Defining and isolating a region of an image internal to a bounding curve

I was recently looking over this very nice question by March Ho: Counting elements which are inside another element on a different colour channel

There are often times where I find myself wanting to draw a curve or define a polytope to cut out and isolate a part of an image for further analysis. One solution, for example, would be to define a set of polytope vertices, e.g.:

poly = {{658., 1224.}, {672., 1054.}, {507., 871.}, {358., 876.}, {344., 1432.}, {483., 1410.}};


And then calculate a winding number for each pixel in the image, or equivalently apply InPolygonQ, to define a "cutout" region. Here's a naive approach:

CutoutRegion = ImageData[pic];
ImageDimX = ImageDimensions[pic][[1]]
ImageDimY = ImageDimensions[pic][[2]]
Length[CutoutRegion[[1]]]

For[y = 1, y <= ImageDimY, y++,
For[x = 1, x <= ImageDimX, x++,
pt = {x, (ImageDimY-y)};
If[GraphicsMeshInPolygonQ[poly, pt] == True,
CutoutRegion[[y, x]] = 1;
,
CutoutRegion[[y, x]] = 0;
];
];
];

ImageMultiply[Image[CutoutRegion], pic]


This works, however very slowly, and it's a little bit clumsy to define a region of interest with a polygon. Really you'd want to freehand draw something.

My question is:

• Is there a trivial way to speedup the above approach?
• Is there a more elegant way to "hand" or "mouse" define a region of interest in an image and isolate it to generate a final product similar to the output of the above approach?
-

Is there a trivial way to speedup the above approach?

You can just rasterize a polygon and use the resulting bitmap as a mask:

cutoutRegion = Binarize[Rasterize[
Graphics[Polygon[poly],
PlotRange -> {{0, imageDimX}, {0, imageDimY}},
ImageSize -> imageDimX]]]


(btw: It's bad style to start variable names with uppercase letters)

Is there a more elegant way to "hand" or "mouse" define a region of interest in an image and isolate it to generate a final product similar to the output of the above approach?

The closest thing I'm aware of is using LocatorPane with LocatorAutoCreate -> True to enter the polygon.

-
I believe you need PlotRangePadding -> 0 as well; I am adding it to the answer. If this is incorrect just revert the edit. –  Mr.Wizard Aug 6 '13 at 23:23

Here's a small program I wrote for practice. It uses nikie's technique, I also used it here.

The code

locatorPositions[dim_, 0] := {};
locatorPositions[dim_, n_] := Module[{r},
r = 0.8 Min[dim/2];
Table[dim/2 + {r Cos[\[Theta]], r Sin[\[Theta]]}, {\[Theta], 0,
2 \[Pi], 2 \[Pi]/n}]
]
locatorConnectingLines[pos_] := Line /@ Partition[pos, 2, 1, {1, 1}];
ColorNegate[
Binarize[Rasterize[
Graphics[Polygon[pos],
PlotRange -> {{0, dim[[1]]}, {0, dim[[2]]}},
ImageSize -> dim]]]]
locatorInterface[image_, n_, f_] :=
DynamicModule[{dim = ImageDimensions[image],
pt = locatorPositions[ImageDimensions[image], n],
background = image},
Panel[
Column[
{
LocatorPane[
Dynamic[pt],
Dynamic[
Show[background,
Graphics[{Green, Dynamic[locatorConnectingLines[pt]]}]]
], Appearance -> Style["*", Large, Green],
LocatorAutoCreate -> Length[pt] == 0],
Button["Apply function",
background =
Show[background,
]
}
]
]
];


Example usage

stu = Import["http://upload.wikimedia.org/wikipedia/commons/6/65/2011_State_of_the_Union.jpg"];
locatorInterface[stu, 10, Blur[#, 12] &]


Before pressing the button:

After pressing the button:

• This is not very fast on my computer. It's probably much faster to manipulate the pixels directly.
• If you set the second argument, the number of locators, to zero, you will be able to create locators by alt-clicking or, on Mac OS, cmd-clicking.
-

Here's another approach, similar to Anon's:

i = ExampleData[{"TestImage", "Girl"}]
Manipulate[
Row[
{Show[
i,
Graphics[{Red,
Opacity[0.6],
Dynamic[Polygon[u]]},
PlotRange -> 2,
Background -> White]],
Panel[result]
}],
{{u, {{100, 100}, {200, 200}, {200, 100}}},
Locator,
LocatorAutoCreate -> True},
Initialization :> (crop[i_, poly_] := Module[{mask},
Rasterize[
Graphics[
Polygon[poly],

result contains the answer for further experiments...
The more the better, yes. There are as well a collection of 3D images---and many more with WolframAlpha["image of ..."] –  Matthias Odisio Aug 14 '13 at 22:30