TY Petr, it works. I will then replace NA by 0. 2010/8/4 Petr PIKAL <[email protected]>: > Hi > > you tried OK > > result <- merge(zz, av, by="DESCRIPTION", all=TRUE) > > and as you did not specify what to do when one value is NA here is one > possible solution > > rowSums(cbind(result$PL.x, result$PL.y), na.rm=T) > > Regards > Petr > > [email protected] napsal dne 04.08.2010 11:52:00: > >> Dear list, >> >> here are my two data frames: >> >> av <- >> structure(list(DESCRIPTION = c("COFFEE C Sep/10", "COPPER Sep/10", >> "CORN Dec/10", "CRUDE OIL miNY Sep/10", "GOLD Aug/10", "HENRY HUB >> NATURAL GAS Sep/10", >> "PALLADIUM Sep/10", "SILVER Sep/10", "SOYBEANS Nov/10", "SPCL HIGH >> GRADE ZINC USD", >> "SUGAR NO.11 Oct/10", "WHEAT Sep/10"), prix = c(-168.3, > -1.60000000000002, >> -773.75, -78.75, -1168.3, -0.0919999999999996, -470.75, 1758.5, >> -975.25, 1964, -19.09, -605.75), pos = c(-1, 0, -2, -1, -1, 0, >> -1, 1, -1, 1, -1, -1), PL = c(-12.03, -31.68, -43.2, -70.49, >> -11.88, -95.04, -3.96, -35.64, -30.24, -12.5, -36.09, -4.32)), .Names >> = c("DESCRIPTION", >> "prix", "pos", "PL"), row.names = c(NA, 12L), class = "data.frame") >> >> zz <- >> structure(list(DESCRIPTION = structure(c(1L, 2L, 3L, 4L, 5L, >> 6L, 7L, 8L, 9L, 10L, 13L, 11L, 12L), .Label = c("COFFEE C Sep/10", >> "COPPER Sep/10", "CORN Dec/10", "CRUDE OIL miNY Sep/10", "GOLD Aug/10", >> "HENRY HUB NATURAL GAS Sep/10", "PALLADIUM Sep/10", "PRM HGH GD > ALUMINIUM USD", >> "SILVER Sep/10", "SOYBEANS Nov/10", "SPCL HIGH GRADE ZINC USD", >> "SUGAR NO.11 Oct/10", "WHEAT Sep/10"), class = "factor"), pl = >> c(-8.20000000000002, >> 2.30000000000001, -47.25, -0.230000000000004, -12.7000000000000, >> -0.236, -2, 11.7100000000000, 14.4000000000001, -34.75, -10.75, >> 55, -0.669999999999998), PL = c(3075.00000000001, -575.000000000003, >> 2362.5, 115.000000000002, 1270.00000000000, 2360, 200, > -292.750000000001, >> -720.000000000005, 1737.5, 537.5, -1375, 750.399999999998), POSITION = > c(1, >> -1, 2, 2, 1, 2, -1, -1, 1, 2, 0, 0, 0), SETTLEMENT = c(167.4, >> 324.55, 390.75, 76.99, 1160.4, 4.718, 468.75, 2067.71, 1744.1, >> 978, 0, 0, 0)), .Names = c("DESCRIPTION", "pl", "PL", "POSITION", >> "SETTLEMENT"), row.names = c(NA, -13L), class = "data.frame") >> >> I am looking for one data frame with the column $PL=zz$PL+av$PL. >> I have been trying using the merge() function and its different >> arguments with no sucess. >> >> Any help is appreciated. >> Thank You. >> >> ______________________________________________ >> [email protected] 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. > >
-- ****************************** Arnaud Gaboury Mobile: +41 79 392 79 56 BBM: 255B488F ______________________________________________ [email protected] 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.
