Ok. Thanks Peter. It was my bad - typo. You caught it. Sorry everyone.
On Wed, Oct 16, 2013 at 10:03 AM, peter dalgaard wrote:
>
> On Oct 16, 2013, at 16:54 , tom soyer wrote:
>
> > Hi,
> >
> > pnorm(-1.53,0,1) under version 3.0.2 gives 0.05155075. I am pretty su
Hi,
pnorm(-1.53,0,1) under version 3.0.2 gives 0.05155075. I am pretty sure it
should be 0.063. Is there something wrong with this version of R?
I am using:
R version 3.0.2 (2013-09-25) -- "Frisbee Sailing"
Copyright (C) 2013 The R Foundation for Statistical Computing
Platform: i686-pc-linux-gnu
Hi,
I have two time series, y and x. Diff(y) and Diff(x) both show no
autocorrelation. But durbin.watson(lm(Diff(y)~lag(Diff(x),k=-4)) gives a DW
value of zero. How come the residule is autocorrelated while Diff(y) and
Diff(x) are not? Does anyone know if in my case a DW of zero indicates
serial c
Hi,
Does anyone know if the RMSE is one of the values provided by the lm model,
or do we have to calculate it by hand from the residuals?
Thanks,
--
Tom
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Hi,
I need to correct for ar(1) behavior of my residuals of my model. I noticed
that there are multiple gls models in R. I am wondering if anyone
has experience in choosing between gls models. For example, how
should one decide whether to use lm.gls in MASS, or gls in nlme for
correcting ar(1)? Do
Hi,
Does anyone know if R has a function that is similar to lag.plot but instead
of auto-correlation, it plots cross-correlation with lags?
Thanks,
--
Tom
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If the
> model I described isn't appropriate then you should look at Ch 5 of
> P&B to learn about the other varFunc classes.
>
> good luck,
>
> Kingsford
>
> ps - would you mind forwarding to r-help in case this others have the
> same question.
>
>
> On T
Hi,
I would like to use a weighted lm model to reduce heteroscendasticity. I am
wondering if the only way to generate the weights in R is through the
laborious process of trial and error by hand. Does anyone know if R has a
function that would automatically generate the weights need for lm?
Thank
Hi
I have a general statistics question on calculating confidence interval of
log transformed data.
I log transformed both x and y, regressed the transformed y on transformed
x: lm(log(y)~log(x)), and I get the following relationship:
log(y) = alpha + beta * log(x) with se as the standard error
Hi,
I am trying to find a solution in R for the following C++ code that allows
one to skip ahead in the loop:
for (x = 0; x <= 13; x++){
x=12;
cout << "Hello World";
}
Note that "Hello World" was printed only twice using this C++ loop. I
tried to do the same in R:
for(i in 1:13){
i=12
print
Hi,
Does anyone know if R has a built-in function that is similar to Excel's
NETWORKDAYS function? i.e., Returns the number of whole working days between
two dates. Working days exclude weekends.
Thanks,
--
Tom
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Hi,
Is it possible to diagonally fill a rectangle with a color gradient? I
noticed that the gradient.rect of plotrix could fill a rect either up and
down or from side to side. I am looking for something similar but fills
diagonally instead, e.g., from the upper left corner to the bottom right.
Doe
Hi,
I have a 2D chart that is divided into four quadrants, I, II, III, IV:
plot(1:10,ylim=c(0,10),xlim=c(0,10),type="n")
abline(v=5,h=5)
text(x=c(7.5,7.5,2.5,2.5),y=c(2.5,7.5,7.5,2.5),labels=c("I","II","III","IV"))
I would like to fill each quadrant with a background color unique to the
quadrant.
Hi,
Does anyone know how one could format numbers using 1000 separator in R? For
example, format 1000 as 1,000 and 10 as 100,000, etc.
Thanks,
--
Tom
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Hi,
I am currently using the following to formate numbers into percentages:
x=0.00112
paste(round(x*100,2),"%",sep="")
I am wondering if there is a built in R function that does the same. Does
anyone know?
Thanks,
--
Tom
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ss affect the global environment please give a
> code example.
>
> On Feb 6, 2008 12:11 PM, tom soyer <[EMAIL PROTECTED]> wrote:
> > Thanks Hardley. I see what you mean. You are right, I am not an expert
> in
> > oop AND I don't really know how R oo works, so certainly I sho
n't affect
the global environment. That was all I meant.
