paths optimizer

al.niessner · November 12, 2015, 3:47am

I have three paths that each define a bounding box. Lets just say I have
these three in (x,y) coordinates and all are moveto, lineto, lineto,
lineto, and closepoly:

1: (0, 0), (1.1, 0), (1.1, 1.1), (0, 1.1), (0, 0)
2: (.9, 0), (2, 0), (2, 1.1), (.9, 1.1), (.9, 0)
3: (.9, .9), (2, .9), (2, 2), (.9, 2) (.9, .9)

While my paths have more nodes and are not as cleanly vertical and
horizontal, it does cover the nature of what I am experiencing.

The first thing to note is that each box has overlap at x = 1 and y = 1
and all three boxes overlap at (1,1).

What I am looking for is an optimizer that would be smart enough to say
hey, because of the overlap, we can trim the boxes to:

1: (0,0) (1,0) (1,1) (0,1) (0,0)
2: (1,0) (2,0) (2,1), (1,1), (1,0)
3: (1,1) (2,1) (2,2), (1,2), (1,1)

I am not sure that optimizer is the right word to use either. I have
poked around matplotlib a bit (over many years) and tried searching but
guessing the search terms is not simple either. Hence, I am asking if
anyone is aware of matplotlib functionality that will help me do this
task. Any other tool outside of matplotlib would be just as nice of a
suggestion. One last bit, it has to work for N paths.

As always, thanks in advance.

···

--
Al Niessner

I have never found the companion that was so companionable as solitude.
- From Walden by Henry David Thoreau

The universe is indifferent, and life is brutal; however, it is man's
choice of behavior that makes them malevolent rather than benevolent.

Some will fall in love with life and drink it from a fountain
That is pouring like an avalanche coming down the mountain.
- From the song Pepper by the Butthole Surfers

ianthomas23 · November 12, 2015, 8:21am

Hello Al,

The search terms you are looking for are 'polygon clipping' and 'boolean
operations on polygons'.

Matplotlib can clip one polygon with respect to another, there is a simple
example at

Alternatively, for more flexibility use shapely. For example:

    from shapely.geometry import Polygon
    a = Polygon([(0, 0), (1.1, 0), (1.1, 1.1), (0, 1.1)])
    b = Polygon([(.9, 0), (2, 0), (2, 1.1), (.9, 1.1)])
    c = Polygon([(.9, .9), (2, .9), (2, 2), (.9, 2)])
    d = a.intersection(b).intersection(c)
    print(d)

which outputs

POLYGON ((0.9 0.9, 0.9 1.1, 1.1 1.1, 1.1 0.9, 0.9 0.9))

i.e. a rectangle from x = 0.9 to 1.1 and y from 0.9 to 1.1.

Ian Thomas

···

On 12 November 2015 at 03:47, Al Niessner <Al.Niessner at gmx.net> wrote:

I have three paths that each define a bounding box. Lets just say I have
these three in (x,y) coordinates and all are moveto, lineto, lineto,
lineto, and closepoly:

1: (0, 0), (1.1, 0), (1.1, 1.1), (0, 1.1), (0, 0)
2: (.9, 0), (2, 0), (2, 1.1), (.9, 1.1), (.9, 0)
3: (.9, .9), (2, .9), (2, 2), (.9, 2) (.9, .9)

While my paths have more nodes and are not as cleanly vertical and
horizontal, it does cover the nature of what I am experiencing.

The first thing to note is that each box has overlap at x = 1 and y = 1
and all three boxes overlap at (1,1).

What I am looking for is an optimizer that would be smart enough to say
hey, because of the overlap, we can trim the boxes to:

1: (0,0) (1,0) (1,1) (0,1) (0,0)
2: (1,0) (2,0) (2,1), (1,1), (1,0)
3: (1,1) (2,1) (2,2), (1,2), (1,1)

I am not sure that optimizer is the right word to use either. I have
poked around matplotlib a bit (over many years) and tried searching but
guessing the search terms is not simple either. Hence, I am asking if
anyone is aware of matplotlib functionality that will help me do this
task. Any other tool outside of matplotlib would be just as nice of a
suggestion. One last bit, it has to work for N paths.

As always, thanks in advance.

