Question

当我手动计算二进制交叉熵时，我采用S形来获得概率，然后使用交叉熵公式并求出结果的平均值：

logits = tf.constant([-1, -1, 0, 1, 2.])
labels = tf.constant([0, 0, 1, 1, 1.])

probs = tf.nn.sigmoid(logits)
loss = labels * (-tf.math.log(probs)) + (1 - labels) * (-tf.math.log(1 - probs))
print(tf.reduce_mean(loss).numpy()) # 0.35197204

cross_entropy = tf.keras.losses.BinaryCrossentropy(from_logits=True)
loss = cross_entropy(labels, logits)
print(loss.numpy()) # 0.35197204

当logits和labels的大小不同时，如何计算分类交叉熵？

logits = tf.constant([[-3.27133679, -22.6687183, -4.15501118, -5.14916372, -5.94609261,
                       -6.93373299, -5.72364092, -9.75725174, -3.15748906, -4.84012318],
                      [-11.7642536, -45.3370094, -3.17252636, 4.34527206, -17.7164974,
                      -0.595088899, -17.6322937, -2.36941719, -6.82157373, -3.47369862],
                      [-4.55468369, -1.07379043, -3.73261762, -7.08982277, -0.0288562477, 
                       -5.46847963, -0.979336262, -3.03667569, -3.29502845, -2.25880361]])
labels = tf.constant([2, 3, 4])

loss_object = tf.keras.losses.SparseCategoricalCrossentropy(from_logits=True,
                                                            reduction='none')
loss = loss_object(labels, logits)
print(loss.numpy()) # [2.0077195  0.00928135 0.6800677 ]
print(tf.reduce_mean(loss).numpy()) # 0.8990229

我的意思是我怎样才能手工获得相同的结果（[2.0077195 0.00928135 0.6800677 ]）？

@OverLordGoldDragon答案是正确的。在TF 2.0中，它看起来像这样：

loss_object = tf.keras.losses.SparseCategoricalCrossentropy(
    from_logits=True, reduction='none')
loss = loss_object(labels, logits)
print(f'{loss.numpy()}\n{tf.math.reduce_sum(loss).numpy()}')

one_hot_labels = tf.one_hot(labels, 10)

preds = tf.nn.softmax(logits)
preds /= tf.math.reduce_sum(preds, axis=-1, keepdims=True)
loss = tf.math.reduce_sum(tf.math.multiply(one_hot_labels, -tf.math.log(preds)), axis=-1)
print(f'{loss.numpy()}\n{tf.math.reduce_sum(loss).numpy()}')
# [2.0077195  0.00928135 0.6800677 ]
# 2.697068691253662
# [2.0077198  0.00928142 0.6800677 ]
# 2.697068929672241

对于语言模型：

vocab_size = 9
seq_len = 6
batch_size = 2

labels = tf.reshape(tf.range(batch_size*seq_len), (batch_size,seq_len)) # (2, 6)
logits = tf.random.normal((batch_size,seq_len,vocab_size)) # (2, 6, 9)

loss_object = tf.keras.losses.SparseCategoricalCrossentropy(
    from_logits=True, reduction='none')
loss = loss_object(labels, logits)
print(f'{loss.numpy()}\n{tf.math.reduce_sum(loss).numpy()}')

one_hot_labels = tf.one_hot(labels, vocab_size)

preds = tf.nn.softmax(logits)
preds /= tf.math.reduce_sum(preds, axis=-1, keepdims=True)
loss = tf.math.reduce_sum(tf.math.multiply(one_hot_labels, -tf.math.log(preds)), axis=-1)
print(f'{loss.numpy()}\n{tf.math.reduce_sum(loss).numpy()}')
# [[1.341706  3.2518263 2.6482694 3.039099  1.5835983 4.3498387]
#  [2.67237   3.3978183 2.8657475       nan       nan       nan]]
# nan
# [[1.341706  3.2518263 2.6482694 3.039099  1.5835984 4.3498387]
#  [2.67237   3.3978183 2.8657475 0.        0.        0.       ]]
# 25.1502742767334

Answer 1

SparseCategoricalCrossentropy是CategoricalCrossentropy，它使用整数标签，而不是 one 。来自source code的示例，以下两个是等效的：

scce = tf.keras.losses.SparseCategoricalCrossentropy() cce = tf.keras.losses.CategoricalCrossentropy() labels_scce = K.variable([[0, 1, 2]]) labels_cce = K.variable([[1, 0, 0], [0, 1, 0], [0, 0, 1]]) preds = K.variable([[.90,.05,.05], [.50,.89,.60], [.05,.01,.94]]) loss_cce = cce(labels_cce, preds, from_logits=False) loss_scce = scce(labels_scce, preds, from_logits=False)

with tf.Session() as sess: sess.run(tf.global_variables_initializer()) sess.run([loss_cce, loss_scce]) print(K.get_value(loss_cce)) print(K.get_value(loss_scce)) # [0.10536055 0.8046684 0.0618754] # [0.10536055 0.8046684 0.0618754]

关于“手动”操作的方法，我们可以参考Numpy backend：

np_labels = K.get_value(labels_cce) np_preds = K.get_value(preds) losses = [] for label, pred in zip(np_labels, np_preds): pred /= pred.sum(axis=-1, keepdims=True) losses.append(np.sum(label * -np.log(pred), axis=-1, keepdims=False)) print(losses) # [0.10536055 0.8046684 0.0618754]

from_logits = True：preds是模型输出之前，将其传递到softmax中（因此我们将其传递到softmax中）

from_logits = False：preds是模型输出之后的结果，将其传递到softmax中（因此我们跳过此步骤）

因此，总而言之，要手动进行计算：

将整数标签转换为一次性标签

如果preds是softmax之前的模型输出，我们将计算其softmax
pred /= ...在计算日志之前归一化预测；这样，高概率。倾向于零标签对一个标签的 penalize 正确预测。如果为from_logits = False，则此步骤被跳过，因为softmax进行了归一化。参见this snippet。 Further reading
对于每个观察值/样本，计算元素式负数log（基于e）仅在 label==1
对所有观察结果取均值

最后，分类交叉熵的数学公式为：

i遍历N个观测值
c遍历C类
1 是indicator function-在这里，像二进制交叉熵一样，除了对长度为C的向量进行操作
p_model [y_i \in C_c]-属于类别i的预测观察概率c

如何手动计算分类交叉熵？

1 个答案: