Pete
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Hello,
Have 15 RedHat EL6 workstations patched to current.
Over the weekend the kernel was patched to 6.7 and the xorg-x11-server
and our R-3.1.2 will
not open a xterm window.
It appears to select a portion of the screen and lock onto it. This
section can be moved
dimnik wrote
thank you for your answer.Yes,that sounds right.I thought the same thing
but the problem is how can i generalize the command for every vector of
numbers not only for the specific example?not only for c(1,2),c(0.1,0.8).
2015-01-04 0:45 GMT+00:00 Pete Brecknock [via R]
ml-node
, the multivariate version of sapply?
Based on your example ...
mapply(function(x,y) rpois(x,y), c(1,2),c(0.1,0.8))
HTH
Pete
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I have 3 xts objects: test, cond1, cond2
You can download here:
https://dl.dropboxusercontent.com/u/102669/obj.rar
My problem is very simple.
test [ cond1 cond2] = NA THIS WORKS
test [ cond1 cond2] = -test [ cond1 cond2] THIS DOESN'T WORKS
Why?
My objective
30 36 2
9 9 21 35 2
HTH
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, ylim=c(min(a,b,d),max(a,b,d)))
lines(c,b, type=o,col=blue)
lines(c,d, type=o,col=green)
HTH
Pete
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Pete Brecknock wrote
Hi
The code below plots a stacked barchart.
I would like to overlay on this chart a circular plotting character at the
sum of the bars for each month. The plotted characters should be joined
with a line.
So, for 1/1/2014, I would like to see a point at 200 (-1000
suggestions?
Thanks in advance
Krishia
Is this what you are looking for?
# Create example matrice
m - matrix(c(1,2,3,4,5,6,7,8,9,10,11,12), nrow=4, byrow=TRUE)
# Create vector
v - c(t(m))
HTH
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-20 2013-10-15
2 2 2013-11-15 2013-11-20 2013-11-17
3 3 2013-12-25 2013-12-30 2013-12-27
HTH
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[,-1])
ind - as.yearqtr(names(df)[-1])
z - zoo(d,ind)
# Plot
plot(z, plot.type=single, col=1:5, lwd=2)
legend(topleft,legend=c(City1,City2,City3,City4,City5),lty=1,
lwd=2, col=1:5)
HTH
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string holding the contents of the different elements
of the list.
Any thoughts greatly appreciated.
Pete
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Thanks Brian. Perfect.
Brian Diggs wrote
On 7/19/2013 12:54 PM, Pete Brecknock wrote:
Hi
I am trying to add the contents of the list myList to a new column z
in
the data frame myDataframe
myList - list(c(A1,B1), c(A2,B2,C2), c(A3,B3))
myDataframe - data.frame(x=c(1,2,3), y=c(R,S,T
,function(x) x[order(orig[2,])]))
orig
[,1] [,2] [,3]
[1,] 10 20 30
[2,]312
new
[,1] [,2] [,3]
[1,] 20 30 10
[2,]123
HTH
Pete
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2.798235e-05 9.17413e-05 9.209191e-05
subdivisions 32 208 91
message OK OKOK
call Expression Expression Expression
HTH
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I have this little data.frame
http://dl.dropbox.com/u/102669/nanotna.rdata
Two column contains NA, so the best thing to do is use na.locf function (with
fromLast = T)
But locf function doesn't work because NA in my data.frame are not recognized as
real NA.
Is there a way to substitute fake NA
Hi to all, i'm new to R
I have an xts object.
Can i find:
a) how many NA are in my object ?
b) eventually where (in which line) they are
Thank you
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PLEASE do read the
tbl[tbl$ShirtSize,]
# Reorder Factor
tbl$ShirtSize = factor(tbl$ShirtSize, levels=c(S,M,L,XL,XXL))
# ShirtSize Order by Size
tbl[tbl$ShirtSize,]
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...
NAM = c(1,2,3,4,5,6,7,8,9)
ifelse(NAM=7,1,0)
# or
ifelse(NAM %in% c(7,8,9),1,0)
HTH
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))
10% 90%
12 29
But this is for the entire 4000 rows, when I need it to be for each YEAR.
Is there no way to use a by argument in the quantile function?
Thanks for any help you can provide.
