-52-7275845
Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) |
www.r-statistics.com (English)
--
On Wed, Jul 7, 2010 at 12:10 PM, Achim Zeileis achim.zeil...@uibk.ac.atwrote:
On Wed, 7 Jul 2010, Tal Galili wrote:
Hello David,
Thanks to your posting I started looking at the function in the arm
package.
It appears this function is quite mature, and offers (for example) the
ability to easily overlap coefficients from several models.
Re: more mature. arm's coefplot() is more flexible in certain respects,
mine is more convenient in others. The overlay functionality is something
arm's coefplot() is better in and it also as some further options (vertical
vs. horizontal etc.). My coefplot() has the advantage that it does not need
any modification as long as coef() and vcov() methods are available.
Furthermore, level can specify the significance level (instead of always
using one and two standard errors, respectively).
But it shouldn't be too hard to create a superset of all options.
I updated the post I published on the subject, so at the end of it I give
an
example of comparing the coef of several models:
http://www.r-statistics.com/2010/07/visualization-of-regression-coefficient
s-in-r/
As Allan pointed out in his reply, something fully reproducible would be
nice. Also, if you keep the example with quasi-complete separation, it would
be worth pointing this out. (Because the maximum likelihood estimator is
Infinity in this case.)
Finally, the Poisson model in comparison with the binomial models does not
make much sense, I guess.
Best,
Z
Thanks again for the pointer.
Best,
Tal
Contact
Details:---
Contact me: tal.gal...@gmail.com | 972-52-7275845
Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) |
www.r-statistics.com (English)
---
---
On Wed, Jul 7, 2010 at 12:02 AM, David Atkins datk...@u.washington.edu
wrote:
FYI, there is already a function coefplot in the arm package;
for example, compare:
library(arm)
Loading required package: MASS
Loading required package: Matrix
[snip]
Attaching package: 'arm'
The following object(s) are masked from 'package:coda':
traceplot
data(Mroz, package = car)
fm - glm(lfp ~ ., data = Mroz, family = binomial)
coefplot(fm)
with version below.
cheeres, Dave
detach(package:arm)
coefplot - function(object, df = NULL, level = 0.95, parm = NULL,
+labels = TRUE, xlab = Coefficient confidence intervals, ylab =
,
+xlim = NULL, ylim = NULL,
+las = 1, lwd = 1, lty = c(1, 2), pch = 19, col = 1,
+length = 0, angle = 30, code = 3, ...)
+ {
+cf - coef(object)
+se - sqrt(diag(vcov(object)))
+if(is.null(parm)) parm - seq_along(cf)
+if(is.numeric(parm) | is.logical(parm)) parm - names(cf)[parm]
+if(is.character(parm)) parm - which(names(cf) %in% parm)
+cf - cf[parm]
+se - se[parm]
+k - length(cf)
+
+if(is.null(df)) {
+ df - if(identical(class(object), lm)) df.residual(object)
else 0
+}
+
+critval - if(df 0 is.finite(df)) {
+ qt((1 - level)/2, df = df)
+} else {
+ qnorm((1 - level)/2)
+}
+ci1 - cf + critval * se
+ci2 - cf - critval * se
+
+lwd - rep(lwd, length.out = 2)
+lty - rep(lty, length.out = 2)
+pch - rep(pch, length.out = k)
+col - rep(col, length.out = k)
+
+if(is.null(xlim)) xlim - range(c(0, min(ci1), max(ci2)))
+if(is.null(ylim)) ylim - c(1 - 0.05 * k, 1.05 * k)
+
+if(isTRUE(labels)) labels - names(cf)
+if(identical(labels, FALSE)) labels -
+labels - rep(labels, length.out = k)
+
+plot(0, 0, xlim = xlim, ylim = ylim, xlab = xlab, ylab = ylab,
+ axes = FALSE, type = n, las = las, ...)
+arrows(ci1, 1:k, ci2, 1:k, lty = lty[1], lwd = lwd[1], col = col,
+ length = length, angle = angle, code = code)
+points(cf, 1:k, pch = pch, col = col)
+abline(v = 0, lty = lty[2], lwd = lwd[2])
+axis(1)
+axis(2, at = 1:k, labels = labels, las = las)
+box()
+ }
coefplot(fm, parm = -1)
Achim Zeileis wrote:
Re: [R] Visualization of coefficients
On Wed, 7 Jul 2010, Tal Galili wrote:
Hi Achim and Allan,I updated the post with Allan's example (thanks Allan).
Thanks!
Achim, you wrote:
Finally, the Poisson model in comparison with the binomial models does not
make much sense, I guess.
I agree. I wanted something to showcase the function on 3 models (with the
same predictors), and that's the easiest I could think of. If you'd think
of a smarter example I'd be happy to incorporate it.
You could generate Poisson data and then fit the binomial model to the
threshold version of the response.
But I guess that would be a bit over the top. Also, one could argue in
that case that a complementary log-log link should be employed.
Hence, I would simply say (verbally) that it works for baysglm, glm, lm,
polr objects and that a default method is available which takes
pre-computed coefficients and associated standard errors from any suitable
model.
Best,
Z
Best,
Tal
Contact
Details:---
Contact me: tal.gal...@gmail.com | 972-52-7275845
Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) |
www.r-statistics.com (English)
---
---
On Wed, Jul 7, 2010 at 12:10 PM, Achim Zeileis achim.zeil...@uibk.ac.at
wrote:
On Wed, 7 Jul 2010, Tal Galili wrote:
Hello David,
Thanks to your posting I started looking at the
function in the arm package.
It appears this function is quite mature, and
offers (for example) the
ability to easily overlap coefficients from several
models.
Re: more mature. arm's coefplot() is more flexible in certain
respects, mine is more convenient in others. The overlay functionality
is something arm's coefplot() is better in and it also as some further
options (vertical vs. horizontal etc.). My coefplot() has the
advantage that it does not need any modification as long as coef() and
vcov() methods are available. Furthermore, level can specify the
significance level (instead of always using one and two standard
errors, respectively).
But it shouldn't be too hard to create a superset of all options.
I updated the post I published on the subject, so at the
end of it I give an
example of comparing the coef of several models:
http://www.r-statistics.com/2010/07/visualization-of-regression-coefficient
s-in-r/
As Allan pointed out in his reply, something fully reproducible would
be nice. Also, if you keep the example with quasi-complete separation,
it would be worth pointing this out. (Because the maximum likelihood
estimator is Infinity in this case.)
Finally, the Poisson model in comparison with the binomial models does
not make much sense, I guess.
Best,
Z
Thanks again for the pointer.
Best,
Tal
Contact
Details:---
Contact me: tal.gal...@gmail.com | 972-52-7275845
Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il
(Hebrew) |
www.r-statistics.com (English)
---
---
On Wed, Jul 7, 2010 at 12:02 AM, David Atkins
datk...@u.washington.edu
wrote:
FYI, there is already a function coefplot in the arm
package;
for example, compare:
library(arm)
Loading required package: MASS
Loading required package: Matrix
[snip]
Attaching package: 'arm'
The following object(s) are masked from 'package:coda':
traceplot
data(Mroz, package = car)
fm - glm(lfp ~ ., data = Mroz, family = binomial)
coefplot(fm)
with version below.
cheeres, Dave
detach(package:arm)
coefplot - function(object, df = NULL, level = 0.95, parm =
NULL,
+ labels = TRUE, xlab = Coefficient confidence intervals,
ylab =
,
+ xlim = NULL, ylim = NULL,
+ las = 1, lwd = 1, lty = c(1, 2), pch = 19, col = 1,
+ length = 0, angle = 30, code = 3, ...)
+ {
+ cf - coef(object)
+ se - sqrt(diag(vcov(object)))
+ if(is.null(parm)) parm - seq_along(cf)
+ if(is.numeric(parm) | is.logical(parm)) parm -
names(cf)[parm]
+ if(is.character(parm)) parm - which(names(cf) %in% parm)
+ cf - cf[parm]
+ se - se[parm]
+ k - length(cf)
+
+ if(is.null(df)) {
+ df - if(identical(class(object), lm))
df.residual(object)
else 0
+ }
+
+ critval - if(df 0 is.finite(df)) {
+ qt((1 - level)/2, df = df)
+ } else {
+ qnorm((1 - level)/2)
+ }
+ ci1 - cf + critval * se
+ ci2 - cf - critval * se
+
+ lwd - rep(lwd, length.out = 2)
+ lty - rep(lty, length.out = 2)
+ pch - rep(pch, length.out = k)
+ col - rep(col, length.out = k)
+
+ if(is.null(xlim)) xlim - range(c(0, min(ci1), max(ci2)))
+ if(is.null(ylim)) ylim - c(1 - 0.05 * k, 1.05 * k)
+
+ if(isTRUE(labels)) labels - names(cf)
+
Re: [R] Visualization of coefficients
I Achim,
I retained the example (so to illustrate the use of the function) - but
pointed out to it's nonsensical nature.
Credit was mentioned to both you and Allan.
Thanks,
Tal
Contact
Details:---
Contact me: tal.gal...@gmail.com | 972-52-7275845
Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) |
www.r-statistics.com (English)
--
On Wed, Jul 7, 2010 at 12:31 PM, Achim Zeileis achim.zeil...@uibk.ac.atwrote:
On Wed, 7 Jul 2010, Tal Galili wrote:
Hi Achim and Allan,I updated the post with Allan's example (thanks Allan).
Thanks!
Achim, you wrote:
Finally, the Poisson model in comparison with the binomial models does
not
make much sense, I guess.
I agree. I wanted something to showcase the function on 3 models (with
the
same predictors), and that's the easiest I could think of. If you'd think
of a smarter example I'd be happy to incorporate it.
You could generate Poisson data and then fit the binomial model to the
threshold version of the response.
But I guess that would be a bit over the top. Also, one could argue in that
case that a complementary log-log link should be employed.
Hence, I would simply say (verbally) that it works for baysglm, glm, lm,
polr objects and that a default method is available which takes pre-computed
coefficients and associated standard errors from any suitable model.
Best,
Z
Best,
Tal
Contact
Details:---
Contact me: tal.gal...@gmail.com | 972-52-7275845
Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) |
www.r-statistics.com (English)
---
---
On Wed, Jul 7, 2010 at 12:10 PM, Achim Zeileis achim.zeil...@uibk.ac.at
wrote:
On Wed, 7 Jul 2010, Tal Galili wrote:
Hello David,
Thanks to your posting I started looking at the
function in the arm package.
It appears this function is quite mature, and
offers (for example) the
ability to easily overlap coefficients from several
models.
Re: more mature. arm's coefplot() is more flexible in certain
respects, mine is more convenient in others. The overlay functionality
is something arm's coefplot() is better in and it also as some further
options (vertical vs. horizontal etc.). My coefplot() has the
advantage that it does not need any modification as long as coef() and
vcov() methods are available. Furthermore, level can specify the
significance level (instead of always using one and two standard
errors, respectively).
But it shouldn't be too hard to create a superset of all options.
I updated the post I published on the subject, so at the
end of it I give an
example of comparing the coef of several models:
http://www.r-statistics.com/2010/07/visualization-of-regression-coefficient
s-in-r/
As Allan pointed out in his reply, something fully reproducible would
be nice. Also, if you keep the example with quasi-complete separation,
it would be worth pointing this out. (Because the maximum likelihood
estimator is Infinity in this case.)
Finally, the Poisson model in comparison with the binomial models does
not make much sense, I guess.
Best,
Z
Thanks again for the pointer.
Best,
Tal
Contact
Details:---
Contact me: tal.gal...@gmail.com | 972-52-7275845
Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il
(Hebrew) |
www.r-statistics.com (English)
---
---
On Wed, Jul 7, 2010 at 12:02 AM, David Atkins
datk...@u.washington.edu
wrote:
FYI, there is already a function coefplot in the arm
package;
for example, compare:
library(arm)
Loading required package: MASS
Loading required package: Matrix
[snip]
Attaching package: 'arm'
The following object(s) are masked from 'package:coda':
traceplot
data(Mroz, package = car)
fm - glm(lfp ~ ., data = Mroz, family = binomial)
coefplot(fm)
with version below.
cheeres, Dave
detach(package:arm)
coefplot - function(object, df = NULL, level = 0.95, parm =
NULL,
+labels = TRUE, xlab = Coefficient confidence intervals,
ylab =
,
+xlim = NULL, ylim = NULL,
+las = 1, lwd = 1, lty = c(1, 2), pch = 19, col = 1,
+length = 0, angle = 30, code = 3, ...)
