# On cross entropy

Cross entropy can be used to define a loss function in machine learning and is usually used when training a classification problem.

In information theory, the cross entropy between two probability distributions $p$ and $q$ over the same underlying set of events measures the average number of bits needed to identify an event drawn from the set if a coding scheme used for the set is optimized for an estimated probability distribution $q$, rather than the true distribution $p$. (source)

This post tries to implement it in pure python to better understand it’s inner workings and then compare it to other popular implementations for cross-validation.

# Our implementation

import numpy as np
import tensorflow as tf
import torch
from matplotlib import pyplot as plt


The crossentropy function is defined as:

This seems simple enough so let’s implement this!

def categorical_crossentropy(y_true, y_pred):
# - SUM(target * log(pred))
return -np.sum(y_true * np.log(y_pred))

categorical_crossentropy([0, 1], [0.5, 0.5])

0.6931471805599453


I don’t trust my code so I need to certify that my implementation is working correctly by comparing it to known and proven implementations.

The first one that comes to mind is the sklearn one. It is not (confusingly) called crossentropy but goes by its other name: log_loss

from sklearn.metrics import log_loss

log_loss([0, 1], [0.5, 0.5])

0.6931471805599453


Ok! The results matched on both (and also match my analytical computation). Time for a few more tests to make sure we’re not missing something with this happy flow.

def certify():
tests = [
[[0, 0, 1], [0.3, 0.7, 0.0]],
[[0, 1, 0, 0], [0.1, 0.2, 0.3, 0.4]],
[[1, 0], [0.4, 0.6]],
]

for [y_true, y_pred] in tests:
my_xent = categorical_crossentropy(y_true, y_pred)
xent = log_loss(y_true, y_pred)
assert my_xent == xent, f"{y_true}\t{y_pred}\n{my_xent} != {xent}"

certify()

/usr/local/lib/python3.6/dist-packages/ipykernel_launcher.py:3: RuntimeWarning: divide by zero encountered in log
This is separate from the ipykernel package so we can avoid doing imports until

---------------------------------------------------------------------------

AssertionError                            Traceback (most recent call last)

<ipython-input-5-f1d99854990a> in <module>()
11         assert my_xent == xent, f"{y_true}\t{y_pred}\n{my_xent} != {xent}"
12
---> 13 certify()

<ipython-input-5-f1d99854990a> in certify()
9         my_xent = categorical_crossentropy(y_true, y_pred)
10         xent = log_loss(y_true, y_pred)
---> 11         assert my_xent == xent, f"{y_true}\t{y_pred}\n{my_xent} != {xent}"
12
13 certify()

AssertionError: [0, 0, 1]	[0.3, 0.7, 0.0]
inf != 12.033141381058451


Hmm.. it crashes on the first example..

categorical_crossentropy([0, 0, 1], [0.3, 0.7, 0.0]), log_loss([0, 0, 1], [0.3, 0.7, 0.0])

/usr/local/lib/python3.6/dist-packages/ipykernel_launcher.py:3: RuntimeWarning: divide by zero encountered in log
This is separate from the ipykernel package so we can avoid doing imports until

(inf, 12.033141381058451)


The problem is of course in our implementation. We have a 0.0 value (the third in the y_pred) on which we are applying the log. You may remember that the log function is undefined on 0.0. The sklearn implementation actually clips the end of the provided y_pred so it will never be 0.0 or 1.0.

Offtopic: log(1.0) is actually 0, it is defined, and I’m unsure why they clip the top as well. I assume is related either to the vanishing gradient problem or to the idea that a prediction is never actually 100% certain of a result (?).

The clipping is performed, employing a sufficiently small epsilon value (sklearn defaults to 1e-15), as:

y_pred = max(eps, min((1-eps), y_pred))


We can use the above or make use of np.clip which will implement the exact formula above, but faster (they claim).

def _clip_for_log(y_pred, eps=1e-15):
# y_pred = np.maximum(eps, np.minimum((1-eps), y_pred)) # equivalent
y_pred = np.clip(y_pred, eps, 1-eps)
return y_pred

_clip_for_log(1), _clip_for_log(np.array([1, 1, 0, 1, 0, 0.5, 0.4, 0.3]))

(0.999999999999999,
array([1.e+00, 1.e+00, 1.e-15, 1.e+00, 1.e-15, 5.e-01, 4.e-01, 3.e-01]))


