similar to: aggregating data with Zoo

Dear List, Thanks to those who helped with my enquiry a few days ago. I have a another question on loops, in this case I am trying to print out the row of a data frame if the previous 3 values (daily values) in col5 are in descending order. I have this loop which works, but ask whether this can be done differently (without conventional loop) in R: flag="T" d= 3 # d represents

Hi all, Please forgive this newbie posting to the list for the first time if I haven't followed proper procedure. I have read through many of the archives and find them most useful in learning R. I have ten years daily data (stock closing prices) which I read in zoo format. I am having problems coding R to run a count of how many days in each month have a price higher than the first day

Hi R-experts, Thanks very much to Jim Holtman and Gabor on my previous question. I am having another problem with data manipulation in zoo. The following is data (Z) for first business day of every month in zoo format. I am trying to get mean of "open" for each year. I subset Z <- Z[,2] then > sapply(split(Z, format(index(Z), "%Y")),mean) I get error message:

Dear R-experts, Sorry if I've overlooked a simple solution here. I have calculated a proportion of the number of observations which meet a criteria, applied to five years of data. How can I break down this proportion statistic for each year? For example (data in zoo format): open high low close hc lc 2004-12-29 4135 4135 4106 4116 8 -21 2004-12-30 4120 4131

Dear List, Sorry if I'm overlooking something simple here but I have gotten a bit tangled. I am trying to print the next five values(with their dates), which occur after a certain condition is met. I have a series of data in zoo format, call it "A". From this series I have created a subset (also in zoo format) based on a certain condition, call this series "B". I

Hi R-experts, I have a data frame (A) , and a subset (B) of this data frame. I am trying to create a new data frame which gives me all the rows of B, plus the 5th next row(occuring in A). I have used the below code, but it gives me all 5 rows after the matching row. I only want the 5th. FiveDaysLater <- A[c(sapply(match(rownames(B),rownames(A)), seq, length=6))),] Any guidance much

Dear List, Thankyou to Jim and Marc for their help on my previous question. I have a data frame of five columns, the first being a list of dates and the other four columns are numeric values. I wanted to list the days where all 4 columns of values are less than in the previous row. I used the following which works fine, except it doesnt show the dates for each row (the values from column 1).

Dear List, This may be the fifth time Ive tried to send this to the list so apologies if there are multiple emails. I need some help getting started with this problem. I have a data frame containing a year of daily stock prices in the following format: Date Open High Low Close 1/15/2000 10 11 8 10 1/16/2000 12 12 10 11 etc.. I want to create a new

All, I have been just recently working with zoo objects for trading systems. Can someone please help with these basic questions? Given a daily time series downloaded using get.hist.quote() from the tseries package, ie...... startDate= as.Date("2000-01-01") endDate= as.Date("2011-01-29") frequency= 'd' s= get.hist.quote('IWF', start= startDate, end=

Here's my code: http://pastebin.com/0yRxEVtm The important parts are uncommented and should be easy to find using the link above. For the following line of code, I plan on looking for a way to offset it up 7 rows so that the 15 minute timestamp would be considered the "median" of the subset being averaged to find the mean: avgCool = aggregate(intCool, trunc(time(intCool),

Hi, Having an zoo object I can subset it to obtain the days where I have the values within some range: is.zoo(z) TRUE subset(z[,1], z[,1]>=5 & z[,1]<= 10) #Yields: Year(day) 1988(13) 1988(14) 1988(16) 1988(20) 1988(21) 1988(22) 1988(25) 1988(26) 7.973946 9.933518 7.978227 7.512960 6.641862 5.667780 5.721358 6.863729 1988(27) 1988(28) 1988(29) 1988(30) 1988(32)

Good morning all! I am attempting to superimpose a moving-average smoother onto a graph of daily plots. These plots (in table[,2] below) span about 350 days and looks very noisy. I'd like for this smoother to plot the average of each group of 7 consecutive days (weekly) and show a line which joins these series of averages. Given the definition of MA, the first and last points will

Dear R People: I'm trying to convert a daily zoo object to a weekly zoo object: xdate <- seq(as.Date("2002-01-01"),as.Date("2010-07-10"),by="day") library(zoo) length(xdate) xt <- zoo(rnorm(3113),order=xdate) xdat2 <- seq(index(xt)[1],index(xt)[3113],by="week") xt.w <- aggregate(xt,by=xdat2,mean) Error: length(time(x)) ==

Hi R users, I have a time series variable that is only available at a monthly level for 1 years that I need to decompose to a weekly time series level - can anyone recommend a R function that I can use to decompose this series? eg. if month1 = 1200 I would to decompose so that the sum of the weeks for month1 equals 1200, etc.. Many thanks in advance for any help. -- View this message in

I would like to run an aggregation on a zoo object that has multiple series in it, with one of more series having NA values. The problem is that by default the aggregate function will produce an NA value in each aggregated period that contains an NA. For instance, if I run aggregate(x, as.yearmon(index(x)), mean) on the example object "x" which is printed below, I will just get a bunch

Hello, i?ve Samba 4.7 with domain controller with 3 servers, 1 master (samba-ad) and two slaves (samba-slave1) and (samba-slave2). The problem is when create user account from "samba-ad? this account not sync to slave, but i create the account on "samba-slave1" or "samba-slave2? this is sync on all server. Samba version [???????.] root at samba-ad:~# samba -V Version

Hi On http://nutwiki.kanonbra.com/wiki/Category:Eaton_Powerware_5110 you can see what is reported. The unreleased part refers to v2.7.1, which is now released. Alf Greg Vickers <daehenoc at iinet.net.au>: >Hi all, > >I've hooked up an Eaton Powerware 5110 to Ubuntu 13.10 server, and >configured nut appropriately. The cgi scripts are working fine, but on

2014-02-13 6:55 GMT+01:00 Alf H?gemark <alf at i100.no>: > Hi > Hi Alf, > On > http://nutwiki.kanonbra.com/wiki/Category:Eaton_Powerware_5110 cool thing! But duplicating the NUT Device Dumps Library [1]. The DDL can also serve both users and developers purposes, by providing these dumps as .dev files (usable with dummy-ups). I'm looking for someone to help me completing

Hi, I'm using a ZaptelBRI card. It works fine. But I have a small problem with call logs. The leading zeroes of the external calling party are not stored (e.g. : 0140302010 will be stored as 140302010). Same for international numbers for which "00" will be stripped out. I would not mind if the cdr record would give me an indication of the call's origin (national or

Hi Folks. May I suggest a small addition to ?factor -- this is not explicit about how to force the ordering of levels in an ordered factor. One can guess it, but I think it would be a good idea to spell it out. For example, one can do: X <- c("Alf","Bert","Chris","Dave") F <-