keras Model 3 共享的層

1 入門html

2 多個輸入和輸出node

3 共享層this

考慮這樣的一個問題:咱們要判斷連個tweet是否來源於同一我的。spa

首先咱們對兩個tweet進行處理,而後將處理的結構拼接在一塊兒,以後跟一個邏輯迴歸,輸出這兩條tweet來自同一我的機率。翻譯

由於咱們對兩條tweet的處理是相同的,因此對第一條tweet的處理的模型,能夠被重用來處理第二個tweet。咱們考慮用LSTM進行處理。code

假設咱們的輸入是兩條 280*256的向量htm

首先定義輸入:blog

import keras
from keras.layers import Input, LSTM, Dense
from keras.models import Model

tweet_a = Input(shape=(280, 256))
tweet_b = Input(shape=(280, 256))

而後咱們共享LSTM。共享層很簡單,只要實例化層一次,而後在你想處理的tensor上調用你想要應用的次數便可(翻譯無力,看代碼)索引

# This layer can take as input a matrix
# and will return a vector of size 64
shared_lstm = LSTM(64)

# When we reuse the same layer instance
# multiple times, the weights of the layer
# are also being reused
# (it is effectively *the same* layer)
encoded_a = shared_lstm(tweet_a)
encoded_b = shared_lstm(tweet_b)

# We can then concatenate the two vectors:
merged_vector = keras.layers.concatenate([encoded_a, encoded_b], axis=-1)

# And add a logistic regression on top
predictions = Dense(1, activation='sigmoid')(merged_vector)

# We define a trainable model linking the
# tweet inputs to the predictions
model = Model(inputs=[tweet_a, tweet_b], outputs=predictions)

model.compile(optimizer='rmsprop',
              loss='binary_crossentropy',
              metrics=['accuracy'])
model.fit([data_a, data_b], labels, epochs=10)

其實,簡單點說,對一個層的屢次調用,就是在共享這個層。這裏有一個層的節點的概念ip

當你在一個輸入tensor上調用一個層時,就會生成一個輸出tensor,就會在這個層上添加一個節點,這個節點鏈接着這兩個tensor(輸入tensor和輸出tensor)。當你屢次調用同一個層的時,

這個層生成的節點就會按照0 ,1, 2, 。。以此類推編號。

那麼當一個層有多個節點的時候,咱們怎麼獲取它的輸出呢?

若是直接經過output獲取會出錯:

a = Input(shape=(280, 256))
b = Input(shape=(280, 256))

lstm = LSTM(32)
encoded_a = lstm(a)
encoded_b = lstm(b)

lstm.output
>> AttributeError: Layer lstm_1 has multiple inbound nodes,
hence the notion of "layer output" is ill-defined.
Use `get_output_at(node_index)` instead.

這時候應該經過索引進行調用:

assert lstm.get_output_at(0) == encoded_a
assert lstm.get_output_at(1) == encoded_b

對於輸入,也是一樣的

a = Input(shape=(32, 32, 3))
b = Input(shape=(64, 64, 3))

conv = Conv2D(16, (3, 3), padding='same')
conved_a = conv(a)

# Only one input so far, the following will work:
assert conv.input_shape == (None, 32, 32, 3)

conved_b = conv(b)
# now the `.input_shape` property wouldn't work, but this does:
assert conv.get_input_shape_at(0) == (None, 32, 32, 3)
assert conv.get_input_shape_at(1) == (None, 64, 64, 3)
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