I am trying to train a machine learning model where the loss function is binary cross entropy, because of gpu limitations i can only do batch size of 4 and i'm having lot of spikes in the loss graph. So I'm thinking to back-propagate after some predefined batch size(>4). So it's like i'll do 10 iterations of batch size 4 store the losses, after 10th iteration add the losses and back-propagate. will it be similar to batch size of 40.
TL;DR
f(a+b) = f(a)+f(b) is it true for binary cross entropy?
f(a+b) = f(a) + f(b) doesn't seem to be what you're after. This would imply that BCELoss is additive which it clearly isn't. I think what you really care about is if for some index i
# false
f(x, y) == f(x[:i], y[:i]) + f([i:], y[i:])
is true?
The short answer is no, because you're missing some scale factors. What you probably want is the following identity
# true
f(x, y) == (i / b) * f(x[:i], y[:i]) + (1.0 - i / b) * f(x[i:], y[i:])
where b is the total batch size.
This identity is used as motivation for the gradient accumulation method (see below). Also, this identity applies to any objective function which returns an average loss across each batch element, not just BCE.
Caveat/Pitfall: Keep in mind that batch norm will not behave exactly the same when using this approach since it updates its internal statistics based on batch size during the forward pass.
We can actually do a little better memory-wise than just computing the loss as a sum followed by backpropagation. Instead we can compute the gradient of each component in the equivalent sum individually and allow the gradients to accumulate. To better explain I'll give some examples of equivalent operations
Consider the following model
import torch
import torch.nn as nn
import torch.nn.functional as F
class MyModel(nn.Module):
def __init__(self):
super().__init__()
num_outputs = 5
# assume input shape is 10x10
self.conv_layer = nn.Conv2d(3, 10, 3, 1, 1)
self.fc_layer = nn.Linear(10*5*5, num_outputs)
def forward(self, x):
x = self.conv_layer(x)
x = F.max_pool2d(x, 2, 2, 0, 1, False, False)
x = F.relu(x)
x = self.fc_layer(x.flatten(start_dim=1))
x = torch.sigmoid(x) # or omit this and use BCEWithLogitsLoss instead of BCELoss
return x
# to ensure same results for this example
torch.manual_seed(0)
model = MyModel()
# the examples will work as long as the objective averages across batch elements
objective = nn.BCELoss()
# doesn't matter what type of optimizer
optimizer = torch.optim.SGD(model.parameters(), lr=0.001)
and lets say our data and targets for a single batch are
torch.manual_seed(1) # to ensure same results for this example
batch_size = 32
input_data = torch.randn((batch_size, 3, 10, 10))
targets = torch.randint(0, 1, (batch_size, 20)).float()
Full batch
The body of our training loop for an entire batch may look something like this
# entire batch
output = model(input_data)
loss = objective(output, targets)
optimizer.zero_grad()
loss.backward()
optimizer.step()
loss_value = loss.item()
print("Loss value: ", loss_value)
print("Model checksum: ", sum([p.sum().item() for p in model.parameters()]))
Weighted sum of loss on sub-batches
We could have computed this using the sum of multiple loss functions using
# This is simpler if the sub-batch size is a factor of batch_size
sub_batch_size = 4
assert (batch_size % sub_batch_size == 0)
# for this to work properly the batch_size must be divisible by sub_batch_size
num_sub_batches = batch_size // sub_batch_size
loss = 0
for sub_batch_idx in range(num_sub_batches):
start_idx = sub_batch_size * sub_batch_idx
end_idx = start_idx + sub_batch_size
sub_input = input_data[start_idx:end_idx]
sub_targets = targets[start_idx:end_idx]
sub_output = model(sub_input)
# add loss component for sub_batch
loss = loss + objective(sub_output, sub_targets) / num_sub_batches
optimizer.zero_grad()
loss.backward()
optimizer.step()
loss_value = loss.item()
print("Loss value: ", loss_value)
print("Model checksum: ", sum([p.sum().item() for p in model.parameters()]))
Gradient accumulation
The problem with the previous approach is that in order to apply back-propagation, pytorch needs to store intermediate results of layers in memory for every sub-batch. This ends up requiring a relatively large amount of memory and you may still run into memory consumption issues.
