问题描述
在使用TensorFlow进行CNN训练时,我将Keras.losses.poisson
用作损失函数.现在,我想与损失函数一起计算许多指标,并且观察到Keras.metrics.poisson
给出的结果不同-尽管两者是相同的函数.
In my CNN training using TensorFlow, I am using Keras.losses.poisson
as a loss function. Now, I like to calculate many metrics alongside that loss function, and I am observing that Keras.metrics.poisson
gives different results - although the two are the same function.
有关某些示例输出,请参见此处:loss
和poisson
输出具有不同的范围,即0.5与0.12:
See here for some example output: loss
and poisson
outputs have different ranges, 0.5 vs. 0.12:
Epoch 1/20
Epoch 00001: val_loss improved from inf to 0.53228, saving model to P:\Data\xyz.h5
- 8174s - loss: 0.5085 - binary_crossentropy: 0.1252 - poisson: 0.1271 - mean_squared_error: 1.2530e-04 - mean_absolute_error: 0.0035 - mean_absolute_percentage_error: 38671.1055 - val_loss: 0.5323 - val_binary_crossentropy: 0.1305 - val_poisson: 0.1331 - val_mean_squared_error: 5.8477e-05 - val_mean_absolute_error: 0.0035 - val_mean_absolute_percentage_error: 1617.8346
Epoch 2/20
Epoch 00002: val_loss improved from 0.53228 to 0.53218, saving model to P:\Data\xyz.h5
- 8042s - loss: 0.5067 - binary_crossentropy: 0.1246 - poisson: 0.1267 - mean_squared_error: 1.0892e-05 - mean_absolute_error: 0.0017 - mean_absolute_percentage_error: 410.8044 - val_loss: 0.5322 - val_binary_crossentropy: 0.1304 - val_poisson: 0.1330 - val_mean_squared_error: 4.9087e-05 - val_mean_absolute_error: 0.0035 - val_mean_absolute_percentage_error: 545.5222
Epoch 3/20
Epoch 00003: val_loss improved from 0.53218 to 0.53199, saving model to P:\Data\xyz.h5
- 8038s - loss: 0.5066 - binary_crossentropy: 0.1246 - poisson: 0.1266 - mean_squared_error: 6.6870e-06 - mean_absolute_error: 0.0013 - mean_absolute_percentage_error: 298.9844 - val_loss: 0.5320 - val_binary_crossentropy: 0.1304 - val_poisson: 0.1330 - val_mean_squared_error: 4.3858e-05 - val_mean_absolute_error: 0.0031 - val_mean_absolute_percentage_error: 452.3541
我在输入此问题时发现了类似的问题: Keras-损失和指标的计算方式不同?但是,我没有使用正则化.
I have found a similar questions while typing this one:Keras - Loss and Metric calculated differently?However, I am not using regularization.
此外,我遇到了这个问题,至少可以帮助我重现此问题:
In addition, I have come across this one, which at least helped me reproduce the issue: Same function in Keras Loss and Metric give different values even without regularization
from tensorflow import keras
layer = keras.layers.Input(shape=(1, 1, 1))
model = keras.models.Model(inputs=layer, outputs=layer)
model.compile(optimizer='adam', loss='poisson', metrics=['poisson'])
data = [[[[[1]]], [[[2]]], [[[3]]]]]
model.fit(x=data, y=data, batch_size=2, verbose=1)
然后我发现基本上是引发此问题的维度.从下面的扩展示例中,您可以看到
What I have found then is that, basically, it's the dimensionality that triggers this issue. From the following extended example, you can see that
- 这个问题可以用很多损失函数来重现(不是以
mean_
开头的) - 将
tensorflow.keras
替换为keras
时,问题消失了,并且 如果数据的维数大于3,则 -
tensorflow.keras
似乎可以按批大小缩放度量标准.至少那是我的拙劣解释.
- the issue can be reproduced with many loss functions (the ones hat don't begin with
mean_
), - the issue goes away when replacing
tensorflow.keras
withkeras
, and tensorflow.keras
seems to scale the metrics by the batch size if the dimensionality of the data is larger than three. At least that is my humble interpretation.
