search for: x2006

...9;2.04%' 0.0065 '0.65%' Feb 0.0145 '1.45%' 0.0082 '0.82%' Mar 0.0027 '0.27%' 0.0122 '1.22%' i.e. > dput(d_final) structure(list(Year = structure(c(2L, 1L, 3L), .Label = c("Feb", "Jan", "Mar"), class = "factor"), X2006 = c(0.0204, 0.0145, 0.0027 ), X2006.Lbl = structure(c(3L, 2L, 1L), .Label = c("'0.27%'", "'1.45%'", "'2.04%'"), class = "factor"), X2007 = c(0.0065, 0.0082, 0.0122), X2007.Lbl = structure(1:3, .Label = c("'0.65%'", &q...

...08 178 19429 0 1221 4919 0 ...  $ X2002: int  386118 0 2728387 680 0 25014 3707 871 2862 0 ...  $ X2003: int  441647 0 3822701 602 0 7293 6343 0 788 0 ...  $ X2004: int  471009 1349411 2700750 1310 215 52840 20119 7 474 0 ...  $ X2005: int  1081143 3662774 3740324 509 4 148102 30044 7 1962 0 ...  $ X2006: int  1670746 4127605 5920870 531 24 224382 27688 27 954 6 ...  $ X2007: int  2346392 3898345 6262784 5076 0 155818 31331 304 1415 0 ...  $ X2008: int  8151345 8119377 2639949 31010 0 202938 15803 104 495 0 ...  $ X2009: int  5257573 3018965 1062246 425 16 119540 12362 8 2096 0 ...  $ X2010: int  6...

Estimados ¿Cómo puede emitirse un comando para leer estos datos así como están? , yo solo lo sé hacer cuando unos están detrás de otros en una sola columna . Saludos José -- Este mensaje le ha llegado mediante el servicio de correo electronico que ofrece Infomed para respaldar el cumplimiento de las misiones del Sistema Nacional de Salud. La persona que envia este correo asume el

...), .Label = c("", "April", "August", "December", "February", "January", "July", "June", "March", "May", "November", "October", "September"), class = "factor"), X2006 = c(5987.387, 6478.592, 8386, 4651.273, 4339.167, 4631.978, 5217.306, 5846.903, 3867.825, 3886.434, 3959.668, 3848.853, NA, NA, NA), X2007 = c(4354.516, 5924.315, 5559.468, 3967.5, 5053.56, 4808.694, 4017.632, 3969.883, 3910.236, 3782.094, 3961.286, 3711.262, NA, NA, NA), X2008 = c(3685.789, NA, NA...

...have 12 chunks of Sweave code that work fine, but this piece: <<>>= library(car) cabbages$Year<-recode(cabbages$Year,"'X2001'='2001';'X2002'='2002';'X2003'='2003';'X2004'='2004';'X2005'='2005';'X2006'='2006';'X2007'='2007';'X2008'='2008';'X2009'='2009';'X2010'='2010';'X2011'='2011';'X2012'='2012'") @ just prints out verbatim as a single line running off the page (No recoding occurs)...

Folks, I have a data frame with columns 200401, 200402, ..., 201207, 201208. These represent years/months. What would be the best way to sum these columns by year? What about by quarter? Thanks for your time, KW -- [[alternative HTML version deleted]]

I have a file, each column of which is a separate year, and each row of each column is mean precipitation for that month. Looks like this (except it goes back to 1964). month X2000 X2001 X2002 X2003 X2004 X2005 X2006 X2007 X2008 X2009 1 1.600 1.010 4.320 2.110 0.925 3.275 3.460 0.675 1.315 2.920 2 2.960 3.905 3.230 2.380 2.720 1.880 2.430 1.380 2.480 2.380 3 1.240 1.815 1.755 1.785 1.250 3.940 10.025 0.420 2.845 2.460 4 3.775 1.350 2.745 0...