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antonio rodriguez

2006-Nov-01 17:51 UTC

[R] matrix manipulation with a for loop

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

Gabor Grothendieck

2006-Nov-01 18:25 UTC

head link

[R] matrix manipulation with a for loop

Try this where m is the matrix:

100 * colMeans(m > 5 & m < 9, na.rm = TRUE)


On 11/1/06, antonio rodriguez <antonio.raju at gmail.com>
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
>
> ______________________________________________
> R-help at stat.math.ethz.ch 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.
>

antonio rodriguez

2006-Nov-01 18:33 UTC

head link

[R] matrix manipulation with a for loop

Phil Spector escribi?:> 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
>>
>> ______________________________________________
>> R-help at stat.math.ethz.ch 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.
>>
>

antonio rodriguez

2006-Nov-01 18:43 UTC

head link

[R] matrix manipulation with a for loop

Phil Spector escribi?:> Antonio -
>    You need the na.rm in the first invocation of sum -- is.na() will 
> never
> return a missing value:
>
>   apply(F.zoo,2,function(x)sum(x >=5 & x <= 
> 9,na.rm=TRUE)/sum(!is.na(x))*100)
>
>                                                           - PhilDear Phil,

I understand now. Many thanks!!

Best regards,

Antonio


>
>
> On Wed, 1 Nov 2006, antonio rodriguez wrote:
>
>> Phil Spector escribi?:
>>> 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
>>>>
>>>> ______________________________________________
>>>> R-help at stat.math.ethz.ch 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.
>>>>
>>>
>>

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