最小化Tensorflow中一个变量的功能

如果要最小化单个参数，则可以执行以下操作（由于要尝试训练参数，因此我避免使用占位符-占位符通常用于超参数和输入，不被视为可训练参数）：

import tensorflow as tf
x = tf.Variable(10.0, trainable=True)
f_x = 2 * x* x - 5 *x + 4
loss = f_x
opt = tf.train.GradientDescentOptimizer(0.1).minimize(f_x)
with tf.Session() as sess:
sess.run(tf.global_variables_initializer())
for i in range(100):
print(sess.run([x,loss]))
sess.run(opt)


这将输出以下对（x，损耗）对的列表：

[10.0, 154.0]
[6.5, 56.0]
[4.4000001, 20.720001]
[3.1400001, 8.0192013]
[2.3840001, 3.4469128]
[1.9304, 1.8008881]
[1.65824, 1.2083197]
[1.494944, 0.99499512]
[1.3969663, 0.91819811]
[1.3381798, 0.89055157]
[1.3029079, 0.88059855]
[1.2817447, 0.87701511]
[1.2690468, 0.87572551]
[1.2614281, 0.87526155]
[1.2568569, 0.87509394]
[1.2541142, 0.87503386]
[1.2524685, 0.87501216]
[1.2514811, 0.87500429]
[1.2508886, 0.87500143]
[1.2505331, 0.87500048]
[1.2503198, 0.875]
[1.2501919, 0.87500024]
[1.2501152, 0.87499976]
[1.2500691, 0.875]
[1.2500415, 0.875]
[1.2500249, 0.87500024]
[1.2500149, 0.87500024]
[1.2500089, 0.875]
[1.2500054, 0.87500024]
[1.2500032, 0.875]
[1.2500019, 0.875]
[1.2500012, 0.87500024]
[1.2500007, 0.87499976]
[1.2500005, 0.875]
[1.2500002, 0.87500024]
[1.2500001, 0.87500024]
[1.2500001, 0.87500024]
[1.2500001, 0.87500024]
[1.2500001, 0.87500024]
[1.2500001, 0.87500024]
[1.2500001, 0.87500024]
[1.2500001, 0.87500024]
[1.2500001, 0.87500024]
[1.2500001, 0.87500024]
[1.2500001, 0.87500024]
[1.2500001, 0.87500024]
[1.2500001, 0.87500024]
[1.2500001, 0.87500024]
[1.2500001, 0.87500024]
[1.2500001, 0.87500024]