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I have a list of graphics

p1 = Plot[x Sin[x] - 1, {x, -2.5, 2.5}]
p2 = Plot[Evaluate[D[x Sin[x] - 1, x]], {x, -2.5, 2.5},PlotStyle -> Red]
ListNew = {p1, p2};

I can use Show to combine them:

Show[p1,p2]

I can also use MapThread if I understand correctly here:

MapThread[f,{x1,x2}] = f[x1,x2]

But I dont understand why using MapThread with Show wont work?

MapThread[Show,ListNew]
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closed as off-topic by b3m2a1, m_goldberg, MarcoB, Bob Hanlon, Alex Trounev Aug 14 at 14:23

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question arises due to a simple mistake such as a trivial syntax error, incorrect capitalization, spelling mistake, or other typographical error and is unlikely to help any future visitors, or else it is easily found in the documentation." – b3m2a1, m_goldberg, MarcoB, Bob Hanlon, Alex Trounev
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ You want Apply not MapThread. $\endgroup$ – b3m2a1 Aug 12 at 23:07
  • $\begingroup$ MapThread is for mapping a multivariate function over multiple lists of arguments, not for applying such a function to a sequence of arguments. $\endgroup$ – m_goldberg Aug 13 at 1:25
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MapThread does not work the way you believe. Run your actual code:

 MapThread[f,{x1,x2}]

You get an error message: "Object x1 at position {2,1} in MapThread[f,{x1,x2}] has only 0 of required 1 dimensions." Basically, you are being told that MapThread wants Lists as the elements in the second argument. Your expression x1 is not a List nor is x2, nor is p1 or p2. As stated in the documentation under Possible Issues: "All arguments must be lists of the same length." (Although, as set forth in the documentation, MapThread can also work with Associations).

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