在model.fit()期间记录Keras中每个纪元的计算时间

Question

我想比较不同模型之间的计算时间.在拟合期间,每个纪元的计算时间被打印到控制台.

Epoch 5/5
160000/160000 [==============================] - **10s** ......

我正在寻找一种方法来存储这些时间,类似于在每个时期保存并通过历史对象可用的模型指标.

Answer 1

尝试以下回调:

class TimeHistory(keras.callbacks.Callback):
    def on_train_begin(self, logs={}):
        self.times = []

    def on_epoch_begin(self, batch, logs={}):
        self.epoch_time_start = time.time()

    def on_epoch_end(self, batch, logs={}):
        self.times.append(time.time() - self.epoch_time_start)

然后:

time_callback = TimeHistory()
model.fit(..., callbacks=[..., time_callback],...)
times = time_callback.times

在这种情况下times应该存储纪元计算时间.

Answer 2

请参阅Marcin Mo？ejko的答案

import time

class TimeHistory(keras.callbacks.Callback):
    def on_train_begin(self, logs={}):
        self.times = []

    def on_epoch_begin(self, epoch, logs={}):
        self.epoch_time_start = time.time()

    def on_epoch_end(self, epoch, logs={}):
        self.times.append(time.time() - self.epoch_time_start)

然后

time_callback = TimeHistory()
model.fit(..., callbacks=[..., time_callback],...)

执行日志

Train on 17000 samples, validate on 8000 samples
Epoch 1/3
17000/17000 [==============================] - 5s 266us/step - loss: 36.7562 - mean_absolute_error: 4.5074 - val_loss: 34.2384 - val_mean_absolute_error: 4.3929
Epoch 2/3
17000/17000 [==============================] - 4s 253us/step - loss: 33.5529 - mean_absolute_error: 4.2956 - val_loss: 32.0291 - val_mean_absolute_error: 4.2484
Epoch 3/3
17000/17000 [==============================] - 5s 265us/step - loss: 31.0547 - mean_absolute_error: 4.1340 - val_loss: 30.6292 - val_mean_absolute_error: 4.1480

然后

print(time_callback.times)

输出

[4.531331300735474, 4.308278322219849, 4.505300283432007]