To alleviate this problem, instead of computing a single loss and performing back-propagation once, we could perform gradient accumulation. This gives equivalent results of the previous version. The difference here is that we instead perform a backward pass on each component of
the loss, only stepping the optimizer once all of them have been backpropagated. This way the computation graph is cleared after each sub-batch which will help with memory usage. Note that this works because .backward() actually accumulates (adds) the newly computed gradients to the existing .grad member of each model parameter. This is why optimizer.zero_grad() must be called only once, before the loop, and not during or after.
# This is simpler if the sub-batch size is a factor of batch_size
sub_batch_size = 4
assert (batch_size % sub_batch_size == 0)
# for this to work properly the batch_size must be divisible by sub_batch_size
num_sub_batches = batch_size // sub_batch_size
# Important! zero the gradients before the loop
optimizer.zero_grad()
loss_value = 0.0
for sub_batch_idx in range(num_sub_batches):
start_idx = sub_batch_size * sub_batch_idx
end_idx = start_idx + sub_batch_size
sub_input = input_data[start_idx:end_idx]
sub_targets = targets[start_idx:end_idx]
sub_output = model(sub_input)
# compute loss component for sub_batch
sub_loss = objective(sub_output, sub_targets) / num_sub_batches
# accumulate gradients
sub_loss.backward()
loss_value += sub_loss.item()
optimizer.step()
print("Loss value: ", loss_value)
print("Model checksum: ", sum([p.sum().item() for p in model.parameters()]))
I think 10 iterations of batch size 4 is same as one iteration of batch size 40, only here the time taken will be more. Across different training examples losses are added before backprop. But that doesn't make the function linear. BCELoss has a log component, and hence it is not a linear function. However what you said is correct. It will be similar to batch size 40.
I'm trying to build a simple regression model using keras and tensorflow. In my problem I have data in the form (x, y), where x and y are simply numbers. I'd like to build a keras model in order to predict y using x as an input.
Since I think images better explains thing, these are my data:
We may discuss if they are good or not, but in my problem I cannot really cheat them.
My keras model is the following (data are splitted 30% test (X_test, y_test) and 70% training (X_train, y_train)):
model = tf.keras.Sequential()
model.add(tf.keras.layers.Dense(32, input_shape=() activation="relu", name="first_layer"))
model.add(tf.keras.layers.Dense(16, activation="relu", name="second_layer"))
model.add(tf.keras.layers.Dense(1, name="output_layer"))
model.compile(loss = "mean_squared_error", optimizer = "adam", metrics=["mse"] )
history = model.fit(X_train, y_train, epochs=500, batch_size=1, verbose=0, shuffle=False)
eval_result = model.evaluate(X_test, y_test)
print("\n\nTest loss:", eval_result, "\n")
predict_Y = model.predict(X)
note: X contains both X_test and X_train.
Plotting the prediction I get (blue squares are the prediction predict_Y)
I'm playing a lot with layers, activation funztions and other parameters. My goal is to find the best parameters to train the model, but the actual question, here, is slightly different: in fact I have hard times to force the model to overfit the data (as you can see from the above results).
Does anyone have some sort of idea about how to reproduce overfitting?
This is the outcome I would like to get:
(red dots are under blue squares!)
