setting bias for multiclass classification python tensorflow keras - tensorflow

Attached model shows how to add bias in case of the unbalanced classification problem initial_bias = np.log([pos/neg]). Is there a way to add bias if you have multi-class classification with unbalanced data, Say 5 classes where classes are have distribution (0.4,0.3,0.2.0.08 and 0.02)
2) also how to calculate and use class weights in such case?
update 1
I found a way to apply weights, still not sure how to use bias
#####adding weights 20 Feb
weight_for_0 = ( 1/ 370)*(370+ 977+ 795)/3
weight_for_1 = ( 1/ 977)*(370+ 977+ 795)/3
weight_for_2 = (1 / 795)*(370+ 977+ 795)/3
#array([0, 1, 2]), array([370, 977, 795])
class_weights_dict = {0: weight_for_0, 1: weight_for_1, 2:weight_for_2}
class_weights_dict
Dcnn.fit(train_dataset,
epochs=NB_EPOCHS,
callbacks=[MyCustomCallback()],verbose=2,validation_data=test_dataset, class_weight=class_weights_dict)

Considering that you're using 'softmax':
softmax = exp(neurons) / sum(exp(neurons))
And that you want the results of the classes to be:
frequency = [0.4 , 0.3 , 0.2 , 0.08 , 0.02]
Biases should be given by the equation (elementwise):
frequency = exp(biases) / sum(exp(biases))
This forms a system of equations:
f1 = e^b1 / (e^b1 + e^b2 + ... + e^b5)
f2 = e^b2 / (e^b1 + e^b2 + ... + e^b5)
...
f5 = e^b5 / (e^b1 + e^b2 + ... + e^b5)
If you can solve this system of equations, you get the biases you want.
I used excel and test-error method to determine that for the frequencies you wanted, your biases should be respectively:
[1.1 , 0.81 , 0.4 , -0.51 , -1.9]
I don't really know how to solve that system easily, but you can keep experimenting with excel or another thing until you reach the solution.
Adding the biases to the layer - method 1.
Use a name when defining the layer, like:
self.last_dense = layers.Dense(units=3, activation="softmax", name='last_layer')
You may need to build the model first, so:
dummy_predictions = model.predict(np.zeros((1,) + input_shape))
Then you get the weights:
weights_and_biases = model.get_layer('last_layer').get_weights()
w, b = weights_and_biases
new_biases = np.array([-0.45752, 0.51344, 0.30730])
model.get_layer('last_layer').set_weights([w, new_biases])
Method 2
def bias_init(bias_shape):
return K.variable([-0.45752, 0.51344, 0.30730])
self.last_dense = layers.Dense(units=3, activation="softmax", bias_initializer=bias_init)

Just in addition to #Daniel Möller's answer, to solve the system of equations
f1 = e^b1 / (e^b1 + e^b2 + ... + e^b5)
...
f5 = e^b5 / (e^b1 + e^b2 + ... + e^b5)
You don't need excel or anything. Just compute bi = ln(fi).

To calculate fi = e^bi / (sum of e^bj), note that fi/fj = e^(bi-bj). Suppose the lowest frequency is fk. You can set bk= 0 and then compute every other class bias with bi = bj + ln(fi/fj).