The improved crossentropy function is now:

def categorical_crossentropy(y_true, y_pred):
y_pred = _clip_for_log(y_pred)
return -np.sum(y_true * np.log(y_pred))

categorical_crossentropy([0, 0, 1], [0.3, 0.7, 0.0])

34.538776394910684

certify()

---------------------------------------------------------------------------

AssertionError                            Traceback (most recent call last)

<ipython-input-13-e627f102689a> in <module>()
----> 1 certify()

<ipython-input-7-b39bd15b1efe> in certify()
7
8     for [y_true, y_pred] in tests:
----> 9         assert categorical_crossentropy(y_true, y_pred) == log_loss(y_true, y_pred), f"{y_true}\t{y_pred}"
10
11 certify()

AssertionError: [0, 0, 1]	[0.3, 0.7, 0.0]


Trying to run the test again shows that (even if the code doesn’t crashes anymore) we are getting different results:

categorical_crossentropy([0, 0, 1], [0.3, 0.7, 0.0]), log_loss([0, 0, 1], [0.3, 0.7, 0.0])

(34.538776394910684, 12.033141381058451)


What happens is that, in reality, calling

log_loss([0, 0, 1], [0.3, 0.7, 0.0])


is interpreted as

log_loss([
0,
0,
1
],
[
0.3,
0.7,
0.0
])


where each list is a batch of predictions. So the log_loss is actually used as a binary_crossentropy on each pair of (target, prediction) and the results (equal to the number of values in the lists) is averaged together.

Explicitly, we have:

(log_loss(, [0.3], labels=[0, 1]) +
log_loss(, [0.7], labels=[0, 1]) +
log_loss(, [0.0], labels=[0, 1])) / 3

12.033141381058451


This means that we need to make the sklearn log_loss think that we’re not having batches but a single prediction to evaluate (so instead of shape (3,) we need a (1, 3)).

categorical_crossentropy([0, 0, 1], [0.3, 0.7, 0.0]), log_loss([[0, 0, 1]], [[0.3, 0.7, 0.0]])

(34.538776394910684, 34.538776394910684)

def certify():
tests = [
[[0, 0, 1], [0.3, 0.7, 0.0]],
[[0, 1, 0, 0], [0.1, 0.2, 0.3, 0.4]],
[[1, 0], [0.4, 0.6]],
]

for [y_true, y_pred] in tests:
my_xent = categorical_crossentropy(y_true, y_pred)
xent = log_loss([y_true], [y_pred])
assert my_xent == xent, f"{y_true}\t{y_pred}\n{my_xent} != {xent}"
print("Success, results are equal!")

certify()

Success, results are equal!


Does this mean that our implementation does not work on batches?

categorical_crossentropy([[0, 0, 1], [0, 1, 0]], [[0.3, 0.7, 0.0], [0.5, 0.2, 0.3]]), log_loss([[0, 0, 1], [0, 1, 0]], [[0.3, 0.7, 0.0], [0.5, 0.2, 0.3]])

(36.14821430734479, 18.074107153672394)


The results of our computation and sklearn’s log_loss with batches is different..

categorical_crossentropy([[0, 1, 0]], [[0.5, 0.2, 0.3]]) + categorical_crossentropy([0, 0, 1], [0.3, 0.7, 0.0])

36.14821430734479


It works but not correctly. Our implementation does a sum over all errors in a batch but we need to return a mean, so we need to divide it by the number of examples in the batch (the batch_size). As such, the new implementation is:

def ensure_ndarray(value):
if not isinstance(value, np.ndarray):
value = np.asarray(value)
return value

def categorical_crossentropy(y_true, y_pred):
"""
Implements the crossentropy function:
Loss = - SUM(target * log(pred))
"""

y_true = ensure_ndarray(y_true)
y_pred = ensure_ndarray(y_pred)

# dimensions must match
assert y_true.shape == y_pred.shape

y_pred = _clip_for_log(y_pred)
batch_size = y_true.shape

return -np.sum(y_true * np.log(y_pred)) / batch_size

categorical_crossentropy([[0, 0, 1], [0, 1, 0]], [[0.3, 0.7, 0.0], [0.5, 0.2, 0.3]]), log_loss([[0, 0, 1], [0, 1, 0]], [[0.3, 0.7, 0.0], [0.5, 0.2, 0.3]])