EDIT:
Here I provide you the data used in the example above: you can copy paste directly to a python interpreter:
X_train = [0.704619794270697, 0.6779457393024553, 0.8207082120250023, 0.8588819357831449, 0.8692320257603844, 0.6878750931810429, 0.9556331888763945, 0.77677964510883, 0.7211381534179618, 0.6438319113259414, 0.6478339581502052, 0.9710222750072649, 0.8952188423349681, 0.6303124926673513, 0.9640316662124185, 0.869691568491902, 0.8320164648420931, 0.8236399177660375, 0.8877334038470911, 0.8084042532069621, 0.8045680821762038]
y_train = [0.7766424210611557, 0.8210846773655833, 0.9996114311913593, 0.8041331063189883, 0.9980525368790883, 0.8164056182686034, 0.8925487603333683, 0.7758207470960685, 0.37345286573743475, 0.9325789202459493, 0.6060269037514895, 0.9319771743389491, 0.9990691225991941, 0.9320002808310418, 0.9992560731072977, 0.9980241561997089, 0.8882905258641204, 0.4678339275898943, 0.9312152374846061, 0.9542371205095945, 0.8885893668675711]
X_test = [0.9749191829308574, 0.8735366740730178, 0.8882783211709133, 0.8022891400991644, 0.8650601322313454, 0.8697902997857514, 1.0, 0.8165876695985228, 0.8923841531760973]
y_test = [0.975653685270635, 0.9096752789481569, 0.6653736469114154, 0.46367666660348744, 0.9991817903431941, 1.0, 0.9111205717076893, 0.5264993912088891, 0.9989199241685126]
X = [0.704619794270697, 0.77677964510883, 0.7211381534179618, 0.6478339581502052, 0.6779457393024553, 0.8588819357831449, 0.8045680821762038, 0.8320164648420931, 0.8650601322313454, 0.8697902997857514, 0.8236399177660375, 0.6878750931810429, 0.8923841531760973, 0.8692320257603844, 0.8877334038470911, 0.8735366740730178, 0.8207082120250023, 0.8022891400991644, 0.6303124926673513, 0.8084042532069621, 0.869691568491902, 0.9710222750072649, 0.9556331888763945, 0.8882783211709133, 0.8165876695985228, 0.6438319113259414, 0.8952188423349681, 0.9749191829308574, 1.0, 0.9640316662124185]
Y = [0.7766424210611557, 0.7758207470960685, 0.37345286573743475, 0.6060269037514895, 0.8210846773655833, 0.8041331063189883, 0.8885893668675711, 0.8882905258641204, 0.9991817903431941, 1.0, 0.4678339275898943, 0.8164056182686034, 0.9989199241685126, 0.9980525368790883, 0.9312152374846061, 0.9096752789481569, 0.9996114311913593, 0.46367666660348744, 0.9320002808310418, 0.9542371205095945, 0.9980241561997089, 0.9319771743389491, 0.8925487603333683, 0.6653736469114154, 0.5264993912088891, 0.9325789202459493, 0.9990691225991941, 0.975653685270635, 0.9111205717076893, 0.9992560731072977]
Where X contains the list of the x values and Y the corresponding y value. (X_test, y_test) and (X_train, y_train) are two (non overlapping) subset of (X, Y).
To predict and show the model results I simply use matplotlib (imported as plt):
predict_Y = model.predict(X)
plt.plot(X, Y, "ro", X, predict_Y, "bs")
plt.show()
Overfitted models are rarely useful in real life. It appears to me that OP is well aware of that but wants to see if NNs are indeed capable of fitting (bounded) arbitrary functions or not. On one hand, the input-output data in the example seems to obey no discernible pattern. On the other hand, both input and output are scalars in [0, 1] and there are only 21 data points in the training set.
Based on my experiments and results, we can indeed overfit as requested. See the image below.
