TensorFlow样例一

假设原函数为 f(x) = 5x^2 + 3，为了估计出这个函数，定义参数未知的函数g(x, w) = w0 x^2 + w1 x + w2，现要找出适合的w使g(x, w) ≈ f(x)。将这个问题转化为求解参数w使得损失函数L(w) = ∑ (f(x) - g(x, w))^2最小，求解过程使用了随机梯度下降（Stochastic Gradient Descent）。求解问题的代码如下：

 import numpy as np
 import tensorflow as tf

 # Placeholders are used to feed values from python to TensorFlow ops. We define
 # two placeholders, one for input feature x, and one for output y.
 x = tf.placeholder(tf.float32)
 y = tf.placeholder(tf.float32)

 # Assuming we know that the desired function is a polynomial of 2nd degree, we
 # allocate a vector of size 3 to hold the coefficients. The variable will be
 # automatically initialized with random noise.
 w = tf.get_variable("w", shape=[3, 1])

 # We define yhat to be our estimate of y.
 f = tf.stack([tf.square(x), x, tf.ones_like(x)], 1)
 yhat = tf.squeeze(tf.matmul(f, w), 1)

 # The loss is defined to be the l2 distance between our estimate of y and its
 # true value. We also added a shrinkage term, to ensure the resulting weights
 # would be small.
 loss = tf.nn.l2_loss(yhat - y) + 0.1 * tf.nn.l2_loss(w)

 # We use the Adam optimizer with learning rate set to 0.1 to minimize the loss.
 train_op = tf.train.AdamOptimizer(0.1).minimize(loss)

 def generate_data():
     x_val = np.random.uniform(-10.0, 10.0, size=100)
     y_val = 5 * np.square(x_val) + 3
     return x_val, y_val

 sess = tf.Session()
 # Since we are using variables we first need to initialize them.
 sess.run(tf.global_variables_initializer())
 for _ in range(1000):
     x_val, y_val = generate_data()
     _, loss_val = sess.run([train_op, loss], {x: x_val, y: y_val})
     print(loss_val)
 print(sess.run([w]))

求解过程如下：

4380421.0
3147655.5
4625718.5
3493661.0
3061016.0
3057624.5
3104206.2
……
103.7392
98.461266
113.29772
104.56809
89.75495
……
17.354445
17.66056
17.716873
18.782757
16.015532
[array([[4.9863739e+00],
       [6.9120852e-04],
       [3.8031762e+00]], dtype=float32)]