Displaying 20 results from an estimated 3000 matches similar to: "lm() on a matrix of zoo series"
2010 Aug 20
1
differecing a zoo series
A quick question
x <- as.yearmon(2000 + seq(0, 23)/12)
x
[1] "Jan 2000" "Feb 2000" "Mar 2000" "Apr 2000" "May 2000" "Jun 2000" "Jul
2000" "Aug 2000" "Sep 2000" "Oct 2000" "Nov 2000" "Dec 2000" "Jan 2001"
[14] "Feb 2001" "Mar 2001" "Apr
2011 Sep 12
5
Hourly data with zoo
I have date data as a numeric and hourly data in 0 to 2300 hours in a dataframe.
d <- rep(20110101,24)
h <- seq(from = 0, to = 2300, by = 100)
df <- data.frame(LST_DATE = d, LST_TIME = h, data = rnorm(24, 0, 1))
S <- chron(dates. = as.character(df$LST_DATE), times. =
paste(as.character(df$LST_TIME/100), ":0:0", sep = ""),
format =
2010 Aug 11
2
Sweeping a zoo series
Given a long zoo matrix, the goal is to "sweep" out a statistic from the
entire length of the
sequences.
longzoomatrix<-zoo(matrix(rnorm(720),ncol=6),as.yearmon(outer(1900,seq(0,length=120)/12,"+")))
cnames<-c(12345,23456,34567,45678,56789,67890)
colnames(longzoomatrix)<-cnames
longzoomatrix[1:24,]
12345 23456 34567 45678
2010 Apr 25
3
Noobie question on aggregate tapply and by
I have a 43MB dataframe ( 5 variables) and I'm trying to summarize subsets
of the data.
I've RTFM ( not very clear) and looked at a variety of samples but cant seem
to figure out
how to make these functions work.
A sample of what I want to do would be this:
ids<-seq(1,50)
years<-c(rep(5,10),rep(6,10),rep(7,10),rep(8,20))
2010 Jun 14
1
recursively Merging a list a zoo objects
The zoo package as a merge function which merges a set of zoo objects
result<-merge(zoo1,zoo2,...)
Assume your zoo objects are already collected in a list
# make a phony list to illustrate the situation. ( hat tip to david W for
constructing a list in a loop)
ddat <- as.list(rep("", 20))
ytd<-seq(3,14)
for(i in 1:20) {
+ ddat[[i]] <- zoo(data,ytd )
+ }
ddat
[[1]]
1 2
2011 May 27
2
help with barplot
Hi,
I'm really struggling with barplot
I have a data.frame with 3 columns. The first column represents an
"incident" type
The second column represents a "month"
The third column represents a "time"
Code for a sample data.frame
incidents <- rep(c('a','b','d','e'), each =25)
months <- rep(c(1,2), each =10)
times
2010 Jul 15
3
Summing over intervals
Given a matrix of MxN
want to take the means of rows in the following fashion
m<-matrix(seq(1,80),ncol=20, nrow=4)
result<-matrix(NA,nrow=4,ncol=20/5)
result[,1]<-apply(m[,1:5],1,mean)
result[,2]<-apply(m[,6:10],1,mean)
result[,3]<-apply(m[,11:15],1,mean)
result[,4]<-apply(m[,16:20],1,mean)
result
[,1] [,2] [,3] [,4]
[1,] 9 29 49 69
[2,] 10 30 50 70
2010 Jun 05
5
Matrix to Vector
Given a matrix of m*n, I want to reorder it as a vector, using a row major
transpose.
so:
> m<-matrix(seq(1,48),nrow=6,byrow=T)
> m
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
[1,] 1 2 3 4 5 6 7 8
[2,] 9 10 11 12 13 14 15 16
[3,] 17 18 19 20 21 22 23 24
[4,] 25 26 27 28 29 30 31 32
[5,] 33 34 35 36 37
2011 Jul 27
2
Elegant way to subtract matrix from array
there are really two related problems here
I have a 2D matrix
A <- matrix(1:100,nrow=20,ncol =5)
S <- matrix(1:10,nrow=2,ncol =5)
#I want to subtract S from A. so that S would be subtracted from the
first 2 rows of
#A, then the next two rows and so on.
#I have a the same problem with a 3D array
# where I want to subtract Q for every layer (1-10) in Z
# I thought I solved this one
2015 Aug 05
1
Samba4 not able to write to group writeable folder???
