TensorFlow2.1入门学习笔记(7)——损失函数

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TensorFlow2.0入门学习笔记(7)——损失函数

损失函数（loss）：

预测值（y）与已知答案（y_）的差距
神经网络的优化目标：loss最小 $\Rightarrow \{\begin{array}{l}mse(MeanAquaredError)\\ 自定义\\ ce(CrossEntropy)\end{array}$

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

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

• 例
  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):
    with tf.GradientTape() as tape:
        y = tf.matmul(x, w1)
        loss_mse = tf.reduce_mean(tf.square(y_ - y))

    grads = tape.gradient(loss_mse, w1)
    w1.assign_sub(lr * grads)

    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
  

自定义损失函数：

如预测产品销量，预测多了，损失成本；预测少了，损失利润。若利润$\ne$成本，则mse产生的loss无法利益最大化。
自定义损失函数：$loss(y\_,y)={\sum }_{n}f(y\_,y)$
$$f(y\_,y)=\{\begin{array}{lr}PROFIT\ast (y\_-y)& y<y\_\\ COST\ast (y\_-y)& y\ge y\_\end{array}$$
loss_zdy=tf.reduce_sum(tf.where(tf.greater(y,y_),COST(y-y_),PROFIT(y_-y)))
如：预测酸奶销量，酸奶成本（COST）1元，酸奶利润（PROFIT）99元
预测少了损失利润99元，预测多了损失1元，希望生成的预测函数往多了预测

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):
    with tf.GradientTape() as tape:
        y = tf.matmul(x, w1)
        loss = tf.reduce_sum(tf.where(tf.greater(y, y_), (y - y_) * COST, (y_ - y) * PROFIT))

    grads = tape.gradient(loss, w1)
    w1.assign_sub(lr * grads)

    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)

$$H(y\_,y)=-\sum y\_\ast lny$$
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)
预测$y_1$=(0.6,0.4) $y_2$=(0.8,0.2)哪个更接近标准答案
$H_1$((1,0),(0.6,0.4))=-(1$\ast$ln0.6 + 0$\ast$ln0.4)$\approx$-(-0.511 + 0) = 0.511
$H_2$((1,0),(0.8,0.2))=-(1$\ast$ln0.8 + 0$\ast$ln0.2)$\approx$-(-0.223 + 0) = 0.223
因为$H_1>H_2$,所以$y_2$预测更准确

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)

输出先过softmax函数，再计算y与y_的交叉熵损失函数

# 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)

• 运行结果：
  
  主要学习的资料，西安科技大学：神经网络与深度学习——TensorFlow2.0实战，北京大学：人工智能实践Tensorflow笔记