Loop over rows of dataframe applying function with if-statement

This operation doesn't require loops, apply statements or if statements. Vectorised operations and subsetting is all you need:

t.d <- within(t.d, V4 <- V1 + V3)t.d[!(t.d$V1>1 & t.d$V3<9), "V4"] <- 0t.d  V1 V2 V3 V41  1  4  7  02  2  5  8 103  3  6  9  0

Why does this work?

In the first step I create a new column that is the straight sum of columns V1 and V4. I use within as a convenient way of referring to the columns of d.f without having to write d.f$V all the time.

In the second step I subset all of the rows that don't fulfill your conditions and set V4 for these to 0.

ifelse is your friend here:

t.d$V4<-ifelse((t.d$V1>1)&(t.d$V3<9), t.d$V1+ t.d$V3, 0)

I'll chip in and provide yet another version. Since you want zero if the condition doesn't mach, and TRUE/FALSE are glorified versions of 1/0, simply multiplying by the condition also works:

t.d<-as.data.frame(matrix(1:9,ncol=3))t.d <- within(t.d, V4 <- (V1+V3)*(V1>1 & V3<9))

...and it happens to be faster than the other solutions ;-)

t.d <- data.frame(V1=runif(2e7, 1, 2), V2=1:2e7, V3=runif(2e7, 5, 10))system.time( within(t.d, V4 <- (V1+V3)*(V1>1 & V3<9)) )         # 3.06 secondssystem.time( ifelse((t.d$V1>1)&(t.d$V3<9), t.d$V1+ t.d$V3, 0) ) # 5.08 secondssystem.time( { t.d <- within(t.d, V4 <- V1 + V3);                t.d[!(t.d$V1>1 & t.d$V3<9), "V4"] <- 0 } )       # 4.50 seconds