When we train neural networks, we typically use gradient descent, which relies on a continuous, differentiable real-valued cost function. The final cost function might, for example, take the mean squared error. Or put another way, gradient descent implicitly assumes the end goal is regression - to minimize a real-valued error measure.
Sometimes what we want a neural network to do is perform classification - given an input, classify it into two or more discrete categories. In this case, the end goal the user cares about is classification accuracy - the percentage of cases classified correctly.
But when we are using a neural network for classification, though our goal is classification accuracy, that is not what the neural network is trying to optimize. The neural network is still trying to optimize the real-valued cost function. Sometimes these point in the same direction, but sometimes they don't. In particular, I've been running into cases where a neural network trained to correctly minimize the cost function, has a classification accuracy worse than a simple hand-coded threshold comparison.
I've boiled this down to a minimal test case using TensorFlow. It sets up a perceptron (neural network with no hidden layers), trains it on an absolutely minimal dataset (one input variable, one binary output variable) assesses the classification accuracy of the result, then compares it to the classification accuracy of a simple hand-coded threshold comparison; the results are 60% and 80% respectively. Intuitively, this is because a single outlier with a large input value, generates a correspondingly large output value, so the way to minimize the cost function is to try extra hard to accommodate that one case, in the process misclassifying two more ordinary cases. The perceptron is correctly doing what it was told to do; it's just that this does not match what we actually want of a classifier. But the classification accuracy is not a continuous differentiable function, so we can't use it as the target for gradient descent.
How can we train a neural network so that it ends up maximizing classification accuracy?
import numpy as np
import tensorflow as tf
sess = tf.InteractiveSession()
tf.set_random_seed(1)
# Parameters
epochs = 10000
learning_rate = 0.01
# Data
train_X = [
[0],
[0],
[2],
[2],
[9],
]
train_Y = [
0,
0,
1,
1,
0,
]
rows = np.shape(train_X)[0]
cols = np.shape(train_X)[1]
# Inputs and outputs
X = tf.placeholder(tf.float32)
Y = tf.placeholder(tf.float32)
# Weights
W = tf.Variable(tf.random_normal([cols]))
b = tf.Variable(tf.random_normal([]))
# Model
pred = tf.tensordot(X, W, 1) + b
cost = tf.reduce_sum((pred-Y)**2/rows)
optimizer = tf.train.GradientDescentOptimizer(learning_rate).minimize(cost)
tf.global_variables_initializer().run()
# Train
for epoch in range(epochs):
# Print update at successive doublings of time
if epoch&(epoch-1) == 0 or epoch == epochs-1:
print('{} {} {} {}'.format(
epoch,
cost.eval({X: train_X, Y: train_Y}),
W.eval(),
b.eval(),
))
optimizer.run({X: train_X, Y: train_Y})
# Classification accuracy of perceptron
classifications = [pred.eval({X: x}) > 0.5 for x in train_X]
correct = sum([p == y for (p, y) in zip(classifications, train_Y)])
print('{}/{} = perceptron accuracy'.format(correct, rows))
# Classification accuracy of hand-coded threshold comparison
classifications = [x[0] > 1.0 for x in train_X]
correct = sum([p == y for (p, y) in zip(classifications, train_Y)])
print('{}/{} = threshold accuracy'.format(correct, rows))
How can we train a neural network so that it ends up maximizing classification accuracy?
I'm asking for a way to get a continuous proxy function that's closer to the accuracy
To start with, the loss function used today for classification tasks in (deep) neural nets was not invented with them, but it goes back several decades, and it actually comes from the early days of logistic regression. Here is the equation for the simple case of binary classification:
The idea behind it was exactly to come up with a continuous & differentiable function, so that we would be able to exploit the (vast, and still expanding) arsenal of convex optimization for classification problems.
It is safe to say that the above loss function is the best we have so far, given the desired mathematical constraints mentioned above.
Should we consider this problem (i.e. better approximating the accuracy) solved and finished? At least in principle, no. I am old enough to remember an era when the only activation functions practically available were tanh and sigmoid; then came ReLU and gave a real boost to the field. Similarly, someone may eventually come up with a better loss function, but arguably this is going to happen in a research paper, and not as an answer to a SO question...
That said, the very fact that the current loss function comes from very elementary considerations of probability and information theory (fields that, in sharp contrast with the current field of deep learning, stand upon firm theoretical foundations) creates at least some doubt as to if a better proposal for the loss may be just around the corner.
There is another subtle point on the relation between loss and accuracy, which makes the latter something qualitatively different than the former, and is frequently lost in such discussions. Let me elaborate a little...
All the classifiers related to this discussion (i.e. neural nets, logistic regression etc) are probabilistic ones; that is, they do not return hard class memberships (0/1) but class probabilities (continuous real numbers in [0, 1]).
