[R] nrow()

[Sandra Stankowski]

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

 > 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)

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.
>>
>> ______________________________________________
>> R-help at r-project.org mailing list
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide 
>> http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>