A complete answer is here:
### To solve that set of nonlinear equations, use scipy fsolve
from scipy.optimize import fsolve
from math import exp
# define the frequency of different classes
f=(0.4, 0.3, 0.2, 0.08, 0.02)
# define the equation
def eqn(x, frequency):
sum_exp = sum([exp(x_i) for x_i in x])
return [exp(x[i])/sum_exp - frequency[i] for i in range(len(frequency))]
# calculate bias init
bias_init = fsolve(func=eqn,
x0=[0]*len(f),
).tolist()
bias_init
To put all things together
def init_imbalanced_class_weight_bias(df:pd.DataFrame, lable:str):
"""To handle imbalanced classification, provide initial bias list and class weight dictionary to 2 places in a tf classifier
1) In the last layer of classifier: tf.keras.layers.Dense(..., bias_initializer = bias_init)
2) model.fit(train_ds, #x=dict(X_train), y=y_train,
batch_size=batch_size,
validation_data= valid_ds, #(dict(X_test), y_test),
epochs=epochs,
callbacks=callbacks,
class_weight=class_weight,
)
Args:
df:pd.DataFrame=train_df
label:str
Returns:
class_weight:dict, e.g. {0: 1.6282051282051282, 1: 0.7604790419161677, 2: 0.9338235294117647}
bias_init:list e.g. [0.3222079660508266, 0.1168690393701237, -0.43907701967633633]
Examples:
class_weight, bias_init = init_imbalanced_class_weight_bias(df=train_df, lable=label)
References:
1. https://www.tensorflow.org/tutorials/structured_data/imbalanced_data
2. https://stackoverflow.com/questions/60307239/setting-bias-for-multiclass-classification-python-tensorflow-keras#new-answer
"""
from scipy.optimize import fsolve
from math import exp
# to deal with imbalance classification, calculate class_weight
d = dict(df[label].value_counts())
m = np.mean(list(d.values()))
class_weight = {k:m/v for (k,v) in d.items()} #e.g. {0: 1.6282051282051282, 1: 0.7604790419161677, 2: 0.9338235294117647}
# define classes frequency list
frequency = list(list(d.values())/sum(d.values()))
# define equations to solve initial bias
def eqn(x, frequency=frequency):
sum_exp = sum([exp(x_i) for x_i in x])
return [exp(x[i])/sum_exp - frequency[i] for i in range(len(frequency))]
# calculate init bias
bias_init = fsolve(func=eqn,
x0=[0]*len(frequency),
).tolist()
return class_weight, bias_init
class_weight, bias_init = init_imbalanced_class_weight_bias(df=train_df, lable=label)
I will post a colab notebook if anyone interested.
In case your tf classifier complains about ValueError: ('Could not interpret initializer identifier:', then add the tf.keras.initializers.Constant() around bias_init:
def init_imbalanced_class_weight_bias(...)
...
return class_weight, tf.keras.initializers.Constant(bias_init)

Related

Convert Tensorflow 1.x code with custom loss into 2.x

Suppose I have the following code written in Tensorflow 1.x where I define custom loss function. I wish to remove .compat.v1., Session, placeholder etc. and convert it into Tensorflow 2.x.
How to do so?
import DGM
import tensorflow as tf
import numpy as np
import scipy.stats as spstats
import matplotlib.pyplot as plt
from tqdm.notebook import trange
# Option parameters
phi = 10
n = 0.01
T = 4
# Solution parameters (domain on which to solve PDE)
t_low = 0.0 - 1e-10
x_low = 0.0 + 1e-10
x_high = 1.0
# neural network parameters
num_layers = 3
nodes_per_layer = 50
# Training parameters
sampling_stages = 2500 # number of times to resample new time-space domain points
steps_per_sample = 20 # number of SGD steps to take before re-sampling
# Sampling parameters
nsim_interior = 100
nsim_boundary_1 = 50
nsim_boundary_2 = 50
nsim_initial = 50
x_multiplier = 1.1 # multiplier for oversampling i.e. draw x from [x_low, x_high * x_multiplier]
def sampler(nsim_interior, nsim_boundary_1, nsim_boundary_2, nsim_initial):
''' Sample time-space points from the function's domain; points are sampled
uniformly on the interior of the domain, at the initial/terminal time points
and along the spatial boundary at different time points.
Args:
nsim_interior: number of space points in the interior of U
nsim_boundary_1: number of space points in the boundary of U
nsim_boundary_2: number of space points in the boundary of U_x
nsim_initial: number of space points at the initial time
'''
# Sampler #1: domain interior
t_interior = np.random.uniform(low=t_low, high=T, size=[nsim_interior, 1])
x_interior = np.random.uniform(low=x_low, high=x_high*x_multiplier, size=[nsim_interior, 1])
# Sampler #2: spatial boundary 1
t_boundary_1 = np.random.uniform(low=t_low, high=T, size=[nsim_boundary_1, 1])
x_boundary_1 = np.ones((nsim_boundary_1, 1))
# Sampler #3: spatial boundary 2
t_boundary_2 = np.random.uniform(low=t_low, high=T, size=[nsim_boundary_2, 1])
x_boundary_2 = np.zeros((nsim_boundary_2, 1))
# Sampler #4: initial condition
t_initial = np.zeros((nsim_initial, 1))
x_initial = np.random.uniform(low=x_low, high=x_high*x_multiplier, size=[nsim_initial, 1])
return (
t_interior, x_interior,
t_boundary_1, x_boundary_1,
t_boundary_2, x_boundary_2,
t_initial, x_initial
)
def loss(
model,
t_interior, x_interior,
t_boundary_1, x_boundary_1,
t_boundary_2, x_boundary_2,
t_initial, x_initial
):
''' Compute total loss for training.
Args:
model: DGM model object
t_interior, x_interior: sampled time / space points in the interior of U
t_boundary_1, x_boundary_1: sampled time / space points in the boundary of U
t_boundary_2, x_boundary_2: sampled time / space points in the boundary of U_x
t_initial, x_initial: sampled time / space points at the initial time
'''
# Loss term #1: PDE
# compute function value and derivatives at current sampled points
u = model(t_interior, x_interior)
u_t = tf.gradients(ys=u, xs=t_interior)[0]
u_x = tf.gradients(ys=u, xs=x_interior)[0]
u_xx = tf.gradients(ys=u_x, xs=x_interior)[0]
diff_u = u_t - u_xx + phi**2 * (tf.nn.relu(u) + 1e-10)**n
# compute average L2-norm for the PDE
L1 = tf.reduce_mean(input_tensor=tf.square(diff_u))
# Loss term #2: First b. c.
u = model(t_boundary_1, x_boundary_1)
bc1_error = u - 1
# Loss term #3: Second b. c.
u = model(t_boundary_2, x_boundary_2)
u_x = tf.gradients(ys=u, xs=x_boundary_2)[0]
