[R] matrix manipulation with a for loop

[antonio rodriguez]

Phil Spector escribió:
> Antonio -
>    When you're operating on each column of a matrix, you really should
> consider the apply() function, which was written for the task.  Also, 
> it's usually easier to count things in R by taking the sum of a logical
> expression, rather than the length of a subsetted vector.
>    Here's code that will solve your problem:
>
>     apply(F.zoo,1,function(x)sum(x >=5 & x <= 9)/sum(!is.na(x))*100)
Dear Phil,

The problem is that the columns have some missing values (NA's) so the 
result for:

apply(F.zoo,2,function(x)sum(x >=5 & x <= 9)/sum(!is.na(x))*100) # yields:

X.1   X.2   X.3   X.4   X.5   X.6   X.7   X.8   X.9  X.10  X.11  X.12  X.13
   NA    NA    NA    NA    NA    NA    NA    NA    NA    NA    NA    
NA    NA
 X.14  X.15  X.16  X.17  X.18  X.19  X.20  X.21  X.22  X.23  X.24  X.25  
X.26
   NA    NA    NA    NA    NA    NA    NA    NA    NA    NA    NA    
NA    NA

So it is supposed that using na.rm=T should do the task, but if I write:

 apply(F.zoo,2,function(x)sum(F.zoo >=5 & F.zoo <= 
9)/sum(!is.na(F.zoo))*100,na.rm=T)

I get:

Erro en FUN(newX[, i], ...) : unused argument(s) (na.rm = TRUE)

Antonio



>
>                                        - Phil Spector
>                      Statistical Computing Facility
>                      Department of Statistics
>                      UC Berkeley
>                      spector at stat.berkeley.edu
>
>
> On Wed, 1 Nov 2006, antonio rodriguez wrote:
>
>> Hi,
>>
>> Having a matrix F.zoo (6575,189) with NA's in some columns I'm trying to
>> extract from each column the percent of days within an specific range,
>> so I've wrote this procedure:
>>
>> length(subset(F.zoo[,86],(F.zoo[,86]>=5) & (F.zoo[,86]<=
>> 9)))/(length(F.zoo[,86])-length(subset(F.zoo[,86],is.na(F.zoo[,86]))))*100 
>>
>>
>> But to do this for each column (189) is pretty hard, so I want to write
>> a function in order to perform this automatically, such I have the
>> percent value corresponding to a specific column. I' tried these two
>> formulas but I can't get it. I think the problem is how to set the
>> initial values for the loop:
>>
>> Formula1:
>>
>> nnn<-function(x){for (i in F.zoo[,i]){
>>    print(length(subset(F.zoo[,i],(F.zoo[,i]>=5) & (F.zoo[,i]<=
>> 9)))/(length(F.zoo[,i])-length(subset(F.zoo[,i],is.na(F.zoo[,i]))))*100)
>> }
>> }
>>
>> Formula 2:
>>
>> H<-t(matrix(1,189))
>>
>> nnn<-function(x){for (i in col(H){
>>    print(length(subset(F.zoo[,i],(F.zoo[,i]>=5) & (F.zoo[,i]<=
>> 9)))/(length(F.zoo[,i])-length(subset(F.zoo[,i],is.na(F.zoo[,i]))))*100)
>> }
>> }
>>
>> Thanks,
>>
>> Antonio
>>
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>>
>