David
check out
?aggregate or ?by should be of help
HTH
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bradleyd wrote
Thanks for your help Pete. I can almost get it to work with;
by(day,year,quantile)
but this only gives me 0% 25% 50% 75% 100%, not the ones I'm looking
for, 10% and 90%.
I have tried;
by(day,year,quantile(c(0.1, 0.9))) but this is rejected by
Error in FUN(X[[1L
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
how about
as.POSIXlt(2008-01-01 02:30, format=%Y-%m-%d %H:%M)
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,data$year,day,c(0.1,0.9))
*
was correct in that it calculated the quantile values as intended, but I
don't know how to then calculate the mean TEMP and IBI values encompasses
within those quantiles.
Thanks again,
David
have a look at trim argument of the mean function
?mean
Pete
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bradleyd wrote
Thanks Pete. The TRIM argument in the MEAN function tells me how to trim
off decimal points, but I am lost as to how to append the mean values of
TEMP and IBI between the 10% and 90% quantiles of DAY in each YEAR.
DAY is the julian date that an event occurred in certain years
(100,200,300,400))
# Ouput Data
outputData - do.call(myfun, inputData)
print(outputData)
HTH
Pete
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?
ticket.ns-c(1,1,1,1,2,5,5,10,10,10)
draw=NULL
for (i in 1:25){
draw[i] - sum(sample(ticket.ns,40,replace=TRUE))
}
print(draw)
HTH
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, order=c(1,1)) to
mod = arima(epsi, order=c(1,0,1))
You can extract the parameters of interest using
a= mod$coef[ar1]
b= mod$coef[ma1]
c= mod$coef[intercept]
HTH
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Sent from
,yellow and then
either 1 or 0 put in the relevant row.
Thanks
maybe model.matrix will help
# d is my understanding of your data
d-factor(c(red,green,red,blue,green,yellow,red))
model.matrix(~d -1)
HTH
Pete
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)
plot(rnorm(100),1:100)
plot(rnorm(100),1:100)
plot(rnorm(100),1:100)
# Title
title(MY TITLE, outer = TRUE, cex = 1.5, adj=0, col=blue, font=2)
Thanks for any pointers
Pete
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David Winsemius wrote
On Jan 15, 2013, at 2:49 PM, Pete Brecknock wrote:
Any recommendations for how I can embed my title below in a single red
strip/box across the plot area in the outer margin?
I would like to avoid the color appearing in any other area defined by
the
oma.
The code
David Winsemius wrote
On Jan 15, 2013, at 3:25 PM, Pete Brecknock wrote:
David Winsemius wrote
On Jan 15, 2013, at 2:49 PM, Pete Brecknock wrote:
Any recommendations for how I can embed my title below in a single red
strip/box across the plot area in the outer margin?
I would like
) c(mean=mean(x), sd=sd(x
# print(a) output
Sepal.Length Sepal.Width Petal.Length Petal.Width
mean5.843 3.057 3.758000 1.199
sd 0.8280661 0.4358663 1.765298 0.7622377
HTH
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'by' must specify uniquely valid column(s)
how about ...
#Data
d1-data.frame(id=c(9,8,6,4,4,3,1))
d2-data.frame(id=c(9,8,6),age=c(46,56,52))
# Left Merge
d-merge(d1,d2,all.x=TRUE)
# Reorder
d[order(d$id,decreasing=TRUE),]
HTH
Pete
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do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
maybe ...
replicate(1, rnorm(50))
could work for you
HTH
Pete
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sampsize=5
i=0
y=matrix(rnorm(nsamples*sampsize,50,3),nrow=nsamples)
s=matrix(NA,10,5)
for(i in 1:10){
s[i,]=sample(y,5,replace=T)
}
HTH
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),
IQR=IQR(x),
Max = max(x)))
HTH
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[,time]),]
Or have I misinterpreted your request?
HTH
Pete (B not D)
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Is there an easy way?
Thanks
How about ...
x - matrix(sample(1:8000),nrow=100)
colnames(x)- paste(Col,1:ncol(x),sep=)
apply(x,2,function(x) c(summary(x), sd=sd(x), IQR=IQR(x)))
HTH
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- rbind(part1, part2)
HTH
Pete
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, and maximum likelihood
estimation for a class of random effects joint models.
Best wishes, Pete.
