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812 lines
45 KiB
Markdown
812 lines
45 KiB
Markdown
# 数组广播机制
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In [1]:
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```py
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import numpy as np
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```
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正常的加法：
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In [2]:
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```py
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a = np.array([[ 0, 0, 0],
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[10,10,10],
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[20,20,20],
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[30,30,30]])
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b = np.array([[ 0, 1, 2],
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[ 0, 1, 2],
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[ 0, 1, 2],
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[ 0, 1, 2]])
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a + b
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```
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Out[2]:
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```py
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array([[ 0, 1, 2],
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[10, 11, 12],
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[20, 21, 22],
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[30, 31, 32]])
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```
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将 `b` 的值变成一维的 `[0,1,2]` 之后的加法：
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In [3]:
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```py
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b = np.array([0,1,2])
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a + b
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```
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Out[3]:
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```py
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array([[ 0, 1, 2],
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[10, 11, 12],
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[20, 21, 22],
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[30, 31, 32]])
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```
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结果一样，虽然两个数组的维数不一样，但是 **Numpy** 检测到 `b` 的维度与 `a` 的维度匹配，所以将 `b` 扩展为之前的形式，得到相同的形状。
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对于更高维度，这样的扩展依然有效。
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如果我们再将 `a` 变成一个列向量呢？
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In [4]:
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```py
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a = np.array([0,10,20,30])
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a.shape = 4,1
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a
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```
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Out[4]:
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```py
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array([[ 0],
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[10],
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[20],
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[30]])
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```
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In [5]:
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```py
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b
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```
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Out[5]:
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```py
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array([0, 1, 2])
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```
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In [6]:
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```py
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a + b
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```
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Out[6]:
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```py
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array([[ 0, 1, 2],
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[10, 11, 12],
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[20, 21, 22],
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[30, 31, 32]])
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```
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可以看到，虽然两者的维度并不相同，但是**Numpy**还是根据两者的维度，自动将它们进行扩展然后进行计算。
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对于 **Numpy** 来说，维度匹配当且仅当：
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* 维度相同
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* 有一个的维度是1
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匹配会从最后一维开始进行，直到某一个的维度全部匹配为止，因此对于以下情况，**Numpy** 都会进行相应的匹配：
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| A | B | Result |
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| --- | --- | --- |
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| 3d array: 256 x 256 x 3 | 1d array: 3 | 3d array: 256 x 256 x 3 |
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| 4d array: 8 x 1 x 6 x 1 | 3d array: 7 x 1 x 5 | 3d array: 8 x 7 x 6 x 5 |
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| 3d array: 5 x 4 x 3 | 1d array: 1 | 3d array: 5 x 4 x 3 |
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| 3d array: 15 x 4 x 13 | 1d array: 15 x 1 x 13 | 3d array: 15 x 4 x 13 |
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| 2d array: 4 x 1 | 1d array: 3 | 2d array: 4 x 3 |
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匹配成功后，**Numpy** 会进行运算得到相应的结果。
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当然，如果相应的维度不匹配，那么**Numpy**会报错：
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In [7]:
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```py
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a = np.array([0,10,20,30])
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a.shape
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```
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Out[7]:
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```py
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(4L,)
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```
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In [8]:
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```py
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b.shape
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```
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Out[8]:
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```py
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(3L,)
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```
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In [9]:
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```py
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a + b
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```
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```py
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---------------------------------------------------------------------------
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ValueError Traceback (most recent call last)
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<ipython-input-9-f96fb8f649b6> in <module>()
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----> 1 a + b
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ValueError: operands could not be broadcast together with shapes (4,) (3,)
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```
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将 `a` 转换为列向量，还是可以计算出结果：
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In [10]:
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```py
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a[:, np.newaxis] + b
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```
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Out[10]:
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```py
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array([[ 0, 1, 2],
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[10, 11, 12],
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[20, 21, 22],
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[30, 31, 32]])
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```
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## 例子
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In [11]:
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```py
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x = np.linspace(-.5,.5, 21)
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```
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In [12]:
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```py
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y = x[:, np.newaxis]
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```
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In [13]:
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```py
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x.shape
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```
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Out[13]:
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```py
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(21L,)
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```
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In [14]:
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```py
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y.shape
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```
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Out[14]:
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```py
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(21L, 1L)
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```
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先形成一个 21 乘 21 的网格，再计算网格到原点的距离：
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In [15]:
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```py
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radius = np.sqrt(x ** 2 + y ** 2)
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```
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In [16]:
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```py
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import matplotlib.pyplot as plt
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%matplotlib inline
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plt.imshow(radius)
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```
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Out[16]:
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```py
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<matplotlib.image.AxesImage at 0xa2cb358>
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```
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