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I don't understand how the first of the basic examples in the documentation on ListPlot3D is generating the plot that it does. 

ListPlot3D[{{1, 1, 1, 1}, {1, 2, 1, 2}, {1, 1, 3, 1}, {1, 2, 1, 4}},  Mesh -> All]

plot

Which of these values is it taking as $x$, $y$ and $z$ values and what is the fourth value in each set used for? I'm trying to learn Mathematica with zero previous knowledge of arrays, so I don't know if that's the problem or if I'm misunderstanding some completely different aspect.

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    $\begingroup$ From the details: In ListPlot3D[array], array must be a rectangular array. Each element can be either a single real number representing a value, or an (x, y, z) triple. There will be holes in the surface corresponding to array elements that do not represent explicit height values. ListPlot3D[array] by default takes the x and y coordinate values for each data point to be successive integers starting at 1. $\endgroup$
    – b3m2a1
    Jan 2, 2018 at 16:50
  • $\begingroup$ @b3m2a1 So then in the example because they aren't triples then each element is a z value: ListPlot3D[{{1, 1, 1, 1}, {1, 2, 1, 2}, {1, 1, 3, 1}, {1, 2, 1, 4}}, Mesh -> All] and it's assigning x and y co-ordinates in integer steps? So would the first element, 1, be assigned x=1,y=1 and the sixth element is x=6,y=6? $\endgroup$
    – user13948
    Jan 2, 2018 at 16:56
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    $\begingroup$ Yeah. The array basically unfurls to this: MapIndexed[Append[#2, #] &, {{1, 1, 1, 1}, {1, 2, 1, 2}, {1, 1, 3, 1}, {1, 2, 1, 4}}, {2}]//Flatten[#, 1]&. But the Interpolation is different. You can see the first uses only the array as the mesh whereas using that array gives a full Mesh (b.c. it interpolates differently, I think) $\endgroup$
    – b3m2a1
    Jan 2, 2018 at 16:59
  • $\begingroup$ @b3m2a1 So, if I've got this right, it ignores subgroupings? It ignores the fact that the numbers come in sets of four and just makes it into one list of z values? $\endgroup$
    – user13948
    Jan 2, 2018 at 17:10
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    $\begingroup$ I think not quite. Unless I misunderstood the docs, it uses the array indexing, i and j, as x and y. $\endgroup$
    – b3m2a1
    Jan 2, 2018 at 17:13

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