gpt4 book ai didi

python - 在联合损失函数上训练多输出 Keras 模型

转载 作者:行者123 更新时间:2023-12-02 19:56:05 25 4
gpt4 key购买 nike

我正在 Keras 中编写两个联合解码器,具有一个公共(public)输入、两个单独的输出以及一个将两个输出都考虑在内的损失函数。我遇到的问题是损失函数。

这是您可以重现错误的最小 Keras 代码:

import tensorflow as tf
from scat import *

from keras.layers import Input, Reshape, Permute, Lambda, Flatten
from keras.layers.core import Dense
from keras.layers.advanced_activations import LeakyReLU
from keras.models import Model
from keras import backend as K

def identity(x):
return K.identity(x)

# custom loss function
def custom_loss():
def my_loss(y_dummy, pred):
fcn_loss_1 = tf.nn.softmax_cross_entropy_with_logits(labels=y_dummy[0], logits=pred[0])
fcn_loss_2 = tf.nn.softmax_cross_entropy_with_logits(labels=y_dummy[1], logits=pred[1])
fcn_loss_2 = tf.matrix_band_part(fcn_loss_2, 0, -1) - tf.matrix_band_part(fcn_loss_2, 0, 0)

fcn_loss = tf.reduce_mean(fcn_loss_1) + 2 * tf.reduce_mean(fcn_loss_2)

return fcn_loss
return my_loss

def keras_version():
input = Input(shape=(135,), name='feature_input')
out1 = Dense(128, kernel_initializer='glorot_normal', activation='linear')(input)
out1 = LeakyReLU(alpha=.2)(out1)
out1 = Dense(256, kernel_initializer='glorot_normal', activation='linear')(out1)
out1 = LeakyReLU(alpha=.2)(out1)
out1 = Dense(512, kernel_initializer='glorot_normal', activation='linear')(out1)
out1 = LeakyReLU(alpha=.2)(out1)
out1 = Dense(45, kernel_initializer='glorot_normal', activation='linear')(out1)
out1 = LeakyReLU(alpha=.2)(out1)
out1 = Reshape((9, 5))(out1)

out2 = Dense(128, kernel_initializer='glorot_normal', activation='linear')(input)
out2 = LeakyReLU(alpha=.2)(out2)
out2 = Dense(256, kernel_initializer='glorot_normal', activation='linear')(out2)
out2 = LeakyReLU(alpha=.2)(out2)
out2 = Dense(512, kernel_initializer='glorot_normal', activation='linear')(out2)
out2 = LeakyReLU(alpha=.2)(out2)
out2 = Dense(540, kernel_initializer='glorot_normal', activation='linear')(out2)
out2 = LeakyReLU(alpha=.2)(out2)
out2 = Reshape((9, 4, 15))(out2)
out2 = Lambda(lambda x: K.dot(K.permute_dimensions(x, (0, 2, 1, 3)),
K.permute_dimensions(x, (0, 2, 3, 1))), output_shape=(4,9,9))(out2)
out2 = Flatten()(out2)
out2 = Dense(324, kernel_initializer='glorot_normal', activation='linear')(out2)
out2 = LeakyReLU(alpha=.2)(out2)
out2 = Reshape((4, 9, 9))(out2)
out2 = Lambda(lambda x: K.permute_dimensions(x, (0, 2, 3, 1)))(out2)

out1 = Lambda(identity, name='output_1')(out1)
out2 = Lambda(identity, name='output_2')(out2)

return Model(input, [out1, out2])

model = keras_version()
model.compile(loss=custom_loss(), optimizer='adam')

model.summary()

feature_final = np.random.normal(0,1,[5000, 9, 15])
train_features_array = np.random.normal(0,1,[5000, 9, 5])
train_adj_array = np.random.normal(0,1,[5000, 9, 9, 4])

feature_final = feature_final.reshape(-1, 135)
model.fit(feature_final, [train_features_array, train_adj_array],
batch_size=50,
epochs=10
)

我得到的错误是:

File "...", line 135, in <module>
epochs=10
File ".../keras/engine/training.py", line 1039, in fit
validation_steps=validation_steps)
File ".../keras/backend/tensorflow_backend.py", line 2675, in _call
fetched = self._callable_fn(*array_vals)
File ".../tensorflow/python/client/session.py", line 1458, in __call__
run_metadata_ptr)
tensorflow.python.framework.errors_impl.InvalidArgumentError: input must be at least 2-dim, received shape: [9]
[[{{node loss/output_1_loss/MatrixBandPart_1}}]]

