tf.layers.dense() 和 tf.contrib.layers.fully_connected() 是否可以互换? [英] Are tf.layers.dense() and tf.contrib.layers.fully_connected() interchangeable?

问题描述

我习惯使用 tf.contrib.layers.fully_connected 来构建全连接层.最近我遇到了 tf.layers.dense 显然用于第一个功能可以使用的地方.是否可以互换，产生相同的输出?

I am used to using tf.contrib.layers.fully_connected to build a fully connected layer. Recently I ran into tf.layers.dense apparently used where the first functioned could be used. Are the interchangeable, producing the same output?

推荐答案

本质上是一样的，后者调用前者.

然而，tf.contrib.fully_connected 在 dense 之上添加了一些功能，特别是在参数中传递规范化和激活的可能性，à la Keras.正如@wordforthewise 所指出的，请注意后者默认为 tf.nn.relu.

However tf.contrib.fully_connected adds a few functionalities on top of dense, in particular the possibility to pass a normalization and an activation in the parameters, à la Keras. As noted by @wordforthewise, mind that the later defaults to tf.nn.relu.

更一般地说，TF API 提出了(并且有些令人困惑地混合)低级和高级 API；在此处了解更多信息.

More generally, the TF API proposes (and mixes somewhat confusingly) low- and hi-level APIs; more on that here.

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