Displaying 20 results from an estimated 10000 matches similar to: "Plotting multiple xts/zoo time series on a single plot."
2011 Jan 22
1
Plotting by factor with xts
Hi all,
I've got an xts time series of stock symbols and closing prices.
> head(x)
symbol close
2010-01-04 "AFB" "13.46"
2010-01-04 "AKP" "12.80"
2010-01-04 "APX" " 8.78"
2010-01-04 "AYN" "13.15"
2010-01-04 "BAF" "13.50"
2010-01-04 "BBF" "12.86"
>
2012 Dec 18
1
How to draw frequency domain plot with xts time series data
Hello,
I'd like to convert the below time-series data with fft or wavelet related function and plot it.
Could you let me know
1. How to convert xts data frame format to list format ?
2. How to plot fft or wavelet diagram ?
Here is the data :
> class(zc)
[1] "xts" "zoo"
> str(zc)
An ‘xts’ object from (10/15/12 09:00:00) to (10/15/12 15:15:00)
2012 Mar 04
1
Store vectors as values in xts time-series object
Hi R programmers,
I have stumbled across what seems a very simple problem. My goal is to
create a xts time series object which contains vectors as values. In
other words, I try to create something like this:
2009-01-01 => c('aa', 'bb', 'dd')
...
2010-02-01 => c('mm')
I have figured out parts of separately. Here's what works (new xts
time-series with
2011 Oct 03
1
xts/time-series and plot questions...
Hello,
I'm a complete newbie to R. Spent this past weekend reading The Art of R Programming, The R Cookbook, the language spec, Wikis and FAQs. I sort-of have my head around R; the dizzying selection of libraries, packages, etc? Not really. I've probably missed or failed to understand something...
I have very a simple data set. Two years (ish) of temperature data, collected and
2017 Oct 06
2
Time series: xts/zoo object at annual (yearly) frequency
Hi,
I'd like to make a time series at an annual frequency.
> a<-xts(x=c(2,4,5), order.by=c("1991","1992","1993"))
Error in xts(x = c(2, 4, 5), order.by = c("1991", "1992", "1993")) :
order.by requires an appropriate time-based object
> a<-xts(x=c(2,4,5), order.by=1991:1993)
Error in xts(x = c(2, 4, 5), order.by =
2011 Mar 02
1
Create a zoo/xts Time Series with Millisecond jumps
Is there a easy way to create the time index for a zoo/xts object for every
100 milliseconds.
eg. time Index would be:
10:00:00:100
10:00:00:200
10:00:00:300
10:00:00:400
I am looking to build an empty zoo/xts object with time index from 10am to
3pm, index jumps by 100ms each row.
Thanks,
Chris
--
View this message in context:
2011 Jul 30
1
Plot.xts - how to change the x-axis labels to show weekly labels.
Dear R-users
I am new to R and struggling not to bother the list with silly questions.
I read the documentation on xts and searched for some examples over the
internet on how to use plot.xts.
The xts object is as follows
dataxts : An 'xts' object from 2010-06-27 to 2010-08-05 containing:
Data: num [1:56161, 1:14] 74 74.2 74.2 74.1 73.9 ...
Indexed by objects of
2011 Jan 04
1
XTS : merge.xts seems to have problem with character vectors
Hi,
Please can you tell me what I am doing wrong. When trying to merge two xts
objects, one of which has multiple character vectors for columns...I am just
getting NAs.
> str(t)
POSIXct[1:1], format: "2011-01-04 11:45:37"
> y2 = xts(matrix(c(letters[1:10]),5), order.by=as.POSIXct(c(t + 1:5)))
> names(y2) = c(1,2)
> y2
1 2
2011-01-04 11:45:38
2012 May 29
2
Converting to XTS loses data.frame structure
Hello,
I noticed something odd when working with data frames and xts objects.
If I read in a CSV file, R creates a nice data.frame. This works well.
If I then convert to an XTS object, I see that all the values in the data are now quoted. My data is a mix of numeric and character. This is usually seen when converting a data.frame to a matrix, as R will treat all the data as the same class.
2013 May 13
1
Math problem with xts objects
Hello,
I coming across a strange problem doing math on an xts object.
If I have an xts object of stock prices (perhaps 5 minute bars of open, high, low,close) and want to do some math, the results fail.
For example:
d$close[10] - d$open[10] works perfectly
d$close[10] - d$open[9] fails. I just get an answer of "numeric(0) Index: numeric(0)".
My guess is that xts is breaking
2011 Jul 17
1
FOMULATING TIME SERIES DATA FROM DATA FRAME
I am estimating Value at Risk using PerfomanceAnalytics package. The?variables are stored in a data frame. I formated the data variables using zoo() and as.xtx() but it is not working. The working example is below.