+ {
+cf - coef(object)
+se - sqrt(diag(vcov(object)))
+if(is.null(parm)) parm - seq_along(cf)
+if(is.numeric(parm) | is.logical(parm)) parm -
names(cf)[parm]
+if(is.character(parm)) parm -
Re: [R] plotmath vector problem; full program enclosed
On 07/07/2010 1:03 AM, Paul Johnson wrote: On Tue, Jul 6, 2010 at 12:41 PM, Duncan Murdoch murdoch.dun...@gmail.com wrote: On 06/07/2010 10:54 AM, Paul Johnson wrote: Here's another example of my plotmath whipping boy, the Normal distribution. You want as.expression(b1), not expression(b1). The latter means the expression consisting of the symbol b1. The former means take the object stored in b1, and convert it to an expression.. It's not perfect, because you'll end up with mu - -1.96sigma (i.e. two minus signs), but it's closer than what you had. Duncan Murdoch Hi, Duncan and David Thanks for looking. I suspect from the comment you did not run the code. And I'm certain from your comment that you didn't run my code, or read the explanation carefully. Duncan Murdoch The expression examples I give do work fine already. But I have to explicitly put in values like 1.96 to make them work. I'm trying to avid that with substitute, which does work for b2, b3, b4, b5, all but b1. Why just one? I'm uploading a picture of it so you can see for yourself: http://pj.freefaculty.org/R/plotmathwrong.pdf please look in the middle axis. Why does only b1 not work, but the rest do? __ 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] sqlQuery not functioning properly
Hello, I am using Rcommander to generate codes and when I go through it to load my excel file, it works. But every time I re-open my script and try to load it directly with the script Rcommander generated, I get an error message. Basically the code works one time and then no more. Here's the script and the error message I get: data - sqlQuery(channel = 2, select * from [Questionnaire$]) [5] ERREUR: ']' inattendu(e) dans pal - sqlQuery(channel = 2, select * from [Questionnaire$] I would be very grateful if someone could help me with that, Thanks! Simon __ 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] Break in the y-axis
On 07/07/2010 02:39 AM, beloitstudent wrote: Thanks for the advice! It has worked for the most part. However, I am still coming up with an error message when placing my break line in the axis that I'm not sure what it means. If you could help me out, that would be fantastic...otherwise I might just have to see if I can add it on powerpoint. Here is the code you gave me and my script that doesn't work. ... axis.break(axis=2, breakpos=50,style=slash,pos=-20,brw=0.02)* This line should be: axis.break(axis=2, breakpos=50,style=slash,pos=-20,brw=0.02) That is, you have passed a character string instead of a number. Jim __ 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] Help With ANOVA (corrected please ignore last email)
Hi: I'd suggest looking at the following plot (data in original post, copied below): library(lattice) stripplot(Intensity ~ Group, data = zzzanova) Some things stand out in this plot that merit attention. As Josh Wiley pointed out in an earlier reply, the concentration of -4.60517 values in this data set should be a concern. Of your 54 responses, 26 of them have this value, spread across all six groups - in addition, this value is concentrated in the first few groups and becomes rarer in the later groups. Consider: with(zzzanova, table(Group)) Group Group1 Group2 Group3 Group4 Group5 Group6 9 8 9 10 9 9 with(subset(zzzanova, Intensity -1), table(Intensity, Group)) Group Intensity Group1 Group2 Group3 Group4 Group5 Group6 -4.60517 8 5 4 4 4 1 Also note that there is one other value with a negative intensity in group 6 (-0.024). (Side note: Is negative intensity meaningful in the context of your scientific problem?) I'm curious about what this -4.60 value is supposed to represent. Is it a missing value code, as Josh inferred, a left-censoring value, or something else? The reason for asking is that its purpose could well have an impact on the type of analysis that is appropriate for these data: * if -4.60 is meant to be a missing value code, its inclusion greatly inflates the degrees of freedom actually present in the data. Moreover, its presence also inflates the variability both within and between groups, which reduces the sensitivity of tests. Also observe from the strip plot above that if -4.60 is a missing value code, then your non-missing data appear to increase in variance with increasing group numbers; moreover, the imbalance in observations between groups is more severe, which in turn makes the p-values of tests less reliable for the non-missing data. (Even worse, the imbalance and increasing variance don't appear to be independent if -4.60 represents a missing value.) * if -4.60 is meant to be something like a lower detection limit or upper bound on a left-censored response, then one is artificially reducing variability within groups and the p-values of tests turn out to be optimistic. There are better ways to handle nondetects. One approach is outlined in Helsel's book 'Nondetects and Data Analysis', which is the reference of the R package NADA. Other approaches to nondetect data are also available outside of R (e.g., the SCOUT software at EPA). * if -4.60 is meant to represent a lower bound on a (censored?) response, then setting -4.60 as the response artificially increases the variation in the data. In this case, one would be artificially left-truncating the data, but for a different reason from that given in the immediately preceding point. Josh also noted the difference in p-values of the between group test when the groups were factors as opposed to numeric. This is a common 'gotcha' in R - you need to pay attention to the classes of your inputs when fitting any statistical model. In the case where the groups comprise a factor, R performs a one-way ANOVA. [FWIW, I got the same p-value as Josh for your 'complete' data (54 obs.)]. When the group variable is numeric, you are fitting a simple linear regression analysis, which implies that the numeric values of the groups are meaningful. There is a big difference in the interpretation of the two types of models. HTH, Dennis On Tue, Jul 6, 2010 at 6:12 AM, Amit Patel amitrh...@yahoo.co.uk wrote: Sorry i had a misprint in the appendix code in the last email Hi I needed some help with ANOVA I have a problem with My ANOVA analysis. I have a dataset with a known ANOVA p-value, however I can not seem to re-create it in R. I have created a list (zzzanova) which contains 1)Intensity Values 2)Group Number (6 Different Groups) 3)Sample Number (54 different samples) this is created by the script in Appendix 1 I then conduct ANOVA with the command zzz.aov - aov(Intensity ~ Group, data = zzzanova) I get a p-value of Pr(F)1 0.9483218 The expected p-value is 0.00490 so I feel I maybe using ANOVA incorrectly or have put in a wrong formula. I am trying to do an ANOVA analysis across all 6 Groups. Is there something wrong with my formula. But I think I have made a mistake in the formula rather than anything else. APPENDIX 1 datalist - c(-4.60517, -4.60517, -4.60517, -4.60517, -4.60517, -4.60517, -4.60517, 3.003749, -4.60517, 2.045314, 2.482557, -4.60517, -4.60517, -4.60517, -4.60517, 1.592743, -4.60517, -4.60517, 0.91328, -4.60517, -4.60517, 1.827744, 2.457795, 0.355075, -4.60517, 2.39127, 2.016987, 2.319903, 1.146683, -4.60517, -4.60517, -4.60517, 1.846162, -4.60517, 2.121427, 1.973118, -4.60517, 2.251568, -4.60517, 2.270724, 0.70338, 0.963816, -4.60517, 0.023703, -4.60517, 2.043382, 1.070586,
Re: [R] grouped logit regression
by grouped data are you saying that you have counts of outcomes and counts of trials? That is how I interpret the glogit in stata. If that is the case you can put your data into glm() like this fit-glm(nevents~xvars, weights=ntrials, family=binomial, data=yourdataset) will fit the binomial regression model summary(fit) will print the coefficients and model fit In the future could you please read the posting guide and put in a data example or some R code you have tried. CS - Corey Sparks, PhD Assistant Professor Department of Demography and Organization Studies University of Texas at San Antonio 501 West Durango Blvd Monterey Building 2.270C San Antonio, TX 78207 210-458-3166 corey.sparks 'at' utsa.edu https://rowdyspace.utsa.edu/users/ozd504/www/index.htm -- View this message in context: http://r.789695.n4.nabble.com/grouped-logit-regression-tp2280763p2280806.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] how to download files from ftp site
Hi all, I'm running R 2.10.1 on Windows XP and I'd like to read files from a ftp site. Does anybody know how to do ? Thanks Benoit -- Benoit Wastine Laboratoire des Sciences du Climat et de l’Environnement (LSCE/IPSL) CEA-CNRS-UVSQ CE Saclay Orme des merisiers Bât 703 - Pte 13A 91191 Gif sur Yvette Cedex France Tel : 33 (0)1 69 08 21 97 Fax : 33 (0)1 69 08 77 16 __ 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] Odp: identifying odd or even number
Hi r-help-boun...@r-project.org napsal dne 01.07.2010 17:40:22: Hi, I run into problem when writing a syntax, I don't know syntax that will return true or false if an integer is odd or even. fff - function(x) as.logical(x%%2) Regards Petr Thanks OYEYEMI, Gafar Matanmi Department of Statistics University of Ilorin P.M.B 1515 Ilorin, Kwara State Nigeria Tel: +2348052278655 Tel: +2348068241885 [[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-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] numerical derivative R help
Which package are you using?
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of Parminder Mankoo
Sent: Tuesday, July 06, 2010 10:50 PM
To: r-help@r-project.org
Subject: [R] numerical derivative R help
Importance: Low
I fit my CDF to sum of exponentials and now I want to take the numerical
derivative of this function to obtain probability density.I will really
appreciate your help reagrding the error messages I am getting which I
don't
understand.
*
*
fitterma - function(xtime) {
a - -0.09144115
b - -0.01335756
c - -2.368057
d - -0.00600052
return(exp(a+b*xtime)+exp(c+d*xtime))
}
numericDeriv(fitterma,xtime)
*Error in numericDeriv(fitterma, xtime) : *
* cannot coerce type 'closure' to vector of type 'double'*
*
*
*Thanks,*
*parmee*
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DISCLAIMER:\ Sample Disclaimer added in a VBScript.\ ...{{dropped:3}}
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[R] Need help in handling date
Dear all, I have a date related question. Suppose I have a character string March-2009, how I can convert it to a valid date object like as.yearmon(2009-01-03) in the zoo package? Is there any possibility there? Ans secondly is there any R function which will give the names of of all months as LETTERS does? Thanks for your time. [[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.
Re: [R] Need help in handling date
?strptime ‘%B’ Full month name in the current locale. (Also matches abbreviated name on input.) On Wed, Jul 7, 2010 at 8:40 AM, Christofer Bogaso bogaso.christo...@gmail.com wrote: Dear all, I have a date related question. Suppose I have a character string March-2009, how I can convert it to a valid date object like as.yearmon(2009-01-03) in the zoo package? Is there any possibility there? Ans secondly is there any R function which will give the names of of all months as LETTERS does? Thanks for your time. [[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-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] Need help in handling date
On Jul 7, 2010, at 7:40 AM, Christofer Bogaso wrote: Dear all, I have a date related question. Suppose I have a character string March-2009, how I can convert it to a valid date object like as.yearmon(2009-01-03) in the zoo package? Is there any possibility there? Ans secondly is there any R function which will give the names of of all months as LETTERS does? Thanks for your time. You may need to append a default day, so something like: as.Date(paste(March-2009, -15, sep = ), format = %B-%Y-%d) [1] 2009-03-15 Otherwise on OSX, I get: as.Date(March-2009, format = %B-%Y) [1] NA The above behavior regarding missing components is system specific, as per the Note in ?as.Date. Also see ?strptime for the formatting options. For the second part: month.name [1] January February March April May [6] June July AugustSeptember October [11] November December See ?LETTERS HTH, Marc Schwartz __ 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 download files from ftp site
?download.file Uwe Ligges On 07.07.2010 14:05, Benoit Wastine wrote: Hi all, I'm running R 2.10.1 on Windows XP and I'd like to read files from a ftp site. Does anybody know how to do ? Thanks Benoit __ 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] Need help in handling date
On Wed, Jul 7, 2010 at 8:40 AM, Christofer Bogaso bogaso.christo...@gmail.com wrote: Dear all, I have a date related question. Suppose I have a character string March-2009, how I can convert it to a valid date object like as.yearmon(2009-01-03) in the zoo package? Is there any possibility there? Try this. The first returns a Date class object and the second returns a character string: d - March-2009 as.Date(as.yearmon(d, %B-%Y)) [1] 2009-03-01 format(as.Date(as.yearmon(d, %B-%Y))) [1] 2009-03-01 See R News 4/1 for more on times and dates. Ans secondly is there any R function which will give the names of of all months as LETTERS does? month.abb [1] Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec month.name [1] January February March April May June [7] July AugustSeptember October November December __ 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] can't open file
I have a text file log2.log encoded Ansi in Windows. When I execute: out - read.zoo(readLines(con - file(log2.log, encoding=UCS-2LE)),FUN = as.chron) have errors: Error en file(file, rt) : no se puede abrir la conexión Además: Mensajes de aviso perdidos 1: In file(file, rt) : sólo fue usado el primer elemento del argumento 'description' 2: In file(file, rt) : no fue posible abrir el archivo '#Software: Microsoft Internet Information Services 5.0': No such file or directory Why? Thks, Sebastián. __ 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] Loading text file from data folder of package
Hello, I am creating a package and in my vignette I would like to load a text file from the data folder of the package. Currently, I am doing the following: filepath - paste(.libPaths(), pkgname, data, sample.txt, sep = /) file(filepath) Is there a better way of doing this? Thanks, Andrew __ 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] Trimming in R
I am looking for a way to trim leading and trailing spaces in a character
string in R. For example:
this is random text
should become:
this is random text.
I have a short function to perform this task as follows:
trim - function(str){
str - sub(^ +, , str)
str - sub( +$, , str)
}
While this function does seem to work, I am curious if there's anything built
into R that I can use instead, as that would be preferable.
Thanks in advance,
Andrew
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Re: [R] Trimming in R
Try this:
gsub(^\\s+|\\s+$, ,this is random text)
On Wed, Jul 7, 2010 at 10:23 AM, Andrew Leeser aml05willi...@yahoo.comwrote:
I am looking for a way to trim leading and trailing spaces in a character
string in R. For example:
this is random text
should become:
this is random text.
I have a short function to perform this task as follows:
trim - function(str){
str - sub(^ +, , str)
str - sub( +$, , str)
}
While this function does seem to work, I am curious if there's anything
built
into R that I can use instead, as that would be preferable.
Thanks in advance,
Andrew
[[alternative HTML version deleted]]
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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
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Re: [R] Loading text file from data folder of package
See ?system.file On Wed, Jul 7, 2010 at 10:20 AM, Andrew Liu andrew.t@williams.eduwrote: Hello, I am creating a package and in my vignette I would like to load a text file from the data folder of the package. Currently, I am doing the following: filepath - paste(.libPaths(), pkgname, data, sample.txt, sep = /) file(filepath) Is there a better way of doing this? Thanks, Andrew __ 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. -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O [[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.
Re: [R] can't open file
You've tried this? out - read.zoo(log2.log, encoding=UCS-2LE, FUN = as.chron) On Wed, Jul 7, 2010 at 10:16 AM, Sebastian Kruk residuo.so...@gmail.comwrote: I have a text file log2.log encoded Ansi in Windows. When I execute: out - read.zoo(readLines(con - file(log2.log, encoding=UCS-2LE)),FUN = as.chron) have errors: Error en file(file, rt) : no se puede abrir la conexión Además: Mensajes de aviso perdidos 1: In file(file, rt) : sólo fue usado el primer elemento del argumento 'description' 2: In file(file, rt) : no fue posible abrir el archivo '#Software: Microsoft Internet Information Services 5.0': No such file or directory Why? Thks, Sebastián. __ 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. -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O [[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.
Re: [R] can't open file
On Wed, Jul 7, 2010 at 9:16 AM, Sebastian Kruk residuo.so...@gmail.com wrote: I have a text file log2.log encoded Ansi in Windows. When I execute: out - read.zoo(readLines(con - file(log2.log, encoding=UCS-2LE)),FUN = as.chron) have errors: Error en file(file, rt) : no se puede abrir la conexión Además: Mensajes de aviso perdidos 1: In file(file, rt) : sólo fue usado el primer elemento del argumento 'description' 2: In file(file, rt) : no fue posible abrir el archivo '#Software: Microsoft Internet Information Services 5.0': No such file or directory Why? The file argument of read.zoo is a character string giving the *name* of the file, not the contents of the file as a vector of character strings. Alternately it can be a connection (undocumented but works) or a data.frame so you likely want one of these: read.zoo(file(log2.log, encoding=UCS-2LE), FUN = as.chron) read.zoo(log2.log, FUN = as.chron) See the examples section of ?read.zoo for more examples. __ 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] can't open file
If I edit(out): structure(logical(0), .Dim = c(707L, 0L), .Dimnames = list(NULL, NULL), index = structure(1:707, format = m/d/y, origin = structure(c(1, 1, 1970), .Names = c(month, day, year)), class = c(dates, times)), class = zoo) I tried: z - read.zoo(textConnection(L), index = 1:2,FUN = function(x) paste(x[,1], x[,2])) I have de following error: Error en x[, 1] : número incorreto de dimensiones 2010/7/7 Henrique Dallazuanna www...@gmail.com: You've tried this? out - read.zoo(log2.log, encoding=UCS-2LE, FUN = as.chron) On Wed, Jul 7, 2010 at 10:16 AM, Sebastian Kruk residuo.so...@gmail.com wrote: I have a text file log2.log encoded Ansi in Windows. When I execute: out - read.zoo(readLines(con - file(log2.log, encoding=UCS-2LE)),FUN = as.chron) have errors: Error en file(file, rt) : no se puede abrir la conexión Además: Mensajes de aviso perdidos 1: In file(file, rt) : sólo fue usado el primer elemento del argumento 'description' 2: In file(file, rt) : no fue posible abrir el archivo '#Software: Microsoft Internet Information Services 5.0': No such file or directory Why? Thks, Sebastián. __ 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. -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O __ 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] Trimming in R
On Jul 7, 2010, at 9:23 AM, Andrew Leeser wrote:
I am looking for a way to trim leading and trailing spaces in a
character
string in R. For example:
this is random text
should become:
this is random text.
I have a short function to perform this task as follows:
trim - function(str){
str - sub(^ +, , str)
str - sub( +$, , str)
}
While this function does seem to work, I am curious if there's
anything built
into R that I can use instead, as that would be preferable.
I have been using the trim function in package gdata. The code is
quite similar to yours.