Limiting the discussion for simplicity to the binary case, when converting a class probability to a (hard) class membership, we are implicitly involving a threshold, usually equal to 0.5, such as if p[i] > 0.5, then class[i] = "1". Now, we can find many cases whet this naive default choice of threshold will not work (heavily imbalanced datasets are the first to come to mind), and we'll have to choose a different one. But the important point for our discussion here is that this threshold selection, while being of central importance to the accuracy, is completely external to the mathematical optimization problem of minimizing the loss, and serves as a further "insulation layer" between them, compromising the simplistic view that loss is just a proxy for accuracy (it is not). As nicely put in the answer of this Cross Validated thread:
the statistical component of your exercise ends when you output a probability for each class of your new sample. Choosing a threshold beyond which you classify a new observation as 1 vs. 0 is not part of the statistics any more. It is part of the decision component.
Enlarging somewhat an already broad discussion: Can we possibly move completely away from the (very) limiting constraint of mathematical optimization of continuous & differentiable functions? In other words, can we do away with back-propagation and gradient descend?
Well, we are actually doing so already, at least in the sub-field of reinforcement learning: 2017 was the year when new research from OpenAI on something called Evolution Strategies made headlines. And as an extra bonus, here is an ultra-fresh (Dec 2017) paper by Uber on the subject, again generating much enthusiasm in the community.
I think you are forgetting to pass your output through a simgoid. Fixed below:
import numpy as np
import tensorflow as tf
sess = tf.InteractiveSession()
tf.set_random_seed(1)
# Parameters
epochs = 10000
learning_rate = 0.01
# Data
train_X = [
[0],
[0],
[2],
[2],
[9],
]
train_Y = [
0,
0,
1,
1,
0,
]
rows = np.shape(train_X)[0]
cols = np.shape(train_X)[1]
# Inputs and outputs
X = tf.placeholder(tf.float32)
Y = tf.placeholder(tf.float32)
# Weights
W = tf.Variable(tf.random_normal([cols]))
b = tf.Variable(tf.random_normal([]))
# Model
# CHANGE HERE: Remember, you need an activation function!
pred = tf.nn.sigmoid(tf.tensordot(X, W, 1) + b)
cost = tf.reduce_sum((pred-Y)**2/rows)
optimizer = tf.train.GradientDescentOptimizer(learning_rate).minimize(cost)
tf.global_variables_initializer().run()
# Train
for epoch in range(epochs):
# Print update at successive doublings of time
if epoch&(epoch-1) == 0 or epoch == epochs-1:
print('{} {} {} {}'.format(
epoch,
cost.eval({X: train_X, Y: train_Y}),
W.eval(),
b.eval(),
))
optimizer.run({X: train_X, Y: train_Y})
# Classification accuracy of perceptron
classifications = [pred.eval({X: x}) > 0.5 for x in train_X]
correct = sum([p == y for (p, y) in zip(classifications, train_Y)])
print('{}/{} = perceptron accuracy'.format(correct, rows))
# Classification accuracy of hand-coded threshold comparison
classifications = [x[0] > 1.0 for x in train_X]
correct = sum([p == y for (p, y) in zip(classifications, train_Y)])
print('{}/{} = threshold accuracy'.format(correct, rows))
The output:
0 0.28319069743156433 [ 0.75648874] -0.9745011329650879
1 0.28302448987960815 [ 0.75775659] -0.9742625951766968
2 0.28285878896713257 [ 0.75902224] -0.9740257859230042
4 0.28252947330474854 [ 0.76154679] -0.97355717420578
8 0.28187844157218933 [ 0.76656926] -0.9726400971412659
16 0.28060704469680786 [ 0.77650583] -0.970885694026947
32 0.27818527817726135 [ 0.79593837] -0.9676888585090637
64 0.2738055884838104 [ 0.83302218] -0.9624817967414856
128 0.26666420698165894 [ 0.90031379] -0.9562843441963196
256 0.25691407918930054 [ 1.01172411] -0.9567816257476807
512 0.2461051195859909 [ 1.17413962] -0.9872989654541016
1024 0.23519910871982574 [ 1.38549554] -1.088881492614746
2048 0.2241383194923401 [ 1.64616168] -1.298340916633606
4096 0.21433120965957642 [ 1.95981205] -1.6126530170440674
8192 0.2075471431016922 [ 2.31746769] -1.989408016204834
9999 0.20618653297424316 [ 2.42539024] -2.1028473377227783
4/5 = perceptron accuracy
4/5 = threshold accuracy
Related
what is the standard way to detect if a model has converged? I was going to record 5 losses with 95 confidence intervals each loss and if they all agreed then I’d halt the script. I assume training until convergence must be implemented already in PyTorch or PyTorch Lightning somewhere. I don’t need a perfect solution, just the standard way to do this automatically - i.e. halt when converged.
