[R] Summing identical IDs

David Winsemius dwinsemius at comcast.net
Fri Oct 30 14:43:58 CET 2009


On Oct 29, 2009, at 5:23 PM, PDXRugger wrote:

>
> Terrific help thank you.
> dupbuild<-aggregate(DF$Acres, list(Bldgid), sum)
> This line worked best.
>
> Now im going to challenge everyone (i think?)
>
> Consider the following:
>
>
> Acres<-c(100,101,100,130,156,.5,293,300,.09,100,12.5)
> Bldgid<-c(1,2,3,4,5,5,6,7,7,8,8)
> Year<-c(1946,1952,1922,1910,1955,1955,1999,1990,1991,2000,2000)
> ImpValue<-c(1000,1400,1300,900,5000,1200,500,1000,300,1000,1000)
> DF=cbind(Acres,Bldgid,Year,ImpValue)
> DF<-as.data.frame(DF)
>
> I would like to do the same, except there are some rules i want to  
> follow.
> I only want to aggregate the Acres if :
> a) The Years are not identical
> b) The ImpValues are not identical
> c) The Years are identical and the ImpValue are not
> d)The ImpValues are identical and the Years are not

As I review your Boolean logic, I run into serious problems.

c) and d) cannot be true if a and b) are true.

So no cases satisfy all 4 specs. In particular both of the pairs you  
say you want aggregated (5+6) and 10+11) violate rule a) and the  
second pair also violates b).

-- 
David
>
> but if the Acres and ImpValues are identical i would still like to  
> add the
> Acres together and form one case.
> If the cases are put together i would also like to add the ImpValues
> together.  So the below
>
>    Acres Bldgid Year ImpValue
> 1  100.00      1 1946     1000
> 2  101.00      2 1952     1400
> 3  100.00      3 1922     1300
> 4  130.00      4 1910      900
> 5  156.00      5 1955     5000
> 6    0.50      5 1955     1200
> 7  293.00      6 1999      500
> 8  300.00      7 1990     1000
> 9    0.09      7 1991      300
> 10 100.00      8 2000     1000
> 11  12.50      8 2000     1000
>
> would become
>
>    Acres Bldgid Year ImpValue
> 1  100.00      1 1946     1000
> 2  101.00      2 1952     1400
> 3  100.00      3 1922     1300
> 4  130.00      4 1910      900
> 5  156.50     5 1955     6200
> 7  293.00      6 1999      500
> 8  300.09     7 1990     1300
> 10 112.50      8 2000     1000
>
> Thanks, i gave it a bunch of shots but nothing worth posting.
>
>
>
>
>
> PDXRugger wrote:
>>
>> Hello All,
>>   I would like to select records with identical IDs then sum an  
>> attribute
>> then and return them to the data frame as a single record.  Please
>> consider
>>
>>
>> Acres<-c(100,101,100,130,156,.5,293,300,.09)
>> Bldgid<-c(1,2,3,4,5,5,6,7,7)
>>
>> DF=cbind(Acres,Bldgid)
>> DF<-as.data.frame(DF)
>>
>> So that:
>>
>>  Acres Bldgid
>> 1 100.00      1
>> 2 101.00      2
>> 3 100.00      3
>> 4 130.00      4
>> 5 156.00      5
>> 6   0.50      5
>> 7 293.00      6
>> 8 300.00      7
>> 9   0.09      7
>>
>> Becomes
>>
>>  Acres Bldgid
>> 1 100.00      1
>> 2 101.00      2
>> 3 100.00      3
>> 4 130.00      4
>> 5 156.50      5
>> 7 293.00      6
>> 8 300.09      7
>>
>> dup<-unique(DF$Bldgid[duplicated(Bldgid)])
>> dupbuild<-DF[DF$Bldgid %in% dup,]
>> dupbuild..dupareasum<-sum(dupbuild$Acres[duplicated(dupbuild 
>> $Bldgid)])
>>
>> This sums the unique Ids of the duplicated records, not whati want.
>> Thanks ahead of time
>>
>> JR
>>
>>
>>
>
> -- 
> View this message in context: http://www.nabble.com/Summing-identical-IDs-tp26118922p26121056.html
> Sent from the R help mailing list archive at Nabble.com.
>
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> and provide commented, minimal, self-contained, reproducible code.

David Winsemius, MD
Heritage Laboratories
West Hartford, CT




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