bc2_error = u_x - 0
# Loss term #3: Initial condition
u = model(t_initial, x_initial)
init_error = u - 1
# compute average L2-norm for the initial/boundary conditions
L2 = tf.reduce_mean(input_tensor=tf.square(bc1_error + bc2_error + init_error))
return L1, L2
# initialize DGM model (last input: space dimension = 1)
model = DGM.DGMNet(nodes_per_layer, num_layers, 1)
# tensor placeholders (_tnsr suffix indicates tensors)
# inputs (time, space domain interior, space domain at initial time)
t_interior_tnsr = tf.compat.v1.placeholder(tf.float32, [None,1])
x_interior_tnsr = tf.compat.v1.placeholder(tf.float32, [None,1])
t_boundary_1_tnsr = tf.compat.v1.placeholder(tf.float32, [None,1])
x_boundary_1_tnsr = tf.compat.v1.placeholder(tf.float32, [None,1])
t_boundary_2_tnsr = tf.compat.v1.placeholder(tf.float32, [None,1])
x_boundary_2_tnsr = tf.compat.v1.placeholder(tf.float32, [None,1])
t_initial_tnsr = tf.compat.v1.placeholder(tf.float32, [None,1])
x_initial_tnsr = tf.compat.v1.placeholder(tf.float32, [None,1])
# loss
L1_tnsr, L2_tnsr = loss(
model,
t_interior_tnsr, x_interior_tnsr,
t_boundary_1_tnsr, x_boundary_1_tnsr,
t_boundary_2_tnsr, x_boundary_2_tnsr,
t_initial_tnsr, x_initial_tnsr
)
loss_tnsr = L1_tnsr + L2_tnsr
# set optimizer
starting_learning_rate = 3e-4
global_step = tf.Variable(0, trainable=False)
lr = tf.compat.v1.train.exponential_decay(
learning_rate=starting_learning_rate,
global_step=global_step,
decay_steps=1e5,
decay_rate=0.96,
staircase=True,
)
optimizer = tf.compat.v1.train.AdamOptimizer(learning_rate=lr).minimize(loss_tnsr)
# initialize variables
init_op = tf.compat.v1.global_variables_initializer()
# open session
sess = tf.compat.v1.Session()
sess.run(init_op)
try:
model.load_weights("checkpoint/")
print("Loading from checkpoint.")
except:
print("Checkpoint not found.")
# for each sampling stage
for i in trange(sampling_stages):
# sample uniformly from the required regions
t_interior, x_interior, \
t_boundary_1, x_boundary_1, \
t_boundary_2, x_boundary_2, \
t_initial, x_initial = sampler(
nsim_interior, nsim_boundary_1, nsim_boundary_2, nsim_initial
)
# for a given sample, take the required number of SGD steps
for _ in range(steps_per_sample):
loss, L1, L2, _ = sess.run(
[loss_tnsr, L1_tnsr, L2_tnsr, optimizer],
feed_dict = {
t_interior_tnsr: t_interior,
x_interior_tnsr: x_interior,
t_boundary_1_tnsr: t_boundary_1,
x_boundary_1_tnsr: x_boundary_1,
t_boundary_2_tnsr: t_boundary_2,
x_boundary_2_tnsr: x_boundary_2,
t_initial_tnsr: t_initial,
x_initial_tnsr: x_initial,
}
)
if i % 10 == 0:
print(f"Loss: {loss:.5f},\t L1: {L1:.5f},\t L2: {L2:.5f},\t iteration: {i}")
model.save_weights("checkpoint/")
I tried searching how to implement custom loss functions with model as an argument, but couldn't implement it.
For model.compile there is a loss argument for which you can pass the Loss function. May be a string (name of loss function), or a tf.keras.losses.Loss instance. For example
Model.compile(optimizer=tf.keras.optimizers.Adam(learning_rate=1e-3),
loss=tf.keras.losses.BinaryCrossentropy())
If you have created your custom loss function you can also pass that loss function to the loss argument by providing the name of that loss function. For example
def my_loss_fn(y_true, y_pred):
squared_difference = tf.square(y_true - y_pred)
return tf.reduce_mean(squared_difference, axis=-1)
model.compile(optimizer='adam', loss=my_loss_fn)
Thank You.

