用数组表达式完成数据操作任务，无需写大量循环。向量化的数组操作会比纯 Python 的等价实现快一到两个数量级。

In [42]:  points = np.arange(-5, 5, 0.01) # 1000 equally spaced points
In [44]: xs, ys = np.meshgrid(points, points) # np.meshgrid 接受两个一维数组,(x,y)对生成一个二维矩阵

In [45]: xs
Out[45]:
array([[-5.  , -4.99, -4.98, ...,  4.97,  4.98,  4.99],
       [-5.  , -4.99, -4.98, ...,  4.97,  4.98,  4.99],
       [-5.  , -4.99, -4.98, ...,  4.97,  4.98,  4.99],
       ...,
       [-5.  , -4.99, -4.98, ...,  4.97,  4.98,  4.99],
       [-5.  , -4.99, -4.98, ...,  4.97,  4.98,  4.99],
       [-5.  , -4.99, -4.98, ...,  4.97,  4.98,  4.99]])

In [46]: ys
Out[46]:
array([[-5.  , -5.  , -5.  , ..., -5.  , -5.  , -5.  ],
       [-4.99, -4.99, -4.99, ..., -4.99, -4.99, -4.99],
       [-4.98, -4.98, -4.98, ..., -4.98, -4.98, -4.98],
       ...,
       [ 4.97,  4.97,  4.97, ...,  4.97,  4.97,  4.97],
       [ 4.98,  4.98,  4.98, ...,  4.98,  4.98,  4.98],
       [ 4.99,  4.99,  4.99, ...,  4.99,  4.99,  4.99]])
In [65]: z = np.sqrt(xs ** 2 + ys ** 2)

In [66]: z
Out[66]:
array([[7.07106781, 7.06400028, 7.05693985, ..., 7.04988652, 7.05693985,
        7.06400028],
       [7.06400028, 7.05692568, 7.04985815, ..., 7.04279774, 7.04985815,
        7.05692568],
       [7.05693985, 7.04985815, 7.04278354, ..., 7.03571603, 7.04278354,
        7.04985815],
       ...,
       [7.04988652, 7.04279774, 7.03571603, ..., 7.0286414 , 7.03571603,
        7.04279774],
       [7.05693985, 7.04985815, 7.04278354, ..., 7.03571603, 7.04278354,
        7.04985815],
       [7.06400028, 7.05692568, 7.04985815, ..., 7.04279774, 7.04985815,
        7.05692568]])

In [67]: import matplotlib.pyplot as plt

In [68]: plt.imshow(z, cmap=plt.cm.gray); plt.colorbar()
Out[68]: <matplotlib.colorbar.Colorbar at 0x1c56b5b3dd8>

In [69]: plt.title("Image plot of $\sqrt{x^2 + y^2}$ for a grid of values")
Out[69]: Text(0.5, 1.0, 'Image plot of $\\sqrt{x^2 + y^2}$ for a grid of values')

In [70]: plt.show()