Displaying 20 results from an estimated 10000 matches similar to: "zoo plot warning messages - I don't know what they mean or how to inspect the data to figure this out"
2009 Dec 15
1
for loop for automatic pdf generation
I know this is not reproducible, but I don't want to clog up mail
boxes with the data frame. I would be happy to send this off list. I
am sure that I am missing something simple. The plotting works if I
just paste the call to qplot into R and replace the [i] with a number.
Thanks for all of your help in advance.
#loop to spit out PDFs
list.names <-
2008 Nov 04
2
Zoo seems to be running slow in R 2.8.0 windows
R version 2.8.0 (2008-10-20)
i386-pc-mingw32
locale:
LC_COLLATE=English_United States.1252;LC_CTYPE=English_United
States.1252;LC_MONETARY=English_United
States.1252;LC_NUMERIC=C;LC_TIME=English_United States.1252
attached base packages:
[1] stats graphics grDevices utils datasets methods base
other attached packages:
[1] StreamMetabolism_0.01 chron_2.3-24 zoo_1.5-4
loaded
2008 Oct 19
1
zoo in ggplot2
library(zoo)
d<-(structure(c(1.39981554315924, 0.89196314359498, 0.407816250252697,
0.823496839063978, 1.14429021220358, 1.23971035967413, 0.960868900583432,
0.927685306209829, 1.22072345292821, 0.249842897450642, 1.00879641624694,
0.925372139878243, 0.317259909172362, 0.382677149697482), index =
structure(c(11808,
11869, 11961, 11992, 12084, 12173, 12265, 12418, 12600, 12631,
12753, 12996,
2010 Sep 16
1
How do I create a plotable zoo object based on weekly data?
Dear readers,
The problem is simple: I have weekly time series data with a maximum
week number of (52,53,52,52) in (2008,2009,2010,2011) respectively. That
means I have a dataset looking like this:
2008-01 value
2008-02 value
.
.
2011-52 value
And I would like to turn that data into a plotable zoo object. Now a
simple example containing 4 data points, each being on the December
28th, would be:
2008 Sep 22
1
as.day() Function (zoo question)
I am was going to look at the as.yearmon function in the zoo package
and write a as.day function to aggregate a time series of 96
observations per day into the mean for each day, but I don't know how
to look at the code so that I can convert it into something I can use.
On top of that I believe that it is probably an S3 method and I
haven't quite gotten that far in my programming
2009 Aug 11
1
merge zoo objects contained in a list
I would like to merge zoo objects that are stored in a list into one
big zoo object with one index for all of the observations.
I have created the list (74 dataframes) with the code below, and have
tried the do.call(merge, foo) in the call and the output is not what I
expected. Any help would be greatly appreciated.
Stephen Sefick
###################################################level logger
2008 Aug 21
1
max and min with the indexes in a zoo object (or anything else that could solve the problem)
library(zoo)
library(chron)
t1 <- chron("1/1/2006", "00:00:00")
t2 <- chron("1/31/2006", "23:45:00")
deltat <- times("00:15:00")
tt <- seq(t1, t2, by = times("00:15:00"))
d <- sample(33:700, 2976, replace=TRUE)
sin.zoo <- zoo(d,tt)
#there are ninety six reading in a day
d.max <- rollapply(sin.zoo, width=96, FUN=max)
2010 May 18
2
Function that is giving me a headache- any help appreciated (automatic read )
note: whole function is below- I am sure I am doing something silly.
when I use it like USGS(input="precipitation") it is choking on the
precip.1 <- subset(DF, precipitation!="NA")
b <- ddply(precip.1$precipitation, .(precip.1$gauge_name), cumsum)
DF.precip <- precip.1
DF.precip$precipitation <- b$.data
part, but runs fine outside of the function:
days=7
2008 Jun 26
1
How to turn Time Series daily values into weekly means (aggregate?)
#this is a daily series of precipitation data. I would like to condense it
into weekly means. How can I do this
#as a side note I would like to do this same thing to two years worth of
fifteen minute interval data and make it into
#a series of daily averages (there are 96 readings per day)
#is aggregate the right function? or...
y <- c(1.23, 0, 0, 0, 0, 0, 0, 0, 0, 0.27, 0, 0.29, 0, 0, 0,
2011 Jan 06
1
[zoo] - Individual zoo or data frames from non-continuous zoo series
#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.
