On Apr 28, 2010, at 5:14 AM, arnaud Gaboury wrote:

Dear group,

Here is my data.frame :

avprix <-
structure(list(DESCRIPTION = c("CORN Jul/10", "CORN May/10",
"ROBUSTA COFFEE (10) Jul/10", "SOYBEANS Jul/10", "SPCL HIGH GRADE ZINC USD
Jul/10",
"STANDARD LEAD USD Jul/10"), prix = c(-1.5, -1082, 11084, 1983.5,
-2464, -118), quantity = c(0, -3, 8, 2, -1, 0)), .Names = c("DESCRIPTION",
"prix", "quantity"), row.names = c(NA, -6L), class = "data.frame")

avprix
                     DESCRIPTION    prix quantity
1                     CORN Jul/10    -1.5        0
2                     CORN May/10 -1082.0       -3
3      ROBUSTA COFFEE (10) Jul/10 11084.0        8
4                 SOYBEANS Jul/10  1983.5        2
5 SPCL HIGH GRADE ZINC USD Jul/10 -2464.0       -1
6        STANDARD LEAD USD Jul/10  -118.0        0

I need to remove the date (i.e. Jul/10 in this example) for each element of the DESCRIPTION column that contains the USD symbol. I am trying to do this
using regular expressions, but must admit I am going nowhere.
My elements in the DESCRIPTION column and the dates can change every day.

This searches for the pattern USD and then replaces any three characters , forward-slash, any two characters:
> sub("USD+.*(.../..)", "", avprix$DESCRIPTION)
[1] "CORN Jul/10" "CORN May/10" "ROBUSTA COFFEE (10) Jul/10" [4] "SOYBEANS Jul/10" "SPCL HIGH GRADE ZINC " "STANDARD LEAD "

This tightens up the matching by requiring that that the characters after the slash be digits:

> sub("USD+.*(.../\\d{2})", "", avprix$DESCRIPTION)
[1] "CORN Jul/10" "CORN May/10" "ROBUSTA COFFEE (10) Jul/10" [4] "SOYBEANS Jul/10" "SPCL HIGH GRADE ZINC " "STANDARD LEAD "

-- David.


>

TY for any help.

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David Winsemius, MD
West Hartford, CT

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