similar to: Plotting multiple xts/zoo time series on a single plot.

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