How to implement custom Keras ordinal loss function with tensor evaluation without disturbing TF>2.0 Model Graph?

I am trying to implement a custom loss function in Tensorflow 2.4 using the Keras backend.
The loss function is a ranking loss; I found the following paper with a somewhat log-likelihood loss: Chen et al. Single-Image Depth Perception in the Wild.
Similarly, I wanted to sample some (in this case 50) points from an image to compare the relative order between ground-truth and predicted depth maps using the NYU-Depth dataset. Being a fan of Numpy, I started working with that but came to the following exception:
ValueError: No gradients provided for any variable: [...]
I have learned that this is caused by the arguments not being filled when calling the loss function but instead, a C function is compiled which is then used later. So while I know the dimensions of my tensors (4, 480, 640, 1), I cannot work with the data as wanted and have to use the keras.backend functions on top so that in the end (if I understood correctly), there is supposed to be a path between the input tensors from the TF graph and the output tensor, which has to provide a gradient.
So my question now is: Is this a feasible loss function within keras?
I have already tried a few ideas and different approaches with different variations of my original code, which was something like:
def ranking_loss_function(y_true, y_pred):
# Chen et al. loss
y_true_np = K.eval(y_true)
y_pred_np = K.eval(y_pred)
if y_true_np.shape[0] != None:
num_sample_points = 50
total_samples = num_sample_points ** 2
err_list = [0 for x in range(y_true_np.shape[0])]
for i in range(y_true_np.shape[0]):
sample_points = create_random_samples(y_true, y_pred, num_sample_points)
for x1, y1 in sample_points:
for x2, y2 in sample_points:
if y_true[i][x1][y1] > y_true[i][x2][y2]:
#image_relation_true = 1
err_list[i] += np.log(1 + np.exp(-1 * y_pred[i][x1][y1] + y_pred[i][x2][y2]))
elif y_true[i][x1][y1] < y_true[i][x2][y2]:
#image_relation_true = -1
err_list[i] += np.log(1 + np.exp(y_pred[i][x1][y1] - y_pred[i][x2][y2]))
else:
#image_relation_true = 0
err_list[i] += np.square(y_pred[i][x1][y1] - y_pred[i][x2][y2])
err_list = np.divide(err_list, total_samples)
return K.constant(err_list)
As you can probably tell, the main idea was to first create the sample points and then based on the existing relation between them in y_true/y_pred continue with the corresponding computation from the cited paper.
Can anyone help me and provide some more helpful information or tips on how to correctly implement this loss using keras.backend functions? Trying to include the ordinal relation information really confused me compared to standard regression losses.
EDIT: Just in case this causes confusion: create_random_samples() just creates 50 random sample points (x, y) coordinate pairs based on the shape[1] and shape[2] of y_true (image width and height)
EDIT(2): After finding this variation on GitHub, I have tried out a variation using only TF functions to retrieve data from the tensors and compute the output. The adjusted and probably more correct version still throws the same exception though:
def ranking_loss_function(y_true, y_pred):
#In the Wild ranking loss
y_true_np = K.eval(y_true)
y_pred_np = K.eval(y_pred)
if y_true_np.shape[0] != None:
num_sample_points = 50
total_samples = num_sample_points ** 2
bs = y_true_np.shape[0]
w = y_true_np.shape[1]
h = y_true_np.shape[2]
total_samples = total_samples * bs
num_pairs = tf.constant([total_samples], dtype=tf.float32)
output = tf.Variable(0.0)
for i in range(bs):
sample_points = create_random_samples(y_true, y_pred, num_sample_points)
for x1, y1 in sample_points:
for x2, y2 in sample_points:
y_true_sq = tf.squeeze(y_true)
y_pred_sq = tf.squeeze(y_pred)
d1_t = tf.slice(y_true_sq, [i, x1, y1], [1, 1, 1])
d2_t = tf.slice(y_true_sq, [i, x2, y2], [1, 1, 1])
d1_p = tf.slice(y_pred_sq, [i, x1, y1], [1, 1, 1])
d2_p = tf.slice(y_pred_sq, [i, x2, y2], [1, 1, 1])
d1_t_sq = tf.squeeze(d1_t)
d2_t_sq = tf.squeeze(d2_t)
d1_p_sq = tf.squeeze(d1_p)
d2_p_sq = tf.squeeze(d2_p)
if d1_t_sq > d2_t_sq:
# --> Image relation = 1
output.assign_add(tf.math.log(1 + tf.math.exp(-1 * d1_p_sq + d2_p_sq)))
elif d1_t_sq < d2_t_sq:
# --> Image relation = -1
output.assign_add(tf.math.log(1 + tf.math.exp(d1_p_sq - d2_p_sq)))
else:
output.assign_add(tf.math.square(d1_p_sq - d2_p_sq))
return output/num_pairs
EDIT(3): This is the code for create_random_samples():
(FYI: Because it was weird to get the shape from y_true in this case, I first proceeded to hard-code it here as I know it for the dataset which I am currently using.)
def create_random_samples(y_true, y_pred, num_points=50):
y_true_shape = (4, 480, 640, 1)
y_pred_shape = (4, 480, 640, 1)
if y_true_shape[0] != None:
num_samples = num_points
population = [(x, y) for x in range(y_true_shape[1]) for y in range(y_true_shape[2])]
sample_points = random.sample(population, num_samples)
return sample_points

