I am trying to build an architecture which will be used for machine language translation (from English to French)
model = Sequential()
model.add(LSTM(256, input_shape =(15,1)))
model.add(RepeatVector(output_sequence_length))
model.add(LSTM(21,return_sequences=True))
model.add(TimeDistributed(Dense(french_vocab_size, activation='sigmoid'))
Max length of English sentence is 15 and that of French is 21. Max number of English words is 199 and that of French is 399. output_sequence_length is 21.
This model throws me an error
Error when checking input: expected lstm_40_input to have shape (None, 15, 1) but got array with shape (137861, 21, 1)
I am stuck with the understanding of the LSTM in keras.
1.The first argument according to documentation must be 'dimensionality of output space'. I did not understand what that means. Also,
what exactly happens return_sequences is set to True
Please let me know.
What Kind of data are you trying to feed your network ? Because it seems to me that you didn't convert your words to vectors (binary vectors or encoded vectors).
Anyway, a LSTM Netword need a 3 dimensional entry, the dimensions correspond to that : (samples , timesteps , features).
In your case, samples correspond to the numebr of your sentences, I guess 137861. Timesteps correspond to the length of each sequence, which In your case is 15, and features is the size of each encoded word ( Depending on which type of encoding you choose. If you choose OneHotEncoding, it will be 199).
The error that you got shows that you fed your network sequences with 21 timesteps instead of 15.
For your second question, when return_sequences is set to False, it returns only one output per LSTM layer, which in your case will be (256, ) for your first LSTM layer. when it's set to True, it will have one output per timestep, giving you an overall output of shape (15 , 256). When you want to stack two or more LSTM layers, you always have to set the first layers to return_sequences = True.
Also, what you are building is called a Many to Many architecture, with different timestep lengths for the input and the output (15 vs 21). As far as I know, it's not that easy to implement in keras.
I'm trying to train a CNN to categorize text by topic. When I use binary cross-entropy I get ~80% accuracy, with categorical cross-entropy I get ~50% accuracy.
I don't understand why this is. It's a multiclass problem, doesn't that mean that I have to use categorical cross-entropy and that the results with binary cross-entropy are meaningless?
model.add(embedding_layer)
model.add(Dropout(0.25))
# convolution layers
model.add(Conv1D(nb_filter=32,
filter_length=4,
border_mode='valid',
activation='relu'))
model.add(MaxPooling1D(pool_length=2))
# dense layers
model.add(Flatten())
model.add(Dense(256))
model.add(Dropout(0.25))
model.add(Activation('relu'))
# output layer
model.add(Dense(len(class_id_index)))
model.add(Activation('softmax'))
Then I compile it either it like this using categorical_crossentropy as the loss function:
model.compile(loss='categorical_crossentropy', optimizer='adam', metrics=['accuracy'])
or
model.compile(loss='binary_crossentropy', optimizer='adam', metrics=['accuracy'])
Intuitively it makes sense why I'd want to use categorical cross-entropy, I don't understand why I get good results with binary, and poor results with categorical.
The reason for this apparent performance discrepancy between categorical & binary cross entropy is what user xtof54 has already reported in his answer below, i.e.:
the accuracy computed with the Keras method evaluate is just plain
wrong when using binary_crossentropy with more than 2 labels
I would like to elaborate more on this, demonstrate the actual underlying issue, explain it, and offer a remedy.
This behavior is not a bug; the underlying reason is a rather subtle & undocumented issue at how Keras actually guesses which accuracy to use, depending on the loss function you have selected, when you include simply metrics=['accuracy'] in your model compilation. In other words, while your first compilation option
model.compile(loss='categorical_crossentropy', optimizer='adam', metrics=['accuracy'])
is valid, your second one:
model.compile(loss='binary_crossentropy', optimizer='adam', metrics=['accuracy'])
will not produce what you expect, but the reason is not the use of binary cross entropy (which, at least in principle, is an absolutely valid loss function).
Why is that? If you check the metrics source code, Keras does not define a single accuracy metric, but several different ones, among them binary_accuracy and categorical_accuracy. What happens under the hood is that, since you have selected binary cross entropy as your loss function and have not specified a particular accuracy metric, Keras (wrongly...) infers that you are interested in the binary_accuracy, and this is what it returns - while in fact you are interested in the categorical_accuracy.
Let's verify that this is the case, using the MNIST CNN example in Keras, with the following modification:
model.compile(loss='binary_crossentropy', optimizer='adam', metrics=['accuracy']) # WRONG way
model.fit(x_train, y_train,
batch_size=batch_size,
epochs=2, # only 2 epochs, for demonstration purposes
verbose=1,
validation_data=(x_test, y_test))
# Keras reported accuracy:
score = model.evaluate(x_test, y_test, verbose=0)
score[1]
# 0.9975801164627075
# Actual accuracy calculated manually:
import numpy as np
y_pred = model.predict(x_test)
acc = sum([np.argmax(y_test[i])==np.argmax(y_pred[i]) for i in range(10000)])/10000
acc
# 0.98780000000000001
score[1]==acc
# False
To remedy this, i.e. to use indeed binary cross entropy as your loss function (as I said, nothing wrong with this, at least in principle) while still getting the categorical accuracy required by the problem at hand, you should ask explicitly for categorical_accuracy in the model compilation as follows:
from keras.metrics import categorical_accuracy
model.compile(loss='binary_crossentropy', optimizer='adam', metrics=[categorical_accuracy])
In the MNIST example, after training, scoring, and predicting the test set as I show above, the two metrics now are the same, as they should be:
# Keras reported accuracy:
score = model.evaluate(x_test, y_test, verbose=0)
score[1]
# 0.98580000000000001
# Actual accuracy calculated manually:
y_pred = model.predict(x_test)
acc = sum([np.argmax(y_test[i])==np.argmax(y_pred[i]) for i in range(10000)])/10000
acc
# 0.98580000000000001
score[1]==acc
# True
System setup:
Python version 3.5.3
Tensorflow version 1.2.1
Keras version 2.0.4
UPDATE: After my post, I discovered that this issue had already been identified in this answer.
It all depends on the type of classification problem you are dealing with. There are three main categories
binary classification (two target classes),
multi-class classification (more than two exclusive targets),
multi-label classification (more than two non exclusive targets), in which multiple target classes can be on at the same time.
In the first case, binary cross-entropy should be used and targets should be encoded as one-hot vectors.
In the second case, categorical cross-entropy should be used and targets should be encoded as one-hot vectors.
In the last case, binary cross-entropy should be used and targets should be encoded as one-hot vectors. Each output neuron (or unit) is considered as a separate random binary variable, and the loss for the entire vector of outputs is the product of the loss of single binary variables. Therefore it is the product of binary cross-entropy for each single output unit.
The binary cross-entropy is defined as
and categorical cross-entropy is defined as
where c is the index running over the number of classes C.
I came across an "inverted" issue — I was getting good results with categorical_crossentropy (with 2 classes) and poor with binary_crossentropy. It seems that problem was with wrong activation function. The correct settings were:
for binary_crossentropy: sigmoid activation, scalar target
for categorical_crossentropy: softmax activation, one-hot encoded target
It's really interesting case. Actually in your setup the following statement is true:
binary_crossentropy = len(class_id_index) * categorical_crossentropy
This means that up to a constant multiplication factor your losses are equivalent. The weird behaviour that you are observing during a training phase might be an example of a following phenomenon:
At the beginning the most frequent class is dominating the loss - so network is learning to predict mostly this class for every example.
After it learnt the most frequent pattern it starts discriminating among less frequent classes. But when you are using adam - the learning rate has a much smaller value than it had at the beginning of training (it's because of the nature of this optimizer). It makes training slower and prevents your network from e.g. leaving a poor local minimum less possible.
That's why this constant factor might help in case of binary_crossentropy. After many epochs - the learning rate value is greater than in categorical_crossentropy case. I usually restart training (and learning phase) a few times when I notice such behaviour or/and adjusting a class weights using the following pattern:
class_weight = 1 / class_frequency
This makes loss from a less frequent classes balancing the influence of a dominant class loss at the beginning of a training and in a further part of an optimization process.
EDIT:
Actually - I checked that even though in case of maths:
binary_crossentropy = len(class_id_index) * categorical_crossentropy
should hold - in case of keras it's not true, because keras is automatically normalizing all outputs to sum up to 1. This is the actual reason behind this weird behaviour as in case of multiclassification such normalization harms a training.