On 2/6/08, hadley wickham <[EMAIL PROTECTED]> wrote:
>
> On Feb 6, 2008 10:13 AM, tom soyer <[EMAIL PROTECTED]> wrote:
> > Thanks Gabor. I guess true oo encapsulation is not possible in R.
>
> Before
Thanks Gabor. I guess true oo encapsulation is not possible in R.
It seems that there is an IDE for S+ in Eclipse...
On 2/6/08, Gabor Grothendieck <[EMAIL PROTECTED]> wrote:
>
> On Feb 6, 2008 9:45 AM, tom soyer <[EMAIL PROTECTED]> wrote:
> > Thanks Gabor for illustra
d
> # mountainBike will inherit the bicycle method by default
> applyBrake <- function(x, decrement) UseMethod("applyBrake")
> applyBrake.bicycle <- function(x, decrement) { x$speed <- x$speed -
> decrement }
>
> # list the applyBrake methods available
> methods(apply
Hi,
I read section 5, oop, of the R lang doc, and I am still not sure I
understand how to build a class in R for oop. I thought that since I
understand the oop syntex of Java and VB, I am wondering if the R programmig
experts could help me out by comparing and contrasting the oop syntex in R
with
7303)), lag=12)
> >>
> >>ARCH test (univariate)
> >>
> >> data: Residual of y1 equation
> >> Chi-squared = 13.1483, df = 12, p-value = 0.3584
> >>
> >> Warning message:
> >> In VAR(s, p = 1, type = "const") :
> &
mn names supplied in y, using: y1, y2, y3, y4, y5, y6, y7, y8,
> y9, y10, y11, y12 , instead.
>
>
> TOM: What can you tell me about the warning message?
>
> Thanks for your help with this.
> Spencer Graves
>
> tom soyer wrote:
> > Spencer,
> &g
OK, it's no good. Here is the result:
> data(m.intc7303)
> archTest(log(1+as.numeric(m.intc7303)), lags=12)
ARCH test (univariate)
data: Residual of y1 equation
Chi-squared = 13.1483, df = 12, p-value = 0.3584
On 2/2/08, tom soyer <[EMAIL PROTECTED]> wrote:
>
because the expected value
> of the F distribution is close to 1 [d2/(d2-2), where d2 = denominator
> degrees of freedom; http://en.wikipedia.org/wiki/F-distribution], while
> the expected value for a chi-square is the number of degrees of freedom
>
> Unfortunately, I don
n
> example on p. 103 that could be used for a reference.
>
> Hope this helps.
> Spencer Graves
>
> tom soyer wrote:
> > Hi,
> >
> > Does anyone know if R has a Lagrange multiplier (LM) test for ARCH
> > effects for univariant time series?
>
Hi,
Does anyone know if R has a Lagrange multiplier (LM) test for ARCH
effects for univariant time series?
Thanks!
--
Tom
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Hi,
Does anyone know if there are formal tests for long vs short memory
processes? i.e., quantitative tests instead of visual examination
of corellograms produced by acf.
Thanks!
--
Tom
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Hi,
does anyone know how to simulate two seasonal data series that are
cointegrated?
Thanks!
--
Tom
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PLEASE do read the pos
Hi,
I was wondering if there is a test that would help one choose whether adf or
pp should be used. Would the shapiro.test work for this purpose?
Thanks!
--
Tom
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There is a "vanilla"
> options to start R, e.g.
>
> Rterm --vanilla
>
> that you might wanna try.
>
> /Henrik
>
>
>
> On Jan 26, 2008 3:54 PM, tom soyer <[EMAIL PROTECTED]> wrote:
> > Hi,
> >
> > My R just froze. I can't get it to do a
Hi,
My R just froze. I can't get it to do anything. It gives "Error: band value"
message to everything I type. Does anyone know if R has a safe mode that I
could check for errors and perform diagnostics? I am using R 2.6.1 on
Windows XP.
> ls()
Error: bad value
> search()
Error: bad value
> ?ls
Er
axt="n",type="n") #1st plot
> par(mar=c(0, 5.1, 0, 5.1))
> plot(y2,xaxt="n",type="n") #2nd plot
>
> #try to draw lines onto each plot on the screen
> lines(y2) #draws a line in the 2nd plot
> par(mfg=c(1, 1))
> lines(y1,col=2) #also
Hi,
Suppose I already have two plots on the same screen, and I want to draw
lines in each of them. Is that possible in R? It seems that once you have
two plots on the screen, you can only draw lines in the the last plot, never
the 1st. Here is what I mean:
#some data
y1=rnorm(1:3)
y2=rnorm(1:3)
Thanks Charles and Gabor! Sorry Charles, the numbers were wrong in my
example. You had the correct one.
