On Apr 15, 2010, at 5:33 AM, arnaud Gaboury wrote:
> Found this solution. It is maybe not the most elegant way, but it
> does the
> job.
>
>> a=as.data.frame(substr(lme$DESCRIPTION,1,14))
>> colnames(a)=c("DESCRIPTION")
>> lme=as.data.frame(c(a,lme[,2:3]))
>
>> lme
>
> DESCRIPTION CLOSING.PRICE POSITION
> 1 PRIMARY NICKEL 25,755.7100 0
> 2 PRIMARY NICKEL 25,760.8600 0
> 3 PRM HGH GD ALU 2,415.9000 0
> 4 SPCL HIGH GRAD 2,420.1000 -1
> 5 SPCL HIGH GRAD 2,420.4100 -1
> 6 SPCL HIGH GRAD 2,420.7300 1
> 7 SPCL HIGH GRAD 2,421.0500 1
> 8 SPCL HIGH GRAD 2,388.4300 0
> 9 SPCL HIGH GRAD 2,389.0000 0
> 10 SPCL HIGH GRAD 2,389.5700 0
> 11 SPCL HIGH GRAD 2,402.2900 0
> 12 SPCL HIGH GRAD 2,402.6400 -2
> 13 SPCL HIGH GRAD 2,391.8600 0
> 14 SPCL HIGH GRAD 2,403.0000 2
> 15 SPCL HIGH GRAD 2,392.4300 0
> 16 SPCL HIGH GRAD 2,393.0000 0
>
>
> If someone has a better idea, it is welcomed.
>
>
> -----Original Message-----
> From: arnaud Gaboury [mailto:arnaud.gaboury at gmail.com]
> Sent: Thursday, April 15, 2010 9:50 AM
> To: 'r-help at r-project.org'
> Subject: sum rows in a data.frame
>
> Dear group,
>
> Here is a data.frame, "lme":
>
>> lme
> DESCRIPTION CLOSING.PRICE POSITION
> 4 PRIMARY NICKEL USD 04/06/10 25,755.7100 0
> 5 PRIMARY NICKEL USD 10/06/10 25,760.8600 0
> 6 PRM HGH GD ALUMINIUM USD 09/07/10 2,415.9000 0
> 8 SPCL HIGH GRADE ZINC USD 06/07/10 2,420.1000 -1
> 9 SPCL HIGH GRADE ZINC USD 07/07/10 2,420.4100 -1
> 10 SPCL HIGH GRADE ZINC USD 08/07/10 2,420.7300 1
> 11 SPCL HIGH GRADE ZINC USD 09/07/10 2,421.0500 1
> 12 SPCL HIGH GRADE ZINC USD 13/04/10 2,388.4300 0
> 13 SPCL HIGH GRADE ZINC USD 14/04/10 2,389.0000 0
> 14 SPCL HIGH GRADE ZINC USD 15/04/10 2,389.5700 0
> 15 SPCL HIGH GRADE ZINC USD 17/05/10 2,402.2900 0
> 16 SPCL HIGH GRADE ZINC USD 18/05/10 2,402.6400 -2
>
> DESCRIPTION and CLOSING.PRICE are factors, POSITION is numeric.
>
> I want to sum POSITION by products, i.e. PRIMARY NICKEL, PRM HGH GD
> ALUMINIUM and SPCL HIGH GRADE ZINC. The problem is that, as you can
> see,
> there is a different date as a part of each product description. Can
> anyone
> tell me how to get rid of these dates so I can sum the position
> column? The
> number of rows is not fixed and will change every day.
See earlier post for one approach to the sum within description
problem. But I did use mean instead of sum.
>
> TY.
>
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David Winsemius, MD
West Hartford, CT