Transforming a time-series into a data frame and back

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Here are two ways. The first way creates dimnames for the matrix about to be created and then strings out the data into a matrix, transposes it and converts it to data frame. The second way creates a by list consisting of year and month variables and uses tapply on that later converting to data frame and adding names.

# create test data
set.seed(123)
tt <- ts(rnorm(12*5, 17, 8), start=c(1981,1), frequency = 12)

1) matrix. This solution requires that we have whole consecutive years

dmn <- list(month.abb, unique(floor(time(tt))))
as.data.frame(t(matrix(tt, 12, dimnames = dmn)))

If we don’t care about the nice names it is just as.data.frame(t(matrix(tt, 12))) .

We could replace the dmn<- line with the following simpler line using @thelatemail’s comment:

dmn <- dimnames(.preformat.ts(tt))

2) tapply. A more general solution using tapply is the following:

Month <-  factor(cycle(tt), levels = 1:12, labels = month.abb)
tapply(tt, list(year = floor(time(tt)), month = Month), c)

Note: To invert this suppose X is any of the solutions above. Then try:

ts(c(t(X)), start = 1981, freq = 12)

Update

Improvement motivated by comments of @latemail below.

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