How to replace a column in a data frame with another one with a different size
On Jul 8, 2012, at 9:31 AM, Stathis Kamperis <ekamperi at gmail.com> wrote:
Hello everyone, I have a dataframe with 1 column and I'd like to replace that column with a moving average. Example:
library('zoo')
mydat <- seq_len(10)
mydat
[1] 1 2 3 4 5 6 7 8 9 10
df <- data.frame("V1" = mydat)
df
V1 1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9 9 10 10
df[df$V1 <- rollapply(df$V1, 3, mean)]
Error in `$<-.data.frame`(`*tmp*`, "V1", value = c(2, 3, 4, 5, 6, 7, 8, : replacement has 8 rows, data has 10
I'm not sure you need the outer df[...] -- I think you just want df$V1 <- rollapply(df$V1,3,mean) However, this will still give you the error message you're seeing because rollapply() only returns 8 values here (you don't get the "endpoints" by default). To get the right number of rows, you want rollapply(df$V1, 3, mean, fill = NA) # Change NA if desired which will put NA's on each end and give you a length 10 result, as needed. Best, Michael
I could use a temporary variable to store the results of rollapply() and then reconstruct the data frame, but I was wondering if there is a one-liner that can achieve the same thing. Best regards, Stathis P.S. If you don't mind, cc me at your reply because I'm not subscribed to the list (but I will check the archive anyway).
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