代码:
import numpy as np
from tensorflow import keras
# import keras
nSamples = 98765
nBatch = 2345
metric = 'poisson'
# metric = 'squared_hinge'
# metric = 'logcosh'
# metric = 'cosine_proximity'
# metric = 'binary_crossentropy'
# example data: always the same samples
np.random.seed(0)
dataIn = np.random.rand(nSamples)
dataOut = np.random.rand(nSamples)
for dataDim in range(1, 10):
# reshape samples into size (1,), ..., (1, 1, ...) according to dataDim
dataIn = np.expand_dims(dataIn, axis=-1)
dataOut = np.expand_dims(dataOut, axis=-1)
# build a model that does absolutely nothing
Layer = keras.layers.Input(shape=np.ones(dataDim))
model = keras.models.Model(inputs=Layer, outputs=Layer)
# compile, fit and observe loss ratio
model.compile(optimizer='adam', loss=metric, metrics=[metric])
history = model.fit(x=dataIn, y=dataOut, batch_size=nBatch, verbose=1)
lossRatio = history.history['loss'][0] / history.history[metric][0]
print(lossRatio)
我发现这种行为至少是不一致的.我应该认为它是错误还是功能?
I find this behavior inconsistent at least. Should I consider it a bug or a feature?
更新:经过进一步调查,我发现度量值被正确计算,而损失值却没有被正确计算.实际上,损失是样本损失的加权总和,其中每个样本的权重是该样本所在批次的大小.这有两个含义:
Update: After further investigation, I have found out that the metrics values seen to be computed correctly, while the loss values are not; in fact, the losses are weighted sums of the sample losses, where the weighting of each sample is the size of the batch that sample is in. This has two implications:
- 如果批次大小除以样品数量,则所有样品的权重都相同,并且损失等于该批次大小的因数.
- 如果批次大小未除以样本数,则由于通常会对批次进行混洗,因此权重(因此计算得出的损失)会从一个时期变化到另一个时期,尽管没有其他变化.这也适用于MSE等指标.
以下代码证明了这些观点:
The following code proves these points:
import numpy as np
import tensorflow as tf
from tensorflow import keras
# metric = keras.metrics.poisson
# metricName = 'poisson'
metric = keras.metrics.mse
metricName = 'mean_squared_error'
nSamples = 3
nBatchSize = 2
dataIn = np.random.rand(nSamples, 1, 1, 1)
dataOut = np.random.rand(nSamples, 1, 1, 1)
tf.InteractiveSession()
layer = keras.layers.Input(shape=(1, 1, 1))
model = keras.models.Model(inputs=layer, outputs=layer)
model.compile(optimizer='adam', loss=metric, metrics=[metric])
h = model.fit(x=dataIn, y=dataOut, batch_size=nBatchSize, verbose=1, epochs=10)
for (historyMetric, historyLoss) in zip(h.history[metricName], h.history['loss']):
# the metric value is correct and can be reproduced in a number of ways
kerasMetricOfData = metric(dataOut, dataIn).eval()
averageMetric = np.mean(kerasMetricOfData)
assert np.isclose(historyMetric, averageMetric), "..."
flattenedMetric = metric(dataOut.flatten(), dataIn.flatten()).eval()
assert np.isclose(historyMetric, flattenedMetric), "..."
if metric == keras.metrics.poisson:
numpyMetric = np.mean(dataIn - np.log(dataIn) * dataOut)
assert np.isclose(historyMetric, numpyMetric), "..."
# the loss value is incorrect by at least a scaling factor (~ batch size).
# also varies *randomly* if the batch size does not divide the # of samples:
if nSamples == 3:
incorrectLoss = np.array([
np.mean(kerasMetricOfData.flatten() * [1, nBatchSize, nBatchSize]),
np.mean(kerasMetricOfData.flatten() * [nBatchSize, 1, nBatchSize]),
np.mean(kerasMetricOfData.flatten() * [nBatchSize, nBatchSize, 1]),
])
elif nSamples == 4:
incorrectLoss = np.mean(kerasMetricOfData) * nBatchSize
assert np.any(np.isclose(historyLoss, incorrectLoss)), "..."