On 1/22/08, Charles C. Berry <[EMAIL PROTECTED]> wrote:
>
> On Tue, 22 Jan 2008, tom soyer wrote:
>
> > Hi,
> >
> > I am trying to reproduce some function
Hi,
I am trying to reproduce some functionalities of Excel pivot table in R,
sadly, I couldn't figure out how to do it. I am wondering if this is even
possible in R. Does anyone know?
Here is an example:
year=rep(2003,16)
quarter=rep(1:4,each=4)
sales=1:16
company=rep(c("a","b","c","d"),4)
df=da
Thanks Marc and Gabor, I got it!
On 1/20/08, Marc Schwartz <[EMAIL PROTECTED]> wrote:
>
> tom soyer wrote:
> > Hi,
> >
> > I am trying to extract data from a ts object by month, e.g., extract
> Jan,
> > Feb, and Aug data from a monthly ts object. I tri
Hi,
I am trying to extract data from a ts object by month, e.g., extract Jan,
Feb, and Aug data from a monthly ts object. I tried the following but it
didn't work:
> xa=1:50
> ta=ts(xa,start=c(1990,1),frequency=12)
> ta[cycle(ta)==c(1,2)] # this method works but it's not what I want
[1] 1 2 13
Thanks Marc!
On 1/19/08, Marc Schwartz <[EMAIL PROTECTED]> wrote:
>
> tom soyer wrote:
> > Hi,
> >
> > I have a plot with type="o", or overstruck. Now I am trying to add the
> > legend, but I couldn't figure out how to show the overstruck ty
Hi,
I have a plot with type="o", or overstruck. Now I am trying to add the
legend, but I couldn't figure out how to show the overstruck type in the
legend. It seems that the legend only allows one to set lty. Does anyone
know how to show overstruck in the legend?
Thanks!
--
Tom
[[alter
Hi,
I have an AR(1) series, so I thought that the order of the series should be
1. A simple lm fit with one period lag predicts the series pretty well. But
when I tried ar, I got different orders: ar.mle selected order 6,
ar.burgselected order 14, and
ar.yw selected order 6. So I am wondering mayb
Thanks Richard. I am just trying to understand exactly what is R's arima
doing, and I am having a hard time. It seems that xreg is necessary to force
arima to include the constant term, but it appears that exactly how this is
done is not documented. If a series is not differenced, e.g. AR(1), then
Hi,
I am trying to understand exactly what xreg does in arima. The documentation
for xreg says:"xreg Optionally, a vector or matrix of external regressors,
which must have the same number of rows as x." What does this mean with
regard to the action of xreg in arima?
Apparently somehow xreg made t
Thanks Achim. Data manipulation in zoo and coerce back to ts. Sounds good!
On 1/10/08, Achim Zeileis <[EMAIL PROTECTED]> wrote:
>
> On Thu, 10 Jan 2008, tom soyer wrote:
>
> > Hi,
> >
> > I have two questions about ts.
> >
> > (1) How do I subset a ts
Hi,
I have two questions about ts.
(1) How do I subset a ts object and still preserve the time index? for
example:
> x=ts(1:10, frequency = 4, start = c(1959, 2)) # the ts object
> x
Qtr1 Qtr2 Qtr3 Qtr4
1959 123
19604567
196189 10
I don't want the 1
Thanks Paul. I will try it.
On 1/9/08, Paul Smith <[EMAIL PROTECTED]> wrote:
>
> On Jan 9, 2008 2:13 PM, tom soyer <[EMAIL PROTECTED]> wrote:
> > Thanks Paul. I thought constrOptim does not do equality. I will check
> again.
>
> Indeed, constrOptim does not do
r the
smoothing constant that instead of minimizes the root mean square error
(RMSE), it targets a particular value of RMSE. Is this possible in R?
On 1/9/08, Paul Smith <[EMAIL PROTECTED]> wrote:
>
> On Jan 9, 2008 1:14 PM, tom soyer <[EMAIL PROTECTED]> wrote:
> > Th
ead of min and max? Is there a function like that in R?
Thanks!