After commenting #Marcin answer, I have more carefully checked one of my students code where I found the same weird behavior, even after only 2 epochs ! (So #Marcin's explanation was not very likely in my case).
And I found that the answer is actually very simple: the accuracy computed with the Keras method evaluate is just plain wrong when using binary_crossentropy with more than 2 labels. You can check that by recomputing the accuracy yourself (first call the Keras method "predict" and then compute the number of correct answers returned by predict): you get the true accuracy, which is much lower than the Keras "evaluate" one.
a simple example under a multi-class setting to illustrate
suppose you have 4 classes (onehot encoded) and below is just one prediction
true_label = [0,1,0,0]
predicted_label = [0,0,1,0]
when using categorical_crossentropy, the accuracy is just 0 , it only cares about if you get the concerned class right.
however when using binary_crossentropy, the accuracy is calculated for all classes, it would be 50% for this prediction. and the final result will be the mean of the individual accuracies for both cases.
it is recommended to use categorical_crossentropy for multi-class(classes are mutually exclusive) problem but binary_crossentropy for multi-label problem.
As it is a multi-class problem, you have to use the categorical_crossentropy, the binary cross entropy will produce bogus results, most likely will only evaluate the first two classes only.
50% for a multi-class problem can be quite good, depending on the number of classes. If you have n classes, then 100/n is the minimum performance you can get by outputting a random class.
You are passing a target array of shape (x-dim, y-dim) while using as loss categorical_crossentropy. categorical_crossentropy expects targets to be binary matrices (1s and 0s) of shape (samples, classes). If your targets are integer classes, you can convert them to the expected format via:
from keras.utils import to_categorical
y_binary = to_categorical(y_int)
Alternatively, you can use the loss function sparse_categorical_crossentropy instead, which does expect integer targets.
model.compile(loss='sparse_categorical_crossentropy', optimizer='adam', metrics=['accuracy'])
when using the categorical_crossentropy loss, your targets should be in categorical format (e.g. if you have 10 classes, the target for each sample should be a 10-dimensional vector that is all-zeros except for a 1 at the index corresponding to the class of the sample).
Take a look at the equation you can find that binary cross entropy not only punish those label = 1, predicted =0, but also label = 0, predicted = 1.
However categorical cross entropy only punish those label = 1 but predicted = 1.That's why we make assumption that there is only ONE label positive.
The main point is answered satisfactorily with the brilliant piece of sleuthing by desernaut. However there are occasions when BCE (binary cross entropy) could throw different results than CCE (categorical cross entropy) and may be the preferred choice. While the thumb rules shared above (which loss to select) work fine for 99% of the cases, I would like to add a few new dimensions to this discussion.
The OP had a softmax activation and this throws a probability distribution as the predicted value. It is a multi-class problem. The preferred loss is categorical CE. Essentially this boils down to -ln(p) where 'p' is the predicted probability of the lone positive class in the sample. This means that the negative predictions dont have a role to play in calculating CE. This is by intention.
On a rare occasion, it may be needed to make the -ve voices count. This can be done by treating the above sample as a series of binary predictions. So if expected is [1 0 0 0 0] and predicted is [0.1 0.5 0.1 0.1 0.2], this is further broken down into:
expected = [1,0], [0,1], [0,1], [0,1], [0,1]
predicted = [0.1, 0.9], [.5, .5], [.1, .9], [.1, .9], [.2, .8]
Now we proceed to compute 5 different cross entropies - one for each of the above 5 expected/predicted combo and sum them up. Then:
CE = -[ ln(.1) + ln(0.5) + ln(0.9) + ln(0.9) + ln(0.8)]
The CE has a different scale but continues to be a measure of the difference between the expected and predicted values. The only difference is that in this scheme, the -ve values are also penalized/rewarded along with the +ve values. In case your problem is such that you are going to use the output probabilities (both +ve and -ves) instead of using the max() to predict just the 1 +ve label, then you may want to consider this version of CE.
