TensorFlow-线性回归三段代码的比较
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目的
省去必要的头与声明， 对关键部分进行比较与理解性说明。 感觉： 虽然这仅仅是一个线性回归的问题，哪怕是我手算也能得到很好的结果。 可是突然间多出来Tensorflow这样的一个大框架，给出的接口实在太多，一时间应接不暇。
Step
训练样本的准备
Basic Usage
x = np.array([1., 2., 3., 4.])
y = np.array([0., -1.,-2.,-3.])
input_fn = tf.contrib.learn.io.numpy_input_fn({"x":x}, y, batch_size=4,num_epochs=1000)
Complete Codes
# training date
x_train = [1, 2, 3, 4]
y_train = [0,-1,-2,-3]
Custom Model
x = np.array([1., 2., 3., 4.])
y = np.array([0.,-1.,-2.,-3.])
input_fn = tf.contrib.learn.io.numpy_input_fn({'x':x}, y, 4, num_epochs=1000)
机器学习的三要素：模型-策略-算法
Basic Usage
features = [tf.contrib.layers.real_valued_column('x', dimension=1)]
estimator = tf.contrib.learn.LinearRegressor(feature_columns=features)
Complete Codes
linear_model = W*x + b
loss = tf.reduce_sum(tf.square(linear_model - y))
optimizer = tf.train.GradientDescentOptimizer(0.01)
train = optimizer.minimize(loss)
Custrom Model
def model(features, labels, mode):
W = tf.get_variable('W', [1], dtype=tf.float64)
b = tf.get_variable('b', [1], dtype=tf.float64)
y = W*features['x'] + b
loss = tf.reduce_sum(tf.square(y - labels))
global_step = tf.train.get_global_step()
optimizer = tf.train.GradientDescentOptimizer(0.01)
train = tf.group(optimizer.minimize(loss), tf.assign_add(global_step, 1))
return tf.contrib.learn.ModelFnOps(mode=mode, predictions=y, loss=loss, train_op=train)
estimator = tf.contrib.learn.Estimator(model_fn=model)
训练过程
Basic Usage
estimator.fit(input_fn=input_fn, steps=1000)
print(estimator.evaluate(input_fn=input_fn))
Complete Codes
for i in range(1000):
sess.run(train, {x:x_train, y:y_train})
# parameters' results and error evaluation
result_W, result_b, result_e = sess.run([W, b, loss], {x:x_train, y:y_train})
print('W: %s b: %s loss: %s' %(result_W, result_b, result_e))
Custom Code
estimator.fit(input_fn=input_fn, steps=1000)
print(estimator.evaluate(input_fn=input_fn, steps=10))
Summary
经过上面简单的横向对比，及纵向梳理，就条理了些许。