My solution is easy to implement. Once create a criterion and changes the reduction to none. Then it will output a tensor of size [B]. Every you log you record that and it's 95 confidence interval (or std if you prefer, but that is much less accuracy). Then every time you add a new loss with it's confidence interval make sure it remains of size 5 (or 10) and that the 5 losses are within a 95 CI of each other. Then if that is true halt.
You can compute the CI with this:
def torch_compute_confidence_interval(data: Tensor,
confidence: float = 0.95
) -> Tensor:
"""
Computes the confidence interval for a given survey of a data set.
"""
n = len(data)
mean: Tensor = data.mean()
# se: Tensor = scipy.stats.sem(data) # compute standard error
# se, mean: Tensor = torch.std_mean(data, unbiased=True) # compute standard error
se: Tensor = data.std(unbiased=True) / (n**0.5)
t_p: float = float(scipy.stats.t.ppf((1 + confidence) / 2., n - 1))
ci = t_p * se
return mean, ci
and you can create the criterion as follow:
loss: nn.Module = nn.CrossEntropyLoss(reduction='none')
so the train loss is now of size [B].
note that I know how to train with a fixed number of epochs, so I am not really looking for that - just the halting criterion for when to stop when models looks converged, what a person would sort of do when they look at their learning curve but automatically.
ref:
https://forums.pytorchlightning.ai/t/what-is-the-standard-way-to-halt-a-script-when-it-has-converged/1415
Set an EarlyStopping (https://pytorch-lightning.readthedocs.io/en/stable/api/pytorch_lightning.callbacks.EarlyStopping.html#pytorch_lightning.callbacks.EarlyStopping) callback in your trainer by
checkpoint_callbacks = [
EarlyStopping(
monitor="val_f1_score",
min_delta=0.01,
patience=10, # NOTE no. val epochs, not train epochs
verbose=False,
mode="min",
),
]
trainer = pl.Trainer(callbacks=callbacks)
This will monitor changes in val_f1_score during training (notice that you have to log this value with self.log("val_f1_score", val_f1) in your pl.LightningModule). And it will stop the training if the minimum change to quantity to qualify as an improvement (min_delta) for more than the number of epoch specified as patience
I am new babie to the Deep Learning field, and I am use log-likelihood method to compare the MSE metrics.Could anyone be able to show how to calculate the following 2 predicted output examples with 3 outputs neurons each. Thanks
yt = [ [1,0,0],[0,0,1]]
yp = [ [0.9, 0.2,0.2], [0.2,0.8,0.3] ]
MSE or Mean Squared Error is simply the expected value of the squared difference between the predicted and the ground truth labels, represented as
\text{MSE}(\hat{\theta}) = E\left[(\hat{\theta} - \theta)^2\right]
where theta is the ground truth labels and theta^hat is the predicted labels
I am not sure what are you referring to exactly, like a theoretical question or a part of code
As a Python implementation
def mean_squared_error(A, B):
return np.square(np.subtract(A,B)).mean()
yt = [[1,0,0],[0,0,1]]
yp = [[0.9, 0.2,0.2], [0.2,0.8,0.3]]
mse = mean_squared_error(yt, yp)
print(mse)
This will give a value of 0.21
If you are using one of the DL frameworks like TensorFlow, then they are already providing the function which calculates the mse loss between tensors
tf.losses.mean_squared_error
where
tf.losses.mean_squared_error(
labels,
predictions,
weights=1.0,
scope=None,
loss_collection=tf.GraphKeys.LOSSES,
reduction=Reduction.SUM_BY_NONZERO_WEIGHTS
)
Args:
labels: The ground truth output tensor, same dimensions as 'predictions'.
predictions: The predicted outputs.
weights: Optional Tensor whose rank is either 0, or the same rank as labels, and must be broadcastable to labels (i.e., all dimensions
must be either 1, or the same as the corresponding losses dimension).
scope: The scope for the operations performed in computing the loss.
loss_collection: collection to which the loss will be added.
reduction: Type of reduction to apply to loss.
Returns:
Weighted loss float Tensor. If reduction is NONE, this has the same
shape as labels; otherwise, it is scalar.
I'm a newbie to machine learning and this is one of the first real-world ML tasks challenged.
Some experimental data contains 512 independent boolean features and a boolean result.
There are about 1e6 real experiment records in the provided data set.
In a classic XOR example all 4 out of 4 possible states are required to train NN. In my case its only 2^(10-512) = 2^-505 which is close to zero.
I have no more information about the data nature, just these (512 + 1) * 1e6 bits.
Tried NN with 1 hidden layer on available data. Output of the trained NN on the samples even from the training set are always close to 0, not a single close to "1". Played with weights initialization, gradient descent learning rate.
