在神经网络入门回顾(感知器、多层感知器)中整理了关于感知器和多层感知器的理论,这里实现关于与门、与非门、或门、异或门的代码,以便对感知器有更好的感觉。
此外,我们使用 pytest 框架进行测试。
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pip install pytest
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与门、与非门、或门
通过一层感知器就可以实现与门、与非门、或门。
先写测试代码 test_perception.py:
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from perception import and_operate, nand_operate, or_operate
def test_and_operate():
"""
测试与门
:return:
"""
assert and_operate( 1 , 1 ) = = 1
assert and_operate( 1 , 0 ) = = 0
assert and_operate( 0 , 1 ) = = 0
assert and_operate( 0 , 0 ) = = 0
def test_nand_operate():
"""
测试与非门
:return:
"""
assert nand_operate( 1 , 1 ) = = 0
assert nand_operate( 1 , 0 ) = = 1
assert nand_operate( 0 , 1 ) = = 1
assert nand_operate( 0 , 0 ) = = 1
def test_or_operate():
"""
测试或门
:return:
"""
assert or_operate( 1 , 1 ) = = 1
assert or_operate( 1 , 0 ) = = 1
assert or_operate( 0 , 1 ) = = 1
assert or_operate( 0 , 0 ) = = 0
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写完测试代码,后面直接输入命令 pytest -v 即可测试代码。
这三个门的权重和偏置是根据人的直觉或者画图得到的,并且不是唯一的。以下是简单的实现,在 perception.py 中写上:
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import numpy as np
def step_function(x):
"""
阶跃函数
:param x:
:return:
"""
if x < = 0 :
return 0
else :
return 1
def and_operate(x1, x2):
"""
与门
:param x1:
:param x2:
:return:
"""
x = np.array([x1, x2])
w = np.array([ 0.5 , 0.5 ])
b = - 0.7
return step_function(np. sum (w * x) + b)
def nand_operate(x1, x2):
"""
与非门
:param x1:
:param x2:
:return:
"""
x = np.array([x1, x2])
w = np.array([ - 0.5 , - 0.5 ])
b = 0.7
return step_function(np. sum (w * x) + b)
def or_operate(x1, x2):
"""
或门
:param x1:
:param x2:
:return:
"""
x = np.array([x1, x2])
w = np.array([ 0.5 , 0.5 ])
b = - 0.3
return step_function(np. sum (w * x) + b)
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运行 pytest -v 确认测试通过。
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= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = test session starts = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
platform darwin - - python 3.6 . 8 , pytest - 5.1 . 2 , py - 1.8 . 0 , pluggy - 0.12 . 0 - - / users / mac / .virtualenvs / work / bin / python3
...
collected 3 items
test_perception.py::test_and_operate passed [ 33 % ]
test_perception.py::test_nand_operate passed [ 66 % ]
test_perception.py::test_or_operate passed [ 100 % ]
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = 3 passed in 0.51s = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
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异或门
如上图所示,由于异或门不是线性可分的,因此需要多层感知器的结构。
使用两层感知器可以实现异或门。
修改 test_perception.py 文件,加入异或门的测试代码 :
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from perception import and_operate, nand_operate, or_operate, xor_operate
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以及
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def test_xor_operate():
"""
测试异或门
:return:
"""
assert xor_operate( 1 , 1 ) = = 0
assert xor_operate( 1 , 0 ) = = 1
assert xor_operate( 0 , 1 ) = = 1
assert xor_operate( 0 , 0 ) = = 0
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在 perception.py 文件里加入异或门的函数:
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def xor_operate(x1, x2):
"""
异或门
:param x1:
:param x2:
:return:
"""
s1 = nand_operate(x1, x2)
s2 = or_operate(x1, x2)
return and_operate(s1, s2)
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我们通过与非门和或门的线性组合实现了异或门。
运行命令 pytest -v 测试成功。
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= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = test session starts = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
platform darwin - - python 3.6 . 8 , pytest - 5.1 . 2 , py - 1.8 . 0 , pluggy - 0.12 . 0 - - / users / mac / .virtualenvs / work / bin / python3
...
collected 4 items
test_perception.py::test_and_operate passed [ 25 % ]
test_perception.py::test_nand_operate passed [ 50 % ]
test_perception.py::test_or_operate passed [ 75 % ]
test_perception.py::test_xor_operate passed [ 100 % ]
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = 4 passed in 0.60s = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
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原文作者:雨先生
原文链接: https://www.cnblogs.com/noluye/p/11465389.html
许可协议:知识共享署名-非商业性使用 4.0 国际许可协议
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原文链接:https://www.cnblogs.com/noluye/p/11465389.html