Thank to both of you. Works great.

Marco -- Interesting concept: the structure of the empty set. Maybe related to 1==={ {} }, 2==={ {},{} } . . . . ;-)

Hi David,

I guess, that the same trick works:

DateListPlot[{FinancialData["GE", "Jan. 1, 2000"], FinancialData["AA", "Jan. 1, 2000"], {{{},}}},  PlotLegends -> Automatic]

The full version would be something like:

DateListPlot[{FinancialData["GE", "Jan. 1, 2000"], FinancialData["AA", "Jan. 1, 2000"], {}} /. {} -> {{{},}}, PlotLegends -> Automatic]

which gives

The main idea is that DateListPlot requires data in a different format from ListPlot. Roughly speaking a valid "empty set" for ListPlot, if the other lists are of the form {x,y}, is

{\[EmptySet]}

whereas DateListPlot requires a different "empty set". I suppose that there is a much nicer argument/way of solving this though.

Cheers,

Marco

Thank you, Marco. That is extremely useful By not stripping lists, I meant I would not what to just delete a list from the list of lists based on it having Length 0. But I can certainly run this rule and that will be much cleaner than handling the 4 cases separately. You rock!