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I can get a colored ListLinePlot by doing something like
ListLinePlot[Range[420, 680, 20], ColorFunction -> "VisibleSpectrum", ColorFunctionScaling -> False]
However, as indicated by the help file ("ColorFunction requires at least one dataset to be Joined"), if I do the equivalent
ListPlot[Range[420, 680, 20], ColorFunction -> "VisibleSpectrum", ColorFunctionScaling -> False]
all my points are blue. Is there a nice way to get ColorFunction to work for ListPlot with Joined -> False?
That is, is there a nicer way to get something like
ListPlot[
List /# Transpose[{Range[(680 - 420)/20 + 1], Range[420, 680, 20]}],
PlotMarkers -> ({Graphics[{#, Disk[]}], 0.05} & /# ColorData["VisibleSpectrum"] /# Range[420, 680, 20])
]
?
(Also, does anyone have an explanation of why Mathematica requires Joined -> True in order to make use of ColorFunction?)
Edit: I'm also looking for a way to do a similar coloring with ErrorListPlot in the ErrorBarPlots package.
The problem is, that Joined->True draws a Line[] which can be given VertexColors for each containing point. I assume doing the same for the points when setting Joined->False leads to situations where it does not work. Nevertheless, Line[] and Point[] work pretty much the same in your case. So what is about
ListLinePlot[Range[420, 680, 20], ColorFunction -> "VisibleSpectrum",
ColorFunctionScaling -> False] /. Line[arg___] :> Point[arg]
And, by the way, if your using a ListLinePlot only, where the only Line[] directives arising are the one from your data, this should work even if you have more datasets and {x,y} coordinates
data = Transpose[Table[{{x, Sin[x]}, {x, Cos[x]}}, {x, 0, 2 Pi, 0.2}]];
ListLinePlot[data, ColorFunction -> Hue] /. Line[arg___] :> Point[arg]
You can use DiscretePlot:
data = Range[420, 680, 20];
DiscretePlot[data[[i]], {i, Length[data]},
ColorFunction -> "VisibleSpectrum", ColorFunctionScaling -> False,
Filling -> None]
If you're plotting a list of x,y points, it gets a little trickier:
data = Transpose[{Range[420, 680, 20], Range[400, 530, 10]}];
mapping = Apply[Rule, data, 2];
DiscretePlot[i/.mapping, {i, data[[;;,1]]},
ColorFunction -> "VisibleSpectrum", ColorFunctionScaling -> False,
Filling -> None]
It does seem rather odd that DiscretePlot will let you color the points differently whereas ListPlot won't. I'm sure it must have something to do with the implementation details, but I can't think of a reason why that would be the case.
I came across this problem in my work too. I assign a colour to each point in the following manner:
data = ...
ListPlot[data] /. Point[args___] :> Point[args, VertexColors -> {c1, c2, ...}]
where c1 is the colour for the first data point, and so on. The colour list may be programmatically generated, eg
ColorData["Rainbow"] /# (Range#Length#data / Length#data)
Here is the result.
The good points of this method are as follows.
It's straightforward: we have a list of pairs, then we create a corresponding list of colours.
Our original ListPlot code needs not be modified (eg, changed into ListLinePlot).
Is there any reason that underlies mathematica's way of presenting this graph
ListPlot[
Table[{x, x*01}, {x, -5, 5, .08}],
PlotStyle -> White,
Filling -> 0,
FillingStyle -> {Dashed, Brown}]
While the dashing is present for the part of the graph above the zero boundary, another part of the graph has the filling that is solid.
Am I doing something wrong?
Not that wrong. Mathematica is interpreting your filling style as being Dashed below zero and Brown above. You just need another pair of braces, like so:
ListPlot[Table[{x, x*01}, {x, -5, 5, .08}], PlotStyle -> White,
Filling -> 0, FillingStyle -> {{Dashed, Brown}}]
Hope that helps.
I would like to draw a rectangle (or more) which printed on paper shows the rectangle in units of cm.
So
Graphics[{Rectangle[{0, 0}, {19, 28}], Orange, Rectangle[{0, 0}, {1, 1}]}]
will print out as two rectangles which can be measured as exactly 1cm x 1cm (orange one) and the black one as 19x28 cm.
It seems that some variables are important:
The ImageSize and of course the AspectRatio.
I used AspectRatio->19/28 and for the ImageSize various settings like ImageSize->{19*27,28*27} but it keeps being not very accurate.
I export the graphics to TIFF and then print out with windows photo gallery to a full page photo. Does anyone have experience with this? There must be a formula instead of trial and error.
