SciPy 创建 2D 多边形蒙版
我需要使用标准 Python 包创建一个 numpy 2D 数组,它表示多边形的二进制掩码。
- 输入:多边形顶点,图像尺寸
- 输出:多边形的二进制掩码(numpy 2D 数组)
(更大的上下文:我想使用 scipy.ndimage.morphology.distance_transform_edt 获得此多边形的距离变换。)
任何人都可以告诉我如何做到这一点?
I need to create a numpy 2D array which represents a binary mask of a polygon, using standard Python packages.
- input: polygon vertices, image dimensions
- output: binary mask of polygon (numpy 2D array)
(Larger context: I want to get the distance transform of this polygon using scipy.ndimage.morphology.distance_transform_edt.)
Can anyone show me how to do this?
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答案其实很简单:
The answer turns out to be quite simple:
作为 @Anil 答案的稍微更直接的替代方案,matplotlib 有
matplotlib.nxutils.points_inside_poly
可用于快速栅格化任意多边形。例如,它产生(一个布尔numpy数组):
您应该能够很好地将
grid
传递给任何 scipy.ndimage.morphology 函数。As a slightly more direct alternative to @Anil's answer, matplotlib has
matplotlib.nxutils.points_inside_poly
that can be used to quickly rasterize an arbitrary polygon. E.g.Which yields (a boolean numpy array):
You should be able to pass
grid
to any of the scipy.ndimage.morphology functions quite nicely.乔评论的更新。
自评论发布以来,Matplotlib API 已发生变化,现在您需要使用子模块
matplotlib.path
提供的方法。工作代码如下。
An update on Joe's comment.
Matplotlib API has changed since the comment was posted, and now you need to use a method provided by a submodule
matplotlib.path
.Working code is below.
作为 @Yusuke N. 答案的一种稍微替代方案,请考虑使用 matplotlib.path,它与 from PIL import Image 的方法一样高效, ImageDraw(不需要安装
Pillow
,不需要考虑integer
或float
。对我有用吗?)工作代码如下:
结果图像如下,其中暗区为
False
,亮区为True
。As a slight alternative to @Yusuke N.'s answer, consider using
matplotlib.path
, which is just as efficient as the one byfrom PIL import Image, ImageDraw
(no need to installPillow
, no need to considerinteger
orfloat
. Useful me?)Working code is below:
And the result image is below, where dark area is
False
, bright area isTrue
.您可以尝试使用python的图像库PIL。首先初始化画布。然后创建一个绘图对象,并开始绘制线条。这是假设多边形位于 R^2 中并且输入的顶点列表的顺序正确。
输入 = [(x1, y1), (x2, y2), ..., (xn, yn)] , (width, height)
这是您正在寻找的东西,还是您在询问不同的东西?
You could try to use python's Image Library, PIL. First you initialize the canvas. Then you create a drawing object, and you start making lines. This is assuming that the polygon resides in R^2 and that the vertex list for the input are in the correct order.
Input = [(x1, y1), (x2, y2), ..., (xn, yn)] , (width, height)
Is this what you were looking for, or were you asking something different?
这是一个实现 @IsaacSutherland 方法(接受的答案)的函数,并进行了一些我认为有用的修改。欢迎评论!
poly_mask()
接受多个多边形作为输入,以便输出蒙版可以由多个最终不连接的多边形区域组成。此外,因为在某些情况下 0 不是一个好的掩码值(例如,如果 0 是要应用掩码的数组的有效值),我添加了一个
value
关键字来设置实际掩码值(例如非常小/很大的数字或 NAN):为了实现此目的,掩码将转换为浮点数组。我更喜欢直接使用
shape
作为输入,而不是 (width, height),这样我就可以像这样使用它:其中
my_array
是掩码必须指向的数组应用(当然,或者具有相同形状的另一个数组)。Here is a function that implements @IsaacSutherland method (the accepted answer) with some modifications I find useful. Comments are welcome!
poly_mask()
accepts multiple polygons as input so that the output mask can be made of multiple, eventually not connected, polygonal regions.Moreover, because in some cases 0 is not a good value for masking (e.g. if 0 is a valid value of the array to which the maskhas to be applied ) I added a
value
keyword that sets the actual masking value (e.g. a very small/big number or NAN): to achieve this the mask is converted to array of float.Instead of (width, height) I prefer to have directly
shape
as input so that I can use it like this:where
my_array
is the array to which the mask has to be applied (or another array with the same shape, of course).这是一个 cv2 版本:
Here is a cv2 version: