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283 lines
14 KiB
Markdown
283 lines
14 KiB
Markdown
# 向量化函数
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自定义的 `sinc` 函数：
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In [1]:
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```py
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import numpy as np
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def sinc(x):
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if x == 0.0:
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return 1.0
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else:
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w = np.pi * x
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return np.sin(w) / w
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```
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作用于单个数值：
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In [2]:
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```py
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sinc(0.0)
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```
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Out[2]:
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```py
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1.0
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```
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In [3]:
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```py
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sinc(3.0)
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```
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Out[3]:
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```py
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3.8981718325193755e-17
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```
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但这个函数不能作用于数组：
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In [4]:
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```py
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x = np.array([1,2,3])
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sinc(x)
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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-4-9d4f36f2aa7a> in <module>()
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1 x = np.array([1,2,3])
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----> 2 sinc(x)
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<ipython-input-1-dffe464e3332> in sinc(x)
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2
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3 def sinc(x):
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----> 4 if x == 0.0:
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5 return 1.0
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6 else:
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ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()
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```
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可以使用 `numpy` 的 `vectorize` 将函数 `sinc` 向量化，产生一个新的函数：
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In [5]:
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```py
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vsinc = np.vectorize(sinc)
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vsinc(x)
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```
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Out[5]:
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```py
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array([ 3.89817183e-17, -3.89817183e-17, 3.89817183e-17])
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```
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其作用是为 `x` 中的每一个值调用 `sinc` 函数：
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In [6]:
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```py
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import matplotlib.pyplot as plt
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%matplotlib inline
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x = np.linspace(-5,5,101)
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plt.plot(x, vsinc(x))
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```
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Out[6]:
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```py
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[<matplotlib.lines.Line2D at 0xa24e4e0>]
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```
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因为这样的用法涉及大量的函数调用，因此，向量化函数的效率并不高。 |