On 1/9/08, Paul Smith <[EMAIL PROTECTED]> wrote:
>
> On Jan 9, 2008 4:01 AM, tom soyer <[EMAIL PROTECTED]> wrote:
> > I noticed that R has a few bound-constrained nonlinear min and max
> solvers,
> > such
Hi,
I noticed that R has a few bound-constrained nonlinear min and max solvers,
such as optim, nlm, etc. But I could not find a constrained min and max
solver that is not LP. Does this mean R do not have this capability? It is
hard to believe that R may not be as advanced as Excel in certain areas
Thanks Gabor!!
On 1/6/08, Gabor Grothendieck <[EMAIL PROTECTED]> wrote:
>
> On Jan 6, 2008 5:17 PM, tom soyer <[EMAIL PROTECTED]> wrote:
> > Hi,
> >
> > I have a ts object with a frequency of 4, i.e., quarterly data, and I
> would
> > like to calcul
Hi,
I have a ts object with a frequency of 4, i.e., quarterly data, and I would
like to calculate the mean for each quarter. So for example:
> ts.data=ts(1:20,start=c(1984,2),frequency=4)
> ts.data
Qtr1 Qtr2 Qtr3 Qtr4
1984 123
19854567
198689 10 11
Hi,
I just discovered decompose() and stl(), both are very nice! I am wondering
if R also has a function that calculates the seasonal index, or make the
seasonal adjustment directly using the results generated from either
decompose() or stl(). It seems that there should be one, but I couldn't find
oops, it should be: rms=(sum((x-mean(x))^2)/(length(x)-1))^(1/2)
On 1/3/08, tom soyer <[EMAIL PROTECTED]> wrote:
>
> Thanks Jim. Yes it does... but I calculated the root mean square (rms),
> and couldn't reproduce the result without multiplying the rms by 2. I don't
[1,] -1.6
> [2,] 0.4
> [3,] -0.6
> [4,] 0.4
> [5,] 1.4
> attr(,"scaled:center")
> [1] 3.6
> >
>
> Default is to performance scaling: "If scale is TRUE then scaling is
> done by dividing the (centered) columns of x by their
> root-mean-square, and if sc
Never mind. I forgot the scale= parameter.
On 1/3/08, tom soyer <[EMAIL PROTECTED]> wrote:
>
> Hi,
>
> The documentation for scale() states:"If center is TRUE then centering is
> done by subtracting the column means (omitting NAs) of x from their
> corresponding col
Hi,
The documentation for scale() states:"If center is TRUE then centering is
done by subtracting the column means (omitting NAs) of x from their
corresponding columns". But it seems that R is subtracting something else
instead of the column mean:
> x=c(2,4,3,4,5)
> mean(x)
[1] 3.6
> x-mean(x)
[1
par(mar=c(4,4,0,2))
> plot(rnorm(1:3))
>
> On Dec 31, 2007 1:16 PM, tom soyer <[EMAIL PROTECTED]> wrote:
> > Thanks Jim! It seems layout() is necessary in addition to mar=. I have a
> > follow up question: is there a way to specify the height of each chart
> so
&
Sorry Gabor, you are right, using mar= alone is enough to do the stacking. I
was wrong.
On 12/31/07, tom soyer <[EMAIL PROTECTED]> wrote:
>
> Thanks Gabor. mar= and oma= by themselves won't be able to do it. layout()
> is necessary per Jim's post. But now I am stuck wi
ECTED]> wrote:
>
> Check out:
>
> http://research.stowers-institute.org/efg/R/Graphics/Basics/mar-oma/index.htm
>
> On Dec 31, 2007 11:53 AM, tom soyer <[EMAIL PROTECTED]> wrote:
> > Hi,
> >
> > I tried to stack two charts on top of each other using the following
> &
))
> plot(rnorm(1:3))
>
> On Dec 31, 2007 11:53 AM, tom soyer <[EMAIL PROTECTED]> wrote:
> > Hi,
> >
> > I tried to stack two charts on top of each other using the following
> > R functions:
> >
> > par(mfrow=c(2,1))
> > plot(rnorm(1:3),xaxt=
Hi,
I tried to stack two charts on top of each other using the following
R functions:
par(mfrow=c(2,1))
plot(rnorm(1:3),xaxt="n",xlab="")
plot(rnorm(1:3))
This created two charts, one on top of the other, but there is too much
space between them. Does anyone know how to elimiate the space in bet
Thanks Gabor!
Also, can you give an example of customizing the major tick marks on the
x-axis and adding minor tick marks between major tick marks?
