search for: 1997

...fit1<-coxph(Surv(Entry, Exit, Fate)~Sex + Agerelease + Dayrelease + strata(Year), data=mydat) coxfit.apc<-survfit(apc.coxfit1) coxfit.apc plot(survfit(apc.coxfit1), conf.int=F, log=T, lty=c(1:2), col=c(1:2), xlim=c(205, 800)) #not run--first entry for this example is day 205 for 1996, 259 for 1997 >mydat ID Year Dayrelease Agerelease Survivorship Entry Exit Fate Sex 16240 1996 205 95 164 205 369 1 0 16319 1996 205 88 140 205 345 1 0 16378 1996 248 108 100 248 348 1 0 20383 1996 241 98 204 241 445 1 0 16324 1996 219 90 227 219 446 1 0 16327 1996 219 90 497 219 716 1 0 20373 1996 241 114 4...

...frame from a long format to a wide format. Unfortunately, I have a continuous date variable which gives me headaches. Consider the following example: > id=c("034","034","016","016","016","340","340") > date=as.Date(c("1997-09-28", "1997-10-06", "1997-11-04", "2000-09-27", > "2003-07-20", "1997-11-08", "1997-11-08")) > ref=c("2","2","1","1","2","1","1") > data1=data.frame(id,d...

Hi list, based on the following data.frame I would like to create a variable that indicates the number of occurrences of A in the 3 years prior to the current year: DF = data.frame(read.table(textConnection(" A B 8025 1995 8026 1995 8029 1995 8026 1996 8025 1997 8026 1997 8025 1997 8027 1997 8026 1999 8027 1999 8028 1995 8029 1998 8025 1997 8027 1997 8026 1999 8027 1999 8028 1995 8029 1998"),head=TRUE,stringsAsFactors=FALSE)) becomes: A B C 8025 1995 0 8026 1995 0 8029 1995 0 8026 1996 1 8025 1997 1 8026 199...

...uot;1996-10-29", "1996-11-02", "1996-11-09", "1996-11-14", "1996-11-19", "1996-11-23", "1996-11-25", "1996-11-29", "1996-12-02", "1996-12-09", "1996-12-15", "1996-12-25", "1997-01-06", "1997-01-10", "1997-01-16", "1997-01-20", "1997-01-22", "1997-01-27", "1997-02-01", "1997-02-04", "1997-02-10", "1997-02-17", "1997-02-26", "1997-03-04", "1997-03-...

...j without observation i and call it "med" 3. substract "med" from the sales of observation i to obtain the group median adjusted sales performance for observation i The dataset, function, and questions are as follows. x1 <- read.table(textConnection("YEAR GROUP SALES 1997 11 0.027 1997 11 0.001 1997 11 0.697 1997 11 0.047 1997 11 0.225 1998 11 0.025 1998 11 0.002 1998 11 0.001 1998 11 0.659 1998 11 0.037 1997 12 0.152 1997 12 0.025 1997 12 0.417 1997 12 0.081 1997 12 0.194 1997 12 0.069 1997 12 0.062 1998 12 0.146 1998 12 0.028 1998 12 0.437 1997...

...than shown), 1 2 3 4 5 Year Total Tus Whi Norw 1994 1.00 1830 0 355 1995 1.00 0 0 0 1995 1.00 0 0 0 1995 1.00 4910 4280 695 1997 1.00 0 0 110 1997 0.58 0 0 0 1997 1.00 0 0 0 1994 1.00 0 0 0 1997 1.00 0 40 70 1998 1.00 0 0 1252 1999 1.04 0...

...a large data frame that is organized by date in a peculiar way. I am seeking advice on how to transform the data into a format that is of more use to me. The data is organized as follows: STN_ID YEAR MM ELEM X1 X2 X3 X4 X5 X6 X7 1 2402594 1997 9 1 *-00233* *-00204* *-00119* -00190 -00251 -00243 -00249 2 2402594 1997 10 1 -00003 -00005 -00001 -00039 -00031 -00036 -00033 3 2402594 1997 11 1 000025 000065 000070 000069 000115 000072 000093 4 2402594 1997 12 1 000160 000...

...uot;, "1996-10-29", "1996-11-02", "1996-11-09", "1996-11-14", "1996-11-19", "1996-11-23", "1996-11-25", "1996-11-29", "1996-12-02", "1996-12-09", "1996-12-15", "1996-12-25", "1997-01-06", "1997-01-10", "1997-01-16", "1997-01-20", "1997-01-22", "1997-01-27", "1997-02-01", "1997-02-04", "1997-02-10", "1997-02-17", "1997-02-26", "1997-03-04", "1997-03-09"...

...uot;, "Congo", "Congo", "Congo", "Congo", "Congo", "Gabon", "Gabon", "Gabon", "Gabon", "Gabon", "Gabon", "Congo", "Congo"), Debut = c("24/01/1995", "01/05/1997", "31/12/1997", "02/02/1995", "28/02/1995", "01/03/1995", "13/03/1995", "01/01/1996", "31/01/1996", "24/01/1995", "01/07/1995", "01/09/1995", "01/07/1997", "01/01/1998", &quo...