##########################################################?
reguire(zoo)
require(PerformanceAnalytics)
reguire(xts)
?
year<- c(1991-12-30, 1992-12-30, 1993-12-30, 1994-12-30)
R1
2011 Mar 04
4
xts POSIXct index format
Hi,
I cannot figure out how to change the index format when displaying POSIXct
objects.
Would like the xts index to display as %H:%M:%OS3 when doing viewing the xts
object.
Think I am missing the obvious.
Cheers,
Chris
--
View this message in context: http://r.789695.n4.nabble.com/xts-POSIXct-index-format-tp3336136p3336136.html
Sent from the R help mailing list archive at Nabble.com.
2010 May 17
1
Isn't aggreate.zoo supposed to work with POSIXct (zoo/TTR/xts issue)?
library(xts)
library(TTR)
ndx = getYahooData("^NDX")
aa = ndx$Close
bb = aggregate(aa, as.yearweek, tail, 1)
The last operation takes forever, and then the bb dates are messed up. The following produces the desired result:
time(aa) = as.Date(time(aa))
bb = aggregate(aa, as.yearweek, tail, 1)
The index of ndx and aa is of POSIXct (as reported by is(time(ndx))) , which apparently
2011 Jan 11
1
Interpolate xts
Hello,
I have a xts object, I would like to fill the NA with linear
interpolated data. Can anyone please help.
> str(zz)
An ‘xts’ object from 2010-11-24 15:59:29 to 2010-11-24 16:00:00 containing:
Data: num [1:23401, 1] 312 312 312 312 312 ...
Indexed by objects of class: [POSIXct,POSIXt] TZ:
xts Attributes:
List of 2
$ src : chr "datafeed"
$ updated: POSIXct[1:1],
2012 Oct 16
1
XTS Subsetting question (noob)
Hi,
How can an xts object be subset using a date in a variable?
E.g. in below, what is the right syntax for Q to get the same value as P?
Obviously the syntax shown below doesnt work...
V is an xts object.
d = "2007-01-01"
P <- V['2007-01-01/']
Q <- V['d/']
Thanks,
J
--
View this message in context:
2011 Oct 08
1
Filling missing days in xts time series
Hi,
I have a bunch of irregularly spaced xts time series (with a POSIX index),
and I'm trying to write a function that fillls the missing days. Using a
solution suggested by Gabor Grothendieck for zoo, I wrote the following:
# FD: Fill missing days
FD<-function(ser) {rng<-range(time(ser))
> temp<-merge(ser,xts(,seq(rng[1],rng[2],"day")))
>
2011 May 28
1
How to do operations on zoo/xts objects with Monthly and Daily periodicities
Is there an elegant way to do operations (+/-/*/ / ) on zoo/xts objects when
one serie is monthly (end of month) and the other daily (weekdays only) -
typically a monthly economic indicator and a stock index price?
Thanks,
TDB
--
View this message in context: http://r.789695.n4.nabble.com/How-to-do-operations-on-zoo-xts-objects-with-Monthly-and-Daily-periodicities-tp3558081p3558081.html
2012 May 22
1
Quantmod, Xts, TTR and Postgresql
Hi Everyone,
I'm currently using the latest build of R and R-Studio server (both are
amazing products)
I'm still very new to this but I came across this issue:
I'm trying to do a select from postgres and put the data into and xts
object like so:
# Libs
library('RPostgreSQL') # http://code.google.com/p/rpostgresql/
library('quantmod')
library('TTR')
2012 Jun 13
1
what does .indexDate() do - R::xts
Dear R experts,
I am learning the very useful XTS package, but cannot figure out the
purpose of some commands.
in particular, the .indexDate() command does not work as expected.
say:
x <- timeBasedSeq('2010-01-01/2010-01-02 12:00')
x <- xts(1:length(x), x)
then i can subset on date as follows:
x['2010-01-01']
however the .indexDate() command does not work as expected;
2012 Dec 02
2
How to calculate mean of every nth time series data with zoo or xts ?
Hello,
I have 1-minute time series stock data and I'd like to calculate mean of every n-th candle data of m-days.
result = c(mean of 1th data, mean of 2nd data, ...)
mean of 1th data = (1th data of 2012-1-1 + 1th data of 2012-1-2 + 1th data of 2012-1-3) / 3
mean of 2nd data = (2nd data of 2012-1-1 + 2nd data of 2012-1-2 + 2nd data of 2012-1-3) / 3
...
Could you let me know the fastest