(18.074107153672394, 18.074107153672394)

certify()

---------------------------------------------------------------------------

AssertionError                            Traceback (most recent call last)

<ipython-input-14-e627f102689a> in <module>()
----> 1 certify()

<ipython-input-12-f328bbe97973> in certify()
9         my_xent = categorical_crossentropy(y_true, y_pred)
10         xent = log_loss([y_true], [y_pred])
---> 11         assert my_xent == xent, f"{y_true}\t{y_pred}\n{my_xent} != {xent}"
12     print("Success, results are equal!")
13

AssertionError: [0, 0, 1]	[0.3, 0.7, 0.0]
11.512925464970229 != 34.538776394910684


Hmm… We’re back to square one. The first example doesn’t fit anymore because we are using a single list (and not batches). Our current implementation assumes we have a single prediction to make, but computes the batch size on the first dimension, which is 3 (but that’s actually the number of classes in our single one-hot-encoded vector).

We need to compute the batch_size a little more carefully (considering we have a batch computation if the inputs have at least 2 dimensions, else if only a single dimensions is used, the inputs are a single prediction)

def ensure_ndarray(value):
if not isinstance(value, np.ndarray):
value = np.asarray(value)
return value

def categorical_crossentropy(y_true, y_pred):
"""
Implements the crossentropy function:
Loss = - SUM(target * log(pred))
"""

y_true = ensure_ndarray(y_true)
y_pred = ensure_ndarray(y_pred)

# dimensions must match
assert y_true.shape == y_pred.shape

y_pred = _clip_for_log(y_pred)
batch_size = y_true.shape if len(y_true.shape) > 1 else 1

return -np.sum(y_true * np.log(y_pred)) / batch_size

certify()

Success, results are equal!


Success!

# Sklearn’s vs Ours discussion

Now, there’s an interesting discussion about our above heuristic:

• form an interface perspective, the implementation acts is two ways, given the shape of the inputs:
• in batch mode (dims >=2, first dim is the batch one)
• in single prediction mode (dims == 1, we only have a single prediction to evaluate)

Since we have a hybrid behavior, we may try to standardize a part of it:

• consider that we always have batches.

This makes the behavior on the: categorical_crossentropy([0, 0, 1], [0.3, 0.7, 0.0]) be interpreted as a batch of [(0, 0.3), (0, 0.7), (1, 0.0)] examples (3 in this case) and the result be a mean of these.

This is actually the behavior of the sklearn implementation. It always assumes you send in batches. It may make sense, since the vast majority of the time you want to use this function is in a stochastic gradient descent (batch based) training loop.

Unfortunately this still adds some uncertainties (or heterogeneous behavior) as the pairs above [(0, 0.3), (0, 0.7), (1, 0.0)] cannot be plainly computed anymore with the initial formula: $Loss = -\sum_{i}{target_i*\log(prediction_i)}$ since this formulation is valid for a one-hot-encoded target variable where there is exactly one value of 1. In the pair (0, 0.3) there is no 1 value in the target, so using this formula yields the result 0 (and it always is 0 for targets equal to 0). This basically leads to the Loss value only represent the errors of the positive (target == 1) samples in the batch, because these are the only ones in which (the single) product is not 0.

## Binary Crossentropy

The sklearn implementation solves this case by assuming that if your input dimension is 1 (you have a list of scalar values) the values will not be computed on the categorical crossentropy function but a simplified version of it where the pairs (0, 0.3) have following internal representation:

• 0 is translated to [1, 0] <=> [1 - 0, 0]
• 0.3 is translated to [0.7, 0.3] <=> [1 - 0.3, 0.3]

This redefinition can be translated as: the correct output is label 0 but the prediction for label 1 is 0.3. So basically I want to say that I predict 0 because I predict a really low label 1 value..

Generically, the pairs $(label, pred)$ where $label \in {0,1}$ and $pred \in (0, 1]$ are equivalenced to target = [1 - label, label] and prediction = [1 - pred, pred]. Now we can compute the regular crossentropy formula: $Loss = - \sum_{i}{target_i} *\log(prediction_i)$ $Loss = - [( 1 - label) * \log(1 - pred) + label * \log(pred)]$

The last formulation is called binary crossentropy.

