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有人可以解释一下，TensorFlow 的 Eager 模式是如何工作的吗？我正在尝试构建一个简单的回归，如下所示：

import tensorflow as tf

tfe = tf.contrib.eager
tf.enable_eager_execution()

import numpy as np


def make_model():
    net = tf.keras.Sequential()
    net.add(tf.keras.layers.Dense(4, activation='relu'))
    net.add(tf.keras.layers.Dense(1))
    return net

def compute_loss(pred, actual):
    return tf.reduce_mean(tf.square(tf.subtract(pred, actual)))

def compute_gradient(model, pred, actual):
    """compute gradients with given noise and input"""
    with tf.GradientTape() as tape:
        loss = compute_loss(pred, actual)
    grads = tape.gradient(loss, model.variables)
    return grads, loss

def apply_gradients(optimizer, grads, model_vars):
    optimizer.apply_gradients(zip(grads, model_vars))

model = make_model()
optimizer = tf.train.AdamOptimizer(1e-4)

x = np.linspace(0,1,1000)
y = x+np.random.normal(0,0.3,1000)
y = y.astype('float32')
train_dataset = tf.data.Dataset.from_tensor_slices((y.reshape(-1,1)))

epochs = 2# 10
batch_size = 25
itr = y.shape[0] // batch_size
for epoch in range(epochs):
    for data in tf.contrib.eager.Iterator(train_dataset.batch(25)):
        preds = model(data)
        grads, loss = compute_gradient(model, preds, data)
        print(grads)
        apply_gradients(optimizer, grads, model.variables)
#         with tf.GradientTape() as tape:
#             loss = tf.sqrt(tf.reduce_mean(tf.square(tf.subtract(preds, data))))
#         grads = tape.gradient(loss, model.variables)
#         print(grads)
#         optimizer.apply_gradients(zip(grads, model.variables),global_step=None)

Gradient output: [None, None, None, None, None, None] 错误如下：

----------------------------------------------------------------------
ValueError                           Traceback (most recent call last)
<ipython-input-3-a589b9123c80> in <module>
     35         grads, loss = compute_gradient(model, preds, data)
     36         print(grads)
---> 37         apply_gradients(optimizer, grads, model.variables)
     38 #         with tf.GradientTape() as tape:
     39 #             loss = tf.sqrt(tf.reduce_mean(tf.square(tf.subtract(preds, data))))

<ipython-input-3-a589b9123c80> in apply_gradients(optimizer, grads, model_vars)
     17 
     18 def apply_gradients(optimizer, grads, model_vars):
---> 19     optimizer.apply_gradients(zip(grads, model_vars))
     20 
     21 model = make_model()

~/anaconda3/lib/python3.6/site-packages/tensorflow/python/training/optimizer.py in apply_gradients(self, grads_and_vars, global_step, name)
    589     if not var_list:
    590       raise ValueError("No gradients provided for any variable: %s." %
--> 591                        ([str(v) for _, v, _ in converted_grads_and_vars],))
    592     with ops.init_scope():
    593       self._create_slots(var_list)

ValueError: No gradients provided for any variable:

编辑

我更新了我的代码。现在，问题在于梯度计算，它返回零。我检查了非零的损失值。

4

1 回答 1

42

第 1 部分：问题确实是您输入的数据类型。默认情况下，您的 keras 模型需要 float32，但您传递的是 float64。您可以更改模型的 dtype 或将输入更改为 float32。

要更改您的模型：

def make_model():
    net = tf.keras.Sequential()
    net.add(tf.keras.layers.Dense(4, activation='relu', dtype='float32'))
    net.add(tf.keras.layers.Dense(4, activation='relu'))
    net.add(tf.keras.layers.Dense(1))
    return net

要更改您的输入： y = y.astype('float32')

第 2 部分：model(data)您需要在 tf.GradientTape() 下调用计算模型（即）的函数。例如，您可以将您的compute_loss方法替换为以下内容：

def compute_loss(model, x, y):
    pred = model(x)
    return tf.reduce_mean(tf.square(tf.subtract(pred, y)))
于 2019-01-18T14:16:04.870 回答