为什么我在 TensorFlow Keras 中的损失函数和指标之间得到不同的值？

Question

在我使用 TensorFlow 进行 CNN 训练时，我将其用作Keras.losses.poisson损失函数。现在，我喜欢与损失函数一起计算许多指标，并且我观察到这Keras.metrics.poisson给出​​了不同的结果 - 尽管两者是相同的函数。

请参阅此处的一些示例输出：loss输出poisson具有不同的范围，0.5 与 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 - Loss and Metriccalculated different? 但是，我没有使用正则化。

另外，我遇​​到过这个，它至少帮助我重现了这个问题：Keras Loss 和 Metric 中的相同函数即使没有正则化也会给出不同的值

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，基本上是维度引发了这个问题。从下面的扩展示例中，您可以看到

• 这个问题可以用许多损失函数来重现（那些不以 开头的函数mean_），
• tensorflow.keras当替换为keras, 时，问题就消失了
• tensorflow.keras 如果数据的维度大于 3 ，似乎会按批量大小缩放指标。至少这是我的拙见。

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不一致。我应该将其视为错误还是功能？

更新：经过进一步调查，我发现指标值计算正确，而损失值则不然；事实上，损失是样本损失的加权总和，其中每个样本的权重是样本所在批次的大小。这有两个含义：

1. 如果批量大小除以样品数量，则所有样品的称重都是相同的，并且损失只需除以等于批量大小的系数即可。
2. 如果批次大小不能除以样本数量，则由于批次通常会被打乱，因此权重以及计算出的损失会从一个时期到下一个时期发生变化，尽管其他方面没有发生任何变化。这也适用于 MSE 等指标。

下面的代码证明了这些观点：

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keras.metrics.mse：最后，使用和之间似乎存在差异'mse'，如本示例所示：

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在这里报告了它。

Answer 1

这已被确认为错误并已修复。有关更多信息，请参阅https://github.com/tensorflow/tensorflow/issues/25970。