So in essence, sklearn.log_loss chooses to assume that we always have batches in the input, and when in doubt (single dimension inputs), doesn’t compute the categorical crossentropy but the binary crossentropy.

For my taste and implementation I’m going to assume that we always compute the categorical crossentropy and relax the batching assumption as the function is called categroical... I always do categorical.

# Keras / Tensorflow crossentropy

from tensorflow.keras.metrics import categorical_crossentropy as keras_cat_xent
keras_cat_xent([0, 1], [0.5, 0.5]).numpy()

0.6931472

targets = [[0, 0, 1], [0, 1, 0]]
predics = [[0.3, 0.7, 0.0], [0.5, 0.2, 0.3]]
categorical_crossentropy(targets, predics), keras_cat_xent(targets, predics).numpy()

(18.074107153672394, array([16.118095,  1.609438], dtype=float32))


OK, using the tensorflow / keras version leads to the following 3 questions:

• why do we get 2 values?
• what do these values mean?
• how can make our numbers match?

The answers to the first and second question are somewhat obvious: we have as many results as samples in the batch, ant they are the results of the per-sample categorical crossentropy function.

This is because the K.categorical_crossentropy function also has a axis=-1 parameter which instructs on which dimensions to do the reduction. Since we’re asking for a reduction on only the last dimension (the dimension of one-hot-encoded values) we are left with the dimension 0 elements (the batch size).

We can demonstrate this by showing that calling the K.categorical_crossentropy function individually for each sample in a batch with size 1 will lead the the same 2 values as above.

keras_cat_xent([[0, 0, 1]], [[0.3, 0.7, 0.0]]), keras_cat_xent([[0, 1, 0]], [[0.5, 0.2, 0.3]])

(<tf.Tensor: shape=(1,), dtype=float32, numpy=array([16.118095], dtype=float32)>,
<tf.Tensor: shape=(1,), dtype=float32, numpy=array([1.609438], dtype=float32)>)


Now for the last question (“how can we make our previous numbers match” / “why don’t they match?”).

Recall that on the first sample, our function returned 34.538, the same for log_loss whereas the keras version returned 16.118..

targets = [[0, 0, 1]]
predics = [[0.3, 0.7, 0.0]]
categorical_crossentropy(targets, predics), log_loss(targets, predics), keras_cat_xent(targets, predics).numpy().sum()

(34.538776394910684, 34.538776394910684, 16.118095)


After reading the source code of the keras implementation and couldn’t find any difference with our implementation, I decided to recompute by hand their answer, when I noticed something strange.

TF_EPSILON = 1e-7
SK_EPSILON = 1e-15

- ((1 * np.log(TF_EPSILON))), - ((1 * np.log(SK_EPSILON)))

(16.11809565095832, 34.538776394910684)


Tensorflow / Keras uses a different epsilon!

We were previously using 1e-15 but they choose 1e-7. It is a bit surprising that the resulting errors are that large, while the change between them is rather small.

Sure, mathematically it make sense that the log of a 10^8 smaller value should result in a bigger error, but from an API point of view, predicting either 1/10^7 or 1/10^15 while the correct answer is 1, should give pretty close errors. These two predictions are after all synonymous to pure wrong.

Now, for the question of “why we have 2 values instead of a single one per on a batch_size == 2?” the answer is both surprising and confusing..

In all, using the vanilla keras.io you can get (depending on what you use):

• per-batch with SUM reduction
• per-batch with MEAN reduction
• per-sample

On the TensorFlow keras port you can get:

• per-batch with MEAN reduction
• per-sample

What about the batching dilemma (the one where sklearn and ours diverged?)

keras_cat_xent([0, 0, 1], [0.3, 0.7, 0.0])

16.118095


Well, keras chooses to always do the crossentropy, like we did.

One last note about Tensorflow / Keras. The categroical_crossentropy has also a parameter from_logits=False that can interpret the values of the predictions as logits, meaning that you can use it for multi-class predictions:

lables = [1, 0, 0, 1]


where the network is expected to produce results for multiple classes at the same time. This is interpreted as if each value of the label represents a binary_crossentropy evaluation.