how to calculate entropy on float numbers over a tensor in python keras

I have been struggling on this and could not get it to work. hope someone can help me with this.
I want to calculate the entropy on each row of the tensor. Because my data are float numbers not integers I think I need to use bin_histogram.
For example a sample of my data is tensor =[[0.2, -0.1, 1],[2.09,-1.4,0.9]]
Just for information My model is seq2seq and written in keras with tensorflow backend.
This is my code so far: I need to correct rev_entropy
class entropy_measure(Layer):
def __init__(self, beta,batch, **kwargs):
self.beta = beta
self.batch = batch
self.uses_learning_phase = True
self.supports_masking = True
super(entropy_measure, self).__init__(**kwargs)
def call(self, x):
return K.in_train_phase(self.rev_entropy(x, self.beta,self.batch), x)
def get_config(self):
config = {'beta': self.beta}
base_config = super(entropy_measure, self).get_config()
return dict(list(base_config.items()) + list(config.items()))
def rev_entropy(self, x, beta,batch):
for i in x:
i = pd.Series(i)
p_data = i.value_counts() # counts occurrence of each value
entropy = entropy(p_data) # get entropy from counts
rev = 1/(1+entropy)
return rev
new_f_w_t = x * (rev.reshape(rev.shape[0], 1))*beta
return new_f_w_t
Any input is much appreciated:)
It looks like you have a series of questions that come together on this issue. I'll settle it here.
You calculate entropy in the following form of scipy.stats.entropy according to your code:
scipy.stats.entropy(pk, qk=None, base=None)
Calculate the entropy of a distribution for given probability values.
If only probabilities pk are given, the entropy is calculated as S =
-sum(pk * log(pk), axis=0).
Tensorflow does not provide a direct API to calculate entropy on each row of the tensor. What we need to do is to implement the above formula.
import tensorflow as tf
import pandas as pd
from scipy.stats import entropy
a = [1.1,2.2,3.3,4.4,2.2,3.3]
res = entropy(pd.value_counts(a))
_, _, count = tf.unique_with_counts(tf.constant(a))
# [1 2 2 1]
prob = count / tf.reduce_sum(count)
# [0.16666667 0.33333333 0.33333333 0.16666667]
tf_res = -tf.reduce_sum(prob * tf.log(prob))
with tf.Session() as sess:
print('scipy version: \n',res)
print('tensorflow version: \n',sess.run(tf_res))
scipy version:
1.329661348854758
tensorflow version:
1.3296613488547582
Then we need to define a function and achieve for loop through tf.map_fn in your custom layer according to above code.
def rev_entropy(self, x, beta,batch):
def row_entropy(row):
_, _, count = tf.unique_with_counts(row)
prob = count / tf.reduce_sum(count)
return -tf.reduce_sum(prob * tf.log(prob))
value_ranges = [-10.0, 100.0]
nbins = 50
new_f_w_t = tf.histogram_fixed_width_bins(x, value_ranges, nbins)
rev = tf.map_fn(row_entropy, new_f_w_t,dtype=tf.float32)
new_f_w_t = x * 1/(1+rev)*beta
return new_f_w_t
Notes that the hidden layer will not produce a gradient that cannot propagate backwards since entropy is calculated on the basis of statistical probabilistic values. Maybe you need to rethink your hidden layer structure.

How do I discover the values for variables of an equation with keras/tensorflow?

I have an equation that describes a curve in two dimensions. This equation has 5 variables. How do I discover the values of them with keras/tensorflow for a set of data? Is it possible? Someone know a tutorial of something similar?
I generated some data to train the network that has the format:
sample => [150, 66, 2] 150 sets with 66*2 with the data something like "time" x "acceleration"
targets => [150, 5] 150 sets with 5 variable numbers.
Obs: I know the range of the variables. I know too, that 150 sets of data are too few sample, but I need, after the code work, to train a new network with experimental data, and this is limited too. Visually, the curve is simple, it has a descendent linear part at the beggining and at the end it gets down "like an exponential".
My code is as follows:
def build_model():
model = models.Sequential()
model.add(layers.Dense(512, activation='relu', input_shape=(66*2,)))
model.add(layers.Dense(5, activation='softmax'))
model.compile(optimizer='rmsprop',
loss='categorical_crossentropy',
metrics=['mae'])
return model
def smooth_curve(points, factor=0.9):
[...]
return smoothed_points
#load the generated data
train_data = np.load('samples00.npy')
test_data = np.load('samples00.npy')
train_targets = np.load('labels00.npy')
test_targets = np.load('labels00.npy')
#normalizing the data
mean = train_data.mean()
train_data -= mean
std = train_data.std()
train_data /= std
test_data -= mean
test_data /= std
#k-fold validation:
k = 3
num_val_samples = len(train_data)//k
num_epochs = 100
all_mae_histories = []
for i in range(k):
val_data = train_data[i * num_val_samples: (i + 1) * num_val_samples]
val_targets = train_targets[i * num_val_samples: (i + 1) * num_val_samples]
partial_train_data = np.concatenate(
[train_data[:i * num_val_samples],
train_data[(i + 1) * num_val_samples:]],
axis=0)
partial_train_targets = np.concatenate(
[train_targets[:i * num_val_samples],
train_targets[(i + 1) * num_val_samples:]],
axis=0)
model = build_model()
#reshape the data to get the format (100, 66*2)
partial_train_data = partial_train_data.reshape(100, 66 * 2)
val_data = val_data.reshape(50, 66 * 2)
history = model.fit(partial_train_data,
partial_train_targets,
validation_data = (val_data, val_targets),
epochs = num_epochs,
batch_size = 1,
verbose = 1)
mae_history = history.history['val_mean_absolute_error']
all_mae_histories.append(mae_history)
average_mae_history = [
np.mean([x[i] for x in all_mae_histories]) for i in range(num_epochs)]
smooth_mae_history = smooth_curve(average_mae_history[10:])
plt.plot(range(1, len(smooth_mae_history) + 1), smooth_mae_history)
plt.xlabel('Epochs')
plt.ylabel('Validation MAE')
plt.show()
Obviously as it is, I need to get the best accuracy possible, but I am getting an "median absolute error(MAE)" like 96%, and this is inaceptable.
I see some basic bugs in this methodology. Your final layer of the network has a softmax layer. This would mean it would output 5 values, which sum to 1, and behave as a probability distribution. What you actually want to predict is true numbers, or rather floating point values (under some fixed precision arithmetic).
If you have a range, then probably using a sigmoid and rescaling the final layer would to match the range (just multiply with the max value) would help you. By default sigmoid would ensure you get 5 numbers between 0 and 1.
The other thing should be to remove the cross entropy loss and use a loss like RMS, so that you predict your numbers well. You could also used 1D convolutions instead of using Fully connected layers.
There has been some work here: https://julialang.org/blog/2017/10/gsoc-NeuralNetDiffEq which tries to solve DEs and might be relevant to your work.