--
David Winsemius, MD
West Hartford, CT
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Re: [R] can't open file
I tried: L - readLines(con - file(log2.log, encoding=UCS-2LE) z - read.zoo(textConnection(L), index = 1:2,FUN = function(x) paste(x[,1], x[,2])) Error: Error en x[, 1] : número incorreto de dimensiones __ 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] can't open file
On Wed, Jul 7, 2010 at 9:43 AM, Sebastian Kruk residuo.so...@gmail.com wrote: I tried: L - readLines(con - file(log2.log, encoding=UCS-2LE) z - read.zoo(textConnection(L), index = 1:2,FUN = function(x) paste(x[,1], x[,2])) Error: Error en x[, 1] : número incorreto de dimensiones Please provide a reproducible example like this: library(zoo) L - 4/4/200 10:10:10 3000 5/5/2000 12:12:12 4000 read.zoo(textConnection(L), index = 1:2, FUN = function(x) paste(x[,1], x[,2])) which gives me 4/4/200 10:10:10 5/5/2000 12:12:12 3000 4000 __ 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] can't open file
On 07/07/2010 9:16 AM, Sebastian Kruk wrote: I have a text file log2.log encoded Ansi in Windows. What Windows calls Ansi is called latin1 in R. You said the encoding was UCS-2LE, which Windows calls Unicode. Part of your problem might be this mismatched encoding. Have you tried using encoding=latin1 when you read the file? Duncan Murdoch When I execute: out - read.zoo(readLines(con - file(log2.log, encoding=UCS-2LE)),FUN = as.chron) have errors: Error en file(file, rt) : no se puede abrir la conexión Además: Mensajes de aviso perdidos 1: In file(file, rt) : sólo fue usado el primer elemento del argumento 'description' 2: In file(file, rt) : no fue posible abrir el archivo '#Software: Microsoft Internet Information Services 5.0': No such file or directory Why? Thks, Sebastián. __ 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-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] error in step.gam
Dear r-helpers, I use function step.gam (package gam, T. Hastie) with several explanatory variables to build a model. Unfortunately, I obviously have too many variables. This message occurs on my 4 core 64bit machine with 8GB RAM in R2.11.1 for Windows (64bit build): Error in array(FALSE, term.lengths) : 'dim' specifies too large an array I read that this message occurs when running out of RAM. So, the question is: Could it maybe help to run step.gam as a parallel process on more then 1 core and is parallel processing possible with this function? I tried to manage it with snowfall, but with no success. Any help would be greatly appreciated! Thanks, Kim This is my step.gam code: fit0 = gam( Vegetation ~ 1, family = binomial, data = d ) gam.scope = list( ~ 1 + ALTITUDE + s(ALTITUDE, 2), ~ 1 + SLOPE + s(SLOPE, 2), ~ 1 + SOUTH_EXPOSEDNESS + s(SOUTH_EXPOSEDNESS, 2), ~ 1 + WEST_EXPOSEDNESS + s(WEST_EXPOSEDNESS, 2), ~ 1 + VCURV + s(VCURV, 2), ~ 1 + HCURV + s(HCURV, 2), ~ 1 + PRECIPITATION_ANNUAL_MEAN + s(PRECIPITATION_ANNUAL_MEAN, 2), ~ 1 + PRECIPITATION_JULY + s(PRECIPITATION_JULY, 2), ~ 1 + PRECIPITATION_JANUARY + s(PRECIPITATION_JANUARY, 2), ~ 1 + DISTANCE_TO_ISOBATH + s(DISTANCE_TO_ISOBATH, 2), ~ 1 + VERTICAL_DISTANCE_TO_ISOBATH + s(VERTICAL_DISTANCE_TO_ISOBATH, 2), ~ 1 + HORIZONTAL_DISTANCE_TO_ISOBATH + s(HORIZONTAL_DISTANCE_TO_ISOBATH, 2), ~ 1 + UTM_EASTING + s(UTM_EASTING, 2), ~ 1 + UTM_NORTHING + s(UTM_NORTHING, 2), ~ 1 + DISTANCE_TO_SETTLEMENT + s(DISTANCE_TO_SETTLEMENT, 2), ~ 1 + Bd_1_Mean + s(Bd_1_Mean, 2), ~ 1 + Bd_2_Mean + s(Bd_2_Mean, 2), ~ 1 + Bd_3_Mean + s(Bd_3_Mean, 2), ~ 1 + Bd_4_Mean + s(Bd_4_Mean, 2), ~ 1 + Bd_5_Mean + s(Bd_5_Mean, 2), ~ 1 + Bd_1t_Mean + s(Bd_1t_Mean, 2), ~ 1 + Bd_2t_Mean + s(Bd_2t_Mean, 2), ~ 1 + Bd_3t_Mean + s(Bd_3t_Mean, 2), ~ 1 + Bd_4t_Mean + s(Bd_4t_Mean, 2), ~ 1 + Bd_5t_Mean + s(Bd_5t_Mean, 2), ~ 1 + NDVI_IR_R_Mean + s(NDVI_IR_R_Mean, 2), ~ 1 + NDVI_IR_Rededge_Mean + s(NDVI_IR_Rededge_Mean, 2), ~ 1 + NDVI_Rededge_R_Mean + s(NDVI_Rededge_R_Mean, 2), ~ 1 + ReTCI_Mean + s(ReTCI_Mean, 2), ~ 1 + NDVI_IR_R_t_Mean + s(NDVI_IR_R_t_Mean, 2), ~ 1 + NDVI_IR_Rededge_t_Mean + s(NDVI_IR_Rededge_t_Mean, 2), ~ 1 + NDVI_Rededge_R_t_Mean + s(NDVI_Rededge_R_t_Mean, 2), ~ 1 + ReTCI_t_Mean + s(ReTCI_t_Mean, 2)) gam.scope fit = step.gam(fit0, scope = gam.scope, direction = both, trace = TRUE) The error occurs with the function step.gam. -- GMX DSL: Internet-, Telefon- und Handy-Flat ab 19,99 EUR/mtl. Bis zu 150 EUR Startguthaben inklusive! http://portal.gmx.net/de/go/dsl __ 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] Visualization of coefficients
Tal Galili wrote: Hello David, Thanks to your posting I started looking at the function in the arm package. It appears this function is quite mature, and offers (for example) the ability to easily overlap coefficients from several models. I updated the post I published on the subject, so at the end of it I give an example of comparing the coef of several models: http://www.r-statistics.com/2010/07/visualization-of-regression-coefficients-in-r/ Thanks again for the pointer. Best, Tal Achim Zeileis wrote: Re: more mature. arm's coefplot() is more flexible in certain respects, mine is more convenient in others. The overlay functionality is something arm's coefplot() is better in and it also as some further options (vertical vs. horizontal etc.). My coefplot() has the advantage that it does not need any modification as long as coef() and vcov() methods are available. Furthermore, level can specify the significance level (instead of always using one and two standard errors, respectively). But it shouldn't be too hard to create a superset of all options. @Tal: For the example using library(arm) and the Mroz data, you posted the wrong image. And loose the intercept in the example. @Achim: It would be worthwhile combining the generality of your version with the overlay capability of the arm version, which is extremely useful for model comparison. However, the arm version uses S4 methods. -Michael -- Michael Friendly Email: friendly AT yorku DOT ca Professor, Psychology Dept. York University Voice: 416 736-5115 x66249 Fax: 416 736-5814 4700 Keele StreetWeb: http://www.datavis.ca Toronto, ONT M3J 1P3 CANADA __ 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] relation in aggregated data
Dear all My question is more on statistics than on R, however it can be demonstrated by R. It is about pros and cons trying to find a relationship by aggregated data. I can have two variables which can be related and I measure them regularly during some time (let say a year) but I can not measure them in a same time - (e.g. I can not measure x and respective value of y, usually I have 3 or more values of x and only one value of y per day). I can make a aggregated values (let say quarterly). My questions are: 1. Is such approach sound? Can I use it? 2. What could be the problems 3. Is there any other method to inspect variables which can be related but you can not directly measure them in a same time? My opinion is, that it is not much sound to inspect aggregated values and there can be many traps especially if there are only few aggregated values. Below you can see my examples. If you have some opinion on this issue, please let me know. Best regards Petr Let us have a relation x/y set.seed(555) x - rnorm(120) y - 5*x+3+rnorm(120) plot(x, y) As you can see there is clear relation which can be seen from plot. Now I make a factor for aggregation. fac - rep(1:4,each=30) xprum - tapply(x, fac, mean) yprum - tapply(y, fac, mean) plot(xprum, yprum) Relationship is completely gone. Now let us make other fake data xn - runif(120)*rep(1:4, each=30) yn - runif(120)*rep(1:4, each=30) plot(xn,yn) There is no visible relation, xn and yn are independent but related to aggregation factor. xprumn - tapply(xn, fac, mean) yprumn - tapply(yn, fac, mean) plot(xprumn, yprumn) Here you can see perfect relation which is only due to aggregation factor. __ 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] Visualization of coefficients
Thanks Michael - I now inserted the correct image for Achim example code. Contact Details:--- Contact me: tal.gal...@gmail.com | 972-52-7275845 Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) | www.r-statistics.com (English) -- On Wed, Jul 7, 2010 at 5:19 PM, Michael Friendly frien...@yorku.ca wrote: Tal Galili wrote: Hello David, Thanks to your posting I started looking at the function in the arm package. It appears this function is quite mature, and offers (for example) the ability to easily overlap coefficients from several models. I updated the post I published on the subject, so at the end of it I give an example of comparing the coef of several models: http://www.r-statistics.com/2010/07/visualization-of-regression-coefficients-in-r/ Thanks again for the pointer. Best, Tal Achim Zeileis wrote: Re: more mature. arm's coefplot() is more flexible in certain respects, mine is more convenient in others. The overlay functionality is something arm's coefplot() is better in and it also as some further options (vertical vs. horizontal etc.). My coefplot() has the advantage that it does not need any modification as long as coef() and vcov() methods are available. Furthermore, level can specify the significance level (instead of always using one and two standard errors, respectively). But it shouldn't be too hard to create a superset of all options. @Tal: For the example using library(arm) and the Mroz data, you posted the wrong image. And loose the intercept in the example. @Achim: It would be worthwhile combining the generality of your version with the overlay capability of the arm version, which is extremely useful for model comparison. However, the arm version uses S4 methods. -Michael -- Michael Friendly Email: friendly AT yorku DOT ca Professor, Psychology Dept. York University Voice: 416 736-5115 x66249 Fax: 416 736-5814 4700 Keele StreetWeb: http://www.datavis.ca Toronto, ONT M3J 1P3 CANADA [[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] help with Date class
I am trying to work with the Date class which is written in S3 and I would like to access to the elements of the class (for example the year). I've tryed to do it for example like this: as.Date(Sys.time)-w w$year #Doesn't work w[year] #is NA I would like to know the correct way to acces to this value. Thank you so much -- Elisenda Vila [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Different goodness of fit tests leads to contradictory conclusions
I am trying to test goodness of fit for my legalistic regression using several
options as shown below. Â Hosmer-Lemeshow test (whose function I borrowed from
a previous post), Hosmerâle Cessie omnibus lack of fit test (also borrowed
from a previous post), Pearson chi-square test, and deviance test. Â All the
tests, except the deviance tests, produced p-values well above 0.05. Â Would
anyone please teach me why deviance test produce p-values such a small value
(0.001687886) that suggest lack of fit, while other tests suggest good fit? Did
I do something wrong?
Â
Thank you for your time and help!
Â
Kiyoshi
Â
Â
mod.fit - glm(formula = no.NA$repcnd ~Â no.NA$svl, family = binomial(link =
logit), data =Â no.NA, na.action = na.exclude, control = list(epsilon =
0.0001, maxit = 50, trace = F))
Â
# Option 1: Hosmer-Lemeshow test
mod.fit - glm(formula = no.NA$repcnd ~Â no.NA$svl, family = binomial(link =
logit), data =Â no.NA, na.action = na.exclude, control = list(epsilon =
0.0001, maxit = 50, trace = F))
           Â
  hosmerlem - function (y, yhat, g = 10)
{
cutyhat - cut(yhat, breaks = quantile(yhat, probs = seq(0, 1, 1/g)),
include.lowest = T)
obs - xtabs(cbind(1 - y, y) ~ cutyhat)
expect - xtabs(cbind(1 - yhat, yhat) ~ cutyhat)
chisq - sum((obs - expect)^2/expect)
P - 1 - pchisq(chisq, g - 2)
c(X^2 = chisq, Df = g - 2, P(Chi) = P)
}
Â
hosmerlem(no.NA$repcnd, fitted(mod.fit))
 X^2                     Df Â
                       P(Chi)
7.8320107 Â Â Â Â Â Â Â Â Â Â 8.000Â Â Â Â Â Â Â Â Â Â Â 0.4500497
Â
Â
# Option 2 - Hosmerâle Cessie omnibus lack of fit test:
library(Design)
lrm.GOF - lrm(formula = no.NA$repcnd ~Â no.NA$svl, data =Â no.NA, y = T, x
= T)
resid(lrm.GOF,type = gof)
Sum of squared errors    Expected value|H0      Â
SDÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
ZÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â P
               48.3487115         Â
              48.3017399          Â
             0.1060826    0.4427829    0.6579228
Â
# Option 3 - Pearson chi-square p-value:
pp - sum(resid(lrm.GOF,typ=pearson)^2)
1-pchisq(pp, mod.fit$df.resid)
[1] 0.4623282
Â
Â
# Option 4 - Deviance (G^2) test:
1-pchisq(mod.fit$deviance, mod.fit$df.resid)
[1] 0.001687886
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Re: [R] can't open file
