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seen Apr 15 at 8:56

Apr
1
comment Image processing: extracting ordered contour
A good solution, thank you! Just one more question: do you see a way how to assure that all the contours are passed in the same direction: clockwise or counterclockwise?
Apr
1
accepted Image processing: extracting ordered contour
Mar
31
comment Image processing: extracting ordered contour
@Szabolcs Agreed. The idea is that you know how the curve goes when you have the original image. And I'm thinking about a way not to lose this information. When you come to the boundary the way it is described, you have already lost it.
Mar
31
comment Image processing: extracting ordered contour
@Szabolcs Indeed, that is a quick solution. Using ListPlot[boundary[[FindCurvePath[boundary][[1]]]], Joined -> True] I get an ordered curve. The question remains about MorphometricComponents[].
Mar
31
revised Image processing: extracting ordered contour
added 84 characters in body
Mar
31
asked Image processing: extracting ordered contour
Mar
31
accepted Plot with plot markers without using ListPlot
Mar
31
comment Plot with plot markers without using ListPlot
Thank you. Choosing this solution as the most natural one, although two others are also good.
Mar
21
asked Plot with plot markers without using ListPlot
Mar
3
comment Numerical Differentiation using 1500 data points
@DanielLichtblau Yeah, you are write. It seems to be a mathematical artifact of converting from Sin and Cos to Exp and losing physical meaning of frequency. Still, it's a very interesting question, but suprisingly it is not discussed here.
Feb
24
comment Numerical Differentiation using 1500 data points
Nice answer! But could someone explain please, why the "freq" are calculated in two parts?
Oct
14
comment Filling a curve to a vertical axis
Right. Still, you know, even if I put all the comments in one string: RegionPlot[y > ArcCot[x], {x, 0, 1.0}, {y, 0.5, 2}, AxesLabel -> {x, y}, PlotRangePadding -> 0, FrameTicks -> {Automatic, Automatic, None, None}, Frame -> {True, True, False, False}, PlotRange -> {{-0.1, 1.1}, {-0.1, 2.1}}] the plot looses the curve. If you evaluate the string, you will obtain a region with a grey boundary, which is not exactly what I would like to have. I mean like that the initial curve is hidden, which is no surprise if one uses RegionPlot.
Oct
11
comment Filling a curve to a vertical axis
Right. That works. Although, I think, it would be nicer if I could fill to the axis itself: you see, now one has a small interval left (which is quite expected if one uses RegionPlot, surely).
Oct
11
comment Filling a curve to a vertical axis
Still there seems to be a strange artefact of filling near x=1...
Oct
11
comment Filling a curve to a vertical axis
To my mind, what you wish to obtain if the extrema do not occur at the endpoints, depends on what the nature of the curve you plot. In my case, the filling to the vertical axis represents the region in which a specific physical assumption is valid. So, if I had a sine curve like you show above, I would desire filling also through the white space under the curve between the minima. Although in my special case, such curves are not possible. Still, as a theoretical question, it is an interesting one.
Oct
11
comment Filling a curve to a vertical axis
Thank you, Michael. It is a nice solution indeed, what you proposed. And the curves you get are really pretty. I suppose that I would chose this one if I needed to fill several curves on the same plot.
Oct
7
comment Filling a curve to a vertical axis
Choosing yours as the first simple answer to what I sought.
Oct
7
comment Filling a curve to a vertical axis
Having thought about it, I should add, that may indeed be an advantage. Suppose filling to x=0 if it is not included may be impossible with ordianary methods. So, thanks again. )
Oct
7
accepted Filling a curve to a vertical axis
Oct
7
comment Filling a curve to a vertical axis
Thank you for the answer. Nice function! But I find that filling to a minimum value is just easier in this case.