感知器
真的好像自控中的負反饋
程序不要隨便刪除
from functools import reduce
#python 3 在Python 3里,reduce()函數(shù)已經(jīng)被
#從全局名字空間里移除了,它現(xiàn)在被放置在
#fucntools模塊里 用的話要 先引入：
class Perceptron(object):
def __init__(self, input_num, activator): #初始化函數(shù)
self.activator = activator
self.weights = [0.0 for _ in range(input_num)]
self.bias = 0.0
def __str__(self):
return 'weights\t:%s\nbias\t:%f\n' % (self.weights, self.bias)
def predict(self,input_vec): #這一步與連接中程序有所不同
L=[i*j for i,j in zip(input_vec,self.weights)]
return self.activator(reduce
(lambda a, b: a + b, L, 0.0) + self.bias)
def train(self, input_vecs, labels, iteration, rate):
for i in range(iteration):
self._one_iteration(input_vecs, labels, rate)
def _one_iteration(self, input_vecs, labels, rate):
samples = zip(input_vecs, labels)
for (input_vec,label) in samples:
output = self.predict(input_vec)
self._update_weights(input_vec, output, label, rate)
def _update_weights(self, input_vec, output, label, rate):
delta = label - output
self.weights =[j + rate * delta * i for i,j in zip(input_vec,self.weights)]
self.bias += rate * delta
#這里同樣也有map 和 zip 的問題
def f(x):
return 1 if x>0 else 0
def get_training_dataset():
input_vecs = [[1,1], [0,0], [1,0], [0,1]]
labels = [1, 0, 0, 0]
return input_vecs, labels
def train_and_perceptron():
p = Perceptron(2, f)
input_vecs, labels = get_training_dataset()
p.train(input_vecs, labels, 10, 0.1)
return p
if __name__ == '__main__':
example = train_and_perceptron()
print (example)
print ('1 and 1 = %d' % example.predict([1, 1]))
print ('1 and 0 = %d' % example.predict([1, 0]))
print ('0 and 0 = %d' % example.predict([0, 0]))
print ('0 and 1 = %d' % example.predict([0, 1]))
在連接中 return self.activator(reduce(lambda a, b: a + b, map(lambda (x, w): x * w,
zip(input_vec, self.weights)), 0.0) + self.bias)
這里就是y = f(w * x +b) 但是在python3中回挽，zip 和 map返回的是對象 而非列表娶牌，支持迭代