Numerical results:
x y_true y_pred error
0 0.704620 0.776642 0.773753 -0.002889
1 0.677946 0.821085 0.819597 -0.001488
2 0.820708 0.999611 0.999813 0.000202
3 0.858882 0.804133 0.805160 0.001026
4 0.869232 0.998053 0.997862 -0.000190
5 0.687875 0.816406 0.814692 -0.001714
6 0.955633 0.892549 0.893117 0.000569
7 0.776780 0.775821 0.779289 0.003469
8 0.721138 0.373453 0.374007 0.000554
9 0.643832 0.932579 0.912565 -0.020014
10 0.647834 0.606027 0.607253 0.001226
11 0.971022 0.931977 0.931549 -0.000428
12 0.895219 0.999069 0.999051 -0.000018
13 0.630312 0.932000 0.930252 -0.001748
14 0.964032 0.999256 0.999204 -0.000052
15 0.869692 0.998024 0.997859 -0.000165
16 0.832016 0.888291 0.887883 -0.000407
17 0.823640 0.467834 0.460728 -0.007106
18 0.887733 0.931215 0.932790 0.001575
19 0.808404 0.954237 0.960282 0.006045
20 0.804568 0.888589 0.906829 0.018240
{'me': -0.00015776709314323828,
'mae': 0.00329163070145315,
'mse': 4.0713782563067185e-05,
'rmse': 0.006380735268216915}
OP's code seems good to me. My changes were minor:
Use deeper networks. It may not actually be necessary to use a depth of 30 layers but since we just want to overfit, I didn't experiment too much with what's the minimum depth needed.
Each Dense layer has 50 units. Again, this may be overkill.
Added batch normalization layer every 5th dense layer.
Decreased learning rate by half.
Ran optimization for longer using the all 21 training examples in a batch.
Used MAE as objective function. MSE is good but since we want to overfit, I want to penalize small errors the same way as large errors.
Random numbers are more important here because data appears to be arbitrary. Though, you should get similar results if you change random number seed and let the optimizer run long enough. In some cases, optimization does get stuck in a local minima and it would not produce overfitting (as requested by OP).
The code is below.
import numpy as np
import pandas as pd
import tensorflow as tf
from tensorflow.keras.layers import Input, Dense, BatchNormalization
from tensorflow.keras.models import Model
from tensorflow.keras.optimizers import Adam
import matplotlib.pyplot as plt
# Set seed just to have reproducible results
np.random.seed(84)
tf.random.set_seed(84)
# Load data from the post
# https://stackoverflow.com/questions/61252785/how-to-overfit-data-with-keras
X_train = np.array([0.704619794270697, 0.6779457393024553, 0.8207082120250023,
0.8588819357831449, 0.8692320257603844, 0.6878750931810429,
0.9556331888763945, 0.77677964510883, 0.7211381534179618,
0.6438319113259414, 0.6478339581502052, 0.9710222750072649,
0.8952188423349681, 0.6303124926673513, 0.9640316662124185,
0.869691568491902, 0.8320164648420931, 0.8236399177660375,
0.8877334038470911, 0.8084042532069621,
0.8045680821762038])
Y_train = np.array([0.7766424210611557, 0.8210846773655833, 0.9996114311913593,
0.8041331063189883, 0.9980525368790883, 0.8164056182686034,
0.8925487603333683, 0.7758207470960685,
0.37345286573743475, 0.9325789202459493,
0.6060269037514895, 0.9319771743389491, 0.9990691225991941,
0.9320002808310418, 0.9992560731072977, 0.9980241561997089,
0.8882905258641204, 0.4678339275898943, 0.9312152374846061,
0.9542371205095945, 0.8885893668675711])
X_test = np.array([0.9749191829308574, 0.8735366740730178, 0.8882783211709133,
0.8022891400991644, 0.8650601322313454, 0.8697902997857514,
1.0, 0.8165876695985228, 0.8923841531760973])
Y_test = np.array([0.975653685270635, 0.9096752789481569, 0.6653736469114154,
0.46367666660348744, 0.9991817903431941, 1.0,
0.9111205717076893, 0.5264993912088891, 0.9989199241685126])
X = np.array([0.704619794270697, 0.77677964510883, 0.7211381534179618,
0.6478339581502052, 0.6779457393024553, 0.8588819357831449,
0.8045680821762038, 0.8320164648420931, 0.8650601322313454,
0.8697902997857514, 0.8236399177660375, 0.6878750931810429,