I'm having trouble un 'tar'ing the file ctest_0_9-3_tar.tar. When using the command "tar -xv" on a UNIX machine I get the "checksum error" message. When trying to use WinZip on a Win95 machine it gives an error reading the header. I'm downloading from the CRAN site at Carnegie Mellon. Could the file there be corrupted? Thanks for any help, and sorry if

I'm having trouble un 'tar'ing the file ctest_0_9-3_tar.tar. When using the command "tar -xv" on a UNIX machine I get the "checksum error" message. When trying to use WinZip on a Win95 machine it gives an error reading the header. I'm downloading from the CRAN site at Carnegie Mellon. Could the file there be corrupted? Thanks for any help, and sorry if

Hi R Users, I'm wondering how can I calculate two (or three) way sum of a variable. A sample data is: State Month Year Value NC Jan 1996 1 NC Jan 1996 2 NC Feb 1997 2 NC Feb 1997 3 NC Mar 1998 3 NC Mar 1998 4 NY Jan 1996 4 NY Jan 1996 5 NY Feb 1997 5 NY Feb 1997 6 NY Mar 1998 6 NY Mar 1998 7 I'm trying to sum up "value" column by State*Month and by State*Month*Year. Also, I may need to calculate mean value along with "sum". Any help w...

Hola, Otra forma, utilizando la función de intervalos y la que comprueba si otro intervalo se solapa del paquete "lubridate": #---------------------- library(lubridate) fe.chas <- data.frame( entra=c('2001-01-01','2001-06-01','2003-01-01') ,sale=c('2002-01-01','2002-06-01','2004-01-01') ) ref <-

...# #Produce a histogram of start dates for a set of field measurements. # I didn't reproduce all the dates, because not sure it's relevant, but here's a sample. # Note the invalid date in the fifth element. > StartDate[sample(1:length(StartDate),20)] [1] "2004-08-26" "1997-09-08" "2004-08-19" "1997-09-08" "0999-07-20" [6] "2001-11-28" "2000-11-02" "1997-09-08" "2004-08-19" "2004-10-28" [11] "1997-09-08" "1997-09-09" "1997-09-08" "1998-09-08" &...

I appreciate if anyone can help me, I have a table as follow, > rate DATE VALUE 1 1997-01-10 5.30 2 1997-01-17 5.30 3 1997-01-24 5.28 4 1997-01-31 5.30 5 1997-02-07 5.29 6 1997-02-14 5.26 7 1997-02-21 5.24 8 1997-02-28 5.26 9 1997-03-07 5.30 10 1997-03-14 5.30 . ...... ... . ...... ... . ...... ... I want to extract the DATE(s) on which...

...ase refere to message posted to linux-security on Mon, 23 Dec 1996 under subject ''Buffer overflow in Linux''s login program'' -- alex] From mail@mail.redhat.com mail2.redhat.com dutecai.et.tudelft.nl by (8.6.10/1.34JP) Received: (qmail 3750 invoked from network); 17 Jan 1997 07:08:37 -0000 Received: from rosie.et.tudelft.nl (130.161.127.248) by mail2.redhat.com with SMTP; 17 Jan 1997 07:08:36 -0000 Received: from cave.et.tudelft.nl (cave.et.tudelft.nl [130.161.127.241]) by rosie.et.tudelft.nl (8.7.4/8.7.3) with ESMTP id RAA26124 for <linux-security@redhat.com>;...

I cannot figure this out... I would like to change the owner of a bunch of folders whose name begins with a dash... # chown Administrator \-BILLED\ JOBS\ -\ 1997-2002 -R chown: invalid option -- B Try `chown --help' for more information. # chown Administrator "\-BILLED\ JOBS\ -\ 1997-2002" -R chown: cannot access `\\-BILLED\\ JOBS\\ -\\ 1997-2002': No such file or directory # chown Administrator "-BILLED JOBS - 1997-2002" -R ch...

Hi, I have two lists (c and h - see below) containing matrices with similar cases but different values. I want to split these matrices into multiple matrices based on the values in h. So, I did the following: years<-c(1997:1999) for (t in 1:length(years)) { year=as.character(years[t]) h[[year]]<-sapply(colnames(h[[year]]), function(var) h[[year]][h[[year]][,var]>0, h[[year]][var,]>0]) } Now that I have created list h (with split matrices), I would like to use these sele...

Hello R-users,   The below is a snippet of my data:   fid crop year value 5_1_1 SWHE 1995 171 5_1_1 SWHE 1997 696 5_1_1 BARL 1996 114 5_1_1 BARL 1997 344 5_2_2 SWHE 1995 120 5_2_2 SWHE 1996 511 5_2_2 BARL 1996 239 5_2_2 BARL 1997 349   Here, I want to create dummy variables with the names of the content of a column 'crop' in a way that the new variable 'S...

...PUREZA_A SEXO NAC TRAZAB_P PUREZA_P NAC_P TRAZAB_M PUREZA_M NAC_M 717 pi M 1995-04-21 NA pi 1991-04-17 NA pi 1990-03-10 1434 pi H 2005-11-18 NA pi 1999-08-05 NA pi 1996-11-06 2151 pi H 2002-04-30 NA pi 1997-08-30 NA pi 1996-09-07 2868 pi M 2012-04-12 NA pi 2006-07-19 NA pi 2006-10-09 PESO 717 76 1434 NA 2151 NA 2868 NA Si coloco toda la orden entre na.omit() no me da resultados porque en algunos campos diferentes de PESO existen NA. No logro dar la...