Pete Philipson
Lecturer in Statistics
School of Computing, Engineering and Information Sciences
Northumbria University
email: pete.philip...@northumbria.ac.uk
[[alternative HTML version
:\\temp\\psw09.xls, :
cannot open URL 'http://ir.eia.gov/wpsr/psw09.xls'
In addition: Warning message:
In download.file(http://ir.eia.gov/wpsr/psw09.xls;, c:\\temp\\psw09.xls,
:
InternetOpenUrl failed: 'The operation timed out'
Thanks for any insights.
Pete Brecknock
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thought, methods with built in functions. Didn't
get any further.
Thank you very much.
How about ...
x - c(0,1,0,1,0,0,0,0)
sum(rle(x)$values)
HTH
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as a
variable, that I believe is what I need to solve this problem. Is that
possible?
Thanks a lot, hope to come back often in this forum,
G.
How about ...
A - c(368,369,370,371,393,394,395)
which.max(diff(A))
HTH
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(paste(Percent of data within 5 SD is ,pData(5),%, sep=)) # 97%
print(paste(Percent of data within 6 SD is ,pData(6),%, sep=)) # 98%
HTH
Pete
Ajata Paul wrote
How do you calculate the percentage of data within 2SD, 3SD, 4SD, 5SD, and
6SD of the mean? I used the following link as the data
How about using the legend function ...
plot(rnorm(100))
legend(60,2,100 Random Normal Draws,cex=.8,text.col=blue,
box.col=red,bg=yellow)
You can customize my effort to fit your specific needs
HTH
Pete
Henry wrote
New to R - rookie question.
I'm a mechanical engineer and enjoying using
Function
updates = v[match(u$name,v$enter),coeff]
u$coe = ifelse(!is.na(updates), updates, u$coe)
HTH
Pete
valerie wrote
Hi,
I have two data frames (u and v).
u
coe nam
1 0 Time
2 0Poten
3 0 AdvExp
4 0Share
5 0 Change
6 0 Accounts
7 0 Work
8
. I am not familiar with the diagram package or the examples you describe.
Why the desire to create a data frame? Why not just use a list?
HTH
Pete
dkStevens wrote
Thanks for the reply. Two things - I must have something missing because
copying and pasting your example gave me an error
2, 3
2 21 4, 5
3 3 NULL 2, 3
4 41 4, 5
HTH
Pete
dkStevens wrote
Group
It's unlikely I'm trying this the best way, but I'm trying to create a
data structure from
4 10 1
5 01 1
6 10 1
7 11 2
8 11 3
9 10 3
10 01 3
HTH
Pete
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)
y = rnorm(100)
plot(x,y)
grid()
HTH
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0.1934
Correlation p-value 0.
Tracking Error 0.7447
Active Premium 0.3694
Information Ratio 0.4960
Treynor Ratio 1.8885
HTH
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the solution by searching.
thanks so much
Jeremy
How about
frame2$age = frame1[match(frame2$ID, frame1$ID),age]
print(frame2)
ID age
1 Guy1 20
2 Guy2 33
HTH
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I am trying to rename column names in a dataframe within a function. I am
seeing an error (listed below) that I don't understand.
Would be grateful of an explanation of what I am doing wrong and how I
should rewrite the function to allow me to be able to rename my variables.
Thanks.
# Test
x
comn-c(abc, abc, abc, abc, abc, abc, xyz, xyz,xyz, xyz)
mi- c(1, 1,1, 2, 2, 2, 1, 1, 3, 3)
x- c(-0.0031, 0.0009, -0.007, 0.1929,0.0087, 0.099,-0.089, 0.005, -0.0078,
0.67)
df- data.frame(comn=comn, mi=mi, x=x)
# Aggregate Function
aggregate(df$x, by=list(df$comn,df$mi),FUN=sd)
HTH
Pete
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),
strptime(2010-12-25 06:00:00,%Y-%m-%d
%H:%M:%S)))
lapply(myListStartEnd,function(x) x[2]-x[1])
# Output
[[1]]
Time difference of 365.1042 days
[[2]]
Time difference of 23.8 days
HTH
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),
PW=c(6,7,8,NA,10,11))
# Extract Any Row Containing an NA
myNAs = temp[apply(temp,1,function(x) any(is.na(x))),]
HTH
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263
3 LTR_Unknown 6210 6423 6.080 4.650 9.081 213
4 LTR_Unknown 9658 10124 0.238 0.117 0.347 466
5 LTR_Unknown 14699 14894 3.545 3.625 2.116 195
6 LTR_Unknown 33201 33474 1.275 1.194 0.591 273
HTH
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=c(0,50))
hist(d2, ylim=c(0,50))
HTH
Pete
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)],]
print(ALL_RECORDS)
# Logical Records
TRUE_FALSE - df$ID==df$ID[duplicated(df$ID)]
print(TRUE_FALSE)
HTH
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Hi!