In both cases I obtain: 01/02/70 01/03/70 01/04/70 01/05/70 01/06/70 01/07/70 01/08/70 01/09/70 01/10/70 01/11/70 01/12/70 01/13/70 01/14/70 01/15/70 01/16/70 01/17/70 01/18/70 01/19/70 01/20/70 01/21/70 01/22/70 01/23/70 01/24/70 01/25/70 01/26/70 01/27/70 01/28/70 01/29/70 01/30/70 01/31/70 02/01/70 02/02/70 02/03/70 02/04/70 02/05/70 02/06/70 02/07/70 02/08/70 02/09/70 02/10/70 02/11/70 02/12/70 02/13/70 02/14/70 02/15/70 02/16/70 02/17/70 02/18/70 02/19/70 02/20/70 02/21/70 02/22/70 02/23/70 02/24/70 02/25/70 02/26/70 02/27/70 02/28/70 03/01/70 03/02/70 03/03/70 03/04/70 03/05/70 03/06/70 03/07/70 03/08/70 03/09/70 03/10/70 03/11/70 03/12/70 03/13/70 03/14/70 03/15/70 03/16/70 03/17/70 03/18/70 03/19/70 03/20/70 03/21/70 03/22/70 03/23/70 03/24/70 03/25/70 03/26/70 03/27/70 03/28/70 03/29/70 03/30/70 03/31/70 04/01/70 04/02/70 04/03/70 04/04/70 04/05/70 04/06/70 04/07/70 04/08/70 04/09/70 04/10/70 04/11/70 04/12/70 04/13/70 04/14/70 04/15/70 04/16/70 04/17/70 04/18/70 04/19/70 04/20/70 04/21/70 04/22/70 04/23/70 04/24/70 04/25/70 04/26/70 04/27/70 04/28/70 04/29/70 04/30/70 05/01/70 05/02/70 05/03/70 05/04/70 05/05/70 05/06/70 05/07/70 05/08/70 05/09/70 05/10/70 05/11/70 05/12/70 05/13/70 05/14/70 05/15/70 05/16/70 05/17/70 05/18/70 05/19/70 05/20/70 05/21/70 05/22/70 05/23/70 05/24/70 05/25/70 05/26/70 05/27/70 05/28/70 05/29/70 05/30/70 05/31/70 06/01/70 06/02/70 06/03/70 06/04/70 06/05/70 06/06/70 06/07/70 06/08/70 2010/7/7 Gabor Grothendieck ggrothendi...@gmail.com: On Wed, Jul 7, 2010 at 9:16 AM, Sebastian Kruk residuo.so...@gmail.com wrote: I have a text file log2.log encoded Ansi in Windows. When I execute: out - read.zoo(readLines(con - file(log2.log, encoding=UCS-2LE)),FUN = as.chron) have errors: Error en file(file, rt) : no se puede abrir la conexión Además: Mensajes de aviso perdidos 1: In file(file, rt) : sólo fue usado el primer elemento del argumento 'description' 2: In file(file, rt) : no fue posible abrir el archivo '#Software: Microsoft Internet Information Services 5.0': No such file or directory Why? The file argument of read.zoo is a character string giving the *name* of the file, not the contents of the file as a vector of character strings. Alternately it can be a connection (undocumented but works) or a data.frame so you likely want one of these: read.zoo(file(log2.log, encoding=UCS-2LE), FUN = as.chron) read.zoo(log2.log, FUN = as.chron) See the examples section of ?read.zoo for more examples. __ 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] help with Date class
On Jul 7, 2010, at 7:25 AM, Elisenda Vila wrote: I am trying to work with the Date class which is written in S3 and I would like to access to the elements of the class (for example the year). I've tryed to do it for example like this: as.Date(Sys.time)-w Throws an error ... since Sys.time is a function w$year #Doesn't work Why should it? Dates are not stored as lists. ?Date w[year] #is NA I would like to know the correct way to acces to this value. as.Date(Sys.time)-w Error in as.Date.default(Sys.time) : do not know how to convert 'Sys.time' to class Date as.Date(Sys.time())-w w [1] 2010-07-07 format(w, %Y) [1] 2010 -- David Winsemius, MD West Hartford, CT __ 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] xyplot: key inside the plot region / lme: confidence bands for predicted
No one replied to my second question: how to get standard errors or confidence intervals for the estimated fixed effects from lme().AFAICS, intervals() only gives CIs for coefficients. My working example is: library(nlme) library(lattice) Ortho - Orthodont Ortho$year - Ortho$age - 8 # make intercept = initial status Ortho.mix1 - lme(distance ~ year * Sex, data=Ortho, random = ~ 1 + year | Subject, method=ML) anova(Ortho.mix1) # get predicted values grid - expand.grid(year=0:6, Sex=c(Male, Female)) grid$age - grid$year+8 # plot vs. age fm.mix1 -cbind(grid, distance = predict(Ortho.mix1, newdata = grid, level=0)) xyplot(distance ~ age, data=fm.mix1, groups=Sex, type=b, par.settings = list(superpose.symbol = list(cex = 1.2, pch=c(15,16))), auto.key=list(text=levels(fm.mix1$Sex), points = TRUE, x=0.05, y=0.9, corner=c(0,1)), main=Linear mixed model: predicted growth) intervals(Ortho.mix1) Approximate 95% confidence intervals Fixed effects: lower est. upper (Intercept)21.607212 22.615625 23.6240383 year0.619660 0.784375 0.9490904 SexFemale -3.041252 -1.406534 0.2281834 year:SexFemale -0.562889 -0.304830 -0.0467701 attr(,label) [1] Fixed effects: Random Effects: Level: Subject lower est.upper sd((Intercept))1.0710615 1.7045122 2.712600 sd(year) 0.0281228 0.1541351 0.844783 cor((Intercept),year) -0.9358505 -0.0311195 0.927652 Within-group standard error: lowerest. upper 1.07081 1.31004 1.60272 Peter Ehlers wrote: On 2010-07-02 9:37, Michael Friendly wrote: I have two questions related to plotting predicted values for a linear mixed model using xyplot: 1: With a groups= argument, I can't seem to get the key to appear inside the xyplot. (I have the Lattice book, but don't find an example that actually does this.) 2: With lme(), how can I generate confidence bands or prediction intervals around the fitted values? Once I get them, I'd like to add them to the plot. Example: library(nlme) library(lattice) Ortho - Orthodont Ortho$year - Ortho$year - 8 # make intercept = initial status Ortho.mix1 - lme(distance ~ year * Sex, data=Ortho, random = ~ 1 + year | Subject, method=ML, # correlation = corAR1 (form = ~ 1 | Subject) ) Ortho.mix1 # predicted values grid - expand.grid(year=0:6, Sex=c(Male, Female)) grid$age - grid$year+8 # plot vs. age fm.mix1 -cbind(grid, distance = predict(Ortho.mix1, newdata = grid, level=0)) xyplot(distance ~ age, data=fm.mix1, groups=Sex, type=b, pch=c(15,16), cex=1.2, auto.key=list(text=c(Male, Female), points = TRUE, x=8, y=26), main=Linear mixed model: predicted growth) Can someone help? -Michael Michael, the x,y location should be specified in the unit square (and you might want to set the 'corner' component). Ortho$year - Ortho$age - 8 ##fix typo xyplot(distance ~ age, data=fm.mix1, groups=Sex, type=b, pch=c(15,16), cex=1.2, auto.key=list(text=c(Male, Female), points = TRUE, x=0.1, y=0.8, corner=c(0,1))) See description of 'key' in ?xyplot. -Peter Ehlers -- Michael Friendly Email: friendly AT yorku DOT ca Professor, Psychology Dept. York University Voice: 416 736-5115 x66249 Fax: 416 736-5814 4700 Keele StreetWeb: http://www.datavis.ca Toronto, ONT M3J 1P3 CANADA __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] help with Date class
On Wed, Jul 07, 2010 at 01:25:43PM +0200, Elisenda Vila wrote: I am trying to work with the Date class which is written in S3 and I would like to access to the elements of the class (for example the year). I've tryed to do it for example like this: as.Date(Sys.time)-w w - Sys.Date() # does the same, but in one step w$year #Doesn't work w[year] #is NA I would like to know the correct way to acces to this value. wp - as.POSIXlt(w)# POSIXlt has the components unclass(wp)# shows you all components wp$year + 1900 # stored as year - 1900, see Unix manuals wp$mon + 1 # stores as mon -1, see Unix manuals Arguably, extractor functions would be of help here. -- Regards, Dirk __ 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] Different goodness of fit tests leads to contradictory conclusions
The first two options are GOF-tests for ungrouped data, the latter two
can only be used for grouped data. According to my experience, the G^2
test is more influenced by this than the X^2 test (gives -wrongly-
significant deviations more easily when used for ungrouped data).
If you started from binary data, you can only use the Hosmer tests.
Cheers
Joris
On Wed, Jul 7, 2010 at 4:00 PM, Kiyoshi Sasaki skiyoshi2...@yahoo.com wrote:
I am trying to test goodness of fit for my legalistic regression using
several options as shown below. Hosmer-Lemeshow test (whose function I
borrowed from a previous post), Hosmer–le Cessie omnibus lack of fit test
(also borrowed from a previous post), Pearson chi-square test, and deviance
test. All the tests, except the deviance tests, produced p-values well above
0.05. Would anyone please teach me why deviance test produce p-values such a
small value (0.001687886) that suggest lack of fit, while other tests suggest
good fit? Did I do something wrong?
Thank you for your time and help!
Kiyoshi
mod.fit - glm(formula = no.NA$repcnd ~ no.NA$svl, family = binomial(link =
logit), data = no.NA, na.action = na.exclude, control = list(epsilon =
0.0001, maxit = 50, trace = F))
# Option 1: Hosmer-Lemeshow test
mod.fit - glm(formula = no.NA$repcnd ~ no.NA$svl, family = binomial(link =
logit), data = no.NA, na.action = na.exclude, control = list(epsilon =
0.0001, maxit = 50, trace = F))
hosmerlem - function (y, yhat, g = 10)
{
cutyhat - cut(yhat, breaks = quantile(yhat, probs = seq(0, 1, 1/g)),
include.lowest = T)
obs - xtabs(cbind(1 - y, y) ~ cutyhat)
expect - xtabs(cbind(1 - yhat, yhat) ~ cutyhat)
chisq - sum((obs - expect)^2/expect)
P - 1 - pchisq(chisq, g - 2)
c(X^2 = chisq, Df = g - 2, P(Chi) = P)
}
hosmerlem(no.NA$repcnd, fitted(mod.fit))
X^2 Df P(Chi)
7.8320107 8.000 0.4500497
# Option 2 - Hosmer–le Cessie omnibus lack of fit test:
library(Design)
lrm.GOF - lrm(formula = no.NA$repcnd ~ no.NA$svl, data = no.NA, y = T, x
= T)
resid(lrm.GOF,type = gof)
Sum of squared errors Expected value|H0 SD
Z P
48.3487115 48.3017399
0.1060826 0.4427829 0.6579228
# Option 3 - Pearson chi-square p-value:
pp - sum(resid(lrm.GOF,typ=pearson)^2)
1-pchisq(pp, mod.fit$df.resid)
[1] 0.4623282
# Option 4 - Deviance (G^2) test:
1-pchisq(mod.fit$deviance, mod.fit$df.resid)
[1] 0.001687886
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--
Joris Meys
Statistical consultant
Ghent University
Faculty of Bioscience Engineering
Department of Applied mathematics, biometrics and process control
tel : +32 9 264 59 87
joris.m...@ugent.be
---
Disclaimer : http://helpdesk.ugent.be/e-maildisclaimer.php
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Re: [R] Gray level mosaic plot with shading_Friendly
Dear Achim and Michael,
Thank you so much. Indeed, mosaic(Titanic, gp = shading_hcl, gp_args = list(lty
= 1:2, c = 0)) does almost what I was looking for, except that for consistency
and clarity, I would have expected the negative values on the legend to be be
outlined with lty = 2.
Michael
On Jul 7, 2010, at 2:13 AM, Achim Zeileis wrote:
On Tue, 6 Jul 2010, Michael Friendly wrote:
Michael Kubovy wrote:
Suppose we start with
data(Titanic)
mosaic(Titanic, shade = TRUE)
How do I combine the dashed box contours of shading_Friendly to indicate
negative residuals, with three levels of gray: dark for abs(Pearson Resid)
4, lighter for 4 abs(Pearson Resid) 2, and lightest for bs(Pearson
Resid) 2 ?
Do you mean [1] you want to plot positive residuals in color and negative in
gray scale?
Or [2] to fold + and - residuals by shading all according to abs(resid), and
distinguishing + from - by the dashed box outlines?
In fact, I designed this coding scheme so that mosaic plots in color (with
my blue - white - red scheme) would approximately do exactly what
you might want under [2], when rendered in B/W, since the fully saturated
red and blue are close in darkness in B/W.
And shading_hcl() has been written to do exactly what you want under [2].
While it is hard to come up with colors of different hues in HSV or HLS space
that have the same brightness (aka lightness/luminance) and the same
colorfulness (aka chroma), this is easy in HCL.
Try
mosaic(Titanic, gp=shading_Friendly)
save as a jpg/png and try converting to B/W with an image program and see if
this is good enough.
mosaic(Titanic, shade = TRUE)
is the same as
mosaic(Titanic, gp = shading_hcl)
which you can then modify to have different line types
mosaic(Titanic, gp = shading_hcl, gp_args = list(lty = 1:2))
If you print that on a grayscale printer you will see the same plot without
any chroma, i.e.,
mosaic(Titanic, gp = shading_hcl, gp_args = list(lty = 1:2, c = 0))
The shading_hcl() function is introduced in Zeileis et al. (2007, JCGS), see
?shading_hcl, which provides more detailed references to HCL colors etc.
Best,
Z
Alternatively, write your own, shading_Kubovy, modeled on
shading_Friendly -
function (observed = NULL, residuals = NULL, expected = NULL,
df = NULL, h = c(2/3, 0), lty = 1:2, interpolate = c(2, 4),
eps = 0.01, line_col = black, ...)
{
shading_hsv(observed = NULL, residuals = NULL, expected = NULL,
df = NULL, h = h, v = 1, lty = lty, interpolate = interpolate,
eps = eps, line_col = line_col, p.value = NA, ...)
}
environment: namespace:vcd
attr(,class)
[1] grapcon_generator
In the defaults, lty=1:2 is what distinguishes + and - for outline line type
hope this helps,
-Michael
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[R] interaction post hoc/ lme repeated measures
Hi Everyone, Im trying to figure out how to get R to analyze this experiment properly. I have a series of subjects each with two legs. Within each leg there are two bones that I am interested in. There are also two treatments that I am interested in. That results in four different combinations of treatments. Obviously, since the subjects only have two legs, they cant receive each combination of treatment. The groups are unbalanced so I cant use aov. As I understand it, lme should work but I am having a tough time figuring out what to use as a random term. Is this correct? x.lme - lme(effect~bone*treatment1*treatment2, random= ~1|subject, data=x) anova(x.lme) Ive tried a number of different random terms and get very similar results (and even get similar results running aov) but Id like to know what the proper way of doing it is. My next question is how to do post hoc tests on this. One website recommends something like this from multcomp: summary(glht(am2,linfct=mcp(myfactor=Tukey))) But apparently this doesnt work with interactions. With my data, I get a significant three-way interaction. If I run TukeyHSD after an aov (ignoring the repeated measures) I would get a series of p values for all the different combinations of treatments and bones. Is there a way to do this with lme? Thank you in advance for any help. Doug Bourne [[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.
Re: [R] how to display the clock time in the loop
Thanks a bunch, it works.
On Thu, Jul 1, 2010 at 4:50 PM, Nikhil Kaza nikhil.l...@gmail.com wrote:
explicit call to print usually works for me.
library(audio)
for (i in 1:5){
wait(60)
print(Sys.time())
}
On Jul 1, 2010, at 4:30 PM, Matt Shotwell wrote:
Try to flush output after printing:
cat(paste(Sys.time()),\n); flush(stdout())
On Thu, 2010-07-01 at 16:17 -0400, Jack Luo wrote:
Hi,
I am doing some computation which is pretty time consuming, I want R to
display CPU time after each iteration using the command Sys.time().
However,
I found that the code only began to display the CPU time after quite a
while
and several iterations have finished. Is there a way to ask R to display
time right after each iteration is finished?