0.8923841531760973, 0.8692320257603844, 0.8877334038470911,
0.8735366740730178, 0.8207082120250023, 0.8022891400991644,
0.6303124926673513, 0.8084042532069621, 0.869691568491902,
0.9710222750072649, 0.9556331888763945, 0.8882783211709133,
0.8165876695985228, 0.6438319113259414, 0.8952188423349681,
0.9749191829308574, 1.0, 0.9640316662124185])
Y = np.array([0.7766424210611557, 0.7758207470960685, 0.37345286573743475,
0.6060269037514895, 0.8210846773655833, 0.8041331063189883,
0.8885893668675711, 0.8882905258641204, 0.9991817903431941, 1.0,
0.4678339275898943, 0.8164056182686034, 0.9989199241685126,
0.9980525368790883, 0.9312152374846061, 0.9096752789481569,
0.9996114311913593, 0.46367666660348744, 0.9320002808310418,
0.9542371205095945, 0.9980241561997089, 0.9319771743389491,
0.8925487603333683, 0.6653736469114154, 0.5264993912088891,
0.9325789202459493, 0.9990691225991941, 0.975653685270635,
0.9111205717076893, 0.9992560731072977])
# Reshape all data to be of the shape (batch_size, 1)
X_train = X_train.reshape((-1, 1))
Y_train = Y_train.reshape((-1, 1))
X_test = X_test.reshape((-1, 1))
Y_test = Y_test.reshape((-1, 1))
X = X.reshape((-1, 1))
Y = Y.reshape((-1, 1))
# Is data scaled? NNs do well with bounded data.
assert np.all(X_train >= 0) and np.all(X_train <= 1)
assert np.all(Y_train >= 0) and np.all(Y_train <= 1)
assert np.all(X_test >= 0) and np.all(X_test <= 1)
assert np.all(Y_test >= 0) and np.all(Y_test <= 1)
assert np.all(X >= 0) and np.all(X <= 1)
assert np.all(Y >= 0) and np.all(Y <= 1)
# Build a model with variable number of hidden layers.
# We will use Keras functional API.
# https://www.perfectlyrandom.org/2019/06/24/a-guide-to-keras-functional-api/
n_dense_layers = 30 # increase this to get more complicated models
# Define the layers first.
input_tensor = Input(shape=(1,), name='input')
layers = []
for i in range(n_dense_layers):
layers += [Dense(units=50, activation='relu', name=f'dense_layer_{i}')]
if (i > 0) & (i % 5 == 0):
# avg over batches not features
layers += [BatchNormalization(axis=1)]
sigmoid_layer = Dense(units=1, activation='sigmoid', name='sigmoid_layer')
# Connect the layers using Keras Functional API
mid_layer = input_tensor
for dense_layer in layers:
mid_layer = dense_layer(mid_layer)
output_tensor = sigmoid_layer(mid_layer)
model = Model(inputs=[input_tensor], outputs=[output_tensor])
optimizer = Adam(learning_rate=0.0005)
model.compile(optimizer=optimizer, loss='mae', metrics=['mae'])
model.fit(x=[X_train], y=[Y_train], epochs=40000, batch_size=21)
# Predict on various datasets
Y_train_pred = model.predict(X_train)
# Create a dataframe to inspect results manually
train_df = pd.DataFrame({
'x': X_train.reshape((-1)),
'y_true': Y_train.reshape((-1)),
'y_pred': Y_train_pred.reshape((-1))
})
train_df['error'] = train_df['y_pred'] - train_df['y_true']
print(train_df)
# A dictionary to store all the errors in one place.
train_errors = {
'me': np.mean(train_df['error']),
'mae': np.mean(np.abs(train_df['error'])),
'mse': np.mean(np.square(train_df['error'])),
'rmse': np.sqrt(np.mean(np.square(train_df['error']))),
}
print(train_errors)
# Make a plot to visualize true vs predicted
plt.figure(1)
plt.clf()
plt.plot(train_df['x'], train_df['y_true'], 'r.', label='y_true')
plt.plot(train_df['x'], train_df['y_pred'], 'bo', alpha=0.25, label='y_pred')
plt.grid(True)
plt.xlabel('x')
plt.ylabel('y')
plt.title(f'Train data. MSE={np.round(train_errors["mse"], 5)}.')
plt.legend()
plt.show(block=False)
plt.savefig('true_vs_pred.png')
A problem you may encountering is that you don't have enough training data for the model to be able to fit well. In your example, you only have 21 training instances, each with only 1 feature. Broadly speaking with neural network models, you need on the order of 10K or more training instances to produce a decent model.
