As said by arun, the code is not clear. Ma Teresa, what is it that you actually want to do? Regards, José
-----Original Message----- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of arun Sent: 29 August 2013 15:12 To: R help Subject: Re: [R] Help R HI, Your code is not clear: for( i in 1: nrow(D)) { for( j in 5:ncol(D)) { D[((D[i,j]/D[i,2])>1.5)]15999)]<-1000000 ## "....1.5)]15999)]" } ^^^ } D<- structure(list(X. = c(1108L, 1591L, 3408L, 3872L, 5823L, 8099L, 10640L, 12600L, 14680L, 14698L), media = c(22, 86.6, 807, 103.25, 73, 125.16667, 67.33333, 2417, 38, 698.16667), IE.2003 = c(32, 92, 166, 237.2, 160.6, 169, 494.8, 1958, 142.2, 505), IE.2004 = c(63, 122.8, 290, 117.8, 73.6, 206, 168.2, 1871, 46, 482), IE.2005 = c(60, 276.6, 302, 154.4, 133.6, 196, 424.8, 1960, 30, 553), IE.2006 = c(39L, 87L, 322L, 113L, 70L, 161L, 476L, 2383L, 61L, 664L), IE.2007 = c(4, 95, 621, 116, 80, 150, 670.6, 2453, 404, 847), IE.2008 = c(8L, 94L, 1071L, 90L, 74L, 94L, 74L, 2506L, 42L, 800L), IE.2009 = c(16L, 81L, 1301L, 94L, 69L, 72L, 77L, 2758L, 19L, 679L), IE.2010 = c(5, 76, 1225, 1911.2, 72, 78, 51, 2442, 243.8, 646)), .Names = c("X.", "media", "IE.2003", "IE.2004", "IE.2005", "IE.2006", "IE.2007", "IE.2008", "IE.2009", "IE.2010"), class = "data.frame", row.names = c("1", "2", "3", "4", "5", "6", "7", "8", "9", "10")) D[,-c(1:4)][D[,-c(1:4)]/D[,2]>1.5] # [1] 60.0 276.6 133.6 196.0 424.8 39.0 476.0 61.0 670.6 404.0 #[11] 1301.0 1225.0 1911.2 243.8 A.K. Hi to everyone, I would like to replace some values in a data.frame (D) > str(D) 'data.frame': 116 obs. of 10 variables: $ X. : int 1108 1591 3408 3872 5823 8099 10640 12600 14680 14698 ... $ media : num 22 86.6 807 103.2 73 ... $ IE.2003: num 32 92 166 237 161 ... $ IE.2004: num 63 122.8 290 117.8 73.6 ... $ IE.2005: num 60 277 302 154 134 ... $ IE.2006: num 39 87 322 113 70 ... $ IE.2007: num 4 95 621 116 80 ... $ IE.2008: num 8 94 1071 90 74 ... $ IE.2009: num 16 81 1301 94 69 ... $ IE.2010: num 5 76 1225 1911 72 ... D X. media IE.2003 IE.2004 IE.2005 IE.2006 IE.2007 IE.2008 IE.2009 IE.2010 1 1108 22.00000 32.0 63.0 60.0 39 4.0 8 16 5.0 2 1591 86.60000 92.0 122.8 276.6 87 95.0 94 81 76.0 3 3408 807.00000 166.0 290.0 302.0 322 621.0 1071 1301 1225.0 4 3872 103.25000 237.2 117.8 154.4 113 116.0 90 94 1911.2 5 5823 73.00000 160.6 73.6 133.6 70 80.0 74 69 72.0 6 8099 125.16667 169.0 206.0 196.0 161 150.0 94 72 78.0 7 10640 67.33333 494.8 168.2 424.8 476 670.6 74 77 51.0 8 12600 2417.00000 1958.0 1871.0 1960.0 2383 2453.0 2506 2758 2442.0 9 14680 38.00000 142.2 46.0 30.0 61 404.0 42 19 243.8 10 14698 698.16667 505.0 482.0 553.0 664 847.0 800 679 646.0 WHat I really want to do is: for( i in 1: nrow(D)) { for( j in 5:ncol(D)) { D[((D[i,j]/D[i,2])>1.5)]15999)]<-1000000 } } ______________________________________________ R-help@r-project.org 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. 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