Thanks,
-Jun
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[R] F# vs. R
Hello, everyone F# is now public. Compiled code should run faster than R. Anyone has opinion on F# vs. R? Just curious Best, S -- --- Kniven skärpes bara mot stenen. __ 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] relation in aggregated data
You examples are pretty extreme... Combining 120 data points in 4 points is off course never going to give a result. Try : fac - rep(1:8,each=15) xprum - tapply(x, fac, mean) yprum - tapply(y, fac, mean) plot(xprum, yprum) Relation is not obvious, but visible. Yes, you lose information. Yes, your hypothesis changes. But in the case you describe, averaging the x-values for every day (so you get an average linked to 1 y value) seems like a possibility, given you take that into account when formulating the hypothesis. Optimally, you should take the standard error on the average into account for the analysis, but this is complicated, often not done and in most cases ignoring this issue is not influencing the results to that extent it becomes important. YMMV Cheers On Wed, Jul 7, 2010 at 4:24 PM, Petr PIKAL petr.pi...@precheza.cz wrote: Dear all My question is more on statistics than on R, however it can be demonstrated by R. It is about pros and cons trying to find a relationship by aggregated data. I can have two variables which can be related and I measure them regularly during some time (let say a year) but I can not measure them in a same time - (e.g. I can not measure x and respective value of y, usually I have 3 or more values of x and only one value of y per day). I can make a aggregated values (let say quarterly). My questions are: 1. Is such approach sound? Can I use it? 2. What could be the problems 3. Is there any other method to inspect variables which can be related but you can not directly measure them in a same time? My opinion is, that it is not much sound to inspect aggregated values and there can be many traps especially if there are only few aggregated values. Below you can see my examples. If you have some opinion on this issue, please let me know. Best regards Petr Let us have a relation x/y set.seed(555) x - rnorm(120) y - 5*x+3+rnorm(120) plot(x, y) As you can see there is clear relation which can be seen from plot. Now I make a factor for aggregation. fac - rep(1:4,each=30) xprum - tapply(x, fac, mean) yprum - tapply(y, fac, mean) plot(xprum, yprum) Relationship is completely gone. Now let us make other fake data xn - runif(120)*rep(1:4, each=30) yn - runif(120)*rep(1:4, each=30) plot(xn,yn) There is no visible relation, xn and yn are independent but related to aggregation factor. xprumn - tapply(xn, fac, mean) yprumn - tapply(yn, fac, mean) plot(xprumn, yprumn) Here you can see perfect relation which is only due to aggregation factor. __ 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. -- Joris Meys Statistical consultant Ghent University Faculty of Bioscience Engineering Department of Applied mathematics, biometrics and process control tel : +32 9 264 59 87 joris.m...@ugent.be --- Disclaimer : http://helpdesk.ugent.be/e-maildisclaimer.php __ 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] Sum vectors and numbers
We want to sum many vectors and numbers together as a vector if there is at least one vector in the arguments. For example, 1 + c(2,3) = c(3,4). Since we are not sure arguments to sum, we are using sum function: sum(v1,v2,...,n1,n2,..). The problem is that sum returns the sum of all the values present in its arguments: sum(1,c(2,3))=6 sum(1,2,3)=6 We do not want to turn sum(v1,v2,...,n1,n2,..) to v1+v2+...+n1+n2+... So do you know easy way to sum vectors (v1,v2,...) and numbers (n1,n2,...) as a vector without using v1+v2+...+n1+n2+...? Thanks, -james __ 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] Sum vectors and numbers
g...@ucalgary.ca wrote: We want to sum many vectors and numbers together as a vector if there is at least one vector in the arguments. For example, 1 + c(2,3) = c(3,4). Since we are not sure arguments to sum, we are using sum function: sum(v1,v2,...,n1,n2,..). The problem is that sum returns the sum of all the values present in its arguments: sum(1,c(2,3))=6 sum(1,2,3)=6 We do not want to turn sum(v1,v2,...,n1,n2,..) to v1+v2+...+n1+n2+... So do you know easy way to sum vectors (v1,v2,...) and numbers (n1,n2,...) as a vector without using v1+v2+...+n1+n2+...? Thanks, I must admit I've read this a couple times and am confused. Can you give one example that shows exactly what you want, using R objects? I guess my question is, what is wrong with 1 + c(2, 3) Is this what you're looking for? lapply(c(1,2,3), `+`, c(8,9,10)) __ 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] Sum vectors and numbers
On Wed, Jul 7, 2010 at 11:35 AM, g...@ucalgary.ca wrote: We want to sum many vectors and numbers together as a vector if there is at least one vector in the arguments. For example, 1 + c(2,3) = c(3,4). Since we are not sure arguments to sum, we are using sum function: sum(v1,v2,...,n1,n2,..). The problem is that sum returns the sum of all the values present in its arguments: sum(1,c(2,3))=6 sum(1,2,3)=6 We do not want to turn sum(v1,v2,...,n1,n2,..) to v1+v2+...+n1+n2+... So do you know easy way to sum vectors (v1,v2,...) and numbers (n1,n2,...) as a vector without using v1+v2+...+n1+n2+...? Thanks, Try this: L - list(1:3, 3, 4:6, 6) Reduce(+, L) [1] 14 16 18 __ 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] F# vs. R
On Jul 7, 2010, at 10:31 AM, Sergey Goriatchev wrote: Hello, everyone F# is now public. Compiled code should run faster than R. Anyone has opinion on F# vs. R? Just curious Best, S The key time critical parts of R are written in compiled C and FORTRAN. Of course, if you want to take the time to code and validate a Cox PH or mixed effects model in F# and then run them against R's coxph() or lme()/lmer() functions to test the timing, feel free... :-) So unless there is a pre-existing library of statistical and related functionality for F#, perhaps you need to reconsider your query. Regards, Marc Schwartz __ 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] Appropriateness of survdiff {survival} for non-censored data
I read through Harrington and Fleming (1982) but it is beyond my statistical comprehension. I have survival data for insects that have a very finite expiration date. I'm trying to test for differences in survival distributions between different groups. I understand that the medical field is most often dealing with censored data and that survival analysis, at least in the package survival, is largely built around these conventions and differs from a classical biological perspective. For example, for lifetable analysis of insects there is often no need to estimate survival using a Kaplan-Meir estimate because it is relatively easy to follow a cohort of individuals through the entire course of life. Thus I question the appropriateness of using survdiff in my analysis; I have exact data yet I would be testing on the Kaplan-Meir estimate of these data in survdiff. Thanks for any help. __ 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] Trimming in R
David Winsemius dwinsemius at comcast.net writes:
On Jul 7, 2010, at 9:23 AM, Andrew Leeser wrote:
I am looking for a way to trim leading and trailing spaces in a
character
string in R. For example:
this is random text
should become:
this is random text.
I have a short function to perform this task as follows:
trim - function(str){
str - sub(^ +, , str)
str - sub( +$, , str)
}
While this function does seem to work, I am curious if there's
anything built
into R that I can use instead, as that would be preferable.
I have been using the trim function in package gdata. The code is
quite similar to yours.
You could also use the substr() function, assuming that you know a priori how
many spaces are before and after the text:
str =this is random text
nchar(str)
substr(str, 8, 26) # 7 spaces before and 8 after the text you want to keep.
HTH,
Ken
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Re: [R] Different goodness of fit tests leads to contradictory conclusions
Dear Joris,
Â
Thank you for your prompt reply! I have a binary dependent variable (whether a
snake is pregnant or not pregnant). Independent/predictor variable is the
snake's body size. Each observation (row) of the data represents each snake.
One column of the data contain '0' or '1' to indicate whether a snake is
pregnant. Another column contain body size for each snake. So, if I understand
correctly, I can use only Hosmer-Lemeshow test. Am I correct?
Â
Thank you very much!
Â
Kiyoshi
--- On Wed, 7/7/10, Joris Meys jorism...@gmail.com wrote:
From: Joris Meys jorism...@gmail.com
Subject: Re: [R] Different goodness of fit tests leads to contradictory
conclusions
To: Kiyoshi Sasaki skiyoshi2...@yahoo.com
Cc: r-help@r-project.org
Date: Wednesday, July 7, 2010, 10:08 AM
The first two options are GOF-tests for ungrouped data, the latter two
can only be used for grouped data. According to my experience, the G^2
test is more influenced by this than the X^2 test (gives -wrongly-
significant deviations more easily when used for ungrouped data).
If you started from binary data, you can only use the Hosmer tests.
Cheers
Joris
On Wed, Jul 7, 2010 at 4:00 PM, Kiyoshi Sasaki skiyoshi2...@yahoo.com wrote:
I am trying to test goodness of fit for my legalistic regression using
several options as shown below. Â Hosmer-Lemeshow test (whose function I
borrowed from a previous post), Hosmerâle Cessie omnibus lack of fit test
(also borrowed from a previous post), Pearson chi-square test, and deviance
test. Â All the tests, except the deviance tests, produced p-values well
above 0.05. Â Would anyone please teach me why deviance test produce p-values
such a small value (0.001687886) that suggest lack of fit, while other tests
suggest good fit? Did I do something wrong?
Thank you for your time and help!
Kiyoshi
mod.fit - glm(formula = no.NA$repcnd ~Â no.NA$svl, family = binomial(link =
logit), data =Â no.NA, na.action = na.exclude, control = list(epsilon =
0.0001, maxit = 50, trace = F))
# Option 1: Hosmer-Lemeshow test
mod.fit - glm(formula = no.NA$repcnd ~Â no.NA$svl, family = binomial(link
= logit), data =Â no.NA, na.action = na.exclude, control = list(epsilon =
0.0001, maxit = 50, trace = F))
  hosmerlem - function (y, yhat, g = 10)
{
cutyhat - cut(yhat, breaks = quantile(yhat, probs = seq(0, 1, 1/g)),
include.lowest = T)
obs - xtabs(cbind(1 - y, y) ~ cutyhat)
expect - xtabs(cbind(1 - yhat, yhat) ~ cutyhat)
chisq - sum((obs - expect)^2/expect)
P - 1 - pchisq(chisq, g - 2)
c(X^2 = chisq, Df = g - 2, P(Chi) = P)
}
hosmerlem(no.NA$repcnd, fitted(mod.fit))
 X^2                     Df Â
                       P(Chi)
7.8320107 Â Â Â Â Â Â Â Â Â Â 8.000Â Â Â Â Â Â Â Â Â Â Â 0.4500497
# Option 2 - Hosmerâle Cessie omnibus lack of fit test:
library(Design)
lrm.GOF - lrm(formula = no.NA$repcnd ~Â no.NA$svl, data =Â no.NA, y = T,
x = T)
resid(lrm.GOF,type = gof)
Sum of squared errors    Expected value|H0      Â
SDÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
ZÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â P
               48.3487115         Â
              48.3017399          Â
             0.1060826    0.4427829    0.6579228
# Option 3 - Pearson chi-square p-value:
pp - sum(resid(lrm.GOF,typ=pearson)^2)
1-pchisq(pp, mod.fit$df.resid)
[1] 0.4623282
# Option 4 - Deviance (G^2) test:
1-pchisq(mod.fit$deviance, mod.fit$df.resid)
[1] 0.001687886
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Statistical consultant
Ghent University
Faculty of Bioscience Engineering
Department of Applied mathematics, biometrics and process control
tel : +32 9 264 59 87
joris.m...@ugent.be
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Re: [R] grayscale wireframe??
With grDevices package, I do the following to generate a greyscale: newcols - colorRampPalette(c(white, black)) #generates palette from white to black #OR newcols - colorRampPalette(c(grey90, grey10)) #generates palette frome light to dark grey for better visibility Then in the wireframe() command type the argument: col.regions=newcols(100) -- View this message in context: http://r.789695.n4.nabble.com/grayscale-wireframe-tp2280281p2281127.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 define a function in R
Chapter 10 of 'An Introduction to R' has the suggestive title Writing your own functions. Isn't that the *first* place you would look? -Peter Ehlers On 2010-07-06 20:57, jd6688 wrote: Thanks Joshua for the example which has been great help me to start. Wish you the best Jason Ding On Tue, Jul 6, 2010 at 9:23 PM, Joshua Wiley-2 [via R] ml-node+2280373-448579502-312...@n4.nabble.comml-node%2b2280373-448579502-312...@n4.nabble.com wrote: Hello, As others have said, its hard to give specific advice without specific needs, but that's okay; I made up some examples needs and some (rather silly) code that might handle it. Depending what you need to do, it may help you get started. I tried to explicitly name all the arguments in any functions I used. When I make gmail use basic text format instead of html, code is sent poorly, so you can trundle off here to see the example, if you like. http://gist.github.com/466164 Cheers, Josh On Tue, Jul 6, 2010 at 3:48 PM, jd6688[hidden email]http://user/SendEmail.jtp?type=nodenode=2280373i=0 wrote: 1. how to write a R script? 2.How to write a SAS like macro/generic process to process multiple files by using the same funstion in R? Thanks in advance -- View this message in context: http://r.789695.n4.nabble.com/how-to-define-a-function-in-R-tp2280290p2280290.htmlhttp://r.789695.n4.nabble.com/how-to-define-a-function-in-R-tp2280290p2280290.html?by-user=t Sent from the R help mailing list archive at Nabble.com. __ [hidden email]http://user/SendEmail.jtp?type=nodenode=2280373i=1mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.htmlhttp://www.r-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Joshua Wiley Ph.D. Student, Health Psychology University of California, Los Angeles http://www.joshuawiley.com/ __ [hidden email]http://user/SendEmail.jtp?type=nodenode=2280373i=2mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.htmlhttp://www.r-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ 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] Adding two files into one and vlookup
I dont get any putput by merging. merge(file1,file2) [1] Date Price 0 rows (or 0-length row.names) file3-merge(file1,file2,by.file1='Date') file3 [1] Date Price 0 rows (or 0-length row.names) The above are the results. Nothing is coming in file3. On 7/7/10, Peter Alspach-2 [via R] ml-node+2280305-1773552212-309...@n4.nabble.com wrote: Tena koe Why is merge() not helpful? From your description I would imagine merge(file1, file2, by='Date') would do what you require. HTH Peter Alspach -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r- project.org] On Behalf Of raghu Sent: Wednesday, 7 July 2010 10:18 a.m. To: r-help@r-project.org Subject: [R] Adding two files into one and vlookup I have two files with dates and prices in each. The number of rows in each of them will differ. How do I create a new file which contains data from both these files? Cbind and merge are not helpful. For cbind because the rows are not the same replication occurs. Also if I have similar data how do I write a vlookup kind of function? I am giving an example below: Say Price1 file contains the following: Date Price 2/3/2010 134.00 3/3/2010 133.90 4/3/2010 135.55 And say price2 contains the following: Date Price 2/3/20102300 3/3/20103200 4/3/20101800 5/3/20101900 I want to take both these data together in a single file, and take the smaller vector (or matrix or dataframe??..i am new to R and still confused with the various objects) which is file1 (because it contains fewer rows ) and vlookup prices in the second file basedon the dates on file1 and write three columns (date, price from 1 and price from2) in a new file. How do i do this please? Many thanks... R -- View this message in context: http://r.789695.n4.nabble.com/Adding-two- files-into-one-and-vlookup-tp2280277p2280277.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-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 message @ http://r.789695.n4.nabble.com/Adding-two-files-into-one-and-vlookup-tp2280277p2280305.html To start a new topic under R help, email ml-node+789696-608741344-309...@n4.nabble.com To unsubscribe from R help, click (link removed) -- 'Raghu' -- View this message in context: http://r.789695.n4.nabble.com/Adding-two-files-into-one-and-vlookup-tp2280277p2281026.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.
