Styjun

I'm using R to prepare a data.frame which will be used in mixed-effects regression later. I'm pretty new to R, I've tried to search and find the solution for my problem but I couldn't find exactly what I want.

My data.frame has 20071 rows and I want to add a new column but it has a length of 1767 and I need to fill the rest with zeros.
Here's the code:

data$M1 <- c(data$M1,(data$Mw[(data$Mw > 6.5)]-6.5))

and I'm getting this error:

Error in `$<-.data.frame`(`*tmp*`, "M1", value = c(0.0999999999999996, : 


replacement has 1767 rows, data has 20071

So I knew that I just need to add zeros of length of empty rows, but I couldn't find an answer in my search.

You could do this:

data$M1 <- pmax(data$Mw - 6.5, 0)

The idea here is the following: You want to create a vector that contains data$Mw - 6.5 unless data$Mw < 6.5, in which case the vector should be zero. This means that you will have zero exactly when data$Mw - 6.5 < 0. So, for each i, your vector will contain the maximum of data$Mw - 6.5 and 0.

This is exactly what the function pmax() is for: it takes multiple vectors as inputs and returns the elementwise maximum. This is easiest seen with an example:

pmax(c(1, 4), c(3, 2))
[1] 3 4

The first value of the output corresponds to max(1, 3) the second value to max(4, 2).

This is considerably faster than using ifelse().

What you are doing is extracting data that lives up to your condition (1767 rows), thus a vector that is shorter than you number of rows in your dataframe.

You should use "ifelse" instead.

data$M1 <- ifelse(data$Mw > 6.5,
 data$Mw - 6.5,
 0)

If the number is above 6.5 you subtract 6.5 from the number, else you return zero.

This should do what you want:

data$M1[1768:20071] <- 0

or, if M1 is a separate vector or column of another data-frame:

data$M1 <- c(M1, rep(0, 20071-1768))

Using cbind.fill() from the rowr library, we can bind different size dataframes/vectors while filling holes with a desired fill.

 > a=as.data.frame(matrix(0,4,2),stringsAsFactors = FALSE)
 > a
 V1 V2
 1 0 0
 2 0 0
 3 0 0
 4 0 0
 > b=c(1,2,3)

 > cbind.fill(a,b,fill=0)
 V1 V2 object
 1 0 0 1
 2 0 0 2
 3 0 0 3
 4 0 0 0

Another base R way is to create the column filled with zeros first and then use a logical index.

data$M1 <- 0
data$M1[data$Mw > 6.5] <- data$Mw[data$Mw > 6.5] - 6.5

This is probably faster.