如何删除圆内的一组网格网格点? [英] How to delete a set of meshgrid points inside a circle?

问题描述

我正在尝试创建一个网格网格，而网格中的某些点没有指定的坐标和半径.我无法减去落入圆内的网格点. 这是我的代码

I am trying to create a meshgrid without some of the points that falls within the circle having specified coordinates and a radius. I am not able to subtract the grid points falling inside the circle. Here is my code

import math
import numpy 
import matplotlib.pyplot as plt


N = 50
x_start, x_end = -2.0, 2.0 
y_start, y_end = -1.0, 1.0


x = numpy.linspace(x_start, x_end, N)
y = numpy.linspace(y_start, y_end, N)

circle_x, circle_y, r= 0.0, 0.0, 0.4

#x.remove((r-circle_x)**2)   #need some help with these two lines
#y.remove((r-circle_y)**2)


X, Y = numpy.meshgrid(x, y)


size = 10
fig = plt.figure()
plt.xlabel('x', fontsize = 16)
plt.ylabel('y', fontsize = 16)
plt.scatter(X, Y)
plt.show()

显示网格点的网格##标题

meshgrid showing the mesh points## Heading

推荐答案

如果您希望散点图仅包含圆外的点，请使用布尔索引来从2D网格划分"数组中仅选择那些点:

If you'd like a scatter plot with just the points outside the circle, use boolean indexing to select only those points from your 2D "meshgridded" array:

import numpy as np
import matplotlib.pyplot as plt

N = 50
x_start, x_end = -2.0, 2.0
y_start, y_end = -1.0, 1.0

x = np.linspace(x_start, x_end, N)
y = np.linspace(y_start, y_end, N)

x0, y0, radius = 0.0, 0.0, 0.4

x, y = np.meshgrid(x, y)
r = np.sqrt((x - x0)**2 + (y - y0)**2)

outside = r > radius

fig, ax = plt.subplots()
ax.set(xlabel='X', ylabel='Y', aspect=1.0)

ax.scatter(x[outside], y[outside])

plt.show()

另一方面，如果使用的是类似imshow的东西，需要2D输入，则需要掩盖内部的值(如@JulienSpronck所述，尽管最好将它们设置为np.nan或使用遮罩数组而不是将其设置为0)或在图像上设置剪切路径.

On the other hand, if you were using something like imshow that requires 2D input, you'd need to either mask the values inside (as @JulienSpronck mentions, though it would be better to set them to np.nan or use a masked array than to set them to 0) or set a clip path on the image.

但是，为了分散，您不需要2D输入.

For scatter, however, you don't need 2D input.

在nD数组上的布尔索引将返回1d结果.例如:

Boolean indexing on a nD array will return a 1d result. For example:

In [9]: x = np.arange(100).reshape(10, 10)

In [10]: x
Out[10]: 
array([[ 0,  1,  2,  3,  4,  5,  6,  7,  8,  9],
       [10, 11, 12, 13, 14, 15, 16, 17, 18, 19],
       [20, 21, 22, 23, 24, 25, 26, 27, 28, 29],
       [30, 31, 32, 33, 34, 35, 36, 37, 38, 39],
       [40, 41, 42, 43, 44, 45, 46, 47, 48, 49],
       [50, 51, 52, 53, 54, 55, 56, 57, 58, 59],
       [60, 61, 62, 63, 64, 65, 66, 67, 68, 69],
       [70, 71, 72, 73, 74, 75, 76, 77, 78, 79],
       [80, 81, 82, 83, 84, 85, 86, 87, 88, 89],
       [90, 91, 92, 93, 94, 95, 96, 97, 98, 99]])

In [11]: x[x > 75]
Out[11]: 
array([76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92,
       93, 94, 95, 96, 97, 98, 99])

因为scatter只是绘制点，所以不关心它们的连接方式，因此我们可以轻松地使用布尔索引的一维结果.

Because scatter is only plotting points, it doesn't care how they're connected, and we can easily use the 1D result of boolean indexing.

另一方面，如果要绘制图像，则需要2D网格.在这种情况下，您需要屏蔽这些值:

On the other hand, if you wanted to plot an image, you'd need a 2D grid. In that case, you'd want to mask the values instead:

In [12]: np.ma.masked_where(x <= 75, x)
Out[12]: 
masked_array(data =
 [[-- -- -- -- -- -- -- -- -- --]
 [-- -- -- -- -- -- -- -- -- --]
 [-- -- -- -- -- -- -- -- -- --]
 [-- -- -- -- -- -- -- -- -- --]
 [-- -- -- -- -- -- -- -- -- --]
 [-- -- -- -- -- -- -- -- -- --]
 [-- -- -- -- -- -- -- -- -- --]
 [-- -- -- -- -- -- 76 77 78 79]
 [80 81 82 83 84 85 86 87 88 89]
 [90 91 92 93 94 95 96 97 98 99]])

请注意，这如何维护输入的2D结构.

Notice how this maintains the 2D structure of the input.

另一方面，如果这是一个浮点数组，则可以很容易地将值设置为np.nan而不是对其进行遮罩.用imshow绘制时，两者的行为相同.在这种情况下，我使用了带掩码的数组，因为x是整数数组，并且不能包含NaN.

On a side note, if this were a floating point array, you could just as easily set the values to np.nan instead of masking them. When plotted with imshow, the two will behave identically. I used a masked array in this case because x was an integer array and can't contain NaN's.

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