Re: [R] Adding two files into one and vlookup
I was able to achieve the desired output. The issue I had was that both the files contained the Date column in the data format. So, i first converted this colum alone into numeric and then did a merge on the files and it worked. The moral is that in any dataframe all variable have to be of similar type and then merge works. Correct me if i am wrong? On 7/7/10, Raghu r.raghura...@gmail.com wrote: Say I have two files file and file2: file1 contains the following: Date Price 02/07/201053.96597903 03/07/201056.92825807 04/07/201039.27408645 05/07/201042.59834151 06/07/201070.68512383 07/07/201010.92505265 08/07/201052.12492249 09/07/201049.88767957 file2 contains the following: Date Price 03/07/20105.312006403 04/07/2010673.0705924 05/07/2010442.4679386 06/07/2010851.9158985 07/07/2010581.8592424 I want to create a new file that should look like: Date Price1Price2 03/07/20105.31200640356.928 04/07/2010673.070592439.274 05/07/2010442.467938642.598 06/07/2010851.915898570.685 07/07/2010581.859242410.925 Thx On 7/7/10, Erik Iverson er...@ccbr.umn.edu wrote: raghu wrote: I have two files with dates and prices in each. The number of rows in each of them will differ. How do I create a new file which contains data from both these files? Cbind and merge are not helpful. For cbind because the rows are not the same replication occurs. Also if I have similar data how do I write a vlookup kind of function? I am giving an example below: Say Price1 file contains the following: Date Price 2/3/2010 134.00 3/3/2010 133.90 4/3/2010 135.55 And say price2 contains the following: Date Price 2/3/20102300 3/3/20103200 4/3/20101800 5/3/20101900 I want to take both these data together in a single file, and take the smaller vector (or matrix or dataframe??..i am new to R and still confused with the various objects) which is file1 (because it contains fewer rows ) and vlookup prices in the second file basedon the dates on file1 and write three columns (date, price from 1 and price from2) in a new file. How do i do this please? I think all this can be accomplished with merge. Can you give reproducible examples as the posting guide suggests? Use read.table to read in your data into R objects, then use ?dput to give us the exact copies of the objects (probably data.frames by your example), and what output you want to have. Being precise with the classes of objects you're working with is key, and ?dput is a great way to make sure we have the same objects as you. Another tip is common terminology. For instance, `vlookup` is not a term used in R, and many people will not know what it means. This way, everything is reproducible for us, and we can offer suggestions and show you what the exact output will be. In short, making sure everyone is on the same page goes a long way when getting help from a mailing list. -- 'Raghu' -- 'Raghu' __ 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] Adding two files into one and vlookup
I tried and this error popped: z1 - read.zoo(textConnection(Lines1), header = TRUE, format = fmt) Error in textConnection(Lines1) : invalid 'text' argument Also library(chron) returns error as it is an invalid library. Could you help? On 7/7/10, Gabor Grothendieck [via R] ml-node+2280348-527941373-309...@n4.nabble.com wrote: On Tue, Jul 6, 2010 at 8:26 PM, Gabor Grothendieck ggrothendi...@gmail.com wrote: On Tue, Jul 6, 2010 at 6:18 PM, raghu r.raghura...@gmail.com wrote: I have two files with dates and prices in each. The number of rows in each of them will differ. How do I create a new file which contains data from both these files? Cbind and merge are not helpful. For cbind because the rows are not the same replication occurs. Also if I have similar data how do I write a vlookup kind of function? I am giving an example below: Say Price1 file contains the following: Date Price 2/3/2010 134.00 3/3/2010 133.90 4/3/2010 135.55 And say price2 contains the following: Date Price 2/3/20102300 3/3/20103200 4/3/20101800 5/3/20101900 I want to take both these data together in a single file, and take the smaller vector (or matrix or dataframe??..i am new to R and still confused with the various objects) which is file1 (because it contains fewer rows ) and vlookup prices in the second file basedon the dates on file1 and write three columns (date, price from 1 and price from2) in a new file. How do i do this please? Try this and for more read the three vignettes (pdf documents) in the zoo package and also read the R News 4/1 article on dates and times: Lines1 - Date Price 2/3/2010 134.00 3/3/2010 133.90 4/3/2010 135.55 Lines2 - Date Price 2/3/20102300 3/3/20103200 4/3/20101800 5/3/20101900 library(zoo) library(chron) z1 - read.zoo(textConnection(Lines1), header = TRUE, FUN = chron) z2 - read.zoo(textConnection(Lines2), header = TRUE, FUN = chron) I originally assumed that the dates were the usual month/day/year but looking at it again I suspect they are day/month/year so lets use Date class instead of chron replacing the last three statements with: fmt - %d/%m/%Y z1 - read.zoo(textConnection(Lines1), header = TRUE, format = fmt) z2 - read.zoo(textConnection(Lines2), header = TRUE, format = fmt) merge(z1, z2) # keep all rows in each merge(z1, z2, all = FALSE) # keep only rows in both merge(z1, z2, all = c(TRUE, FALSE)) # keep all rows in z1 merge(z1, z2, all = c(FALSE, TRUE)) # keep all rows in z2 __ 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 message @ http://r.789695.n4.nabble.com/Adding-two-files-into-one-and-vlookup-tp2280277p2280348.html To start a new topic under R help, email ml-node+789696-608741344-309...@n4.nabble.com To unsubscribe from R help, click (link removed) -- 'Raghu' -- View this message in context: http://r.789695.n4.nabble.com/Adding-two-files-into-one-and-vlookup-tp2280277p2281017.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.
Re: [R] Adding two files into one and vlookup
Say I have two files file and file2: file1 contains the following: DatePrice 02/07/2010 53.96597903 03/07/2010 56.92825807 04/07/2010 39.27408645 05/07/2010 42.59834151 06/07/2010 70.68512383 07/07/2010 10.92505265 08/07/2010 52.12492249 09/07/2010 49.88767957 file2 contains the following: DatePrice 03/07/2010 5.312006403 04/07/2010 673.0705924 05/07/2010 442.4679386 06/07/2010 851.9158985 07/07/2010 581.8592424 I want to create a new file that should look like: Date Price1Price2 03/07/2010 5.31200640356.928 04/07/2010 673.070592439.274 05/07/2010 442.467938642.598 06/07/2010 851.915898570.685 07/07/2010 581.859242410.925 Thx On 7/7/10, Erik Iverson er...@ccbr.umn.edu wrote: raghu wrote: I have two files with dates and prices in each. The number of rows in each of them will differ. How do I create a new file which contains data from both these files? Cbind and merge are not helpful. For cbind because the rows are not the same replication occurs. Also if I have similar data how do I write a vlookup kind of function? I am giving an example below: Say Price1 file contains the following: Date Price 2/3/2010 134.00 3/3/2010 133.90 4/3/2010 135.55 And say price2 contains the following: Date Price 2/3/20102300 3/3/20103200 4/3/20101800 5/3/20101900 I want to take both these data together in a single file, and take the smaller vector (or matrix or dataframe??..i am new to R and still confused with the various objects) which is file1 (because it contains fewer rows ) and vlookup prices in the second file basedon the dates on file1 and write three columns (date, price from 1 and price from2) in a new file. How do i do this please? I think all this can be accomplished with merge. Can you give reproducible examples as the posting guide suggests? Use read.table to read in your data into R objects, then use ?dput to give us the exact copies of the objects (probably data.frames by your example), and what output you want to have. Being precise with the classes of objects you're working with is key, and ?dput is a great way to make sure we have the same objects as you. Another tip is common terminology. For instance, `vlookup` is not a term used in R, and many people will not know what it means. This way, everything is reproducible for us, and we can offer suggestions and show you what the exact output will be. In short, making sure everyone is on the same page goes a long way when getting help from a mailing list. -- 'Raghu' __ 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] Gray level mosaic plot with shading_Friendly
On Wed, 7 Jul 2010, Michael Kubovy wrote:
Dear Achim and Michael,
Thank you so much. Indeed, mosaic(Titanic, gp = shading_hcl, gp_args =
list(lty = 1:2, c = 0)) does almost what I was looking for, except that
for consistency and clarity, I would have expected the negative values
on the legend to be be outlined with lty = 2.
In the continuous legend, that is employed by default (legend_resbased),
it is visually not very compelling to show line types as well. But you can
set legend = legend_fixed which displays this information (but is less
intuitive concerning the interval ranges).
Best,
Z
Michael
On Jul 7, 2010, at 2:13 AM, Achim Zeileis wrote:
On Tue, 6 Jul 2010, Michael Friendly wrote:
Michael Kubovy wrote:
Suppose we start with
data(Titanic)
mosaic(Titanic, shade = TRUE)
How do I combine the dashed box contours of shading_Friendly to indicate negative
residuals, with three levels of gray: dark for abs(Pearson Resid) 4, lighter for 4
abs(Pearson Resid) 2, and lightest for bs(Pearson Resid) 2 ?
Do you mean [1] you want to plot positive residuals in color and negative in
gray scale?
Or [2] to fold + and - residuals by shading all according to abs(resid), and
distinguishing + from - by the dashed box outlines?
In fact, I designed this coding scheme so that mosaic plots in color (with my
blue - white - red scheme) would approximately do exactly what
you might want under [2], when rendered in B/W, since the fully saturated red
and blue are close in darkness in B/W.
And shading_hcl() has been written to do exactly what you want under [2]. While
it is hard to come up with colors of different hues in HSV or HLS space that
have the same brightness (aka lightness/luminance) and the same
colorfulness (aka chroma), this is easy in HCL.
Try
mosaic(Titanic, gp=shading_Friendly)
save as a jpg/png and try converting to B/W with an image program and see if
this is good enough.
mosaic(Titanic, shade = TRUE)
is the same as
mosaic(Titanic, gp = shading_hcl)
which you can then modify to have different line types
mosaic(Titanic, gp = shading_hcl, gp_args = list(lty = 1:2))
If you print that on a grayscale printer you will see the same plot without any
chroma, i.e.,
mosaic(Titanic, gp = shading_hcl, gp_args = list(lty = 1:2, c = 0))
The shading_hcl() function is introduced in Zeileis et al. (2007, JCGS), see
?shading_hcl, which provides more detailed references to HCL colors etc.
Best,
Z
Alternatively, write your own, shading_Kubovy, modeled on
shading_Friendly -
function (observed = NULL, residuals = NULL, expected = NULL,
df = NULL, h = c(2/3, 0), lty = 1:2, interpolate = c(2, 4),
eps = 0.01, line_col = black, ...)
{
shading_hsv(observed = NULL, residuals = NULL, expected = NULL,
df = NULL, h = h, v = 1, lty = lty, interpolate = interpolate,
eps = eps, line_col = line_col, p.value = NA, ...)
}
environment: namespace:vcd
attr(,class)
[1] grapcon_generator
In the defaults, lty=1:2 is what distinguishes + and - for outline line type
hope this helps,
-Michael
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[R] use sliding window to count substrings found in large string
Hello together, I'm looking for advice on how to do some tests on strings. What I want to do is the following: (just an example, real strings/sequence are about 200-400 characters long) given set of Strings: String1 abcdefgh String2 bcdefgop use a sliding window of size x to create an vector of all subsequences of size x found in the set (order matters! ). Now create, for every string in the set, an vector containing the counts on how often each subsequence was found in this particular string. It would be great if someone could give me a vague outline on how to start and which methods to work. I did read through the man pages and goggled a lot, but still don't know how to approach this. best regards, Immanuel __ 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] HELP - four.nines.cartesian.probability.grid
My routine (below) works OK but misbehaves if the on-screen plot is made
wider using the mouse.
The problem is caused by using
par(usr)[1] - 0.07 * (par(usr)[2] - par(usr)[1])
to locate two items on the y-axis. The rest of the labeling is controlled
by the line=0 parameter setting. Of course resizing changes the absolute
plot width, and I'd rather have things with respect to the line= setting.
(Try it: Cut and paste the code, then resize the screen plot to be wider.)
Is there another way to do what I'm trying to do?
I'm using R version 2.11.1 (2010-05-31), running on an HP 64 bit Windows 7
machine.
Thanks
four.nines.cartesian.probability.grid -
function (X.data, x.title = X, y.title = Theoretical Normal CDF,
X.start, X.end)
{
probs - c(0.0001, 0.0003, 0.001, 0.002, 0.005, 0.01, 0.02, 0.05, 0.1,
0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 0.95, 0.98, 0.99, 0.995, 0.998,
0.999, 0.9997, 0.)
z.vals - qnorm(probs)
y.min - z.vals[1]
y.max - z.vals[length(z.vals)]
npts - length(X.data)
X.mean - mean(X.data)
X.sd - sd(X.data)
X.001 - X.mean + X.sd * qnorm(0.001)
X.999 - X.mean + X.sd * qnorm(0.999)
plot(NA, axes = FALSE, xlim = c(X.start, X.end), ylim = c(qnorm(0.0001),
qnorm(0.)), xlab = , ylab = )
axis(side = 1, labels = TRUE, at = pretty(c(X.001, X.data, X.999), n=8),
line = 0, tick = TRUE, outer = FALSE)
mtext(text = x.title, side = 1, line = 2.3, adj = 0.5)
axis(side = 2, at = z.vals, labels = c(NA, NA,
as.character(probs)[-(1:2)]),
line = 0, tck = -0.01, las = 1, cex.axis = 0.8)
x.loc - par(usr)[1] - 0.07 * (par(usr)[2] - par(usr)[1]) ## If plot
is resized wider this may be troublesome.
text(x = x.loc, y = z.vals[1], bquote(1* X *10^-4), cex = 0.75, xpd =
TRUE)
text(x = x.loc, y = z.vals[2], bquote(3* X *10^-4), cex = 0.75, xpd =
TRUE)
abline(h = qnorm(c(0.90, 0.95, 0.99, 0.999, 0.)), lty = 5, col =
gray)
abline(h = qnorm(c(0.10, 0.05, 0.01, 0.001, 0.0001)), lty = 5, col =
gray)
mtext(text = y.title, side = 2, line = 3, at = (y.max + y.min)/2, adj =
0.5)
mtext(www.StatisticalEngineering.com, side = 1, line = 2.9, adj = 1,
cex = 0.7)
box()
}
windows(width = 6, height = 6, pointsize = 12)
par(mar = c(4, 4, 0.5, 1) + 0.1)
X.data - rnorm(n=100, mean = 8000, sd = 300)
four.nines.cartesian.probability.grid(X.data, X.start = 7000, X.end = 9000)
points(x = sort(X.data), y = qnorm1:length(X.data)) -
0.5)/length(X.data
Charles Annis, P.E.
charles.an...@statisticalengineering.com
561-352-9699
http://www.StatisticalEngineering.com
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Re: [R] Calculate area under a curve
The mean values theorem of integration (which I think typifies the differences in thinking between mathematicians and statisticians) says that the integral from a to b is equal to the average value of the curve between a and b times the distance from a to b (b - a). I would be interested in how similar the areas computed using the other methods suggested compare to just averaging your data and multiplying by the distance. -- Gregory (Greg) L. Snow Ph.D. Statistical Data Center Intermountain Healthcare greg.s...@imail.org 801.408.8111 -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r- project.org] On Behalf Of suse Sent: Thursday, July 01, 2010 8:57 AM To: r-help@r-project.org Subject: [R] Calculate area under a curve Hi, I want to know the area under a curve, which is not given as a function, but as values in a time series. It is not a smooth curve, but switches often between positive values and zero (the values describe the moisture in the soil over a year, one entry is one day). I already tried area.between.curves, but got only 0 as result. I guess, it doesn't work because of these multiple changes between 0 and positive values (most of the time, the values are 0 and in the certain case I tested, the positive values were only on single days; but for other values positive values last longer) I hope, someone can help me! THank you! -- View this message in context: http://r.789695.n4.nabble.com/Calculate- area-under-a-curve-tp2275283p2275283.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-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] Visualization of coefficients
On Jul 7, 2010, at 10:19 AM, Michael Friendly wrote: Tal Galili wrote: Hello David, Thanks to your posting I started looking at the function in the arm package. It appears this function is quite mature, and offers (for example) the ability to easily overlap coefficients from several models. I updated the post I published on the subject, so at the end of it I give an example of comparing the coef of several models: http://www.r-statistics.com/2010/07/visualization-of-regression-coefficients-in-r/ Thanks again for the pointer. Best, Tal Achim Zeileis wrote: Re: more mature. arm's coefplot() is more flexible in certain respects, mine is more convenient in others. The overlay functionality is something arm's coefplot() is better in and it also as some further options (vertical vs. horizontal etc.). My coefplot() has the advantage that it does not need any modification as long as coef() and vcov() methods are available. Furthermore, level can specify the significance level (instead of always using one and two standard errors, respectively). But it shouldn't be too hard to create a superset of all options. @Tal: For the example using library(arm) and the Mroz data, you posted the wrong image. And loose the intercept in the example. @Achim: It would be worthwhile combining the generality of your version with the overlay capability of the arm version, which is extremely useful for model comparison. However, the arm version uses S4 methods. To my reading coefplot somehow collects parameters from a list of model objects using S4 methods, but then passes these to coefplot.default which uses base graphics. There seems to be an implicit loop, ... perhaps some sort of S4 magic? ... that accumulates like-named coefficients in a .local(...) (function?) object until they all get reduced to class numeric, at which point they are passed to coefplot.default() which does not appear to be an S4 method. (I am not particularly knowledgeable about S4 functions and dispatching, so any corrections or applifications to this account would be welcome.) -- David. -Michael -- Michael Friendly Email: friendly AT yorku DOT ca Professor, Psychology Dept. York University Voice: 416 736-5115 x66249 Fax: 416 736-5814 4700 Keele StreetWeb: http://www.datavis.ca Toronto, ONT M3J 1P3 CANADA __ 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. David Winsemius, MD West Hartford, CT __ 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] use sliding window to count substrings found in large string
On Wed, Jul 7, 2010 at 12:25 PM, Immanuel mane.d...@googlemail.com wrote: Hello together, I'm looking for advice on how to do some tests on strings. What I want to do is the following: (just an example, real strings/sequence are about 200-400 characters long) given set of Strings: String1 abcdefgh String2 bcdefgop use a sliding window of size x to create an vector of all subsequences of size x found in the set (order matters! ). Now create, for every string in the set, an vector containing the counts on how often each subsequence was found in this particular string. It would be great if someone could give me a vague outline on how to start and which methods to work. I did read through the man pages and goggled a lot, but still don't know how to approach this. Try this: # generate an input string n long set.seed(123) n - 300 lets - paste(sample(letters[1:5], n, replace = TRUE), collapse = ) # get rolling k-length sequences and count k - 3 table(substring(lets, 1:(n-k+1), k:n)) __ 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] F# vs. R
Hello, Marc No, I do not want to validate Cox PH. :-) I do use R daily, though right now I do not use the statistical part that much. I just generally wonder if any R-user tried F# and his/her opinions. Regards, Sergey On Wed, Jul 7, 2010 at 17:56, Marc Schwartz marc_schwa...@me.com wrote: On Jul 7, 2010, at 10:31 AM, Sergey Goriatchev wrote: Hello, everyone F# is now public. Compiled code should run faster than R. Anyone has opinion on F# vs. R? Just curious Best, S The key time critical parts of R are written in compiled C and FORTRAN. Of course, if you want to take the time to code and validate a Cox PH or mixed effects model in F# and then run them against R's coxph() or lme()/lmer() functions to test the timing, feel free... :-) So unless there is a pre-existing library of statistical and related functionality for F#, perhaps you need to reconsider your query. Regards, Marc Schwartz -- --- Kniven skärpes bara mot stenen. __ 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] nls + quasi-poisson distribution
Dear Suresh, The gnm package for generalized nonlinear models might be what you want here. This allows you to specify nonlinear models with family=quasipoisson. For an introduction to the package see the article in R News: http://cran.r-project.org/doc/Rnews/Rnews_2007-2.pdf If your model requires a custom nonlin function and you get stuck on this, or you have any other queries about the package, I'm happy to be contacted directly. Best regards, Heather On 06/07/10 07:04, Suresh Krishna wrote: Hello R-helpers, I would like to fit a non-linear function to data (Discrete X axis, over-dispersed Poisson values on the Y axis). I found the functions gnlr in the gnlm package from Jim Lindsey: this can handle nonlinear regression equations for the parameters of Poisson and negative binomial distributions, among others. I also found the function nls2 in the software package accompanying the book Statistical tools for nonlinear regression by Huet et al: this can handle nonlinear regression with Poisson distributed Y-axis values. I was wondering if there was any other option: specifically, any option that handled nonlinear fitting with quasi-Poisson distributions (to handle the overdispersion). This is a very new area for me, and I am still trying to figure out the best way to do this, so I would appreciate any and all pointers. Thanks much, Suresh __ 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-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] possible to plot number line in R?