2009 Oct 06
1
ggplot2 applying a function based on facet
Look at the bottom of the message for my question
#here is a little function that I wrote
USGS <- function(input="discharge", days=7){
library(chron)
library(gsubfn)
#021973269 is the Waynesboro Gauge on the Savannah River Proper (SRS)
#02102908 is the Flat Creek Gauge (ftbrfcms)
#02133500 is the Drowning Creek (ftbrbmcm)
#02341800 is the Upatoi Creek Near Columbus (ftbn)
#02342500 is
2008 Jul 08
1
making zoo objects with zoo without format argument?
#this is a subset of a larger data frame and I am okay with subsetting it as
there are redundant time stamps, but I would like to create a zoo object out
of this and I am having a hard #time figuring out how to do this the date
structure is year and then month
x <- structure(list(Yearmonth = structure(c(12L, 24L, 1L, 13L, 14L,
3L, 15L, 4L, 16L, 5L, 17L, 6L, 18L, 7L, 19L, 8L, 20L, 9L, 21L,
2008 Mar 03
1
write csv file from zoo object
# chron
library(chron)
fmt.chron <- function(x) {
chron(sub(" .*", "", x), gsub(".* (.*)", "\\1:00", x))
}
z1 <- read.zoo(SC3.csv, sep = ",", header = TRUE, FUN = fmt.chron)
z2 <- read.zoo(SC2.csv, sep = ",", header = TRUE, FUN = fmt.chron)
z3<-c(z1, z2)
write.table(z3, sep="," , "SC.csv")
How do you
2008 Apr 02
1
zoo plot not showing whole date
z1 = read.zoo("chemmgL.csv", sep=",", header=T, format="%m/%d/%y")
I would like to the entire date field in the plot - 1/1/07 all that I
get now is 2007
I do not include data because it is to large of a data set for an email
--
Let's not spend our time and resources thinking about things that are
so little or so large that all they really do for us is puff us
2008 Jul 02
1
Zoo plotting behavior
I have a matrix with data that runs from 1/1/06 00:01:00-1/31/08 23:46:00.
I have read in the data with this
fmt.chron <- function(x) {
chron(sub(" .*", "", x), gsub(".* (.*)", "\\1:00", x))
}
x <- read.zoo(file.choose(), sep=",", header=T, FUN=fmt.chron)
plotted with this
plot(x[,(seq(3, by=9, length.out=12))],
2012 Jan 21
2
4th corner analysis ade4 - what do the colors mean
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
2008 Oct 20
2
error message when plotting survival curves
I am trying to plot survival curves using the following code as an example:
>rs1799964.coxph<-(coxph(Surv(sassurvmonths,status)~age+stage+rs1799964_TNFA,method="efron"))
>plot(rs1799964.coxph,lyt=c(1,3),xlab="Survival in Months",ylab="Proportion
Surviving")
I am gettingthe following error message:
>Error in xy.coords(x, y, xlabel, ylabel, log) :
2010 Jun 28
1
Zoo series to a date time stamp that is regular
NOTE: I will provide data if necessary, but I didn't want clutter
everyones mailbox
All:
I have a time series with level and temperature data for 11 sites for
each of three bases. I will have to do this more than once is what I
am saying here. OK, The time series are zoo objects with index
values in chron format. The problem is that the date and times should
be at even 15 min intervals,
2008 Mar 05
1
plotting big zoo object memory problem
the comma seperated file is 37Mb, and I get the below message:
it is zoo object read in this way:
# chron
> library(chron)
> fmt.chron <- function(x) {
+ chron(sub(" .*", "", x), gsub(".* (.*)", "\\1:00", x))
+ }
> z1 <- read.zoo("all.csv", sep = ",", header = TRUE, FUN = fmt.chron)
and then the plot is done with:
2008 Apr 22
2
bootstrap for confidence intervals of the mean
d = c(0L, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, 0L, 0L, 7375L,
NA, NA, 17092L, 0L, 0L, 32390L, 2326L, 22672L, 13550L, 18285L)
boot.out <-boot(d, mean, R=1000, sim="permutation")
Error in mean.default(data, original, ...) :
'trim' must be numeric of length one
I know that I am missing something but I can't figure it out.
thanks
stephen
--
Let's not spend our