Considering a DateObject d and TimeObject t:
In[1]:= t = TimeObject[{14, 1, 45.`}]
In[2]:= d = DateObject[{2012, 06, 11}]
How can I achieve this result:
In[3]:= dt = DateObject[{2012, 06, 11, 14, 1, 45.`}]
Thanks!
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4 Answers
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Another, shorter way is
Append[d, t]
From the docs,
DateObject[date,time]represents the specified date list andTimeObjecttime.
If you need the list you mention in the question, just convert the DateObject using DateList:
DateList@Append[d, t]
(* {2012, 6, 11, 14, 1, 45.} *)
answered Jun 17, 2015 at 11:56
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A little late to the party here but this also works:
In[28]:= DateObject[d, t]
as Szabolcs noted, the documentation references this, but doesn't give an explicit example(which I will make sure gets added).
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6
Strip off their heads. replace them with List and then Flatten them:
dt=DateObject@@(List @@@ {d, t} // Flatten)
DateObject[{2012, 6, 11, 14, 1, 45.}]
Or ....
dt=DateObject @@ Join @@ {d[[1]], t[[1]]}
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$\begingroup$ Why is the result no quantity/object, but only text? $\endgroup$– SuTronJun 17, 2015 at 11:06
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$\begingroup$ Why not
Join[]the lists instead? :) $\endgroup$ Jun 17, 2015 at 11:21 -
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1$\begingroup$ Well, I had
DateObject @@ Join[d[[1]], t[[1]]], myself… :) $\endgroup$ Jun 17, 2015 at 11:24
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Well...
d~d[[0]]~t
...seems to work.
