In order to emphasize data, I try to draw a box from top to down of the graphpane. This works well as long I stay with a linear axis.
When I change to a log axis the height of the box will not fit. How do I calculate the right height of such a box for log scales?
Change the coordinate frame for your box object. Use the XAxisYChartFraction, and (0,1) as the Y coordinates. This should do the trick.
Related
When you make a line profile of all x-values or all y-values the extraction from each pixel is clear. But when you take a line profile along a diagonal, how does DM choose which pixels to use in the one dimensional readout?
Not really a scripting question, but I'm rather certain that it uses bi-linear interpolation between the grid-points along the drawn line. (And if perpendicular integration is enabled, it does so in an integral.) It's the same interpolation you would get for a "rotate" image.
In fact, you can think of it as a rotate-image (bi-linearly interpolated) with a 'cut-out' afterwards, potentially summed/projected onto the new X-axis.
Here is an example
Assume we have a 5 x 4 image, which gives the grid as shown below.
I'm drawing top-left corners to indicate the coordinates system pixel convention used in DigitalMicrgraph, where
(x/y)=(0/0) is the top-left corner of the image
Now extract a LineProfile from (1/1) to (4/3). I have highlighted the pixels for those coordinates.
Note, that a Line drawn from the corners seems to be shifted by half-a-pixel from what feels 'natural', but that is the consequence of the top-left-corner convention. I think, this is why a LineProfile-Marker is shown shifted compared to f.e. LineAnnotations.
In general, this top-left corner convention makes schematics with 'pixels' seem counter-intuitive. It is easier to think of the image simply as grid with values in points at the given coordinates than as square pixels.
Now the maths.
The exact profile has a length of:
As we can only have profiles with integer channels, we actually extract a LineProfile of length = 4, i.e we round up.
The angle of the profile is given by the arc-tangent of dX and dY.
So to extract the profile, we 'rotate' the grid by that angle - done by bilinear interpolation - and then extract the profile as grid of size 4 x 1:
This means the 'values' in the profile are from the four points:
Which are each bi-linearly interpolated values from four closest points of the original image:
In case the LineProfile is averaged over a certain width W, you do the same thing but:
extract a 2D grid of size L x W centered symmetrically over the line.i.e. the grid is shifted by (W-1)/2 perpendicular to the profile direction.
sum the values along W
I am plotting meteor observation data from a sky camera, sometimes using right ascension and declination for my x and y axes, at other times azimuth and elevation. The problem I have in both cases is with the x axis when my observations span the 360 degree mark. Sometimes I get a batch of observations on the left of my plot (near zero degrees, and a batch on the right hand side (near 360 degrees), with a big expanse of nothing in the middle. Is there any easy way I can change the x axis so that the 360/0 degree wrap over is in the centre of the plot? I would still want to show the true azimuth (or right ascension) in the axis labels.
PS. Pointing the camera elsewhere is not an option ]1
PPS So in the image shown the plots on the left hand side should be to the right of those on the right hand side with x axis from 250 (via 360/0) to 100.
PPPS So the second image shows what I am after - but I got to that by doctoring the data - as is obvious from the scale of the x axis in this plot
In blender, I know there is a way to flip a group of keyframes by the x axis by copying and pasting (ctrl+shift+v) but how can I flip them upside down?
Move-rotate-scale operators can be used in 2D space too, so you can scale animation curves with S Y -1 (scale along Y axis -1 times), probably adjusting pivot point to origin first (e.g. set pivot to "2d cursor" and set 2d cursor somewhere at Y=0. The same can be done for X axis.
I have a horizontal bar chart where I decided to hide the Y axis and display the axis labels inside the chart:
Now I'd like to know how I can tell Dimple to use the free space (see red rectangle) available from hiding the Y axis.
You set the bounds of the plot area with setMargins or setBounds. Dimple doesn't automatically size to accommodate axes. So just set the left margin to about 10px.
I have create a Quartz composition for use in MAC OS program as part of my interface.
I am relying on the fact that when you have composition sprite movement (a text bullet point in my case) is limited both in the X plane and Y plane to minimum -1 and maximum +1.
When I scale up the window / make my window full screen, I find that the horizontal plane (X axis) remains the same, with -1 being my far left point and +1 being my far right point. However the vertical plane (Y axis) changes, in full screen mode it goes from -0.7 to +0.7.
This scaling is screwing with my calculations. Is there anyway to get the application to keep the scale as -1 to +1 for both horizontal and vertical planes? Or is there a way of determining the upper and lower limits?
Appreciate any help/pointers
Quartz Composer viewer Y limits are usually -0.75 -> 0.75 but it's only a matter of aspect ratio. X limits are allways -1 -> 1, Y ones are dependents on them.
You might want to assign dynamically customs width and heigth variables, capturing the context bounds size. For example :
double myWidth = context.bounds.size.width;
double myHeight = context.bounds.size.height;
Where "context" is your viewer context object.
If you're working directly with the QC viewer : you should use the Rendering Destination Dimensions patch that will give you the width and the height. Divide Height by 2, then multiply the result by -1 to have the other side.