Lec-02 Simple Liner Regression LAB

>> Hypothesis and Cost

Hypothesis

Cost

で最もコストの低いWとBの値を学習と呼ぶことができます.

cost関数をtesorflowに移行するにはどうすればいいですか?
x_data = [1,2,3,4,5]
y_data = [1,2,3,4,5]

W = tf.Variable(2.9)
b = tf.Variable(0.5)

hypothesis = W * x_data + b

cost(W,b)
cost = tf.reduce_mean(tf.square(hypothesis - y_data))

:入力と出力xとyの値が同じ場合、W値は1、b値は0である.

tf.reduce_mean()
v = [1., 2., 3., 4.] 
tf.reduce_mean(v) #2.5

tf.square()
tf.square(3) #9

Gradient降下(=傾斜降下アルゴリズム)
: minimize cost(W,b)
: learning_rate = 0.01

# 변수(W,b)에 대한 정보 tape에 기록_
with tf.GradientTape() as tape: 	 
    hypothesis = W * x_data + b
    cost = tf.reduce_mean(tf.square(hypothesis - y_data))

#기울기 값 즉 미분값을 구함._
W_grad, b_grad = tape.gradient(cost, [W,b]) 

W.assign_sub(learning_rate * W_grad)
b.assign_sub(learning_rate * b_grad)

A.assign_sub(B)
: A = A - B
: A-=B

Full Code

import tensorflow as tf
#Data
x_data = [1,2,3,4,5]
y_data = [1,2,3,4,5]
  
# W, b initialize
W = tf.Variable(2.9)
b = tf.Variable(0.5)
  
learning_rate = 0.01
  
for i in range(100+1): # W, b update
# Gradient descent
    with tf.GradientTape() as tape:
        hypothesis = W * x_data + b
        cost = tf.reduce_mean(tf.square(hypothesis - y_data))
    W_grad, b_grad = tape.gradient(cost, [W,b])
    W.assign_sub(learning_rate * W_grad)
    b.assign_sub(learning_rate * b_grad)
    if i % 10 == 0:
        print("{:5}|{:10.4f}|{:10.4}|{:10.6f}".format(i, W.numpy(), b.numpy(), cost))

Collabを使用して実行されたコード結果

すべての人のために準備した第2四半期-Tensorflow