Simple method for converting float array to DateTime (or similar)?

New to Julia
6

@rafael.guerra I like this one! Nice that you can easily get millisecond precision with DateTime.(t), or preserve microsecond precision with Date.(t) and Time.(t). So the full solution becomes:

using Dates, CompoundPeriods

T = [   2017 09 06 00 00 04.911360;
        2017 09 06 04 15 55.193727;
        2017 09 06 22 55 12.256655]

tcp = Year.(T[:,1]) + Month.(T[:,2]) + Day.(T[:,3]) + Hour.(T[:,4]) +
      + Minute.(T[:,5]) + Nanosecond.(round.(Int,1e9*T[:,6]))

t = canonical.(tcp)

sdt = DateTime.(t)  # millisecond precision
sdate = Date.(t)    # date only
stime = Time.(t)    # time only with microsecond precision