Displaying 5 results from an estimated 5 matches similar to: "time series line plot: Error in plot.window(...) : invalid 'xlim' value"
2012 Aug 03
5
replacement has length zero. In addition: Warning message: In max(i) : no non-missing arguments to max; returning -Inf
Hi,
Here is my data, the first 10 rows
> u=regCond_all[1:10,]
> dput(u)
structure(c(999, 999, 999, 999, 999, 999, 999, 999, 999, 999,
99, 99, 99, 99, 99, 99, 99, 99, 99, 99, 99, 99, 99, 99, 99, 99,
99, 99, 99, 99, 1.9, 2, 1.97, 1.99, 1.83, 1.78, 1.6, 1.52, 1.52,
1.36, 10.53, 9.88, 9.88, 10.53, 10.53, 10.53, 5.26, 9.88, 10.53,
10.53, 5.4, 5.57, 5.46, 5.34, 5.5, 5.59, 5.62, 5.76, 6.23, 6.19,
2012 Jul 16
4
Error in as.xts
Hi
I got the following error using as.xts
Error in xts(x, order.by = order.by, frequency = frequency, ...) :
NROW(x) must match length(order.by)
Here is how the data looks like
> d1 <- read.csv(file.path(dataDir,"AppendixA-FishCountsTable-2009.csv"),
as.is=T)
> d1[1:3,]
dive_id date time species count size site depth level
TRANSECT VIS_M
1 62 10/12/2009
2012 Jul 18
2
loop searching the id corresponding to the given index (timestamp)
Hello,
I have the following loop for two data sets: diveData_2008 and
diveData_2009. It uses two other data: diveCond_all and fishTable. The
problem is at the point to identify the dive_id for the given index (index
is timestamp). It keeps on saying
for the1st loop
Error in fishReport$dive_id[i] <- dive_id : replacement has length zero
for the 2nd loop
Error in fishReport$dive_id[i + j] <-
2006 Apr 17
0
difference of means as response?
Dear R users,
I am looking for some advice on the proper construction of a mixed model
in R, using the difference in means as the response and treating
within-means residuals as a random effect.
I have a dataframe (my own, a snippet of which is given below) that is
composed of observations of pollen viability in flowers along tree
branches. Flowers (1 to 3 per position) were collected from
2006 May 20
1
intervals from cut() as numerics?
Hi,
Given some example data:
dat <- seq(4, 7, by = 0.05)
x <- sample(dat, 30)
y <- sample(dat, 30)
error <- x - y
I have broken the rage of x into 10 groups and I can calculate the bias
(mean(error)) for each of these 10 groups:
groups <- cut(x, breaks = 10)
max.bias <- aggregate(error, list(group = groups), mean)
max.bias
group x
1 (4,4.3] -0.7750000
2