--
Al Niessner

I have never found the companion that was so companionable as solitude.
- From Walden by Henry David Thoreau

The universe is indifferent, and life is brutal; however, it is man's
choice of behavior that makes them malevolent rather than benevolent.

Some will fall in love with life and drink it from a fountain
That is pouring like an avalanche coming down the mountain.
- From the song Pepper by the Butthole Surfers

_______________________________________________
Matplotlib-users mailing list
Matplotlib-users at python.org
Matplotlib-users Info Page

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al.niessner · November 12, 2015, 4:34pm

If I understand them both correctly (will need more time to fully
appreciate what the documentation is telling me) it seems I will need to
use both. I can use shapely to find the intersections in order to form
my new path and then use matplotlib patches to clip the original
polygons. Maybe shapely can clip too but I did not find it in the first
30 seconds that I was looking at the documentation.

Thanks for the help.

···

On Thu, 2015-11-12 at 08:21 +0000, Ian Thomas wrote:

Hello Al,

The search terms you are looking for are 'polygon clipping' and
'boolean operations on polygons'.

Matplotlib can clip one polygon with respect to another, there is a
simple example at
clipping a triagle with a circle in matplotlib - Stack Overflow

Alternatively, for more flexibility use shapely. For example:

    from shapely.geometry import Polygon
    a = Polygon([(0, 0), (1.1, 0), (1.1, 1.1), (0, 1.1)])
    b = Polygon([(.9, 0), (2, 0), (2, 1.1), (.9, 1.1)])
    c = Polygon([(.9, .9), (2, .9), (2, 2), (.9, 2)])

    d = a.intersection(b).intersection(c)

    print(d)

which outputs

POLYGON ((0.9 0.9, 0.9 1.1, 1.1 1.1, 1.1 0.9, 0.9 0.9))

i.e. a rectangle from x = 0.9 to 1.1 and y from 0.9 to 1.1.

Ian Thomas

On 12 November 2015 at 03:47, Al Niessner <Al.Niessner at gmx.net> wrote:

        I have three paths that each define a bounding box. Lets just
        say I have
        these three in (x,y) coordinates and all are moveto, lineto,
        lineto,
        lineto, and closepoly:

        1: (0, 0), (1.1, 0), (1.1, 1.1), (0, 1.1), (0, 0)
        2: (.9, 0), (2, 0), (2, 1.1), (.9, 1.1), (.9, 0)
        3: (.9, .9), (2, .9), (2, 2), (.9, 2) (.9, .9)

        While my paths have more nodes and are not as cleanly vertical
        and
        horizontal, it does cover the nature of what I am
        experiencing.

        The first thing to note is that each box has overlap at x = 1
        and y = 1
        and all three boxes overlap at (1,1).

        What I am looking for is an optimizer that would be smart
        enough to say
        hey, because of the overlap, we can trim the boxes to:

        1: (0,0) (1,0) (1,1) (0,1) (0,0)
        2: (1,0) (2,0) (2,1), (1,1), (1,0)
        3: (1,1) (2,1) (2,2), (1,2), (1,1)

        I am not sure that optimizer is the right word to use either.
        I have
        poked around matplotlib a bit (over many years) and tried
        searching but
        guessing the search terms is not simple either. Hence, I am
        asking if
        anyone is aware of matplotlib functionality that will help me
        do this
        task. Any other tool outside of matplotlib would be just as
        nice of a
        suggestion. One last bit, it has to work for N paths.

        As always, thanks in advance.

        --
        Al Niessner

        I have never found the companion that was so companionable as
        solitude.
        - From Walden by Henry David Thoreau

        The universe is indifferent, and life is brutal; however, it
        is man's
        choice of behavior that makes them malevolent rather than
        benevolent.

        Some will fall in love with life and drink it from a fountain
        That is pouring like an avalanche coming down the mountain.
        - From the song Pepper by the Butthole Surfers

        _______________________________________________
        Matplotlib-users mailing list
        Matplotlib-users at python.org
        Matplotlib-users Info Page

--
Al Niessner

I have never found the companion that was so companionable as solitude.
- From Walden by Henry David Thoreau

The universe is indifferent, and life is brutal; however, it is man's
choice of behavior that makes them malevolent rather than benevolent.

Some will fall in love with life and drink it from a fountain
That is pouring like an avalanche coming down the mountain.
- From the song Pepper by the Butthole Surfers