Loop over rows of dataframe applying function with if-statement
Solution 1
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"] <- 0
t.d
V1 V2 V3 V4
1 1 4 7 0
2 2 5 8 10
3 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$V
all the time.
In the second step I subset all of the rows that don't fulfill your conditions and set V4 for these to 0.
Solution 2
ifelse
is your friend here:
t.d$V4<-ifelse((t.d$V1>1)&(t.d$V3<9), t.d$V1+ t.d$V3, 0)
Solution 3
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 seconds
system.time( ifelse((t.d$V1>1)&(t.d$V3<9), t.d$V1+ t.d$V3, 0) ) # 5.08 seconds
system.time( { t.d <- within(t.d, V4 <- V1 + V3);
t.d[!(t.d$V1>1 & t.d$V3<9), "V4"] <- 0 } ) # 4.50 seconds
Elinka
Updated on April 10, 2020Comments
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Elinka about 4 years
I'm new to R and I'm trying to sum 2 columns of a given dataframe, if both the elements to be summed satisfy a given condition. To make things clear, what I want to do is:
> t.d<-as.data.frame(matrix(1:9,ncol=3)) > t.d V1 V2 V3 1 4 7 2 5 8 3 6 9 > t.d$V4<-rep(0,nrow(t.d)) > for (i in 1:nrow(t.d)){ + if (t.d$V1[i]>1 && t.d$V3[i]<9){ + t.d$V4[i]<-t.d$V1[i]+t.d$V3[i]} + } > t.d V1 V2 V3 V4 1 4 7 0 2 5 8 10 3 6 9 0
I need an efficient code, as my real dataframe has about 150000 rows and 200 columns. This gives an error:
t.d$V4<-t.d$V1[t.d$V1>1]+ t.d$V3[t.d$V3>9]
Is "apply" an option? I tried this:
t.d<-as.data.frame(matrix(1:9,ncol=3)) t.d$V4<-rep(0,nrow(t.d)) my.fun<-function(x,y){ if(x>1 && y<9){ x+y} } t.d$V4<-apply(X=t.d,MAR=1,FUN=my.fun,x=t.d$V1,y=t.d$V3)
but it gives an error as well. Thanks very much for your help.