Here is some code to get you started: plot.new() plot.window( c(0,10), c(-1, 1) ) axis(1, at=0:10, pos=0) lines( c(2,2,5,5), c( -0.25, -0.5, -0.5, -0.25 ) ) text( 3, -0.6, 'Interval 1' ) # the plotrix package has a function for text in a box lines( c(3,3,6,6), c( 0.1, 0.3, 0.3, 0.1) ) text( 4.5, 0.4, 'Interval 2' ) lines( c(4,4,7,7), c( -0.4, -0.6, -0.6, -0.4 ) ) text( 5.5, -0.7, 'Interval 3' ) pieces could be placed in functions to automate the parts that you want. hope this helps, -- Gregory (Greg) L. Snow Ph.D. Statistical Data Center Intermountain Healthcare greg.s...@imail.org 801.408.8111 -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r- project.org] On Behalf Of Kroepfl, Julia (julia.kroe...@uni-graz.at) Sent: Friday, July 02, 2010 2:40 AM To: 'r-help@r-project.org' Subject: [R] possible to plot number line in R? Thank you very much for your answers, but I think I did not explain thoroughly enough what I needed. I attached a demo of the plot. I need the number line between 2 and 3, both values being shown on the line, interval values should be printed next to the dashes and lines should connect the intervals. Is this possible to do in R? Best Regards, Julia __ 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] use sliding window to count substrings found in large string
Hey, big help, thanks! One little question remains, if I create more then one string and table ... - # generate an input string n long set.seed(123) n - 300 lets_1 - paste(sample(letters[1:5], n, replace = TRUE), collapse = ) lets_2 - paste(sample(letters[1:5], n, replace = TRUE), collapse = ) # get rolling k-length sequences and count k - 3 table_1 -table(substring(lets_1, 1:(n-k+1), k:n)) table_2 -table(substring(lets_2, 1:(n-k+1), k:n)) --- is it possible to manipulate table_1 so that it contains zero entries for all the substrings found in table_2 but not in table_1? best regards Immanuel Try this: # generate an input string n long set.seed(123) n - 300 lets - paste(sample(letters[1:5], n, replace = TRUE), collapse = ) # get rolling k-length sequences and count k - 3 table(substring(lets, 1:(n-k+1), k:n)) [[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.
Re: [R] help on bar chart
fortune(197) If anything, there should be a Law: Thou Shalt Not Even Think Of Producing A Graph That Looks Like Anything From A Spreadsheet. -- Ted Harding (in a discussion about producing graphics) R-help (August 2007) Filling graphics objects with lines dates back to the days when the only way to get decent graphics was with a pen plotter (device that actually had a mechanical arm that would move a pen over paper (or paper under the pen)) which could not easily fill areas (well you could draw the lines really densely, but the usual outcome was that you annoyed other users by taking a long time and using all the ink and often ended up with a hole in the paper instead of a nice fill. Technology has progressed quite a bit since then, it is best if your graphs reflect that. Tufte also discusses that such lines actually make graphs harder to read, can actually distort features, and cause other problems. The fact that lattice graphics do not include a simple way to do this is a feature. -- Gregory (Greg) L. Snow Ph.D. Statistical Data Center Intermountain Healthcare greg.s...@imail.org 801.408.8111 -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r- project.org] On Behalf Of ppcrystal Sent: Friday, July 02, 2010 11:47 PM To: r-help@r-project.org Subject: [R] help on bar chart Hey guys, This is the bar chart that I am working on: library(lattice); data - data.frame( X1 = c(2300, 1300, 1300, 450), X2 = c(2110, 2220, 1100, 660), Y = factor(c(sample1, sample2, sample3, sample4)) ); barchart( Y ~ X1 + X2, data, stack = TRUE, horiz = TRUE, lwd = 1.5, xlab = expression(bold(Sample size)), col = colors()[c(24,1)], xlim = c(0,5000), xat = seq(0,5000,1000) ); I wanted to make a bar chart that has hatching lines inside the bar: with sample 2 and 4 having vertical lines and sample 1 and 3 having horizontal lines, like the following (I kind of photoshopped the image to demonstrate what I wanted it to look like): http://r.789695.n4.nabble.com/file/n2277107/test.png Anyone knows how I can add hatching to the bar charts? Thanks very much for your time!!! -- View this message in context: http://r.789695.n4.nabble.com/help-on- bar-chart-tp2277107p2277107.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-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] use sliding window to count substrings found in large string
On Wed, Jul 7, 2010 at 1:15 PM, Immanuel mane.d...@googlemail.com wrote: Hey, big help, thanks! One little question remains, if I create more then one string and table ... - # generate an input string n long set.seed(123) n - 300 lets_1 - paste(sample(letters[1:5], n, replace = TRUE), collapse = ) lets_2 - paste(sample(letters[1:5], n, replace = TRUE), collapse = ) # get rolling k-length sequences and count k - 3 table_1 -table(substring(lets_1, 1:(n-k+1), k:n)) table_2 -table(substring(lets_2, 1:(n-k+1), k:n)) --- is it possible to manipulate table_1 so that it contains zero entries for all the substrings found in table_2 but not in table_1? best regards Immanuel Turn them into factors with the appropriate levels before counting them with table: # generate an input string n long set.seed(123) n - 300 lets_1 - paste(sample(letters[1:5], n, replace = TRUE), collapse = ) lets_2 - paste(sample(letters[1:5], n, replace = TRUE), collapse = ) # get rolling k-length sequences and count k - 3 s1 - substring(lets_1, 1:(n-k+1), k:n) s2 - substring(lets_2, 1:(n-k+1), k:n) levs - sort(unique(union(s1, s2))) table(factors(s1, levs)) table(factors(s2, levs)) __ 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] use sliding window to count substrings found in large string
On Wed, Jul 7, 2010 at 1:25 PM, Gabor Grothendieck ggrothendi...@gmail.com wrote: On Wed, Jul 7, 2010 at 1:15 PM, Immanuel mane.d...@googlemail.com wrote: Hey, big help, thanks! One little question remains, if I create more then one string and table ... - # generate an input string n long set.seed(123) n - 300 lets_1 - paste(sample(letters[1:5], n, replace = TRUE), collapse = ) lets_2 - paste(sample(letters[1:5], n, replace = TRUE), collapse = ) # get rolling k-length sequences and count k - 3 table_1 -table(substring(lets_1, 1:(n-k+1), k:n)) table_2 -table(substring(lets_2, 1:(n-k+1), k:n)) --- is it possible to manipulate table_1 so that it contains zero entries for all the substrings found in table_2 but not in table_1? best regards Immanuel Turn them into factors with the appropriate levels before counting them with table: # generate an input string n long set.seed(123) n - 300 lets_1 - paste(sample(letters[1:5], n, replace = TRUE), collapse = ) lets_2 - paste(sample(letters[1:5], n, replace = TRUE), collapse = ) # get rolling k-length sequences and count k - 3 s1 - substring(lets_1, 1:(n-k+1), k:n) s2 - substring(lets_2, 1:(n-k+1), k:n) levs - sort(unique(union(s1, s2))) table(factors(s1, levs)) table(factors(s2, levs)) That should be factor, not factors: table(factor(s1, levs)) table(factor(s2, levs)) __ 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] using objects from different workspaces
Check out the high performance computing task view on CRAN. -- Gregory (Greg) L. Snow Ph.D. Statistical Data Center Intermountain Healthcare greg.s...@imail.org 801.408.8111 -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r- project.org] On Behalf Of Martin Dózsa Sent: Sunday, July 04, 2010 5:12 PM To: r-help@r-project.org Subject: [R] using objects from different workspaces Hi all, I have the following problem: I need to run a large number of simulations, therefore I use many computers. After the computations I need to make some operations with the obtained results (such as weighted average, sum, etc.). My question is, how is it possible to combine the output of several R sessions? My objects are quite complex (multi-dimensional arrays), therefore export to some simple csv file seems not to be a solution. I also need to do this operation several times, so it would be good if it could be automatized. Thank you in advance, Martin [[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-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] F# vs. R
On Wed, 2010-07-07 at 17:31 +0200, Sergey Goriatchev wrote: Hello, everyone F# is now public. Compiled code should run faster than R. Anyone has opinion on F# vs. R? Just curious Best, S Sergey, F# is public, but is not open source. F# run in windows but run in AIX, linux, MAC, UNIX etc? Compiled code should run faster than R, but is precise? Compiled code should run faster than R, but is reliable? Compiled code should run faster than R, but have 2.440 packages for extend your capacities? Compiled code should run faster than R, but critical code is in C o FORTRAN So I think the F# is not a good alternative, if your concern is velocity dou you look Littler http://code.google.com/p/littler/ -- Bernardo Rangel Tura, M.D,MPH,Ph.D National Institute of Cardiology Brazil __ 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] Data Labels in a barchart (Lattice or otherwise)
fortune(197) If anything, there should be a Law: Thou Shalt Not Even Think Of Producing A Graph That Looks Like Anything From A Spreadsheet. -- Ted Harding (in a discussion about producing graphics) R-help (August 2007) Also read the discussion started with: http://tolstoy.newcastle.edu.au/R/e2/help/07/08/22858.html -- Gregory (Greg) L. Snow Ph.D. Statistical Data Center Intermountain Healthcare greg.s...@imail.org 801.408.8111 -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r- project.org] On Behalf Of RaoulD Sent: Sunday, July 04, 2010 9:44 PM To: r-help@r-project.org Subject: [R] Data Labels in a barchart (Lattice or otherwise) Hi, Can anyone please help me with how I could add labels with the value for each bar in a barchart? (similar to how data labels can be added in Excel) I have done a lot of searching but havent been lucky. Thanks, Raoul -- View this message in context: http://r.789695.n4.nabble.com/Data- Labels-in-a-barchart-Lattice-or-otherwise-tp2278027p2278027.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-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] use sliding window to count substrings found in large string
Hey, saved my day. Now can watch the football semi-final thanks Turn them into factors with the appropriate levels before counting them with table: # generate an input string n long set.seed(123) n - 300 lets_1 - paste(sample(letters[1:5], n, replace = TRUE), collapse = ) lets_2 - paste(sample(letters[1:5], n, replace = TRUE), collapse = ) # get rolling k-length sequences and count k - 3 s1 - substring(lets_1, 1:(n-k+1), k:n) s2 - substring(lets_2, 1:(n-k+1), k:n) levs - sort(unique(union(s1, s2))) table(factor(s1, levs)) table(factor(s2, levs)) __ 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] subscripts for panel.superpose in lattice
Hi,
I am trying to superimpose (overlay) regression lines to scatter plots
by groups with xyplot (dysfunctional code below). However, my call of
panel.superpose breaks down because of the subscripts requirement. I
tried to research the documentation and examples, but I cannot figure
out how to make xyplot plug subscripts to a panel... call. Could you
have a look? It would be greatly appreciated.
Thank you,
Laszlo
scatter_contrast - function(depvar,bins,cutvar,cutvarname = NULL,
yvarlab = NULL,xvarlab =
NULL,nbins=20,maxbins=100,yrange=c(0,9),plottitle=NULL,legendtitle=NULL)
{
library('lattice')
library('grid')
trellis.par.set(
plot.symbol = list(cex = 1.5,col=rgb(26,71,111,max=255)),
superpose.symbol = list(cex = rep(1,
times=7),pch=c(15:21),col=c(rgb(26,71,111,max=255),
rgb(144,53,59,max=255),rgb(85,117,47,max=255),#ff,orange,#00ff00,brown),fill=c(rgb(26,71,111,max=255),
rgb(144,53,59,max=255),rgb(85,117,47,max=255),#ff,orange,#00ff00,brown)),
plot.line = list(cex = 1,lwd=2,col=rgb(26,71,111,max=255)),
superpose.line = list(cex = rep(1,
times=7),lwd=rep(2,times=7),col=c(rgb(26,71,111,max=255),
rgb(144,53,59,max=255),rgb(85,117,47,max=255),#ff,orange,#00ff00,brown)),
reference.line = list(col=rgb(234,242,243,max=255)),
add.line = list(rgb(85,117,47,max=255),lwd=2),
grid.pars = list(col=black),#rgb(234,242,243,max=255)),
superpose.polygon =
list(col=black),#rgb(234,242,243,max=255)),
fontsize = list(text=16),
par.xlab.text = list(cex = 0.8),
par.ylab.text = list(cex = 0.8),
)
if (length(unique(bins))maxbins) bins - binning(bins,nbins)
temp - summary(cutvar)
cut - 1*(cutvar temp[3])
if (length(unique(!is.na(cutvar)))6) cut - cutvar
legval1 - names(data.frame(cutvar))
xl - names(data.frame(bins))
leg - paste(unique(sort(cut)))
legval2 - leg[1:(length(unique(leg))-1)]
if (length(na.omit(cutvar)) == length(cutvar)) legval2 - leg
ht - depvar[[1]]
if (is.na(ht)) ht - 0
if (ht == hist) {
bins -
replace(bins,binsquantile(bins,0.99),quantile(bins,0.99))
bins -
replace(bins,binsquantile(bins,0.05),quantile(bins,0.05))
mes -
aggregate(matrix(1,length(bins),1),list(bins,cut),sum,na.rm = TRUE)
cnt - aggregate(matrix(1,length(bins),1),list(cut),sum,na.rm =
TRUE)
mes$x - mes$V1
mes$V1 - NULL
mes - merge(mes,cnt,by.x = Group.2,by.y = Group.1)
mes$x - mes$x/mes$V1 }
else {mes - aggregate(depvar,list(bins,cut),mean,na.rm = TRUE)}
if (yrange[2] == 9) {
ran - (max(mes$x)- min(mes$x))*0.1
yrange = c(min(mes$x)-ran,max(mes$x)+ran)}
xyplot(x ~ Group.1,groups = Group.2,data = mes, subscripts =TRUE,type =
p,
auto.key = list(cex=0.7,cex.title=0.7,title=legendtitle,space =
bottom,points = FALSE,lines=TRUE,columns=2),
# key = simpleKey(paste(cutvarname,leg),points = FALSE,lines=TRUE
),
xlab = xvarlab, ylab = yvarlab,ylim=yrange,
aspect=fill,
panel = function(...) {
panel.grid(h = -1, v = 0)
panel.xyplot(...)