I need to perform this simple sampling function several hundred times:
x1=as.character(rnorm(1000, 100, 15))
x2=as.character(rnorm(1000, 150, 10))
y1=as.data.frame(x1,x2)
sample1=as.data.frame(sample(y1$x1, 12, replace = FALSE, prob = NULL))
sample1
write.table(sample1, sample1.txt, sep=
can any body help me ? Appreciate your help and thanks in advance.
Reza
Not pretty but this works ...
lst1 = list(c(0,1,2,3),c(0,1,5),c(2,3,4))
t(sapply(lst1, function(x) c(x,rep(0,4-length(x)
HTH
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region.
This is a simple problem, but I can't find anything wrong. How could I fix
it?
Maybe something like ...
# example data
d = data.frame(age=1:20, names=letters[1:20])
# 1. subset using [
d[d$age10 d$age16,]
# 2. subset using subset function
subset(d,d$age10 d$age16)
HTH
Pete
Thanks, Josh!
The index variable (time) was my problem. My R skills are too low! :)
Problem solved!
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Hi,
I need to reshape my dataframe from a long format to a wide format.
Unfortunately, I have a continuous date variable which gives me headaches.
Consider the following example:
id=c(034,034,016,016,016,340,340)
date=as.Date(c(1997-09-28, 1997-10-06, 1997-11-04, 2000-09-27,
2003-07-20,
Dear R-help,
I am using the chull function to create a convex
hull of a series of about 20,000 data points.
A
pain is temporary, glory is forever!
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Dear R-help,
I am using the chull function to create a convex hull of a series of about
20,000 data points.
A
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PLEASE
a1- c(1, 5, 8, 8, 9, 9, 14, 20, 3, 10, 10, 12, 6, 16, 7, 11, 13, 13, 17,
18, 2, 4, 15, 19)
Does anyone have a suggestion how to deal with this? Thank you in advance.
Lisa
is a1 = b[a] what you are looking for?
HTH
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d.z$V2 = d.z[,V1] *100
# recreate col names
names(d.z) = c(names(d),V2)
#-
Thanks
Pete
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Gabor Grothendieck wrote:
On Thu, May 26, 2011 at 7:35 PM, Pete Brecknock
lt;peter.breckn...@bp.comgt; wrote:
I have a zoo object that contains 2 time series named A-B and V1.
When I create a third series V2, the name of the A-B series is
changed
to A.B.
Although I could recreate
)
There is a nice document on CRAN that you may find useful.
http://cran.r-project.org/doc/contrib/Ricci-refcard-ts.pdf
HTH
Pete
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, stage=d$stage), FUN=mean)
# 2. using the by function
by(d$wage_accepted,list(d$period,d$stage),FUN=mean)
As for tutorial resources, I would recommend visiting CRAN at
http://cran.r-project.org/ ... follow the Manuals link on the left hand
side of the page.
HTH
Pete
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Gabor Grothendieck wrote:
On Tue, Dec 7, 2010 at 11:30 AM, Pete Pete lt;noxyp...@gmail.comgt;
wrote:
Hi,
consider the following two dataframes:
x1=c(232,3454,3455,342,13)
x2=c(1,1,1,0,0)
data1=data.frame(x1,x2)
y1=c(232,232,3454,3454,3455,342,13,13,13,13)
y2=c(E1,F3,F5,E1,E2,H4,F8,G3
... is the apply function what you are looking for?
A=matrix(1,2,4)
apply(A,1,sum)
HTH
Pete
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You could try the urca package
Also, I would maybe have a look a the CRAN Task View on computational
econometrics at http://cran.r-project.org/web/views/Econometrics.html
HTH
Pete
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)
x1 x2 x3 x4 x5 x6
[1,] NA NA 3 7 NA NA
[2,] 5 8 12 NA NA NA
[3,] 7 10 14 18 NA NA
[4,] 11 14 18 23 NA NA
[5,] 67 71 75 NA NA NA
HTH
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)
HTH
Pete
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to generate them on the fly.