它输出:
Epoch 1/10
2/3 [===================>..........] - ETA: 0s - loss: 0.0044 - mean_squared_error: 0.0022
3/3 [==============================] - 0s 5ms/sample - loss: 0.0099 - mean_squared_error: 0.0084
Epoch 2/10
2/3 [===================>..........] - ETA: 0s - loss: 0.0238 - mean_squared_error: 0.0119
3/3 [==============================] - 0s 2ms/sample - loss: 0.0163 - mean_squared_error: 0.0084
Epoch 3/10
2/3 [===================>..........] - ETA: 0s - loss: 0.0238 - mean_squared_error: 0.0119
3/3 [==============================] - 0s 2ms/sample - loss: 0.0163 - mean_squared_error: 0.0084
Epoch 4/10
2/3 [===================>..........] - ETA: 0s - loss: 0.0238 - mean_squared_error: 0.0119
3/3 [==============================] - 0s 2ms/sample - loss: 0.0163 - mean_squared_error: 0.0084
Epoch 5/10
2/3 [===================>..........] - ETA: 0s - loss: 0.0238 - mean_squared_error: 0.0119
3/3 [==============================] - 0s 2ms/sample - loss: 0.0163 - mean_squared_error: 0.0084
Epoch 6/10
2/3 [===================>..........] - ETA: 0s - loss: 0.0222 - mean_squared_error: 0.0111
3/3 [==============================] - 0s 2ms/sample - loss: 0.0158 - mean_squared_error: 0.0084
Epoch 7/10
2/3 [===================>..........] - ETA: 0s - loss: 0.0222 - mean_squared_error: 0.0111
3/3 [==============================] - 0s 2ms/sample - loss: 0.0158 - mean_squared_error: 0.0084
Epoch 8/10
2/3 [===================>..........] - ETA: 0s - loss: 0.0238 - mean_squared_error: 0.0119
3/3 [==============================] - 0s 2ms/sample - loss: 0.0163 - mean_squared_error: 0.0084
Epoch 9/10
2/3 [===================>..........] - ETA: 0s - loss: 0.0222 - mean_squared_error: 0.0111
3/3 [==============================] - 0s 2ms/sample - loss: 0.0158 - mean_squared_error: 0.0084
Epoch 10/10
2/3 [===================>..........] - ETA: 0s - loss: 0.0044 - mean_squared_error: 0.0022
3/3 [==============================] - 0s 2ms/sample - loss: 0.0099 - mean_squared_error: 0.0084
更新:最后,使用keras.metrics.mse
和'mse'
之间似乎有所不同,如以下示例所示:
Update: Finally, there seems to be a difference between using keras.metrics.mse
and 'mse'
, as this example shows:
import numpy as np
from tensorflow import keras
# these three reproduce the issue:
# metric = keras.metrics.poisson
# metric = 'poisson'
# metric = keras.metrics.mse
# this one does not:
metric = 'mse'
nSamples = 3
nBatchSize = 2
dataIn = np.random.rand(nSamples, 1, 1, 1)
dataOut = np.random.rand(nSamples, 1, 1, 1)
layer = keras.layers.Input(shape=(1, 1, 1))
model = keras.models.Model(inputs=layer, outputs=layer)
model.compile(optimizer='adam', loss=metric, metrics=[metric])
model.fit(x=dataIn, y=dataOut, batch_size=2, verbose=1, epochs=10)
我开始相信这一定是一个错误,并且在这里报告了它.
I begin to believe that this must be a bug and reported it here.
推荐答案
已被确认为错误并已修复.有关更多信息,请参见 https://github.com/tensorflow/tensorflow/issues/25970.
This has been confirmed as a bug and fixed.For more information, see https://github.com/tensorflow/tensorflow/issues/25970.
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