On 12/26/07, Gabor Grothendieck <[EMAIL PROTECTED]> wrote:
>
> On Dec 26, 2007 10:55 PM, tom soyer <[EMAIL PROTECTED]> wrot
Hi,
I have been having very good results using plot.zoo to chart time series
data. But I have three questions about plot.zoo and I am wondering if anyone
knows the answers.
(1) when I tried to use semi-log scale, via log="y", R issued a warning,
although it looked like plot.zoo plotted in semi-lo
t;
> > agg2=aggregate(df[,2],list(year=years(df[,1]),month=fmonth),sum)
> > levels(agg2$month) # even if a factor with levels in the correct order
> is supplied, aggregate(), sortsthe levels by alphabet regardless.
> [1] "Jan" "Feb" "Mar" "Apr&
Hi,
I am constructing a contingency table using xtabs. The function works great:
mo
yr Sep Oct Nov Dec
1950 -7.164486e-02 3.152674e-02 -1.283389e-02 1.570382e-01
1951 3.054293e-02 4.665234e-02 -2.445499e-04 8.720204e-02
1952 3.937034e-0
Hi,
I am using aggregate() to add up groups of data according to year and month.
It seems that the function aggregate() automatically sorts the levels of
factors of the grouping elements, even if the order of the levels of factors
is supplied. I am wondering if this is a bug, or if I missed someth
Hi,
I am trying to count weekday of the month using R. For example, 1/4/2001
is the 4th weekday of Jan, and 1/5/2001 is the 5th weekday of the month, and
1/8/2001 is the 6th weekday of the month, etc. I get as far as extracting
the weekdays from a sequence of dates (see below). But I have not yet
>
> --
> Mango Solutions
> data analysis that delivers
> Tel: +44(0) 1249 467 467
> Mob: +44(0) 1249 467 468
> Fax: +44(0) 7813 526 123
>
>
>
> -Original Message-
> From: [EMAIL PROTECTED] on behalf of tom soyer
> Sent: Tue 11/12/2007 11:17
> To: r-
Hi,
I am doing a calculation on a long series using a For Loop. Here is an
example of the calculation:
accumulate=function(x){
y=0
z=0
for(i in 1:length(x)){
y=y+x[i]
z=c(z,y)
}
return(z[2:length(z)])
}
> x=c(1:10)
> x
[1] 1 2 3 4 5 6 7 8 9 10
> accumulate(x)
[1] 1 3 6 10
m I
> missunderstanding what Excel's percentrank does ?
>
> aa <- rnorm(25); aa # data vector
> percentrank <- function(x) {
> var <- sort(x)
> p.rank <- 1:length(var)/length(var)*100
> dd <- cbind(var,p.rank)
> }
> pr <- percentrank(aa); pr
>
&
Hi,
Does anyone know if R has a built-in function that is equvalent to Excel's
percentrank, i.e., returns the rank of a value in a data set as a percentage
of the data set?
Thanks,
--
Tom
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Thanks Peter!
On 11/30/07, Peter Dalgaard <[EMAIL PROTECTED]> wrote:
>
> tom soyer wrote:
> > Hi,
> >
> > I am trying to find a function in R that would calculate the critical
> value
> > of t for a given probability and df. For example, if p=5% and df=2
Hi,
I am trying to find a function in R that would calculate the critical value
of t for a given probability and df. For example, if p=5% and df=20, then
does R have a function to calculate the t value? I tried:
> dt(0.05,df=20)
[1] 0.3934718
That doesn't look right to me. Could someone tell me
Hi,
I am using lm() for regression analysis of my data set. My regression
results look pretty good, i.e., the coefficient is significant and the p
value is much less than 0.05. But when I checked the residuals, both using
qqnorm() and hist(), the distribution does not look normal. It looks like
t
Merge worked! Thanks!!!
On 11/28/07, Matthew Keller <[EMAIL PROTECTED]> wrote:
>
> Tom,
>
> Check out ?merge. Does exactly what you need
>
> Matt
>
> On Nov 28, 2007 11:27 AM, tom soyer <[EMAIL PROTECTED]> wrote:
> > Hi,
> >
> > I have two s
Hi,
I have two sets of data that I would like to put into a data frame. But
since they have different length, I am not sure how to do this. Here is an
example of my data:
data set one:
date growth
1/1/2007 10
1/2/2007 10.2
1/3/2007 10.4
1/4/2007 10.6
data set two:
date g
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