How to get column mean for specific rows only? - r

Question

I need to get the mean of one column (here: score) for specific rows (here: years). Specifically, I would like to know the average score for three periods:
period 1: year <= 1983
period 2: year >= 1984 & year <= 1990
period 3: year >= 1991
This is the structure of my data:
country year score
Algeria 1980 -1.1201501
Algeria 1981 -1.0526943
Algeria 1982 -1.0561565
Algeria 1983 -1.1274560
Algeria 1984 -1.1353926
Algeria 1985 -1.1734330
Algeria 1986 -1.1327666
Algeria 1987 -1.1263586
Algeria 1988 -0.8529455
Algeria 1989 -0.2930265
Algeria 1990 -0.1564207
Algeria 1991 -0.1526328
Algeria 1992 -0.9757842
Algeria 1993 -0.9714060
Algeria 1994 -1.1422258
Algeria 1995 -0.3675797
...
The calculated mean values should be added to the df in an additional column ("mean"), i.e. same mean value for years of period 1, for those of period 2 etc.
This is how it should look like:
country year score mean
Algeria 1980 -1.1201501 -1.089
Algeria 1981 -1.0526943 -1.089
Algeria 1982 -1.0561565 -1.089
Algeria 1983 -1.1274560 -1.089
Algeria 1984 -1.1353926 -0.839
Algeria 1985 -1.1734330 -0.839
Algeria 1986 -1.1327666 -0.839
Algeria 1987 -1.1263586 -0.839
Algeria 1988 -0.8529455 -0.839
Algeria 1989 -0.2930265 -0.839
Algeria 1990 -0.1564207 -0.839
...
Every possible path I tried got easily super complicated - and I have to calculate the mean scores for different periods of time for over 90 countries ...
Many many thanks for your help!

Answer 1

datfrm$mean <-
with (datfrm, ave( score, findInterval(year, c(-Inf, 1984, 1991, Inf)), FUN= mean) )
The title question is a bit different than the real question and would be answered by using logical indexing. If one wanted only the mean for a particular subset say year >= 1984 & year <= 1990 it would be done via:
mn84_90 <- with(datfrm, mean(score[year >= 1984 & year <= 1990]) )

Answer 2

Since findInterval requires year to be sorted (as it is in your example) I'd be tempted to use cut in case it isn't sorted [proved wrong, thanks #DWin]. For completeness the data.table equivalent (scales for large data) is :
require(data.table)
DT = as.data.table(DF) # or just start with a data.table in the first place
DT[, mean:=mean(score), by=cut(year,c(-Inf,1984,1991,Inf))]
or findInterval is likely faster as DWin used :
DT[, mean:=mean(score), by=findInterval(year,c(-Inf,1984,1991,Inf))]

Answer 3

If the rows are ordered by year, I think the easiest way to accomplish this would be:
m80_83 <- mean(dataframe[1:4,3]) #Finds the mean of the values of column 3 for rows 1 through 4
m84_90 <- mean(dataframe[5:10,3])
#etc.
If the rows are not ordered by year, I would use tapply like this.
list.of.means <- c(tapply(dataframe$score, cut(dataframe$year, c(0,1983.5, 1990.5, 3000)), mean)
Here, tapply takes three parameters:
First, the data you want to do stuff with (in this case, datafram$score).
Second, a function that cuts that data up into groups. In this case, it will cut the data into three groups based on the dataframe$year values. Group 1 will include all rows with dataframe$year values from 0 to 1983.5, Group 2 will include all rows with dataframe$year values from 1983.5 to 1990.5, and Group 3 will include all rows with dataframe$year values from 1983.5 to 3000.
Third, a function that is applied to each group. This function will apply to the data you selected as your first parameter.
So, list.of.means should be a list of the 3 values you are looking for.