Consider the following code that generates a noisy sine wave and tries to train a densely-connected feed-forward neural network to fit the data. My model has two linear layers, each with 50 hidden units and a ReLU activation function. The experiments are parameterized with the variable num_points which I will increase.
import tensorflow as tf
from tensorflow import keras
from tensorflow.keras import layers
import numpy as np
import matplotlib.pyplot as plt
np.random.seed(7)
def generate_data(num_points=100):
X = np.linspace(0.0 , 2.0 * np.pi, num_points).reshape(-1, 1)
noise = np.random.normal(0, 1, num_points).reshape(-1, 1)
y = 3 * np.sin(X) + noise
return X, y
def run_experiment(X_train, y_train, X_test, batch_size=64):
num_points = X_train.shape[0]
model = keras.Sequential()
model.add(layers.Dense(50, input_shape=(1, ), activation='relu'))
model.add(layers.Dense(50, activation='relu'))
model.add(layers.Dense(1, activation='linear'))
model.compile(loss = "mse", optimizer = "adam", metrics=["mse"] )
history = model.fit(X_train, y_train, epochs=10,
batch_size=batch_size, verbose=0)
yhat = model.predict(X_test, batch_size=batch_size)
plt.figure(figsize=(5, 5))
plt.plot(X_train, y_train, "ro", markersize=2, label='True')
plt.plot(X_train, yhat, "bo", markersize=1, label='Predicted')
plt.ylim(-5, 5)
plt.title('N=%d points' % (num_points))
plt.legend()
plt.grid()
plt.show()
Here is how I invoke the code:
num_points = 100
X, y = generate_data(num_points)
run_experiment(X, y, X)
Now, if I run the experiment with num_points = 100, the model predictions (in blue) do a terrible job at fitting the true noisy sine wave (in red).
Now, here is num_points = 1000:
Here is num_points = 10000:
And here is num_points = 100000:
As you can see, for my chosen NN architecture, adding more training instances allows the neural network to better (over)fit the data.
If you do have a lot of training instances, then if you want to purposefully overfit your data, you can either increase the neural network capacity or reduce regularization. Specifically, you can control the following knobs:
increase the number of layers
increase the number of hidden units
increase the number of features per data instance
reduce regularization (e.g. by removing dropout layers)
use a more complex neural network architecture (e.g. transformer blocks instead of RNN)
You may be wondering if neural networks can fit arbitrary data rather than just a noisy sine wave as in my example. Previous research says that, yes, a big enough neural network can fit any data. See:
Universal approximation theorem. https://en.wikipedia.org/wiki/Universal_approximation_theorem
Zhang 2016, "Understanding deep learning requires rethinking generalization". https://arxiv.org/abs/1611.03530
As discussed in the comments, you should make a Python array (with NumPy) like this:-
Myarray = [[0.65, 1], [0.85, 0.5], ....]
Then you would just call those specific parts of the array whom you need to predict. Here the first value is the x-axis value. So you would call it to obtain the corresponding pair stored in Myarray
There are many resources to learn these types of things. some of them are ===>
https://www.geeksforgeeks.org/python-using-2d-arrays-lists-the-right-way/
https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=video&cd=2&cad=rja&uact=8&ved=0ahUKEwjGs-Oxne3oAhVlwTgGHfHnDp4QtwIILTAB&url=https%3A%2F%2Fwww.youtube.com%2Fwatch%3Fv%3DQgfUT7i4yrc&usg=AOvVaw3LympYRszIYi6_OijMXH72
I've been implementing VAE and IWAE models on the caltech silhouettes dataset and am having an issue where the VAE outperforms IWAE by a modest margin (test LL ~120 for VAE, ~133 for IWAE!). I don't believe this should be the case, according to both theory and experiments produced here.
