Converting a data frame to xts

This is clearly documented --- xts and zoo objects are formed by supplying two arguments, a vector or matrix carrying data and Date, POSIXct, chron, ... type supplying the time information (or in the case of zoo the ordering).

So do something like

 qxts <- xts(q[,-1], order.by=q[,1])

and you should be set.

Well, as.xts assumes by default that the dates are stored in the rownames of the data.frame. Hence the error message. A quick and dirty fix is:

rownames(q) = q[1]
as.xts(q)

But you get an extra column with the dates string. Ideally you would construct the data.frame with the dates as rownames to start with.

Here's a solution using the tidyquant package, which contains a function as_xts() that coerces a data frame to an xts object. It also contains as_tibble() to coerce xts objects to tibbles ("tidy" data frames).

Recreate the data frame (note that the date-time class is used in "tidy" data frames, but any unambiguous date or date time class can be used):

> q
# A tibble: 3 × 2
                    t     x
               <dttm> <dbl>
1 2006-01-01 00:00:00     1
2 2006-01-01 01:00:00     2
3 2006-01-01 02:00:00     3

Use as_xts() to convert to "xts" class. Specify the argument, date_col = t, to designate the "t" column as the dates to use as row names:

> library(tidyquant)
> as_xts(q, date_col = t)
                    x
2006-01-01 00:00:00 1
2006-01-01 01:00:00 2
2006-01-01 02:00:00 3

The return is an xts object with the proper date or date-times as row names.

Here is a posible solution:

library(timetk)
q <- xts::xts(q[,-1], order.by = q$t)