TensorFlow2.1入门学习笔记(7)——损失函数

TensorFlow2.0入门学习笔记(7)——损失函数

损失函数（loss）：

均方误差mse：loss_mse = tf.reduce_mean(tf.square(y_-y))

$MSE(y\_,y)=\frac{\sum_{i=1}^{n}(y-y\_)^2}{n}$

• 预测酸奶日销量y，x1、x2是影响日销量的因素。
建模前，应预先采集的数据有：每日x1、x2和销量y_(即已知答案，最佳的情况：产量=销量)
拟造数据集X，Y：y_=x1+x2
噪声：-0.05~+0.05
拟合可以预算销量的函数
import tensorflow as tf
import numpy as np

SEED = 23455

rdm = np.random.RandomState(seed=SEED)  # 生成[0,1)之间的随机数
x = rdm.rand(32, 2)
y_ = [[x1 + x2 + (rdm.rand() / 10.0 - 0.05)] for (x1, x2) in x]  # 生成噪声[0,1)/10=[0,0.1); [0,0.1)-0.05=[-0.05,0.05)
x = tf.cast(x, dtype=tf.float32)

w1 = tf.Variable(tf.random.normal([2, 1], stddev=1, seed=1))

epoch = 15000
lr = 0.002

for epoch in range(epoch):
y = tf.matmul(x, w1)
loss_mse = tf.reduce_mean(tf.square(y_ - y))

if epoch % 500 == 0:
print("After %d training steps,w1 is " % (epoch))
print(w1.numpy(), "\n")
print("Final w1 is: ", w1.numpy())



自定义损失函数：

$f(y\_,y)=\left\{\begin{array}{lr}PROFIT*(y\_-y)&y<y\_ \\COST*(y\_-y)&y\geq y\_\end{array}\right.$

loss_zdy=tf.reduce_sum(tf.where(tf.greater(y,y_),COST(y-y_),PROFIT(y_-y)))

import tensorflow as tf
import numpy as np

SEED = 23455
COST = 1
PROFIT = 99

rdm = np.random.RandomState(SEED)
x = rdm.rand(32, 2)
y_ = [[x1 + x2 + (rdm.rand() / 10.0 - 0.05)] for (x1, x2) in x]  # 生成噪声[0,1)/10=[0,0.1); [0,0.1)-0.05=[-0.05,0.05)
x = tf.cast(x, dtype=tf.float32)

w1 = tf.Variable(tf.random.normal([2, 1], stddev=1, seed=1))

epoch = 10000
lr = 0.002

for epoch in range(epoch):
y = tf.matmul(x, w1)
loss = tf.reduce_sum(tf.where(tf.greater(y, y_), (y - y_) * COST, (y_ - y) * PROFIT))

if epoch % 500 == 0:
print("After %d training steps,w1 is " % (epoch))
print(w1.numpy(), "\n")
print("Final w1 is: ", w1.numpy())


• 运行结果

交叉熵损失函数：tf.losses.categorical_crossentropy(y_,y)

$H(y\_,y)=-\sum y\_*lny$

CE(Cross Entropy):表示两个概率分布之间的距离
eg.二分类 已知答案y_=(1,0)

$H_1$((1,0),(0.6,0.4))=-(1$*$ln0.6 + 0$*$ln0.4)$\approx$-(-0.511 + 0) = 0.511
$H_2$((1,0),(0.8,0.2))=-(1$*$ln0.8 + 0$*$ln0.2)$\approx$-(-0.223 + 0) = 0.223

import tensorflow as tf
loss_ce1 = tf.losses.categorical_crossentropy([1, 0], [0.6, 0.4])
loss_ce2 = tf.losses.categorical_crossentropy([1, 0], [0.8, 0.2])
print("loss_ce1:", loss_ce1)
print("loss_ce2:", loss_ce2)

• 运行结果
softmax与交叉熵解和：tf.nn.softmax_cross_entropy_with_logits(y_,y)

# softmax与交叉熵损失函数的结合
import tensorflow as tf
import numpy as np

y_ = np.array([[1, 0, 0], [0, 1, 0], [0, 0, 1], [1, 0, 0], [0, 1, 0]])
y = np.array([[12, 3, 2], [3, 10, 1], [1, 2, 5], [4, 6.5, 1.2], [3, 6, 1]])
y_pro = tf.nn.softmax(y)
loss_ce1 = tf.losses.categorical_crossentropy(y_,y_pro)
loss_ce2 = tf.nn.softmax_cross_entropy_with_logits(y_, y)

print('分步计算的结果:\n', loss_ce1)
print('结合计算的结果:\n', loss_ce2)

posted @ 2020-05-20 09:48  Moonx5  阅读(264)  评论(0编辑  收藏  举报