similar to: Zoo series to a date time stamp that is regular

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

Below is the code that I am using in a much larger function. I would expect a bankfull measure at zero to be between 0.6 and 0.8 approxfun is returning 0.8136986. I am sure that I am missing something. measure_bkf <- (structure(list(measurment_num = c(0, 0.2, 0.4, 0.6, 0.8, 1, 1.2, 1.4, 1.6, 1.8, 2, 2.2, 2.4, 2.6, 2.8, 3, 3.2, 3.4), bankfull_depths_m = c(-0.48, -0.48, -0.42, -0.26, 0.58,

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

#Is there a way to break the below zoo object into non-NA data frames algorithmically #this is a small example of a much larger problem. #It is really no even necessary to have the continuous chunks #end up as zoo objects but it is important to have them end #up with the index column. #thanks for all of your help in advance, and #if you need anything else please let me know library(zoo) ind.

Another newbie question I got the 1 minute spine interpolation and 15 mean aggregation working with many thanks to Gabor Grothendieck using Zoo functions. I got a tip from Hasan Diwan to look at xts but it seemed I would make better progress using code from Gabor. Now I'm having trouble plotting this zoo object. I'm thinking I want a function to "split" the zoo object back to

Folks, If I have a vector such as the following: x <- c(0, -1, -1, -1, 0, 0, 1, -1, 1, 0) and I want to replace the zeroes by the nearest non-zero number to the left, is there a more elegant way to do this than the following loop? y <- x for (i in 2 : length(x)) { if (y[i] == 0) { y[i] <- y[i - 1] } } > y [1] 0 -1 -1 -1 -1 -1 1 -1 1 1 You can see the

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

s <- 1.00 max(s) returns 1 is there anyway that I can get it to return 1.00. I am using the results of this max statement in a grep statement and it returns the wrong numbers, I will provide more information and code if it would make more sense in context. -- Stephen Sefick ____________________________________ | Auburn University? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?? | | Department of

Is there a way to use a continuous variable to pch in qplot? I believe this worked in previous version. I need to specify certain values of a shape for particular points so that multiple graphs all show the same shapes for the same streams. I have gone to the original data and added a pch column that I would like to use to specify the shapes to pch in qplot. Any help would be greatly

Hello, I'm trying to use na.spline (package zoo) to fill some missing data in a time series. this works fine, however, if I apply the 'maxgap' argument, I always get the error: <------ Error in na.spline.vec(x., coredata(object.), xout = xout., ...) : attempt to apply non-function ------> I couldn't find a similar error for this case in the mailing lists and zoo vignette,

I am thinking about submitting a package to CRAN that contains some units conversion functions that I use on a regular basis. Would this be helpful to the community, or would it be better to keep this as a personal package? I don't want to clutter CRAN. many thanks, -- Stephen Sefick ************************************************** Auburn University Biological Sciences 331 Funchess

R 2.12.2 on Scientific Linux 6.4 #works chron(times.="15:00:00", format=c(times="h:m:s")) #doesn't work chron(times.="15:00", format=c(times="h:m")) From chron Manual: The times format can be any permutation of "h", "m", and "s" separated by any one non-special character. The default is "h:m:s". what am I

I have used the fourthcorner function as suggest by dray and legendre (model 2 and 4 then combine). I plot the combined value with plot(four.comb, type="G"). What do the colors mean? I have both grey and black bars. many thanks, Stephen -- Stephen Sefick ************************************************** Auburn University Biological Sciences 331 Funchess Hall Auburn, Alabama

I know this is a double post, but the subject line was really misleading. Sorry again. This is the first time that I have tried to update packages with a tinkered around with .Rprofile. I start R with R --vanilla and it does not load my .Rprofile, but when I issue the command update.packages() R downloads the packages as expected, but then seems to load .Rprofile before compiling the packages

row <- c("a","b","c","d","e","f","g") #rows from larger data frame row.1 <- c("a","b","c","g") #rows of smaller data frame because d, e, and f don't contain any of the species, but the zeros are important x <- data.frame(sp1=rnorm(4), sp2=rnorm(4), sp3=rnorm(4),

I would like to be able to grab x and y columns out of a dataframe and then grab all of the columns that are not equal to x or y. I am sure that I am missing something easy. ftbr_UTM_downstream <- (structure(list(site = c("Jennie_Creek_Main_Stem", "Wolf_Pit_Creek_Main_Stem", "Little_Rockfish_Main_Stem_North", "Big_Muddy_Creek_Main_Stem",

These data represent stream channel cross-sectional surveys. I would like to be able to find the measurement on the tape (measure) where the Bank Full Depth (bkf_depths) is 0. This will happen twice because the channel has two sides. I thought fitting a loess line to these data and then predicting the measurment number would do it. I was wrong. Below is my failed attempt. My naive thought is

All, 1. I will try and make this clear and concise. Please let me know any information that would be helpful in figuring out this problem (I don't know the relevant information to post). I am on linux- see below for session information. 2. Problem: working directory: home an old version of a function is sourced into the R session and doesn't work working directory: Desktop the

I am not entirely sure what is going on. I am sure that I am missing something. maptools won't detach... Thanks for all of your help. R 2.11.1 OS Ubuntu 10.04 R --vanilla install.packages("sp", dep=TRUE) Loading required package: reshape Loading required package: plyr Loading required package: grid Loading required package: proto This is vegan 1.17-3 Loading required package:

the date were created with chron with this argument format=c(dates="Y/m/d", times="H:M:S")) so I have the dates being displayed as (10/06/22 12:00:00) I would like to have them displayed as "2010-06-22 12:00:00" or "%Y-%m-%d %H:%M:%S" and then I can convert these for mergeing with another data frame x <- (structure(c(14464, 14464.0104166667,