Setting from_logits=True redirects you to using the tensorflow.nn.softmax_cross_entropy_with_logits_v2 function. This function has a few caveats to understand:

NOTE: While the classes are mutually exclusive, their probabilities need not be. All that is required is that each row of labels is a valid probability distribution. If they are not, the computation of the gradient will be incorrect.

This means that even I’ve stated you could have labels=[1, 0, 0, 1] this function actually requires that you send it something like labels = [0.5, 0, 0, 0.5] for it to be a valid probability distribution. You can convert [1, 0, 0, 1] to [0.5, 0, 0, 0.5] by passing it through a softmax or through a simples scaling method:

def scale(values):
return values / np.sum(values)


WARNING: This op expects unscaled logits, since it performs a softmax on logits internally for efficiency. Do not call this op with the output of softmax, as it will produce incorrect results.

This means that while we are required to scale the labels we are required NOT to scale the logits (i.e. predictions)

# PyTorch crossentropy

NLLLoss is the negative log likelihood implementation:

• uses the format (y_pred, y_true) instead of the common (y_true, y_pred) found in sklearn, keras, tensorflow
• y_pred is expected to have log values (i.e. y_pred == log(orig_y_pred)
• y_true should contain class indexes (i.e. ordinal values not one-hot-encoded values). This is equivalent to the sparse_categorical_crossentropy class of modules in keras and TensorFlow
• expects certain types:
• torch.Long for the y_true
• torch.Float for the y_pred

Observation: Because the function requires the y_pred values to be in log format that means that is up to the caller to do the clipping with whatever values he wishes to use.

def f(values):

def l(values):

from torch import nn
nn.NLLLoss()(f([np.log([0.5, 0.5])]), l())

tensor(0.6931)


So let’s respect these documented assumptions and try to check that we can correctly match the results of the sklearn.metrics.log_loss and torch.nn.NLLLoss

orig_targets = [[0, 0, 1], [0, 1, 0]]
orig_predics = [[0.3, 0.7, 0.0], [0.5, 0.2, 0.3]]

targets = np.argmax(orig_targets, axis=-1)
predics = np.log(np.clip(orig_predics, SK_EPSILON, 1-SK_EPSILON)) # same clipping type

nn.NLLLoss()(f(predics), l(targets)).numpy(), log_loss(orig_targets, orig_predics)

(array(18.074108, dtype=float32), 18.074107153672394)


Notice that by default, calling .float() on a PyTorch tensor yields a float32 values which leads to a reduction in precision of the results.

Let’s try to make the tensor a float64 value and notice what happens

def f(values):

def l(values):

orig_targets = [[0, 0, 1], [0, 1, 0]]
orig_predics = [[0.3, 0.7, 0.0], [0.5, 0.2, 0.3]]

targets = np.argmax(orig_targets, axis=-1)
predics = np.log(np.clip(orig_predics, SK_EPSILON, 1-SK_EPSILON))   # same clipping type
nn.NLLLoss()(f(predics), l(targets)).numpy(), log_loss(orig_targets, orig_predics)

(array(18.07410715), 18.074107153672394)


We get more decimal points but the results are still a bit off compared to the sklearn implementation. I’m not sure why that is.

There is also the CrossEntropyLoss layer, which is read it correctly only does a softmax on the predictions, before computing the log_loss.

This means that either we need to invert the softmax before calling it, or we apply the softmax on the sklearn one if we wish to compare the results.

It’s easier to do the second option.

nn.CrossEntropyLoss()(f(orig_predics), l(targets)), log_loss(orig_targets, softmax(orig_predics))

(tensor(1.3566, dtype=torch.float64), 1.3565655522346258)


It worked!!

# Conclusions

• I wasn’t expected things to be so nuanced when I started writing this!
• keras in bit of a mess. There are multiple confusing ways to compute the crossentropy.
• small details (the epsilon) matter
• if not careful we may sometime get to see the results of a binary_crossentropy rather than a categorical_crossentropy
• PyTorch makes you to explicitly do stuff (like the applying the log, the clipping or the softmax) in order to make you aware of the subtle details that if made implicit (like keras and sklearn superbly do) might make you shoot yourself in the foot (without even noticing it)

A relay nice article about the cross-entropy loss can also be found here

### My main takeaways are these:

• implement everything yourself (or read the source-code). I’m afraid of how many details I’ve missed until now in other more convoluted (get it?!) layers / concepts.
• details matter

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