# panel.abline(lm(depvar ~ bins))
panel.superpose(bins,depvar,subscripts,groups =
cutvar,type=nr)
panel.axis(side = left, outside = TRUE,tck =
-1, line.col = 1)
panel.axis(side = bottom, outside = TRUE,tck
= -1, line.col = 1)},
main = plottitle,
par.settings = list(axis.line = list(col =
transparent)),
axis = function(side, ...) {
if (side == left) {
grid.lines(x = c(0, 0), y = c(0,
1),default.units =
npc,gp=gpar(col='black')) }
else if (side == bottom) {
grid.lines(x = c(0, 1), y = c(0,
0),default.units =
npc,gp=gpar(col='black')) }
axis.default(side = side, ...)
}
)
}
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] export VTK from R : impossible to write data as float
So, the problem was that R exports only double sized floats (double), and Paraview requires single sized floats. The solution was just to write : writeBin(data,bfile_celldata,endian=swap,size=4) Special Thanks to Sebastian Gibb : http://www.mail-archive.com/r-help@r-project.org/msg100994.html alex -- View this message in context: http://r.789695.n4.nabble.com/export-VTK-from-R-impossible-to-write-data-as-float-tp2278756p2281217.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] [R-pkgs] ggplot2 version 0.8.8
ggplot2 ggplot2 is a plotting system for R, based on the grammar of graphics, which tries to take the good parts of base and lattice graphics and avoid bad parts. It takes care of many of the fiddly details that make plotting a hassle (like drawing legends) as well as providing a powerful model of graphics that makes it easy to produce complex multi-layered graphics. To install or update, run: install.packages(c(ggplot2, plyr)) Find out more at http://had.co.nz/ggplot2, and check out the nearly 500 examples of ggplot in use. If you're interested, you can also sign up to the ggplot2 mailing list at http://groups.google.com/group/ggplot2, or track development at http://github.com/hadley/ggplot2 ggplot2 0.8.8 (2010-07-02) This version fixes the following 23 bugs: * coord_equal finally works as expected (thanks to continued prompting from Jean-Olivier Irisson) * coord_equal renamed to coord_fixed to better represent capabilities * coord_polar and coord_polar: new munching system that uses distances (as defined by the coordinate system) to figure out how many pieces each segment should be broken in to (thanks to prompting from Jean-Olivier Irisson) * fix ordering bug in facet_wrap (thanks to bug report by Frank Davenport) * geom_errorh correctly responds to height parameter outside of aes * geom_hline and geom_vline will not impact legend when used for fixed intercepts * geom_hline/geom_vline: intercept values not set quite correctly which caused a problem in conjunction with transformed scales (reported by Seth Finnegan) * geom_line: can now stack lines again with position = stack (fixes #74) * geom_segment: arrows now preserved in non-Cartesian coordinate system (fixes #117) * geom_smooth now deals with missing values in the same way as geom_line (thanks to patch from Karsten Loesing) * guides: check all axis labels for expressions (reported by Benji Oswald) * guides: extra 0.5 line margin around legend (fixes #71) * guides: non-left legend positions now work once more (thanks to patch from Karsten Loesing) * label_bquote works with more expressions (factors now cast to characters, thanks to Baptiste Auguie for bug report) * scale_color: add missing US spellings * stat: panels with no non-missing values trigged errors with some statistics. (reported by Giovanni Dall'Olio) * stat: statistics now also respect layer parameter inherit.aes (thanks to bug report by Lorenzo Isella and investigation by Brian Diggs) * stat_bin no longer drops 0-count bins by default * stat_bin: fix small bug when dealing with single bin with NA position (reported by John Rauser) * stat_binhex: uses range of data from scales when computing binwidth so hexes are the same size in all facets (thanks to Nicholas Lewin-Koh for the bug report) * stat_qq has new dparam parameter for specifying distribution parameters (thanks to Yunfeng Zhang for the bug report) * stat_smooth now uses built-in confidence interval (with small sample correction) for linear models (thanks to suggestion by Ian Fellows) * stat_spoke: correctly calculate stat_spoke (cos and sin were flipped, thanks to Jean-Olivier Irisson for bug report and fix) -- Assistant Professor / Dobelman Family Junior Chair Department of Statistics / Rice University http://had.co.nz/ ___ R-packages mailing list r-packa...@r-project.org https://stat.ethz.ch/mailman/listinfo/r-packages __ 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 draw the legend about color from 3d picture
It may not help the original poster, but here's a solution based on what Greg said above: # Load plotrix library(plotrix) # Create a new layout to divide the graphics in 2, the first one (displaying the persp() graph) being 4 times larger than the second one (displying the legend) layout(matrix(c(1,2),1,2,byrow=T),widths=c(4,1)) # Setup the color palette, here for exemple a 20-step heatmap from white through yellow to red colors grey.colors - colorRampPalette( c(white, yellow, black) ) color - grey.colors(20) # Run the persp() command with right definition of facets and colors zfacet - z[-1,-1] + z[-1,-ncol(z)] + z[-nrow(z),-1] + z[-nrow(z),-ncol(z)] persp(x, y, z, col=color[cut(zfacet, nbcol)], theta=...) # Stetup and display the legend, col.labels - c(0.0,0.2,0.4,0.6,0.8,1.0) # For z values between 0 and 1 plot(0:10,type=n,axes=FALSE,xlab=,ylab=) # Blank plot required for the legend to be added to color.legend(1,1.5,9,8,col.labels,color,gradient=y,align=rb) -- View this message in context: http://r.789695.n4.nabble.com/how-to-draw-the-legend-about-color-from-3d-picture-tp866475p2281321.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] plotmath vector problem; full program enclosed
On Wed, Jul 7, 2010 at 5:41 AM, Duncan Murdoch murdoch.dun...@gmail.com wrote:
You want as.expression(b1), not expression(b1). The latter means
the
expression consisting of the symbol b1. The former means take the
object
stored in b1, and convert it to an expression..
Thanks to Duncan and Allen, who pointed out that I was not even
reading my own code carefully. I apologize for trying your patience.
Before I stop swinging at this one, I still want to bother everybody
about one thing, which really was the original question, but I did not
know the words to ask it.
The full code below is a working example that works, but I don't
understand why. Focus on these two commands that produce 2 axes. Both
produce the desired output, but, as far as I can see, they should not!
1:
axis(1, line=6, at=mu+dividers*sigma,
labels=as.expression(c(b1,b2,b3,b4,b5), padj=-1))
2:
axis(1, line=9, at=mu+dividers*sigma,
labels=c(as.expression(b1),b2,b3,b4,b5), padj=-1)
This second one shouldn't work, I think.
It has as.expression on only the first element, and yet they all come
out right. Is there a spill over effect?
My original question should not have asked why b1 does not print
correctly when I do this:
axis(1, line=9, at=mu+dividers*sigma,
labels=c(expression(b1),b2,b3,b4,b5), padj=-1)
but the correct question should have been why do b2, b3, b4 , and b5
get processed properly into plot math even though they are not
expressions??
???
pj
### Filename: plotMathProblem.R
### Paul Johnson July 7, 2010
### email me paulj...@ku.edu
sigma - 10.0
mu - 4.0
myx - seq( mu - 3.5*sigma, mu+ 3.5*sigma, length.out=500)
myDensity - dnorm(myx,mean=mu,sd=sigma)
### xpd needed to allow writing outside strict box of graph
### Need big bottom margin to add several x axes
par(xpd=TRUE, ps=10, mar=c(18,2,2,2))
plot(myx, myDensity, type=l, xlab=, ylab=Probability Density ,
main=myTitle1, axes=FALSE)
axis(2, pos= mu - 3.6*sigma)
axis(1, pos=0)
lines(c(myx[1],myx[length(myx)]),c(0,0)) ### closes off axes
addInteriorLine - function(x, m, sd){
for (i in 1:(length(x))){
lines( c(x[i],x[i]), c(0, dnorm(x[i],m=m,sd=sd)), lty= 14, lwd=.2)
}
}
dividers - c(qnorm(0.025), -1, 0, 1, qnorm(0.975))
addInteriorLine(mu+sigma*dividers, mu,sigma)
# bquote creates an expression that text plotters can use
t1 - bquote( mu== .(mu))
mtext(bquote( mu == .(mu)), 1, at=mu, line=-1)
addInteriorLabel - function(pos1, pos2, m, s){
area - abs(100*( pnorm(m+pos1*s,m,s)-pnorm(m+pos2*s, m,s)))
mid - m+0.5*(pos1+pos2)*s
text(mid, 0.5*dnorm(mid,m,s),label=paste(round(area,2),%))
}
addInteriorLabel(dividers[1],dividers[2], mu, sigma)
addInteriorLabel(dividers[2],dividers[3], mu, sigma)
addInteriorLabel(dividers[3],dividers[4], mu, sigma)
addInteriorLabel(dividers[4],dividers[5], mu, sigma)
b1 - substitute( mu - d*sigma, list(d=round(dividers[1],2)) )
b2 - substitute( mu - sigma )
b3 - substitute( mu )
b4 - substitute( mu + sigma )
b5 - substitute( mu + d*sigma, list(d=round(dividers[5],2)) )
axis(1, line=4, at=mu+dividers*sigma, labels=c(b1,b2,b3,b4,b5), padj=-1)
axis(1, line=6, at=mu+dividers*sigma,
labels=as.expression(c(b1,b2,b3,b4,b5), padj=-1))
axis(1, line=9, at=mu+dividers*sigma,
labels=c(as.expression(b1),b2,b3,b4,b5), padj=-1)
--
Paul E. Johnson
Professor, Political Science
1541 Lilac Lane, Room 504
University of Kansas
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[R] LondonR July Meeting
I am pleased to announce to agenda for next weeks LondonR meeting:
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Date: Tuesday 13th July 2010
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Agenda
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November 2010.
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[R] Weired problem when passing arguments using ...?
Hello All,
I'm trying to pass the argument col.names to write.csv using '...'.
But I got the following warnings. Maybe it is very simple. But I'm not
sure what I am wrong. Could you please help point to me what the
problem is?
#
fun=function(x, ...) {
fr=parent.frame()
tmp=get(x, envir=fr)
write.csv(
tmp
, file=paste(x, '.csv', sep='')
, ...
)
}
f=data.frame(x=1:10,y=letters[1:10])
fun('f', col.names=F)
fun('f', col.names=F)
Warning message:
In write.csv(tmp, file = paste(x, .csv, sep = ), ...) :
attempt to set 'col.names' ignored
--
Tom
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[R] how to process this in R
Here are what i am going to accomplish: I have 400 files named as xxx.txt. the content of the file looks like the following: namecount 1. aaa 100 2. bbb2000 3. ccc300 4. ddd 3000 more that 1000 rows in each files. these are the areas i need help: 1. how can i only read in the files with the string patterns ggg or fff as part of the file names? for instance, I only need the file names with the ggg or fff in it x_ggg_y_1.txt _fff__xxx.txt i don't need to read in the files, such as _aaa_.txt 2.how cam rename the files: for instance: x_ggg_y_1.txt==changed to ggg1a.txt 3.after the files read in, how can i only keep the rows with the aaa and bbb, everything elses show be removed from the files, but the files still remain the same file name? for instance, in the x_ggg_y_1.txt file, it shouls looks like: namecount 1. aaa100 2. bbb2000 3. aaa300 4. bbb400 Thanks so lot, I am very new to R, I am looking forward to any helps from you. -- View this message in context: http://r.789695.n4.nabble.com/how-to-process-this-in-R-tp2281283p2281283.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] Batch files process and String parsing
Here are what i am going to accomplish: I have 400 files named as xxx.txt. the content of the file looks like the following: namecount 1. aaa 100 2. bbb2000 3. ccc300 4. ddd 3000 more that 1000 rows in each files. these are the areas i need help: 1. how can i only read in the files with the string patterns ggg or fff as part of the file names? for instance, I only need the file names with the ggg or fff in it x_ggg_y_1.txt _fff__xxx.txt i don't need to read in the files, such as _aaa_.txt 2.how cam rename the files: for instance: x_ggg_y_1.txt==changed to ggg1a.txt 3.after the files read in, how can i only keep the rows with the aaa and bbb, everything elses show be removed from the files, but the files still remain the same file name? for instance, in the x_ggg_y_1.txt file, it shouls looks like: namecount 1. aaa100 2. bbb2000 3. aaa300 4. bbb400 Thanks so lot, I am very new to R, I am looking forward to any helps from you. -- View this message in context: http://r.789695.n4.nabble.com/Batch-files-process-and-String-parsing-tp2281208p2281208.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] forcing a zero level in contr.sum
I need to use contr.sum and observe that some levels are not statistically different from the overall mean of zero. What is the proper way of forcing the zero estimate? It seems the column corresponding to that level should become a column of zeros. Is there a way to achieve that without me constructing the design matrix? Thank you. Stephen Bond [[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] xyplot of function only shows diagonal plots
I expected this script to show nine panels, each with a plot of the
function.
But when I run it, only the diagonal panels have content.
Executing print(trel_1) shows what I expected in the upper left corner.
I am using R ver. 2.11.0 and lattice ver. 0.18-8 on Windows XP under
Eclipse and StatET.
I have searched the documentation, but found no answer. What am I missing?
--- simple.R ---
# Test xyplot layout for a function
#
require(lattice)
f - function(x,a,b){return( a * sin(x)+ b)}
trellis.par.set(theme = canonical.theme(Windows))
x - seq(-4,4, by=0.1)
# case 1x1
a - c( 1 )
b - c( 3 )
trel_1 - xyplot(f(x,a,b) ~ x | a * b, , type = l,
main = 1x1, ylab = f(x), xlab = x,
xlim = c(-6,6), ylim = c(-6,6), )
# case 3x3
a - c(1,2,3)
b - c(1,2,3)
trel_3 - xyplot(f(x,a,b) ~ x | a * b, , type = l,
main = 3x3, ylab = f(x), xlab = x,
xlim = c(-6,6), ylim = c(-6,6), )
print(trel_3)
#print(trel_1)
--- end of simple.R ---
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