Hi...
With samba4 I sometimes feel like a bloody beginner even I use samba
since ages.
Miracle of the day: Users connecting using samba4 cannot write to group
writeable folders even they should be able to.
I upgraded one 3.6 fileserver to 4.2.3. PDC is always a samba 4.2.3
instance. I am using samba4 in classic PDC mode for a couple of reasons.
On my fileserver I have a folder called
2011 Mar 18
2
Understanding tryCatch
I've read the help and the archives on tryCatch but I'm still stuggling
trying to understand how it
works exactly and how I can use it to get the result I need.
I have a data.frame of urls which point to 11 .zip files. Basically I use
RCurl to get the list of
files from a ftp and then reduce that directory dump to the 11 zip files I
want to download.
Its easy enough to do that in a loop
2010 Aug 13
1
Bug in t.test?
Hello all,
due to unexplained differences between statistical results from
collaborators and our lab that arose in the same large proteomics
dataset we reevaluated the t.test() function. Here, we found a weird
behaviour that is also reproducible in the following small test
dataset:
Suppose, we have two vectors with numbers and some missing values
that refer to the same individuals and that
2012 Mar 20
2
Reshaping data from long to wide without a "timevar"
Hello All,
I was wondering if it's possible to reshape data from long to wide in R without using a "timevar". I've pasted some sample data below along with some code. The data are sorted by Subject and Drug. I want to transpose the Drug variable into multiple columns in alphabetical order.
My data have a variable called "RowNo" that functions almost like a
2005 Apr 22
1
Need help arranging the plot in different fashion than the default format
I want to change the way the plot is appearing in the barchart ..
testdata <-
2003 Sep 23
2
confusion about what to expect?
In playing around with data.frames (and wanting a simple, cheap way to
use the variable and case names in plots; but I've solved that with
some hacks, yech), I noticed the following behavior with subsetting.
testdata <- data.frame(matrix(1:20,nrow=4,ncol=5))
names(testdata) ## expect labels, get them
names(testdata[2,]) ## expect labels, get them
names(testdata[,2]) ## expect labels, but
2005 Apr 25
1
Need help with panel.segment..
Hi All,
For the following code, I'm not sure why the error bars are appearing horizontal. Can someone please tell me how to fix the problem.
testdata <-
2007 Sep 20
1
Bug with Cor(..., method='spearman") and by() (PR#9921)
I posted this on R help, and a few others responded indicating they too
were able to replicate the error as a function of missing data. I
believe this should not be the case and hence and reporting it here.
### Code provided on R-Help by Ivar Herfindal
# Simulate data
testdata <- cbind.data.frame(gr=3Drep(letters[1:4], each=3D5), =
aa=3Drnorm(20),
bb=3Drnorm(20))
# Introduce some missingness
2005 May 05
6
Need some quick help with lattice - barchart
For the following code below, the x-axis ticks are 1,2,3,4,5,6,7 when I was expection them to be 1,2,8,9,10,11,12. Please help me figure out where is the mistake.
library(lattice)
testdata <- as.data.frame(t(structure(c(
1,2005,9.24,6.18,634,
2,2005,8.65,6.05,96,
8,2004,6.81,6.51,16,
9,2004,9.0,7.29,8,
10,2004,8.84,6.18,524,
11,2004,8.54,6.35,579,
12,2004,9.97,6.3,614,
12,2005,8.75,5.84,32,
2007 May 13
2
extracting text contained in brackets ("[ ... ]") from a character string?
I have a text string that contains text within two brackets.
e.g. "testdata[3]" "testdata[-4]", "testdata[-4g]",
I wish to "extract" the string enclosed in brackets?
What is a good way to do this?
e.g.
fun(testdata[3]) = '3'
fun(testdata[-4g]) = '-4g'
---------------------------------
Moody friends. Drama queens. Your life?
2010 Mar 12
1
Length as fun.aggregate in cast function of reshape package: unexpected error
Dear Everyone,
I am having problems with use of the reshape package's cast function using length as an aggregating function.
Unexpectedly, I receive the error: 2 arguments passed to 'length' which requires 1
I don't understand this at all - the data I'm using is very simple, and appears almost identical to that used in the
ChickWeight example in the package. The problem can