Hi All,
I need to do the following operation from data.frame:
df <- data.frame(Year = c("2001", "2002",
"2003", "2004", "2005", "2006",
"2007"), Amount = c(150, 120, 175, 160, 120, 105, 135))
df[which.max(df$Amount),] #to extract row with max Amount.
Now I need to do 3 years average around the max Amount value (ie:
mean(120,175,160))
Thanks!
N
[[alternative HTML version deleted]]
Nico, I expect there are many better ways to do this, but this does work: max_row <- (1:nrow(df))[which.max(df$Amount)] mean(df$Amount[max_row + c(-1, 0, 1)])> max_row <- (1:nrow(df))[which.max(df$Amount)] > mean(df$Amount[max_row + c(-1, 0, 1)])[1] 151.6667 R. Mark Sharp, Ph.D. Director of Primate Records Database Southwest National Primate Research Center Texas Biomedical Research Institute P.O. Box 760549 San Antonio, TX 78245-0549 Telephone: (210)258-9476 e-mail: msharp at TxBiomed.org> On Sep 21, 2015, at 9:52 AM, Nico Gutierrez <nico.gutierrezo at gmail.com> wrote: > > Hi All, > > I need to do the following operation from data.frame: > > df <- data.frame(Year = c("2001", "2002", "2003", "2004", "2005", "2006", > "2007"), Amount = c(150, 120, 175, 160, 120, 105, 135)) > df[which.max(df$Amount),] #to extract row with max Amount. > > Now I need to do 3 years average around the max Amount value (ie: > mean(120,175,160)) > > Thanks! > N > > [[alternative HTML version deleted]] > > ______________________________________________ > R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see > 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.
On Mon, Sep 21, 2015 at 9:52 AM, Nico Gutierrez <nico.gutierrezo at gmail.com> wrote:> Hi All, > > I need to do the following operation from data.frame: > > df <- data.frame(Year = c("2001", "2002", "2003", "2004", "2005", "2006", > "2007"), Amount = c(150, 120, 175, 160, 120, 105, 135)) > df[which.max(df$Amount),] #to extract row with max Amount. > > Now I need to do 3 years average around the max Amount value (ie: > mean(120,175,160)) > > Thanks! > N > >?The simplistic answer is something like: df <- structure(list(Year = structure(1:7, .Label = c("2001", "2002", "2003", "2004", "2005", "2006", "2007"), class = "factor"), Amount = c(150, 120, 175, 160, 120, 105, 135)), .Names = c("Year", "Amount"), row.names c(NA, -7L), class = "data.frame"); wdf <- which.max(df$Amount); adf3 <- mean(df$Amount[adf-1:adr+1]); But that ignores the boundry condition where the maximum is at either end. What do you want to do in that case?? -- Schrodinger's backup: The condition of any backup is unknown until a restore is attempted. Yoda of Borg, we are. Futile, resistance is, yes. Assimilated, you will be. He's about as useful as a wax frying pan. 10 to the 12th power microphones = 1 Megaphone Maranatha! <>< John McKown [[alternative HTML version deleted]]
Note the following problems: 1. " max_row <- (1:nrow(df))[which.max(df$Amount)]" This is a bit silly. max_row <- which.max(df$Amount) will do. See ?which.max 2. What happens if the max is the first or last row? e.g.> dat <- data.frame(a=runif(5),b=1:5) > max_row<- which.max(dat$b) > mean(dat[max_row+c(-1,0,1),"b"]) ## 2-d indexing[1] NA Cheers, Bert Gunter "Data is not information. Information is not knowledge. And knowledge is certainly not wisdom." -- Clifford Stoll On Mon, Sep 21, 2015 at 10:51 AM, Mark Sharp <msharp at txbiomed.org> wrote:> Nico, > > I expect there are many better ways to do this, but this does work: > max_row <- (1:nrow(df))[which.max(df$Amount)] > mean(df$Amount[max_row + c(-1, 0, 1)]) > >> max_row <- (1:nrow(df))[which.max(df$Amount)] >> mean(df$Amount[max_row + c(-1, 0, 1)]) > [1] 151.6667 > > R. Mark Sharp, Ph.D. > Director of Primate Records Database > Southwest National Primate Research Center > Texas Biomedical Research Institute > P.O. Box 760549 > San Antonio, TX 78245-0549 > Telephone: (210)258-9476 > e-mail: msharp at TxBiomed.org > > > > >> On Sep 21, 2015, at 9:52 AM, Nico Gutierrez <nico.gutierrezo at gmail.com> wrote: >> >> Hi All, >> >> I need to do the following operation from data.frame: >> >> df <- data.frame(Year = c("2001", "2002", "2003", "2004", "2005", "2006", >> "2007"), Amount = c(150, 120, 175, 160, 120, 105, 135)) >> df[which.max(df$Amount),] #to extract row with max Amount. >> >> Now I need to do 3 years average around the max Amount value (ie: >> mean(120,175,160)) >> >> Thanks! >> N >> >> [[alternative HTML version deleted]] >> >> ______________________________________________ >> R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see >> https://stat.ethz.ch/mailman/listinfo/r-help >> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html >> and provide commented, minimal, self-contained, reproducible code. > > ______________________________________________ > R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code.