在第二次尝试中,我尝试编写两个损失函数并使用损失权重将它们组合起来。

# custom loss function
def custom_loss_1():
def my_loss_1(y_dummy, pred):
fcn_loss_1 = tf.nn.softmax_cross_entropy_with_logits(labels=y_dummy[0], logits=pred[0])

return tf.reduce_mean(fcn_loss_1)
return my_loss_1

def custom_loss_2():
def my_loss_2(y_dummy, pred):
fcn_loss_2 = tf.nn.softmax_cross_entropy_with_logits(labels=y_dummy[1], logits=pred[1])
fcn_loss_2 = tf.matrix_band_part(fcn_loss_2, 0, -1) - tf.matrix_band_part(fcn_loss_2, 0, 0)

return tf.reduce_mean(fcn_loss_2)
return my_loss_2

model.compile(loss={'output_1':custom_loss_1(), 'output_2':custom_loss_2()},
loss_weights={'output_1':1.0, 'output_2':2.0}, optimizer='adam')

但是我收到了

tensorflow.python.framework.errors_impl.InvalidArgumentError: Matrix size-incompatible: In[0]: [20,25920], In[1]: [324,324]
[[{{node dense_9/BiasAdd}}]]

在这种情况下,问题实际上可能出在模型本身。这是model.summary:

__________________________________________________________________________________________________
Layer (type) Output Shape Param # Connected to
==================================================================================================
feature_input (InputLayer) (None, 135) 0
__________________________________________________________________________________________________
dense_5 (Dense) (None, 128) 17408 feature_input[0][0]
__________________________________________________________________________________________________
leaky_re_lu_5 (LeakyReLU) (None, 128) 0 dense_5[0][0]
__________________________________________________________________________________________________
dense_6 (Dense) (None, 256) 33024 leaky_re_lu_5[0][0]
__________________________________________________________________________________________________
leaky_re_lu_6 (LeakyReLU) (None, 256) 0 dense_6[0][0]
__________________________________________________________________________________________________
dense_7 (Dense) (None, 512) 131584 leaky_re_lu_6[0][0]
__________________________________________________________________________________________________
leaky_re_lu_7 (LeakyReLU) (None, 512) 0 dense_7[0][0]
__________________________________________________________________________________________________
dense_1 (Dense) (None, 128) 17408 feature_input[0][0]
__________________________________________________________________________________________________
dense_8 (Dense) (None, 540) 277020 leaky_re_lu_7[0][0]
__________________________________________________________________________________________________
leaky_re_lu_1 (LeakyReLU) (None, 128) 0 dense_1[0][0]
__________________________________________________________________________________________________
leaky_re_lu_8 (LeakyReLU) (None, 540) 0 dense_8[0][0]
__________________________________________________________________________________________________
dense_2 (Dense) (None, 256) 33024 leaky_re_lu_1[0][0]
__________________________________________________________________________________________________
reshape_2 (Reshape) (None, 9, 4, 15) 0 leaky_re_lu_8[0][0]
__________________________________________________________________________________________________
leaky_re_lu_2 (LeakyReLU) (None, 256) 0 dense_2[0][0]
__________________________________________________________________________________________________
lambda_1 (Lambda) (None, 4, 9, 9) 0 reshape_2[0][0]
__________________________________________________________________________________________________
dense_3 (Dense) (None, 512) 131584 leaky_re_lu_2[0][0]
__________________________________________________________________________________________________
flatten_1 (Flatten) (None, 324) 0 lambda_1[0][0]
__________________________________________________________________________________________________
leaky_re_lu_3 (LeakyReLU) (None, 512) 0 dense_3[0][0]
__________________________________________________________________________________________________
dense_9 (Dense) (None, 324) 105300 flatten_1[0][0]
__________________________________________________________________________________________________
dense_4 (Dense) (None, 45) 23085 leaky_re_lu_3[0][0]
__________________________________________________________________________________________________
leaky_re_lu_9 (LeakyReLU) (None, 324) 0 dense_9[0][0]
__________________________________________________________________________________________________
leaky_re_lu_4 (LeakyReLU) (None, 45) 0 dense_4[0][0]
__________________________________________________________________________________________________
reshape_3 (Reshape) (None, 4, 9, 9) 0 leaky_re_lu_9[0][0]
__________________________________________________________________________________________________
reshape_1 (Reshape) (None, 9, 5) 0 leaky_re_lu_4[0][0]
__________________________________________________________________________________________________
lambda_2 (Lambda) (None, 9, 9, 4) 0 reshape_3[0][0]
__________________________________________________________________________________________________
output_1 (Lambda) (None, 9, 5) 0 reshape_1[0][0]
__________________________________________________________________________________________________
output_2 (Lambda) (None, 9, 9, 4) 0 lambda_2[0][0]
==================================================================================================
Total params: 769,437
Trainable params: 769,437
Non-trainable params: 0
__________________________________________________________________________________________________

如果您认为模型有问题,请检查"model" 。这个问题与this question不同它在损失中仅使用一个输出。这也是用 Tensorflow 编写的类似模型的损失函数:

# -- loss function
Y_1 = tf.placeholder(tf.float32, shape=[None, 9, 9, 4])
Y_2 = tf.placeholder(tf.float32, shape=[None, 9, 5])

loss_1 = tf.nn.softmax_cross_entropy_with_logits(labels=Y_2, logits=fcn(X)[0])
loss_2 = tf.nn.softmax_cross_entropy_with_logits(labels=Y_1, logits=fcn(X)[1])
loss_2 = tf.matrix_band_part(loss_2, 0, -1) - tf.matrix_band_part(loss_2, 0, 0)

loss = tf.reduce_mean(loss_1) + 2 * tf.reduce_mean(loss_2)

