Tensorflow之损失函数与交叉熵

损失函数：预测值与已知答案之间的差距

NN优化目标：loss最小{mse， 自定义， ce)

均方误差tensorflow实现，loss_mse = tf.reduce_mean(tf.sqrue(y_-y)

预测酸奶日销量，y，x1, x2是影响日销量的因素

建模前，应预先采集每日x1,x2,和效率y

拟造数据集x，y:y_=x1 + x2 ，噪声 -0.05-+0.05

import tensorflow as tf
import numpy as np
 
SEED = 2345
 
rdm = np.random.RandomState()
x = rdm.rand(32,2) # 生成32行两列之间的数字
y_ = [[x1 + x2 + (rdm.rand()/10.0 - 0.05)] for (x1, x2) in x] #0.1-0.05=0.005
x = tf.cast(x, dtype=tf.float32)
# 随机初始化w1（2，1）
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) #更新参数

使用均方误差，预测多和预测少是一样的

预测多了，损失成本，预测少了，损失利润，利润不等于成本

自定义损失函数 loss(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):
    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())
 
# 自定义损失函数
# 酸奶成本1元， 酸奶利润99元
# 成本很低，利润很高，人们希望多预测些，生成模型系数大于1，往多了预测

 交叉熵

交叉熵可以表示两个概率分布之间的距离

例如 二分类，已知答案y_(1, 0) 预测 y1(0.6, 0.4), y2=(0.8, 0.2),  那个答案接近标准答案

代码实现， tf.losses.categorical_crossentropy(y_,y)

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)

sotfmax与交叉熵结合

tf.nn.sotfmax_cross_entropy_with_logits(y_, y)

例子：

# 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)
 
 
# 输出的结果相同