I want to create a new column with a running sum of every value greater than 0. I have a dataframe:

df=data.frame(year=c('2007-04-01','2007-04-02','2007-04-03','2007-04-04','2007-04-05','2007-04-06'),air.temp=c(1,2,-1,3,1,0)

and I want to create:

df=data.frame(year=c('2007-04-01','2007-04-02','2007-04-03','2007-04-04','2007-04-05','2007-04-06'),air.temp=c(1,2,-1,3,1,0),temp.sum=c(1,3,3,6,7,7)) 

So far I have tried:

df$temp.sum <- if_else(df$air.temp > 0, cumsum(df$air.temp), 0)

Which resulted in

temp.sum=c(1,3,0,5,6,0))  

How do I not count values at or below 0, without changing the running sum? My dataset is 100,000+ observations, so simple suggestions are helpful!

I want to create a new column with a running sum of every value greater than 0. I have a dataframe:

df=data.frame(year=c('2007-04-01','2007-04-02','2007-04-03','2007-04-04','2007-04-05','2007-04-06'),air.temp=c(1,2,-1,3,1,0)

and I want to create:

df=data.frame(year=c('2007-04-01','2007-04-02','2007-04-03','2007-04-04','2007-04-05','2007-04-06'),air.temp=c(1,2,-1,3,1,0),temp.sum=c(1,3,3,6,7,7)) 

So far I have tried:

df$temp.sum <- if_else(df$air.temp > 0, cumsum(df$air.temp), 0)

Which resulted in

temp.sum=c(1,3,0,5,6,0))  

How do I not count values at or below 0, without changing the running sum? My dataset is 100,000+ observations, so simple suggestions are helpful!

• r
• if-statement
• cumsum

3 Answers 3

Use a parallel maximum to make negative values 0, then continue to do the cumulative sum.

cumsum(pmax(df$air.temp, 0))
#[1] 1 3 3 6 7 7

Seems very quick on 1.2M values:

x <- rep(df$air.temp, 2e5)
system.time(cumsum(pmax(x, 0)))
##   user  system elapsed 
##      0       0       0 

Another solution would be:

cumsum(df$air.temp*(df$air.temp > 0))
[1] 1 3 3 6 7 7

If you are using if_else/ifelse:

cumsum(if_else(df$air.temp > 0,df$air.temp, 0))
[1] 1 3 3 6 7 7

This answer uses the most brute-force while loop. The result is in summation.

a=c(1,2,-1,3,1,0)
n=length(a)
i=1
sum=0
summation=0
while(i<=n){
    if (a[i]>=0){
        sum=sum+a[i]
    }
    summation[i]=sum
    i=i+1
}