Any help would be gratefully received.
Kind regards
Pete
###
# FUNCTION MYPLOT
###
myplot=function(i,j){
pushViewport(viewport
Apologies
I forgot to include that the reference lines should be for the y axis only.
Thanks.
Pete
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Thanks Gabor. I owe you again.
Kind regards
Pete
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Hi,
I am getting the following error when I try to run import rpy from the the
python IDE:
Traceback (most recent call last):
File stdin, line 1, in module
File /usr/lib/python2.6/dist-packages/rpy.py, line 134, in module
% RVERSION)
RuntimeError: No module named _rpy2122
RPy
For 5 years
set.seed = 1
d=data.frame(year=rep(2007:2011,each=12), month=rep(1:12,5), meanTemp =
rnorm(60,10,5))
meanByMonth = ave(d$meanTemp, d$month, FUN = mean)[7:9]
HTH
Pete
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You might want to take a look at Bob Muenchen's book R for SAS and SPSS
Users
There is an 80 page preview at .. http://rforsasandspssusers.com/
Additionally, there is lots of documentation for getting started with R on
the CRAN website http://cran.r-project.org/
HTH
Pete
)
There will be many other approaches.
HTH
Pete
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]])
ret = mapply(cbind,one,two)
colnames(ret) = paste(C,1:length(combs),sep=)
You will need to change seq_len(4) to seq_len(269) in the second line.
HTH
Pete
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Sent
After ordering the table of membership degrees , i must get the difference
between the first and second coloumns , between the first and second largest
membership degree of object i. This for K=2,K=3,to K.max=6.
This difference is multiplyed by the Crisp silhouette index vector (si). Too
it
After ordering the table of membership degrees , i must get the difference
between the first and second coloumns , between the first and second largest
membership degree of object i. This for K=2,K=3,to K.max=6.
This difference is multiplyed by the Crisp silhouette index vector (si). Too
it
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try
subset(D, D$x 5|D$y 5)
HTH
Pete
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After ordering the table of membership degrees , i must get the difference
between the first and second coloumns , between the first and second largest
membership degree of object i. This for K=2,K=3,to K.max=6.
This difference is multiplyed by the Crisp silhouette index vector (si). Too
it
[,sapply(d,is.numeric)],1,sum)
HTH
Pete
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) = Counts
print(counts)
HTH
Pete
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After ordering the table of membership degrees , i must get the difference
between the first and second coloumns , between the first and second largest
membership degree of object i. This for K=2,K=3,to K.max=6.
This difference is multiplyed by the Crisp silhouette index vector (si). Too
it
Dieter is correct, the lengths of the 2 series are different
Try
s = merge(s1,s2)
corr = cor(s[,Close.s1],s[,Close.s2],use=pairwise.complete.obs)
print(corr)
HTH
Pete
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Typing ? (no quotes) followed by a topic of interest will throw up the R
Help documentation.
1. Have a look at ?runif
2. Try ?subset and ?cumsum
3. Look at ?rle. Use in conjunction with cumsum and maybe ifelse.
HTH
Pete
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=zoo(d[,-1],order.by=as.Date(d$Date))
# generate rolling std devs
sd.z = rollapply(d.z,5,sd,align=right)
# you wanted a data frame
df = as.data.frame(merge(d.z,sd.z,all=TRUE))
HTH
Pete
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how about ...
j=NULL
for(i in 4: length(xyz$Close)) {
j[i] = sd(xyz$Close[i-3:i])
}
print(j)
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I believe there are two reasons why your code doesn't work
1. You should replace
sd(Close[i]:Close[(i-3)])
with
sd(Close[(i-3):i]).
This will ensure you select the appropriate obsevations to feed in the sd
function.
2. Per Ray's point above, you need to output the calculated value of sd
.
nums = 1:10
nums +c(1,2)
HTH
Pete
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, .). Making things easy for myself (it's
late), if you wish to simply ignore an NA the following would work
# sample data
y2=c(NA,1,2,3,4,5)
# ignore NA
ave(y2,is.na(y2),FUN=cumprod)
HTH
Pete
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