UPDATE:
I tried the suggestion of #Szabolcs and I used the following code:
g = Graphics[{White, EdgeForm[Directive[Thick, Black]],
Rectangle[{0, 0}, {18, 28}], Orange, Rectangle[{0, 0}, {10, 10}]}]
final = Show[g, AspectRatio -> Automatic,
PlotRange -> {{-0.5, 18.5}, {-0.5, 28.5}}]
cm = 72/2.54
Export["final.pdf", Show[final, ImageSize -> {19 cm, 29 cm}]]
This works great. The orange rectangle of 10x10cm is when measured exactly 10x10cm
the cm 72/2.54 value was not what I expected since I though Windows uses 96dpi and Mac 72dpi (reading from the www). However 72 is the value that works.
I've also beenn playing with the frames but then it gets ugly. Haven't found a way to get the right results dispite playing with all possible settings. What should work is create the frames/ticks etc myself inside the selected boundaries but that's not the path I would like to pursue..
g = Graphics[{Rectangle[{0, 0}, {19, 28}], Orange, Rectangle[{0, 0}, {1, 1}]}]
Okay, first thing you need to do is set the x and y directions to use the same units, which means
Show[g, AspectRatio -> Automatic]
But this is already the default.
Second thing you need to do is choose a size and range for your plot area. Let's make it 21 by 30 with your rectangles centred:
plotArea = {{0, 21}, {0, 30}} - {1, 1}
Show[g, AspectRatio -> Automatic, PlotRange -> plotArea]
Third thing you need to do is turning off adding any padding/margins that make the actual size of your figure larger than your plot range:
final = Show[g, AspectRatio -> Automatic, PlotRange -> plotArea, PlotRangePadding -> 0, ImagePadding -> 0]
I believe ImageMargins does not make a difference, but if it does, set that to 0 as well.
The final thing you need to do is export this to a printable format that preserves the image dimensions, and set the size of the image so that 1 cm will be 1 unit on your plot. Mathematica accepts image sizes in printer's points, so let's define:
cm = 72/2.54
Export["final.pdf", Show[final, ImageSize -> 21 cm]]
We want the plot to be 21 cm wide because it's 21 units wide. Use PDF as export format, not TIFF. The ImageSize needs to be used inside Show to work around some problems with Export ...
Now open your PDF in Adobe Reader, open the print dialogue, and make sure that Page Scaling is set to None! I don't know how to do this in other readers ... Also make sure your figure fits the paper (21 by 30 cm is too large for A4 ...)
I'm not going to do a test print, so let me know if this works for you :-) The size of the PDF generated this way is exactly 21 by 30 cm, so if something goes wrong, it must happen at the printing stage.
I believe you need to add PlotRangePadding -> None and set image dimensions appropriately.
In this case, the "bounding box" size is the same as your larger rectangle: {19, 28}
The robust way to do this is to set ImageSize to the actual required dimensions, and make use of ImageResolution, which will embed this value into the TIFF file for proper printing:
cm = 72 / 2.54;
g = Graphics[{Rectangle[{0, 0}, {19, 28}], Orange,
Rectangle[{0, 0}, {1, 1}]}, PlotRangePadding -> None,
ImageSize -> {19, 28}*cm];
Export["print.tif", g, ImageResolution -> 300]
This assumes that you want to print from a raster format (TIFF) but you can also export to other formats such as PDF with the same method.
Graph[] has a tendency to cut off vertex labels in Mathematica. I am looking for a robust workaround.
Example:
Graph[{1 -> 2, 2 -> 3, 3 -> 1}, VertexLabels -> "Name"]
My present workaround:
SetOptions[Graph, ImagePadding -> 12]
This is not robust because the value of ImagePadding needs to be manually adjusted depending on the label size.
Apparently using FullGraphics on the Graph object will fix the clipping for the purpose of display, at the expense of interactivity.
Per the comment below, Show[] works as well, and avoids modifying the graphics.
Here are two possible workarounds.
Enlarge the vertex size and place the labels within the vertex. Of course, this also depends on the length of the labels, but for shortish labels it works well, whereas your example above clips off any label of more than one character for vertex 1.
ex:
Table[Graph[{1 -> 2, 2 -> 3, 3 -> 1}, VertexSize -> 0.3,
VertexLabels -> Table[i ->
Placed["vertex" <> ToString[i], p], {i, 3}],
VertexShapeFunction -> "Square", PlotLabel -> p],
{p, {Left, Top, Right, Bottom, Center}}]
Use tooltips to store the labels instead of displaying them on the graphic. [Edit: Center probably looks the best, and then you can wrap labels by putting \n in your string if you need to, but again, depends on the label length.]
ex:
Graph[{1 -> 2, 2 -> 3, 3 -> 1}, VertexLabels -> Placed["Name", Tooltip]]
While this stops you from being able to see all the labels at the same time, you never have any clipping.
