Thanks.
I made it!
Best wishes,
S.
Am 22.02.2011 17:41, schrieb Erik Iverson:
Sandra Stankowski wrote:
is.na function does'nt seem to work, but maybe I'm just dealing with
it in a wrong way.
here's an example
> m <- c(2, 3, 5, 6, 3, 7, -99, -99, 6)
> n <- c(1,1,1,1,1,2,2,2,2)
so my matrix contains certain missing values
Thank you for the example.
You're constructing a data.frame, not a matrix.
Those are two separate classes in R.
> m[m==-99] <- NA
> o <- data.frame(m, n)
> o
m n
1 2 1
2 3 1
3 5 1
4 6 1
5 3 1
6 7 2
7 NA 2
8 NA 2
9 6 2
"2" stands for february
> february <- which(o[,2]==2, arr.ind = TRUE)
> prec_feb <- sum(o[february,1], na.rm = TRUE)
> prec_feb
[1] 13
And now I need to know the exact number of rows, where "m" contains
a value. to know how many days a month give any information. (to
create monthly means and stuff)
You might find ?complete.cases useful.
Also try:
sum(!is.na(o$m))
?tapply may also be useful in general: e.g.,
tapply(o$m, o$n, mean, na.rm = TRUE)
None of this is tested...
hope this explains, what I need to know.
Thanks,
S.
Am 22.02.2011 16:50, schrieb Erik Iverson:
Sandra,
Please provide a small, reproducible example of this issue.
You probably want to use ?is.nan and not the inequality
operator.
Similar example, contrast:
x <- NA
is.na(x)
x == NA
Sandra Stankowski wrote:
Hey there,
I tried to count the number of rows, where my data isn't NaN in a
certain column.
this was my guess:
(given is a data frame with 2069 rows and 17 cols)
NROW(data[jan,16] != NaN)
("jan" is defined this way: jan <- which(data[,2]==1, arr.ind= TRUE))
but I only get the number of columns where my data is "1" in the
second col. R isn't removing the NaN.
na.rm isn't working here.
I would appreciate your help.
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