similar to: merging and adding time series

down vote favorite Hello I have a zoo series. It lasts 10 years and its frequency is 15min. I'd like to get a new zoo series (or vector) with the same number of elements, whith each element equal to the first element of the day. That's, The first element everyday is repeated throughout the wole day. This is not same as aggregate(originalseries,as.Date,head,1) because this gives a

Hello How can I substitute all NA values by zero in a R zoo series? I've been reading about na.locf and na.omit but I think none of them do what I need. thanks. -- View this message in context: http://r.789695.n4.nabble.com/Substitute-NAs-by-zero-tp2546715p2546715.html Sent from the R help mailing list archive at Nabble.com.

Hello How can I select several not continuous rows ? If I wanted to select rows 1 to 7 I'll write mydata[,1:7] But what if I need to select rows 1 to 5 and 10 to 15? -- View this message in context: http://r.789695.n4.nabble.com/How-to-select-not-continous-rows-tp3003840p3003840.html Sent from the R help mailing list archive at Nabble.com.

Hello Where could I find examples on how to work with the time index in a timeseries or zoo series? Let say I've got this series DATA 1990-01-01 10:00:00 0.900 1990-01-01 10:01:00 0.910 1990-01-01 10:03:00 0.905 1990-01-01 10:04:00 0.905 1990-01-01 10:05:00 0.890 ....................... 2000-12-31 20:00:00 0.992 How do I make simple calculations such as ... ? Calculate the

Hello Imagine I have a vector with ones and zeroes I write it compactly: 1111111100001111111111110000000001111111111100101 I need to get a new vector replacing the "N" ones following the zeroes to new zeroes. For example for N = 3 1111111100001111111111110000000001111111111100101 becomes 1111111100000001111111110000000000001111111100000 I can do it with a for loop but I've read

Hello What's the differente betwen using "plot" and using "print" in order to plot a graph? For example in order to plot the result of a histogram. cheers -- View this message in context: http://r.789695.n4.nabble.com/plot-vs-print-tp3045256p3045256.html Sent from the R help mailing list archive at Nabble.com.

Hello. I don't uderstant when to use textConnection and when not. Some examples do it, some not. I've even seen something like con <- textConnection(rev(rev(ReadLines('data.txt'))[-(1:2])) data <- read.table(con) close(con) -- View this message in context: http://r.789695.n4.nabble.com/when-to-use-textConnection-tp2327132p2327132.html Sent from the R help mailing list

Hello How do you compare R to SAS in terms of speed and management of large datasets? What about Revolution R? I've seen on their site, they claim that Revolution R is much faster than R and it's multithread... Can you really notice the difference?. What dissadvantage does it have? I think it's based on R 2.10. but R already issued the version 2.12 Regards What alternative

Hello Why this works: ncota <- 1 nslope <- 29 resul <- matrix(rep(0,ncota*nslope*4),ncota*nslope,4) But this doesn't? ncota <- 1 sini <- 0.1; sfin <- 1.5; spaso <- 0.05; nslope <- 1+((sfin-sini)/spaso) resul <- matrix(rep(0,ncota*nslope*4),ncota*nslope,4) I guess the problem is that the division gives a noninteger number. How can I get the second one work? I

Hello I've used read.table to read a file that contains numbers such as 0.00001 when I write them back with write.table those numbers appear as 1e-5 How can I keep the old format? thanks -- View this message in context: http://r.789695.n4.nabble.com/number-format-writing-1e-5-instead-of-0-00001-tp3003831p3003831.html Sent from the R help mailing list archive at Nabble.com.

Hi Usually "aggregate" is used to calculate things such as the sum of all data on the first day, the sum next day, and so on. But how can I calculate the mean of the first hour of all days, the mean of the second hour of all days, and so on. ??? That's Most examples: today at 1am + today at 2am + today at 3am +.... -> sum today tomorrow at 1am + tomorrow at

hello I think I've found a bug I don't know if it's a chron bug or a R one. (05/12/05 23:00:00) +1/24 gives (05/12/05 24:00:00) instead of (05/13/05 00:00:00) it looks like the same but it's not because when you get the date of this datetime it says day 12 instead of 13. Please, forward it to the place where this bugs are supposed to be posted. cheers -- View this message

Hello, I have a series of intraday (high-frequency) price data in the form of POSIX timestamp followed by the value. I sucesfuly loaded that into "its" package object. I would like to create from it a regularly spaced time series of prices (for example 1min, 5min, etc apart) so i could calcualte returns. There is an interpolation function locf() that for timestamp with value NA uses last

Hi everybody, I have a small question about the function "na.locf" from the package "zoo". I saw in the help that this function is able to fill NA gaps with the last value before the NA gap (or with the next value). But it is possible to fill my NA gaps according to the last AND the next value at the same time? Actually, I want R to fill my gaps with the method of

Hello I have a file with this format 2005-01-03 09:05 0.00 2005-01-03 09:10 0.01 2005-01-03 09:15 0.02 2005-01-03 09:20 0.03 2005-01-03 09:25 0.04 2005-01-03 09:30 0.05 2005-01-03 09:35 0.06 2005-01-03 09:40 0.07 2005-01-03 09:45 0.08 2005-01-03 09:50 0.09 2005-01-03 09:55 0.10 2005-01-03 10:00 0.00 2005-01-03 10:05 0.00 .... some

Hi, A quick beginner's question. I have two time series, A with daily data, and another B with data at varying frequencies, but mostly annual. Both the series are sorted ascending. I need to merge these two series together in the following way: For any entry of A, the lookup should match with B until we find an entry of B that's larger than A's. For all A[i], i = 1,...,N and B[j],

Hi R-helpers I'm looking for a vectorised function which does missing value replacement as in last observation carried forward in the zoo package but instead of a locf, I would like the locf function to add +1 to each time a missing value occurred. See below for an example. > require(zoo) > x <- 5:15 > x[4:7] <- NA > coredata(na.locf(zoo(x))) [1] 5 6 7 7 7 7 7 12 13

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 with real NA? In this case na.locf function should work Thank you

I'm sure I'm being stupid so flame away... R2.2.1 on Windoze (boohoo) latest updates of packages. I'm exploring a dataset (land) with three variables looking at an narrowly unbalanced two group (GROUP) ANCOVA of a randomised controlled trial analysing endpoint score (SFQ.LOCF.ENDPOINT) entering the baseline score (SFQ.BASELINE) as covariate and the following work fine: > res.same

Hi! Is there a possibilty in R to carry out LOCF (Last Observation Carried Forward) analysis or to create a new data frame (array, matrix) with LOCF? Or some helpful functions, packages? Karl --------------------------------- Gesendet von http://mail.yahoo.de Schneller als Mail - der neue Yahoo! Messenger. [[alternative HTML version deleted]]