How does Tensorflow Batch Normalization work?

I'm using tensorflow batch normalization in my deep neural network successfully. I'm doing it the following way:
if apply_bn:
with tf.variable_scope('bn'):
beta = tf.Variable(tf.constant(0.0, shape=[out_size]), name='beta', trainable=True)
gamma = tf.Variable(tf.constant(1.0, shape=[out_size]), name='gamma', trainable=True)
batch_mean, batch_var = tf.nn.moments(z, [0], name='moments')
ema = tf.train.ExponentialMovingAverage(decay=0.5)
def mean_var_with_update():
ema_apply_op = ema.apply([batch_mean, batch_var])
with tf.control_dependencies([ema_apply_op]):
return tf.identity(batch_mean), tf.identity(batch_var)
mean, var = tf.cond(self.phase_train,
mean_var_with_update,
lambda: (ema.average(batch_mean), ema.average(batch_var)))
self.z_prebn.append(z)
z = tf.nn.batch_normalization(z, mean, var, beta, gamma, 1e-3)
self.z.append(z)
self.bn.append((mean, var, beta, gamma))
And it works fine both for training and testing phases.
However I encounter problems when I try to use the computed neural network parameters in my another project, where I need to compute all the matrix multiplications and stuff by myself. The problem is that I can't reproduce the behavior of the tf.nn.batch_normalization function:
feed_dict = {
self.tf_x: np.array([range(self.x_cnt)]) / 100,
self.keep_prob: 1,
self.phase_train: False
}
for i in range(len(self.z)):
# print 0 layer's 1 value of arrays
print(self.sess.run([
self.z_prebn[i][0][1], # before bn
self.bn[i][0][1], # mean
self.bn[i][1][1], # var
self.bn[i][2][1], # offset
self.bn[i][3][1], # scale
self.z[i][0][1], # after bn
], feed_dict=feed_dict))
# prints
# [-0.077417567, -0.089603029, 0.000436493, -0.016652612, 1.0055743, 0.30664611]
According to the formula on the page https://www.tensorflow.org/versions/r1.2/api_docs/python/tf/nn/batch_normalization:
bn = scale * (x - mean) / (sqrt(var) + 1e-3) + offset
But as we can see,
1.0055743 * (-0.077417567 - -0.089603029)/(0.000436493^0.5 + 1e-3) + -0.016652612
= 0.543057
Which differs from the value 0.30664611, computed by Tensorflow itself.
So what am I doing wrong here and why I can't just calculate batch normalized value myself?
Thanks in advance!
The formula used is slightly different from:
bn = scale * (x - mean) / (sqrt(var) + 1e-3) + offset
It should be:
bn = scale * (x - mean) / (sqrt(var + 1e-3)) + offset
The variance_epsilon variable is supposed to scale with the variance, not with sigma, which is the square-root of variance.
After the correction, the formula yields the correct value:
1.0055743 * (-0.077417567 - -0.089603029)/((0.000436493 + 1e-3)**0.5) + -0.016652612
# 0.30664642276945747