为什么在评估TensorFlow Keras中作为度量的损失函数时会得到不同的值?

Keras.losses.poisson 作为损失函数。现在,我喜欢计算损失函数旁边的许多指标,我观察到了这一点 Keras.metrics.poisson 给出不同的结果-尽管两者是相同的函数。

loss 和 poisson 输出具有不同的范围,分别为0.5和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

我在键入此问题时发现了类似的问题: Keras - Loss and Metric calculated differently? 但是,我没有使用正则化。

此外,我还遇到了这个问题,它至少帮助我重现了这个问题: 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)

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

• 这个问题可以通过许多损失函数(不是以 mean_ ),
• tensorflow.keras 具有 keras ,及
• 如果数据的维度大于三,则似乎可以通过批量大小来缩放度量

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。我应该认为它是一个bug还是一个特性?

更新 :经过进一步调查,我发现度量值计算正确,而损失值计算不正确;事实上,损失是样本损失的加权和,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. 如果批次大小除以样本数量,则所有样本的称重相同,损失仅通过与批次大小相等的系数进行抵消。
2. 如果批次大小不划分样本数量,因为批次通常会被洗牌,那么权重以及由此计算的损失从一个历元到下一个历元都会发生变化,尽管其他历元没有发生任何变化。这也适用于MSE等指标。

以下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 和 'mse' ,如本例所示:

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和错误 reported it here .