How about a multi-label situation where expected = [1 0 0 0 1]? Conventional approach is to use one sigmoid per output neuron instead of an overall softmax. This ensures that the output probabilities are independent of each other. So we get something like:
expected = [1 0 0 0 1]
predicted is = [0.1 0.5 0.1 0.1 0.9]
By definition, CE measures the difference between 2 probability distributions. But the above two lists are not probability distributions. Probability distributions should always add up to 1. So conventional solution is to use same loss approach as before - break the expected and predicted values into 5 individual probability distributions, proceed to calculate 5 cross entropies and sum them up. Then:
CE = -[ ln(.1) + ln(0.5) + ln(0.9) + ln(0.9) + ln(0.9)] = 3.3
The challenge happens when the number of classes may be very high - say a 1000 and there may be only couple of them present in each sample. So the expected is something like: [1,0,0,0,0,0,1,0,0,0.....990 zeroes]. The predicted could be something like: [.8, .1, .1, .1, .1, .1, .8, .1, .1, .1.....990 0.1's]
In this case the CE =
- [ ln(.8) + ln(.8) for the 2 +ve classes and 998 * ln(0.9) for the 998 -ve classes]
= 0.44 (for the +ve classes) + 105 (for the negative classes)
You can see how the -ve classes are beginning to create a nuisance value when calculating the loss. The voice of the +ve samples (which may be all that we care about) is getting drowned out. What do we do? We can't use categorical CE (the version where only +ve samples are considered in calculation). This is because, we are forced to break up the probability distributions into multiple binary probability distributions because otherwise it would not be a probability distribution in the first place. Once we break it into multiple binary probability distributions, we have no choice but to use binary CE and this of course gives weightage to -ve classes.
One option is to drown the voice of the -ve classes by a multiplier. So we multiply all -ve losses by a value gamma where gamma < 1. Say in above case, gamma can be .0001. Now the loss comes to:
= 0.44 (for the +ve classes) + 0.105 (for the negative classes)
The nuisance value has come down. 2 years back Facebook did that and much more in a paper they came up with where they also multiplied the -ve losses by p to the power of x. 'p' is the probability of the output being a +ve and x is a constant>1. This penalized -ve losses even further especially the ones where the model is pretty confident (where 1-p is close to 1). This combined effect of punishing negative class losses combined with harsher punishment for the easily classified cases (which accounted for majority of the -ve cases) worked beautifully for Facebook and they called it focal loss.
So in response to OP's question of whether binary CE makes any sense at all in his case, the answer is - it depends. In 99% of the cases the conventional thumb rules work but there could be occasions when these rules could be bent or even broken to suit the problem at hand.
For a more in-depth treatment, you can refer to: https://towardsdatascience.com/cross-entropy-classification-losses-no-math-few-stories-lots-of-intuition-d56f8c7f06b0
The binary_crossentropy(y_target, y_predict) doesn't need to apply to binary classification problem.
In the source code of binary_crossentropy(), the nn.sigmoid_cross_entropy_with_logits(labels=target, logits=output) of tensorflow was actually used.
And, in the documentation, it says that:
Measures the probability error in discrete classification tasks in which each class is independent and not mutually exclusive. For instance, one could perform multilabel classification where a picture can contain both an elephant and a dog at the same time.
I have my input state with shape = (84,84,4)
state = Input(shape=(84,84,4), dtype="float")
So I want to pass this to some TimeDistributed layer with time steps size=1..5 (in range of 1 to 5) and I don't know exactly which it equals.
My next layer is something like this:
conv1 = TimeDistributed(Convolution2D(16, 8, 8, subsample=(4, 4), border_mode='valid',
activation='relu', dim_ordering='tf'))(state)
And I've got an error at this layer:
IndexError: tuple index out of range
I just want to pass an unknown time-series size to TimeDistributed and then to LSTM also.
So basically in Keras - you need to provide the sequence length because during computations Keras layers accepts as an input numpy array with a specified shape - what makes compulsory for all inputs (at least in one batch) to have a length fixed. But - you still can deal with varying input size by 0-padding (making all sequence equal size by adding all zero dummy timesteps at the beginning) and then masking what makes your network equivalent to a varying length input network.
You can give a variable sequence-length, like this :
classifier.add(TimeDistributed(Convolution2D(64,(3,3)),input_shape=(None,None,None,3)))
But now you will have to adjust the length of the vector when it flattens or un-rolls at the time prediction.