What function defines accuracy in Keras when the loss is mean squared error (MSE)?

Question

How is Accuracy defined when the loss function is mean square error? Is it mean absolute percentage error?

The model I use has output activation linear and is compiled with loss= mean_squared_error

model.add(Dense(1))
model.add(Activation('linear'))  # number

model.compile(loss='mean_squared_error', optimizer='adam', metrics=['accuracy'])

and the output looks like this:

Epoch 99/100
1000/1000 [==============================] - 687s 687ms/step - loss: 0.0463 - acc: 0.9689 - val_loss: 3.7303 - val_acc: 0.3250
Epoch 100/100
1000/1000 [==============================] - 688s 688ms/step - loss: 0.0424 - acc: 0.9740 - val_loss: 3.4221 - val_acc: 0.3701

So what does e.g. val_acc: 0.3250 mean? Mean_squared_error should be a scalar not a percentage - shouldnt it? So is val_acc - mean squared error, or mean percentage error or another function?

From definition of MSE on wikipedia:https://en.wikipedia.org/wiki/Mean_squared_error

The MSE is a measure of the quality of an estimator—it is always non-negative, and values closer to zero are better.

Does that mean a value of val_acc: 0.0 is better than val_acc: 0.325?

edit: more examples of the output of accuracy metric when I train - where the accuracy is increase as I train more. While the loss function - mse should decrease. Is Accuracy well defined for mse - and how is it defined in Keras?

lAllocator: After 14014 get requests, put_count=14032 evicted_count=1000 eviction_rate=0.0712657 and unsatisfied allocation rate=0.071714
1000/1000 [==============================] - 453s 453ms/step - loss: 17.4875 - acc: 0.1443 - val_loss: 98.0973 - val_acc: 0.0333
Epoch 2/100
1000/1000 [==============================] - 443s 443ms/step - loss: 6.6793 - acc: 0.1973 - val_loss: 11.9101 - val_acc: 0.1500
Epoch 3/100
1000/1000 [==============================] - 444s 444ms/step - loss: 6.3867 - acc: 0.1980 - val_loss: 6.8647 - val_acc: 0.1667
Epoch 4/100
1000/1000 [==============================] - 445s 445ms/step - loss: 5.4062 - acc: 0.2255 - val_loss: 5.6029 - val_acc: 0.1600
Epoch 5/100
783/1000 [======================>.......] - ETA: 1:36 - loss: 5.0148 - acc: 0.2306

Answer 1

There are at least two separate issues with your question.

The first one should be clear by now from the comments by Matias and the other answer: accuracy is not meaningful in a regression problem, such as yours; see also the comment by patyork in this Keras thread. For good or bad, the fact is that Keras will not "protect" you or any other user from putting not-meaningful requests in your code, i.e. you will not get any error, or even a warning, that you are attempting something that does not make sense, such as requesting the accuracy in a regression setting.

Having clarified that, the other issue is:

Since Keras does indeed return an "accuracy", even in a regression setting, what exactly is it and how is it calculated?

To shed some light here, let's revert to a public dataset (since you do not provide any details about your data), namely the Boston house price dataset (saved locally as housing.csv), and run a simple experiment as follows:

import numpy as np
import pandas
import keras

from keras.models import Sequential
from keras.layers import Dense

# load dataset
dataframe = pandas.read_csv("housing.csv", delim_whitespace=True, header=None)
dataset = dataframe.values
# split into input (X) and output (Y) variables
X = dataset[:,0:13]
Y = dataset[:,13]

model = Sequential()
model.add(Dense(13, input_dim=13, kernel_initializer='normal', activation='relu'))
model.add(Dense(1, kernel_initializer='normal'))
# Compile model asking for accuracy, too:
model.compile(loss='mean_squared_error', optimizer='adam', metrics=['accuracy'])

model.fit(X, Y,
     batch_size=5,
     epochs=100,
     verbose=1)

As in your case, the model fitting history (not shown here) shows a decreasing loss, and an accuracy roughly increasing. Let's evaluate now the model performance in the same training set, using the appropriate Keras built-in function:

score = model.evaluate(X, Y, verbose=0)
score
# [16.863721372581754, 0.013833992168483997]

The exact contents of the score array depend on what exactly we have requested during model compilation; in our case here, the first element is the loss (MSE), and the second one is the "accuracy".

At this point, let us have a look at the definition of Keras binary_accuracy in the metrics.py file:

def binary_accuracy(y_true, y_pred):
    return K.mean(K.equal(y_true, K.round(y_pred)), axis=-1)

So, after Keras has generated the predictions y_pred, it first rounds them, and then checks to see how many of them are equal to the true labels y_true, before getting the mean.

Let's replicate this operation using plain Python & Numpy code in our case, where the true labels are Y:

y_pred = model.predict(X)
l = len(Y)
acc = sum([np.round(y_pred[i])==Y[i] for i in range(l)])/l
acc
# array([0.01383399])

Well, bingo! This is actually the same value returned by score[1] above...

To make a long story short: since you (erroneously) request metrics=['accuracy'] in your model compilation, Keras will do its best to satisfy you, and will return some "accuracy" indeed, calculated as shown above, despite this being completely meaningless in your setting.

---

There are quite a few settings where Keras, under the hood, performs rather meaningless operations without giving any hint or warning to the user; two of them I have happened to encounter are:

• Giving meaningless results when, in a multi-class setting, one happens to request loss='binary_crossentropy' (instead of categorical_crossentropy) with metrics=['accuracy'] - see my answers in Keras binary_crossentropy vs categorical_crossentropy performance? and Why is binary_crossentropy more accurate than categorical_crossentropy for multiclass classification in Keras?

• Disabling completely Dropout, in the extreme case when one requests a dropout rate of 1.0 - see my answer in Dropout behavior in Keras with rate=1 (dropping all input units) not as expected