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.gabo...@gmail.com]
Sent: Thursday, April 15, 2010 9:50 AM
To: 'r-help@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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