问题
I am trying to clean some data and would like to replace zeros with values from the previous date. I was hoping the following code works but it doesn't
temp = c(1,2,4,5,0,0,6,7)
temp[which(temp==0)]=temp[which(temp==0)-1]
returns
1 2 4 5 5 0 6 7
instead of
1 2 4 5 5 5 6 7
Which I was hoping for. Is there a nice way of doing this without looping?
回答1:
The operation is called "Last Observation Carried Forward" and usually used to fill data gaps. It's a common operation for time series and thus implemented in package zoo:
temp = c(1,2,4,5,0,0,6,7)
temp[temp==0] <- NA
library(zoo)
na.locf(temp)
#[1] 1 2 4 5 5 5 6 7
回答2:
You could use essentially your same logic except you'll want to apply it to the values vector that results from using rle
temp = c(1,2,4,5,0,0,6,0)
o <- rle(temp)
o$values[o$values == 0] <- o$values[which(o$values == 0) - 1]
inverse.rle(o)
#[1] 1 2 4 5 5 5 6 6
来源：https://stackoverflow.com/questions/17591949/recursive-replacement-in-r