I'm hoping someone can find some issue in how I'm implementing that's causing this to be the case.
The network I'm using to approximate q and p is the same as that detailed in the appendix of the paper above. The calculation part of the model is below:
data_k_vec = data.repeat_interleave(K,0) # Generate K samples (in my case K=50 is producing this behavior)
mu, log_std = model.encode(data_k_vec)
z = model.reparameterize(mu, log_std) # z = mu + torch.exp(log_std)*epsilon (epsilon ~ N(0,1))
decoded = model.decode(z) # this is the sigmoid output of the model
log_prior_z = torch.sum(-0.5 * z ** 2, 1)-.5*z.shape[1]*T.log(torch.tensor(2*np.pi))
log_q_z = compute_log_probability_gaussian(z, mu, log_std) # Definitions below
log_p_x = compute_log_probability_bernoulli(decoded,data_k_vec)
if model_type == 'iwae':
log_w_matrix = (log_prior_z + log_p_x - log_q_z).view(-1, K)
elif model_type =='vae':
log_w_matrix = (log_prior_z + log_p_x - log_q_z).view(-1, 1)*1/K
log_w_minus_max = log_w_matrix - torch.max(log_w_matrix, 1, keepdim=True)[0]
ws_matrix = torch.exp(log_w_minus_max)
ws_norm = ws_matrix / torch.sum(ws_matrix, 1, keepdim=True)
ws_sum_per_datapoint = torch.sum(log_w_matrix * ws_norm, 1)
loss = -torch.sum(ws_sum_per_datapoint) # value of loss that gets returned to training function. loss.backward() will get called on this value
Here are the likelihood functions. I had to fuss with the bernoulli LL in order to not get nan during training
def compute_log_probability_gaussian(obs, mu, logstd, axis=1):
return torch.sum(-0.5 * ((obs-mu) / torch.exp(logstd)) ** 2 - logstd, axis)-.5*obs.shape[1]*T.log(torch.tensor(2*np.pi))
def compute_log_probability_bernoulli(theta, obs, axis=1): # Add 1e-18 to avoid nan appearances in training
return torch.sum(obs*torch.log(theta+1e-18) + (1-obs)*torch.log(1-theta+1e-18), axis)
In this code there's a "shortcut" being used in that the row-wise importance weights are being calculated in the model_type=='iwae' case for the K=50 samples in each row, while in the model_type=='vae' case the importance weights are being calculated for the single value left in each row, so that it just ends up calculating a weight of 1. Maybe this is the issue?
Any and all help is huge - I thought that addressing the nan issue would permanently get me out of the weeds but now I have this new problem.
EDIT:
Should add that the training scheme is the same as that in the paper linked above. That is, for each of i=0....7 rounds train for 2**i epochs with a learning rate of 1e-4 * 10**(-i/7)
The K-sample importance weighted ELBO is
$$ \textrm{IW-ELBO}(x,K) = \log \sum_{k=1}^K \frac{p(x \vert z_k) p(z_k)}{q(z_k;x)}$$
For the IWAE there are K samples originating from each datapoint x, so you want to have the same latent statistics mu_z, Sigma_z obtained through the amortized inference network, but sample multiple z K times for each x.
So its computationally wasteful to compute the forward pass for data_k_vec = data.repeat_interleave(K,0), you should compute the forward pass once for each original datapoint, then repeat the statistics output by the inference network for sampling:
mu = torch.repeat_interleave(mu,K,0)
log_std = torch.repeat_interleave(log_std,K,0)
Then sample z_k. And now repeat your datapoints data_k_vec = data.repeat_interleave(K,0), and use the resulting tensor to efficiently evaluate the conditional p(x |z_k) for each importance sample z_k.