If I do a Plot with Frame->True is there a way I can find the coordinates of the corners of the Frame in the absolute coordinates of the image? I have the numerical values of PlotRange and PlotRangePadding but note that I don't want to tamper with the actual plot in any way, just find out where in the full display area Mathematica chooses to place the frame/axes of the plot.
As pointed out by Brett Champion, I'm looking for the coordinates {x,y} such that Scaled[{0,0}] == ImageScaled[{x,y}].
[Note that I edited this question to remove my confusing misuse of the term "scaled coordinates".]
The corners of the frame are at Scaled[{0,0}] and Scaled[{1,1}].
The corners of the full graphic (including labels) are at ImageScaled[{0,0}] and ImageScaled[{1,1}].
Converting between them is hard, although in theory it's possible to convert Scaled and user (unscaled) coordinates if you know the actual, numeric, settings for PlotRange and PlotRangePadding.
Depending on your application, you might also be able to use MousePosition, which knows these things as well.
Rasterize (and HTML export) also know how to find bounding boxes of annotations, in a bitmap/pixel coordinate system:
In[33]:= Rasterize[
Plot[Sin[x], {x, 0, 10}, Frame -> True,
Prolog -> {LightYellow,
Annotation[Rectangle[Scaled[{0, 0}], Scaled[{1, 1}]], "One",
"Region"]}], "Regions"]
Out[33]= {{"One", "Region"} -> {{22., 1.33573}, {358.9, 209.551}}}
Here's how dreeves used that Rasterize trick to make a function to return exactly what he was looking for (note the assumption of a global variable imgsz which gives the ImageSize option for rasterizing the plot -- the coordinates of the frame depend on that value):
(* Returns the geometry of the frame of the plot:
{width, height, x offset, y offset, total width, total height}. *)
geom[p_Graphics] := Module[{q, x1, y1, x2, y2, xmax, ymax},
q = Show[p, Prolog->{Annotation[Rectangle[Scaled[{0,0}], Scaled[{1,1}]],
"MAGIC00","MAGIC11"]}];
{{x1,y1}, {x2,y2}} = Rasterize[q, "Regions", ImageSize->imgsz][[1,2]];
{xmax,ymax} = Rasterize[p, "RasterSize", ImageSize->imgsz];
{x2-x1, y2-y1, x1, y1, xmax, ymax}]
The coordinates of the upper left corner of the frame are always Scaled[{0,1}].
The coordinates of the lower right corner of the frame are always Scaled[{1,0}].
Let's place large points at the upper left and lower right corners:
Plot[Cos[x], {x, 0, 10}, Frame -> True,
Epilog -> {PointSize[.08], Point[Scaled[{0, 1}]], Point[Scaled[{1, 0}]]} ]
When I click on the graph (see below) , it is obvious that there is no padding around the frame of the plot.
Now, with ImagePadding on, let's place Points in the same corners:
Plot[Cos[x], {x, 0, 10}, Frame -> True,
ImagePadding -> {{37, 15}, {20, 48}},
Epilog -> {PointSize[.08], Point[Scaled[{0, 1}]], Point[Scaled[{1, 0}]]} ]
The Points stay at the corners of the graph frame.
There is ImagePadding around the graph frame.
EDIT: Based on the clarification of the question by dreeves.
Plot[Cos[x], {x, 1, 9}, ImageSize -> 300, AspectRatio -> 1,
Frame -> True, ImagePadding -> 30,
FrameTicks -> {Range[9], Automatic},
Epilog -> {PointSize[.08], Point[Scaled[{0, 1}]], Point[Scaled[{1, 0}]]}]
I've drawn the plot as 300x300 to simplify the numbers.
Here's the analysis.
Documentation states that ImagePadding "is defined within ImageSize".
The image shown above has a width and height of 300 pixels.
There is a 30 pixel margin drawn around the frame; this corresponds to 10% of the width and height.
So the frame corners should be, starting from the origin, at ImageScaled[{.1,.1}], ImageScaled[{.9,.1}, ImageScaled[{.9,.9}] & ImageScaled[{.1,.9}].
It's easy to work out the value for other AspectRatios and ImageSizes.
One possibility is to take manual control of ImagePadding:
Plot[Sin[x], {x, 0, 10}, Frame -> True,
ImagePadding -> {{30, 5}, {20, 5}}]
ImageTake[Rasterize[%], {5, -20}, {30, -5}]