I have the following toy data.table
x = data.table(dt = c('2015-01-01','2015-02-01','2016-04-01'))
I'm trying to add an extra column to the table which has the three letter abbreviation of the date. I tried
x$dt = strptime(x$dt,'%b')
but it gave back the following, a warning and an empty field.
Warning messages:
1: In `[<-.data.table`(x, j = name, value = value) :
Supplied 9 items to be assigned to 3 items of column 'dt' (6 unused)
2: In `[<-.data.table`(x, j = name, value = value) :
Coerced 'list' RHS to 'character' to match the column's type. Either change the target column to 'list' first (by creating a new 'list' vector length 3 (nrows of entire table) and assign that; i.e. 'replace' column), or coerce RHS to 'character' (e.g. 1L, NA_[real|integer]_, as.*, etc) to make your intent clear and for speed. Or, set the column type correctly up front when you create the table and stick to it, please.
> x
dt mnth
1: c(NA, NA, NA)
2: c(NA, NA, NA)
3: c(NA, NA, NA)
I found prior Q&A on SO that said that strptime
returned a list but haven't been able to accomplish this relatively simple task. Could someone provide some pointers?
We can use format
format(strptime(x$dt, '%Y-%m-%d'), '%b')
Or as @Frank mentioned, the format.Date
can be applied to Date
class
format(as.Date(x$dt), "%b")
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