No. On Mon, Sep 21, 2015 at 10:58 AM, John McKown <john.archie.mckown at gmail.com> wrote:> On Mon, Sep 21, 2015 at 9:52 AM, Nico Gutierrez <nico.gutierrezo at gmail.com> > wrote: > >> Hi All, >> >> I need to do the following operation from data.frame: >> >> df <- data.frame(Year = c("2001", "2002", "2003", "2004", "2005", "2006", >> "2007"), Amount = c(150, 120, 175, 160, 120, 105, 135)) >> df[which.max(df$Amount),] #to extract row with max Amount. >> >> Now I need to do 3 years average around the max Amount value (ie: >> mean(120,175,160)) >> >> Thanks! >> N >> >> > The simplistic answer is something like: > > df <- structure(list(Year = structure(1:7, .Label = c("2001", "2002", > "2003", "2004", "2005", "2006", "2007"), class = "factor"), Amount = c(150, > 120, 175, 160, 120, 105, 135)), .Names = c("Year", "Amount"), row.names > c(NA, > -7L), class = "data.frame"); > wdf <- which.max(df$Amount); > adf3 <- mean(df$Amount[adf-1:adr+1]);Typos?! But it won't work anyway. See ?Syntax for operator precedence and Example:> a <- 1:5 > mid <- 3 > a[mid-1:mid+1][1] 3 2 1> a[(mid-1):(mid+1)][1] 2 3 4 Cheers, Bert> > But that ignores the boundry condition where the maximum is at either end. > What do you want to do in that case? > > > -- > > Schrodinger's backup: The condition of any backup is unknown until a > restore is attempted. > > Yoda of Borg, we are. Futile, resistance is, yes. Assimilated, you will be. > > He's about as useful as a wax frying pan. > > 10 to the 12th power microphones = 1 Megaphone > > Maranatha! <>< > John McKown > > [[alternative HTML version deleted]] > > ______________________________________________ > R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see > 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.
year <- df$Year[ which.max( df$Amount)] df[ df$Year %in% (as.numeric( as.character( year)) + -1:1), ] Year Amount 2 2002 120 3 2003 175 4 2004 160 On Mon, Sep 21, 2015 at 04:52:46PM +0200, Nico Gutierrez wrote:> Hi All, > > I need to do the following operation from data.frame: > > df <- data.frame(Year = c("2001", "2002", "2003", "2004", "2005", "2006", > "2007"), Amount = c(150, 120, 175, 160, 120, 105, 135)) > df[which.max(df$Amount),] #to extract row with max Amount. > > Now I need to do 3 years average around the max Amount value (ie: > mean(120,175,160)) > > Thanks! > N > > [[alternative HTML version deleted]] > > ______________________________________________ > R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see > 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. >
better ( if year is an vector of more than 1 element): df[ df$Year %in% outer(as.numeric( as.character( year)), -1:1, FUN='+'), ] Year Amount 2 2002 120 3 2003 175 4 2004 160 On Mon, Sep 21, 2015 at 10:49:34PM +0200, Frank Schwidom wrote:> > year <- df$Year[ which.max( df$Amount)] > df[ df$Year %in% (as.numeric( as.character( year)) + -1:1), ] > Year Amount > 2 2002 120 > 3 2003 175 > 4 2004 160 > > > On Mon, Sep 21, 2015 at 04:52:46PM +0200, Nico Gutierrez wrote: > > Hi All, > > > > I need to do the following operation from data.frame: > > > > df <- data.frame(Year = c("2001", "2002", "2003", "2004", "2005", "2006", > > "2007"), Amount = c(150, 120, 175, 160, 120, 105, 135)) > > df[which.max(df$Amount),] #to extract row with max Amount. > > > > Now I need to do 3 years average around the max Amount value (ie: > > mean(120,175,160)) > > > > Thanks! > > N > > > > [[alternative HTML version deleted]] > > > > ______________________________________________ > > R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see > > 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. > > > > ______________________________________________ > R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see > 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. >