Note you may also want to use the logsumexp operation when calculating the IW-ELBO for numerical stability. I can't quite figure out what's going on with the log_w_matrix calculation in your post, but this is what I would do:
log_pz = ...
log_qzCx = ....
log_pxCz = ...
log_iw = log_pxCz + log_pz - log_qzCx
log_iw = log_iw.reshape(-1, K)
iwelbo = torch.logsumexp(log_iw, dim=1) - np.log(K)
EDIT: Actually after thinking about it a bit and using the score function identity, you can interpret the IWAE gradient as an importance weighted estimate of the standard single-sample gradient, so the method in the OP for calculation of the importance weights is equivalent (if a bit wasteful), provided you place a stop_gradient operator around the normalized importance weights, which you call w_norm. So I the main problem is the absence of this stop_gradient operator.
I read this thread about the difference between SVC() and LinearSVC() in scikit-learn.
Now I have a data set of binary classification problem(For such a problem, the one-to-one/one-to-rest strategy difference between both functions could be ignore.)
I want to try under what parameters would these 2 functions give me the same result. First of all, of course, we should set kernel='linear' for SVC()
However, I just could not get the same result from both functions. I could not find the answer from the documents, could anybody help me to find the equivalent parameter set I am looking for?
Updated:
I modified the following code from an example of the scikit-learn website, and apparently they are not the same:
import numpy as np
import matplotlib.pyplot as plt
from sklearn import svm, datasets
# import some data to play with
iris = datasets.load_iris()
X = iris.data[:, :2] # we only take the first two features. We could
# avoid this ugly slicing by using a two-dim dataset
y = iris.target
for i in range(len(y)):
if (y[i]==2):
y[i] = 1
h = .02 # step size in the mesh
# we create an instance of SVM and fit out data. We do not scale our
# data since we want to plot the support vectors
C = 1.0 # SVM regularization parameter
svc = svm.SVC(kernel='linear', C=C).fit(X, y)
lin_svc = svm.LinearSVC(C=C, dual = True, loss = 'hinge').fit(X, y)
# create a mesh to plot in
x_min, x_max = X[:, 0].min() - 1, X[:, 0].max() + 1
y_min, y_max = X[:, 1].min() - 1, X[:, 1].max() + 1
xx, yy = np.meshgrid(np.arange(x_min, x_max, h),
np.arange(y_min, y_max, h))
# title for the plots
titles = ['SVC with linear kernel',
'LinearSVC (linear kernel)']
for i, clf in enumerate((svc, lin_svc)):
# Plot the decision boundary. For that, we will assign a color to each
# point in the mesh [x_min, m_max]x[y_min, y_max].
plt.subplot(1, 2, i + 1)
plt.subplots_adjust(wspace=0.4, hspace=0.4)
Z = clf.predict(np.c_[xx.ravel(), yy.ravel()])
# Put the result into a color plot
Z = Z.reshape(xx.shape)
plt.contourf(xx, yy, Z, cmap=plt.cm.Paired, alpha=0.8)
# Plot also the training points
plt.scatter(X[:, 0], X[:, 1], c=y, cmap=plt.cm.Paired)
plt.xlabel('Sepal length')
plt.ylabel('Sepal width')
plt.xlim(xx.min(), xx.max())
plt.ylim(yy.min(), yy.max())
plt.xticks(())
plt.yticks(())
plt.title(titles[i])
plt.show()
Result:
Output Figure from previous code
In mathematical sense you need to set:
SVC(kernel='linear', **kwargs) # by default it uses RBF kernel
and
LinearSVC(loss='hinge', **kwargs) # by default it uses squared hinge loss
Another element, which cannot be easily fixed is increasing intercept_scaling in LinearSVC, as in this implementation bias is regularized (which is not true in SVC nor should be true in SVM - thus this is not SVM) - consequently they will never be exactly equal (unless bias=0 for your problem), as they assume two different models
SVC : 1/2||w||^2 + C SUM xi_i
LinearSVC: 1/2||[w b]||^2 + C SUM xi_i
Personally I consider LinearSVC one of the mistakes of sklearn developers - this class is simply not a linear SVM.
After increasing intercept scaling (to 10.0)
However, if you scale it up too much - it will also fail, as now tolerance and number of iterations are crucial.
To sum up: LinearSVC is not linear SVM, do not use it if do not have to.