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How to calculate Categorical Cross-Entropy by hand?

When I calculate Binary Crossentropy by hand I apply sigmoid to get probabilities, then use Cross-Entropy formula and mean the result:

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

How to calculate Categorical Cross-Entropy when logits and labels have different sizes?

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

I mean how can I get the same result ([2.0077195 0.00928135 0.6800677 ]) by hand?

@OverLordGoldDragon answer is correct. In TF 2.0 it looks like this:

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

For language models:

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
Solution

  • SparseCategoricalCrossentropy is CategoricalCrossentropy that takes integer labels as opposed to one-hot. Example from source code, the two below are equivalent:

    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]

    As to how to do it 'by hand', we can refer to the 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 is model output before passing it into softmax (so we pass it into softmax)
    • from_logits = False: preds is model output after passing it into softmax (so we skip this step)

    So in summary, to compute it by hand:

    1. Convert integer labels to one-hot labels
    2. If preds are model outputs before softmax, we compute their softmax
    3. pred /= ... normalizes predictions before computing logs; this way, high-probab. preds on zero-labels penalize correct predictions on one-labels. If from_logits = False, this step is skipped, since softmax does the normalization. See this snippet. Further reading
    4. For each observation / sample, compute element-wise negative log (base e) only where label==1
    5. Take mean of losses for all the observations

    Lastly, the mathematical formula for categorical crossentropy is:

    • i iterates over N observations
    • c iterates over C classes
    • 1 is the indicator function - here, like binary crossentropy, except operates on length-C vectors
    • p_model [y_i \in C_c] - predicted probability of observation i belonging to class c