编辑:我使用实际数据集尝试了答案中的代码,损失函数显示了与代码的 Tensorflow 实现不同的行为。答案中建议的损失函数很快收敛并变为 nan。我同意这个答案,它说output_1应该是绝对的。基于此,我编写了以下损失函数,它仍然不如 Tensorflow 收敛得那么快,但绝对不会崩溃:

def custom_loss_1(model, output_1):
""" This loss function is called for output2
It needs to fetch model.output[0] and the output_1 predictions in
order to calculate fcn_loss_1
"""
def my_loss(y_true, y_pred):
fcn_loss_1 = tf.nn.softmax_cross_entropy_with_logits(labels=model.targets[0], logits=output_1)

return tf.reduce_mean(fcn_loss_1)

return my_loss

def custom_loss_2():
""" This loss function is called for output2
It needs to fetch model.output[0] and the output_1 predictions in
order to calculate fcn_loss_1
"""
def my_loss(y_true, y_pred):
fcn_loss_2 = tf.nn.softmax_cross_entropy_with_logits(labels=y_true, logits=y_pred)
fcn_loss_2 = tf.matrix_band_part(fcn_loss_2, 0, -1) - tf.matrix_band_part(fcn_loss_2, 0, 0)
return tf.reduce_mean(fcn_loss_2)

return my_loss

output_layer_1 = [layer for layer in model.layers if layer.name == 'output_1'][0]
losses = {'output_1': custom_loss_1(model, output_layer_1.output), 'output_2': custom_loss_2()}
model.compile(loss=losses, optimizer='adam', loss_weights=[1.0, 2.0])

最佳答案

您的代码中有两个问题:

首先是 Lambda 内的 K.dot 操作需要是 K.batch_dot

我用过:

def output_mult(x):
a = K.permute_dimensions(x, (0, 2, 1, 3))
b = K.permute_dimensions(x, (0, 2, 3, 1))
return K.batch_dot(a, b)


out2 = Lambda(output_mult)(out2)

它实际上有助于让 Keras 计算输出维度。这是检查代码的简单方法。为了调试它,我首先用现有损失 (mse) 替换了自定义损失,这很容易检测到。

第二个问题是自定义损失函数采用一对目标/输出而不是列表。损失函数的参数不是您最初和编辑中假设的张量列表。所以我将你的损失函数定义为

def custom_loss(model, output_1):
""" This loss function is called for output2
It needs to fetch model.output[0] and the output_1 predictions in
order to calculate fcn_loss_1
"""
def my_loss(y_true, y_pred):
fcn_loss_1 = tf.nn.softmax_cross_entropy_with_logits(labels=model.targets[0], logits=output_1)
fcn_loss_2 = tf.nn.softmax_cross_entropy_with_logits(labels=y_true, logits=y_pred)
fcn_loss_2 = tf.matrix_band_part(fcn_loss_2, 0, -1) - tf.matrix_band_part(fcn_loss_2, 0, 0)
return tf.reduce_mean(fcn_loss_2)

return my_loss

并将其用作

output_layer_1 = [layer for layer in model.layers if layer.name == 'output_1'][0]
losses = {'output_1': 'categorical_crossentropy', 'output_2': custom_loss(model, output_layer_1.output)}
model.compile(loss=losses, optimizer='adam', loss_weights=[1.0, 2.0])

编辑:我最初误读了 output2 的自定义损失,因为需要 fcn_loss_1 的值,但情况似乎并非如此,您可以将其写为:

def custom_loss():
def my_loss(y_true, y_pred):
fcn_loss_2 = tf.nn.softmax_cross_entropy_with_logits(labels=y_true, logits=y_pred)
fcn_loss_2 = tf.matrix_band_part(fcn_loss_2, 0, -1) - tf.matrix_band_part(fcn_loss_2, 0, 0)
return tf.reduce_mean(fcn_loss_2)

return my_loss

并将其用作:

losses = {'output_1': 'categorical_crossentropy', 'output_2': custom_loss()}
model.compile(loss=losses, optimizer='adam', loss_weights=[1.0, 2.0])

我假设output_1的损失是categorical_crossentropy。但即使你需要改变它,最简单的方法就是拥有 2 个独立的损失函数。当然,您也可以选择定义一个返回 0 的损失函数和一个返回全部成本的损失函数...但是将“loss(output1) + 2 * loss(output2)”分成两个损失加上权重,恕我直言。

完整笔记本: https://colab.research.google.com/drive/1NG3uIiesg-VIt-W9254Sea2XXUYPoVH5

关于python - 在联合损失函数上训练多输出 Keras 模型,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/56996864/

25 4 0
Copyright 2021 - 2024 cfsdn All Rights Reserved 蜀ICP备2022000587号
广告合作:1813099741@qq.com 6ren.com