Replace rows with 0s in dataframe with preceding row values diverse than 0

Question

Here an example of my dataframe:

df = read.table(text = 'a  b
120 5
120 5
120 5
119 0
118 0
88 3
88 3
87 0  
10 3
10 3
10 3
7 4
6 0
5 0
4 0', header = TRUE)

I need to replace the 0s within col b with each preceding number diverse than 0.

Here my desired output:

 a  b
120 5
120 5
120 5
119 5
118 5
88 3
88 3
87 3  
10 3
10 3
10 3
7 4
6 4
5 4
4 4

Until now I tried:

df$b[df$b == 0] = (df$b == 0) - 1

But it does not work. Thanks

Answer 1

na.locf from zoo can help with this:

library(zoo)
#converting zeros to NA so that na.locf can get them
df$b[df$b == 0] <- NA
#using na.locf to replace NA with previous value
df$b <- na.locf(df$b)

Out:

> df
     a b
1  120 5
2  120 5
3  120 5
4  119 5
5  118 5
6   88 3
7   88 3
8   87 3
9   10 3
10  10 3
11  10 3
12   7 4
13   6 4
14   5 4
15   4 4

Answer 2

Performing this task in a simple condition seems pretty hard, but you could also use a small for loop instead of loading a package.

for (i in which(df$b==0)) {
  df$b[i] = df$b[i-1]
}

Output:

> df
     a b
1  120 5
2  120 5
3  120 5
4  119 5
5  118 5
6   88 3
7   88 3
8   87 3
9   10 3
10  10 3
11  10 3
12   7 4
13   6 4
14   5 4
15   4 4

I assume that this could be slow for large data.frames

Answer 3

Here is a base R method using rle.

# get the run length encoding of variable
temp <- rle(df$b)
# fill in 0s with previous value
temp$values[temp$values == 0] <- temp$values[which(temp$values == 0) -1]
# replace variable
df$b <- inverse.rle(temp)

This returns

df
     a b
1  120 5
2  120 5
3  120 5
4  119 5
5  118 5
6   88 3
7   88 3
8   87 3
9   10 3
10  10 3
11  10 3
12   7 4
13   6 4
14   5 4
15   4 4

Note that the replacement line will throw an error if the first element of the vector is 0. You can fix this by creating a vector that excludes it.

For example

replacers <- which(temp$values == 0)
replacers <- replacers[replacers > 1]