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Re: ListPlot3D
*To*: mathgroup at smc.vnet.net
*Subject*: [mg131199] Re: ListPlot3D
*From*: "djmpark" <djmpark at comcast.net>
*Date*: Mon, 17 Jun 2013 06:25:51 -0400 (EDT)
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I'll just discuss the first example. For that the entries in the array are
the z values. The coordinates of each z value are the row and column
position indices in the matrix. However if we use the DataRange Option when
plotting then the displayed x and y values will correspond to the DataRange
values, which is usually what we want.
This is a difficult plot command for beginners or people who do not often
use it. One has to figure it out each time. The maxim that if a command is
difficult to explain then it may be ill-designed may apply here. The
confusion is that the form of the matrix is not the same as that normally
produced by the Table statement with the x iterator first and the y iterator
second, nor that used in the Plot3D statement that uses the same iterator
order. Nor does it correspond to the values in the plot if you view it from
above. (It corresponds to the plot arrangement if you view it from below.)
It all depends on whether you want the plot to look like what you would
consider to be the function, or what you consider to be a specific matrix.
As an example let's take a nonsymmetrical array so we can distinguish x and
y variation.
array1 = Table[x y, {x, 0, 4}, {y, -2, 2}];
The following three displays show various forms of the matrix with row and
column labels:
1) The array as it would be produced by a Table command with x and y
iterators in that order.
2) The transpose of the original array.
3) The transpose of the original array with the rows reversed. This
corresponds to looking at the plot from above.
Grid[{{"x\\y", -2, -1, 0, 1, 2}, {0, 0, 0, 0, 0, 0},
{1, -2, -1, 0, 1, 2}, {2, -4, -2, 0, 2, 4},
{3, -6, -3, 0, 3, 6}, {4, -8, -4, 0, 4, 8}},
Dividers -> {{2 -> GrayLevel[0]}, {2 -> GrayLevel[0]}},
Spacings -> {{None, {}}, {None, {}}},
Background -> {{}, {}, {}}, Frame -> {{}, {}, {}},
ItemStyle -> {{}, {}, {}}, ItemSize ->
{{Automatic, {}}, {Automatic, {}}},
Alignment -> {{Right, {}}, {Center, {}}, {}},
BaseStyle -> {14, Plain, FontFamily -> "Times"}]
Grid[{{"y\\x", 0, 1, 2, 3, 4}, {-2, 0, -2, -4, -6, -8},
{-1, 0, -1, -2, -3, -4}, {0, 0, 0, 0, 0, 0},
{1, 0, 1, 2, 3, 4}, {2, 0, 2, 4, 6, 8}},
Dividers -> {{2 -> GrayLevel[0]}, {2 -> GrayLevel[0]}},
Spacings -> {{None, {}}, {None, {}}},
Background -> {{}, {}, {}}, Frame -> {{}, {}, {}},
ItemStyle -> {{}, {}, {}}, ItemSize ->
{{Automatic, {}}, {Automatic, {}}},
Alignment -> {{Right, {}}, {Center, {}}, {}},
BaseStyle -> {14, Plain, FontFamily -> "Times"}]
Grid[{{"y\\x", 0, 1, 2, 3, 4}, {2, 0, 2, 4, 6, 8},
{1, 0, 1, 2, 3, 4}, {0, 0, 0, 0, 0, 0},
{-1, 0, -1, -2, -3, -4}, {-2, 0, -2, -4, -6, -8}},
Dividers -> {{2 -> GrayLevel[0]}, {2 -> GrayLevel[0]}},
Spacings -> {{None, {}}, {None, {}}},
Background -> {{}, {}, {}}, Frame -> {{}, {}, {}},
ItemStyle -> {{}, {}, {}}, ItemSize ->
{{Automatic, {}}, {Automatic, {}}},
Alignment -> {{Right, {}}, {Center, {}}, {}},
BaseStyle -> {14, Plain, FontFamily -> "Times"}]
ListPlot3D and ListDensityPlot require the Transpose of the array as it
would be produced by Table.
ListPlot3D[array1 // Transpose,
InterpolationOrder -> 2,
Mesh -> 4,
DataRange -> {{0, 4}, {-2, 2}},
AxesLabel -> {x, y, z},
BoxRatios -> {1, 1, 3/4}]
ListDensityPlot[array1 // Transpose,
DataRange -> {{0, 4}, {-2, 2}},
ColorFunctionScaling -> False,
ColorFunction -> (ColorData["SolarColors"][Rescale[#, {-8, 8}]] &),
Frame -> True, FrameTicks -> True]
Plot3D requires the same iterator order as Table, i.e. it corresponds to the
original array.
Plot3D[x y, {x, 0, 4}, {y, -2, 2},
Mesh -> 4,
BoxRatios -> {1, 1, 3/4}]
But an ArrayPlot requires the third form of matrix if we want it to
correspond to the other two plots. (But I suppose one would usually want an
ArrayPlot to correspond to a specific matrix and not a function.)
ArrayPlot[array1 // Transpose // Reverse,
DataRange -> {{0, 4}, {-2, 2}},
ColorFunctionScaling -> False,
ColorFunction -> (ColorData["SolarColors"][Rescale[#, {-8, 8}]] &),
Frame -> True, FrameTicks -> True]
And if we want the ListPlot3D to look like the original matrix we would use:
ListPlot3D[array1 // Reverse,
InterpolationOrder -> 2,
Mesh -> 4,
DataRange -> Reverse[{{0, 4}, {-2, 2}}],
AxesLabel -> {y, x, z},
BoxRatios -> {1, 1, 3/4},
ViewPoint -> {1/2, -2, 0.5} 5]
Got that?
David Park
djmpark at comcast.net
http://home.comcast.net/~djmpark/index.html
From: amannucci [mailto:Anthony.J.Mannucci at jpl.nasa.gov]
I could use some help with ListPlot3D. These are examples from the
documentation. My preconceived notion of what this should do is place a data
point at coordinates {x,y,z} and connect a surface along the z_i. Here is an
example from the Mathematica documentation:
ListPlot3D[{{1, 1, 1, 1}, {1, 2, 1, 2}, {1, 1, 3, 1}, {1, 2, 1, 4}}, Mesh
-> All] I cannot figure out how to read these data. What are the x,y values?
They are apparently "the x and y coordinate values for each data point to be
successive integers starting at 1." How does this thread through the data?
For the first point, x=1, y=1, z=1. For the last point, x=4,y=4, z= 4. What
about intermediate points? I can't figure it out. Is it x=1, y= 1..4,
x=2,y=1..4,x=3,y=1..4,x=4,y=1..4? The documentation does not say.
There is also the data triplets. E.g. from the documentation,
ListPlot3D[{{0, 0, 1}, {1, 0, 0}, {0, 1, 0}}, Mesh -> All]
This I understand. It is a triplet of {x,y,z}. However, when I feed in the
following data set, the graph is completely blank:
{{0., 0., 0.}, {0.1, 0.1, 0.248514}, {0.2, 0.2, 0.329812}, {0.3, 0.3,
0.324989}, {0.4, 0.4, 0.275382}, {0.5, 0.5, 0.207606}, {0.6, 0.6,
0.138975}, {0.7, 0.7, 0.0799098}, {0.8, 0.8, 0.0358315}, {0.9, 0.9,
0.008981}, {1., 1., 0.}}
x and y span 0->1. There are z-values, but nothing is plotted. If I manually
change the first three data points to be (mimicing the docs):
{0., 0., 0.}, {1.0, 0.0, 0.248514}, {0., 1., 0.329812}
I then see a surface. I cannot figure this out.
Thanks for any help you can provide.
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