year
1999 1999 1999 2003 2003 2005 2005 2005 2005 2007 2009 2009 2009
A1
15 7 24 6 65 5 89 56 21 15 19 7 23
Above table shows a data frame. I want to have a vector, lets say "median1" which has the median of those values in a1 corresponds to each year. And I know that with a for loop it is easy but I am trying to find a 'vectorized' based solution.
with data.table package if your data.frame is called DF
library(data.table)
DT = data.table(DF)
DT[,median(a1),by='year']
Use ave which is an R base function. Combining ave with transform you'll get a pretty nice output. Consider dat is your data.frame
> transform(dat, Median= ave(a1, year, FUN=median))
year a1 Median
1 1999 20 15.0
2 1999 15 15.0
3 1999 11 15.0
4 2003 11 7.0
5 2003 3 7.0
6 2007 89 40.5
7 2007 25 40.5
8 2007 56 40.5
9 2007 12 40.5
If you only want a vector consisting of medians per each year you can do:
> with(dat, ave(a1, year, FUN=median))
[1] 15.0 15.0 15.0 7.0 7.0 40.5 40.5 40.5 40.5
In base R, you can do this:
foo <- data.frame(
year=c(1999,1999,1999,2003,2003,2005,2005,2005,2005,2007,2009,2009,2009),
A1=c(15,7,24,6,65,5,89,56,21,15,19,7,23))
by(foo$A1,foo$year,median)
Strictly speaking, the result will not be a vector, but you can fix that:
as.vector(by(foo$A1,foo$year,median))
by() is always helpful when you want to do an operation by groups.
It's not clear to me, but it seems like you want the median of each year? If so...
## set up the data
> year <- c(1999,1999,1999,2003,2003,2005,2005,2005,2005,2007,2009,2009,2009)
> A1 <- c(15, 7, 24, 6, 65, 5, 89, 56, 21, 15, 19, 7, 23)
> dd <- data.frame(year, A1)
## solution
> xx <- c(do.call(cbind, lapply(split(dd, dd$year), function(x) median(x$A1))))
> names(xx) <- unique(dd$year)
> xx
1999 2003 2005 2007 2009
15.0 35.5 38.5 15.0 19.0
