dataframe - Compute values relative to specific factor in R data frame -


i have data frame in r of following form:

        bc solvopt istrng tsolv      epb     1   10       1      0  0.10 -78.1450     2   10       1      1  0.15 -78.7174     3   10       1     10  0.14 -78.7175     4   10       1    100  0.12 -78.7184     5   10       1   1000  0.09 -78.7232     6   10       1      2  0.15 -78.7175     7   10       1     20  0.14 -78.7176     8   10       1    200  0.12 -78.7192     30  10       2      0  0.10 -78.1450     31  10       2      1  0.11 -78.7174     32  10       2     10  0.11 -78.7175     33  10       2    100  0.10 -78.7184     34  10       2   1000  0.13 -78.7232     35  10       2      2  0.11 -78.7174     36  10       2     20  0.10 -78.7176     37  10       2    200  0.10 -78.7192     59  10       3      0  0.16 -78.1450     60  10       3      1  0.23 -78.7174     61  10       3     10  0.21 -78.7175     62  10       3    100  0.19 -78.7184     63  10       3   1000  0.17 -78.7232     64  10       3      2  0.22 -78.7175     65  10       3     20  0.21 -78.7176     66  10       3    200  0.18 -78.7192     88  10       4      0  0.44 -78.1450     89  10       4      1 14.48 -78.7162     90  10       4     10 12.27 -78.7175     91  10       4    100  1.23 -78.7184     92  10       4   1000  0.44 -78.7232     93  10       4      2 14.52 -78.7172     94  10       4     20  6.16 -78.7176     95  10       4    200  0.62 -78.7192

i want add column frame shows relative error in epb each value of bc , istrng relative solvopt=3.

for example, compute relative difference in epb @ each row subtract epb value of corresponding row same value of bc , istrng solvopt=3.

is there easy way short of splitting multiple data frames (for each solvopt) , remunging together?

the end goal generate plots of relative error vs istrng each value of bc using qplot.

if merge subset solvopt==3 against main data on both bc , istrong, , subtract difference, should result want, e.g.:

newdat <- merge(dat,dat[dat$solvopt==3,c("bc","istrng","epb")], by=c("bc","istrng")) newdat$diff <- with(newdat, epb.x - epb.y)

...or in 1 fell swoop using match , interaction:

dat$diff <- dat$epb - dat[dat$solvopt==3,"epb"][match(   with(dat,                  interaction(bc,istrng) ),   with(dat[dat$solvopt==3,], interaction(bc,istrng) )  )]

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