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times is a list of times i.e. matching {{hour_, min_, sec_Real}..}.

I want to call f[Dynamic@times].

The following works for a single argument to the function; f[Dynamic@(times: {{_, _, _Real}..})]

Edited; The single argument version above works. The multiple argument version below which is closer to the actual code I am using doesn't.

timestamps = {{2008, 1, 2, 14., 31., 0.}, {2008, 1, 2, 14., 32., 0.}};
bars = {{36.67, 36.69, 36.61, 36.65, 20800.}, {36.68, 36.71, 36.68, 36.68, 
  2600.}};
dates = {{2008, 1, 2}, {2008, 1, 2}};
times = {{14., 31., 0.}, {14., 32., 0.}};
indicators = {36.69, 36.71};

In[203]:= Clear@FormatCoordinates;
FormatCoordinates[pt_, timestamps_, bars_, indicators_] := 
 Block[{}, "Matched non-dynamic version 1."]
FormatCoordinates[pt_, dates_, times : {{_, _, _Real} ..}, 
  indicators_] := Block[{}, "Matched non-dynamic version 2."]
FormatCoordinates[pt_, Dynamic@timestamps_, Dynamic@bars_, 
  Dynamic@indicators_] := Block[{}, "Matched dynamic version 1."]
FormatCoordinates[pt_, Dynamic@dates_, 
  Dynamic@(times : {{_, _, _Real} ..}), Dynamic@indicators_] := 
 Block[{}, "Matched dynamic version 2."]

In[208]:= FormatCoordinates[dummy, timestamps, bars, indicators]
Out[208]= "Matched non-dynamic version 1."

In[209]:= FormatCoordinates[dummy, Dynamic@timestamps, Dynamic@bars, 
 Dynamic@indicators]
Out[209]= "Matched dynamic version 1."

In[210]:= FormatCoordinates[dummy, dates, times, indicators]
Out[210]= "Matched non-dynamic version 2."

In[211]:= FormatCoordinates[dummy, Dynamic@dates, Dynamic@times, 
 Dynamic@indicators]
Out[211]= "Matched dynamic version 1."

UPDATE: Read @Kuba's comment in the answer below for the solution.

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4
  • 1
    $\begingroup$ Would you please add a usage example: an input along with a desired output. It is not clear what you require. If it helps, please add more than one example. $\endgroup$
    – Syed
    Commented Nov 11, 2021 at 16:52
  • $\begingroup$ @Syed - thanks, I've added an example. $\endgroup$
    – RobertNathaniel
    Commented Nov 11, 2021 at 19:10
  • $\begingroup$ Could you provide complete code that shows the issue? Looks like it ought to work to me. $\endgroup$
    – Carl Woll
    Commented Nov 11, 2021 at 20:00
  • $\begingroup$ @Carl - provided more complete code. $\endgroup$
    – RobertNathaniel
    Commented Nov 12, 2021 at 11:09

1 Answer 1

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To match f[Dynamic@times] you do not need the double brackets and the Repeated. E.g. you could define:

ClearAll[f];
t = {hour_, min_, sec_Real};
f[Dynamic[{hour_, min_, sec_Real} ]] = "Worked";

If you now say:

f[Dynamic[{11, 11, 11.}]]

MMA responds with "Worked"

It is more complex with repeated dates. Names should not be used with "Repeated" as this implies that the corresponding values are the same:

ClearAll[f];
t = {hour_, min_, sec_Real};
f[Dynamic[{{_, _, _Real} .. }]] = "Worked";
f[Dynamic[{{11, 11, 11.}, {12, 12, 12.}}]]
(*Worked*)

If you want to get at the individual values you can give the whole input a name and get the values by indexing. E.g. to get the first value:

ClearAll[f];
t = {hour_, min_, sec_Real};
f[Dynamic[x : {{_, _, _Real} ..}]] := x[[1, 1]];
f[Dynamic[{{11, 12, 12.}, {12, 12, 12.}}]]
(*11*)
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  • $\begingroup$ thanks for your answer but it looks like I was premature in approving it. It doesn't work for a list of times as stated in the original question. f[Dynamic[{{11, 11, 11.}, {12, 12, 12.}}]] will fail to match. $\endgroup$
    – RobertNathaniel
    Commented Nov 11, 2021 at 18:44
  • $\begingroup$ This is obvious because f[Dynamic[{{11, 11, 11.}, {12, 12, 12.}}]] is not the same as f[Dynamic[{11, 11, 11.}]] I added this case to my answer. $\endgroup$
    – Daniel Huber
    Commented Nov 11, 2021 at 20:24
  • $\begingroup$ I amended the question to provide a more complete example. If what you say is true about using names with Repeated, then how come it works with the non-dynamic example but not with Dynamic. Also, how would I then be able to refer to the repeated argument in the function definition without a name? $\endgroup$
    – RobertNathaniel
    Commented Nov 12, 2021 at 11:13
  • $\begingroup$ You gave a name to the whole list and get elements by indexing. $\endgroup$
    – Daniel Huber
    Commented Nov 12, 2021 at 11:48
  • $\begingroup$ I still don't understand. Please could you look at the amended question again and provide me a working example if possible. Thanks. $\endgroup$
    – RobertNathaniel
    Commented Nov 13, 2021 at 11:23

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