My code utilizing TensorFlow 1.3, Python 3. Model excerpt:
with tf.name_scope("Layer1"):
#W1 = tf.Variable(tf.random_uniform([512, innerN], minval=-2/512, maxval=2/512), name="Weights_1")
W1 = tf.Variable(tf.zeros([512, innerN]), name="Weights_1")
b1 = tf.Variable(tf.zeros([1]), name="Bias_1")
Out1 = tf.sigmoid( tf.matmul(x, W1) + b1)
with tf.name_scope("Layer2"):
W2 = tf.Variable(tf.random_uniform([innerN, 1], minval=-2/512, maxval=2/512), name="Weights_2")
#W2 = tf.Variable(tf.zeros([innerN, 1]), name="Weights_2")
b2 = tf.Variable(tf.zeros([1]), name="Bias_2")
y = tf.nn.sigmoid( tf.matmul(Out1, W2) + b2)
with tf.name_scope("Training"):
y_ = tf.placeholder(tf.float32, [None,1])
cross_entropy = tf.reduce_mean(
tf.nn.softmax_cross_entropy_with_logits(
labels = y_, logits = y)
)
train_step = tf.train.GradientDescentOptimizer(0.005).minimize(cross_entropy)
with tf.name_scope("Testing"):
# Test trained model
correct_prediction = tf.equal( tf.round(y), tf.round(y_))
# ...
# Train
for step in range(500):
batch_xs, batch_ys = Datasets.train.next_batch(300, shuffle=False)
_, my_y, summary = sess.run([train_step, y, merged_summaries],
feed_dict={x: batch_xs, y_: batch_ys})
I suspect two cases:
my fault – bad NN implementation, wrong architecture;
bad data. Compared to XOR example, incomplete training data would result in a failing NN. However, the training examples fed to the trained NN are supposed to give right predictions, aren't they?
How to evaluate if it is possible at all to train a neural network (a 2-layer perceptron) on the provided data to forecast the result? A case of aceptable set would be the XOR example. Opposed to some random noise.
There are only ad hoc ways to know if it is possible to learn a function with a differentiable network from a dataset. That said, these ad hoc ways do usually work. For example, the network should be able to overfit the training set without any regularisation.
A common technique to gauge this is to only fit the network on a subset of the full dataset. Check that the network can overfit to that, then increase the size of the subset, and increase the size of the network as well. Unfortunately, deciding whether to add extra layers or add more units in a hidden layer is an arbitrary decision you'll have to make.
However, looking at your code, there are a few things that could be going wrong here:
Are your outputs balanced? By that I mean, do you have the same number of 1s as 0s in the dataset targets?
Your initialisation in the first layer is all zeros, the gradient to this will be zero, so it can't learn anything (although, you have a real initialisation above it commented out).
Sigmoid nonlinearities are more difficult to optimise than simpler nonlinearities, such as ReLUs.
I'd recommend using the built-in definitions for layers in Tensorflow to not worry about initialisation, and switching to ReLUs in any hidden layers (you need sigmoid at the output for your boolean target).
Finally, deep learning isn't actually very good at most "bag of features" machine learning problems because they lack structure. For example, the order of the features doesn't matter. Other methods often work better, but if you really want to use deep learning then you could look at this recent paper, showing improved performance by just using a very specific nonlinearity and weight initialisation (change 4 lines in your code above).
I am attempting to train an ANN on time series data in Keras. I have three vectors of data that are broken into scrolling window sequences (i.e. for vector l).
np.array([l[i:i+window_size] for i in range( len(l) - window_size)])
The target vector is similarly windowed so the neural net output is a prediction of the target vector for the next window_size number of time steps. All the data is normalized with a min-max scaler. It is fed into the neural network as a shape=(nb_samples, window_size, 3). Here is a plot of the 3 input vectors.
The only output I've managed to muster from the ANN is the following plot. Target vector in blue, predictions in red (plot is zoomed in to make the prediction pattern legible). Prediction vectors are plotted at window_size intervals so each one of the repeated patterns is one prediction from the net.
I've tried many different model architectures, number of epochs, activation functions, short and fat networks, skinny, tall. This is my current one (it's a little out there).
Conv1D(64,4, input_shape=(None,3)) ->
Conv1d(32,4) ->
Dropout(24) ->
LSTM(32) ->
Dense(window_size)
But nothing I try will affect the neural net from outputting this repeated pattern. I must be misunderstanding something about time-series or LSTMs in Keras. But I'm very lost at this point so any help is greatly appreciated. I've attached the full code at this repository.
https://github.com/jaybutera/dat-toy
I played with your code a little and I think I have a few suggestions for getting you on the right track. The code doesn't seem to match your graphs exactly, but I assume you've tweaked it a bit since then. Anyway, there are two main problems:
The biggest problem is in your data preparation step. You basically have the data shapes backwards, in that you have a single timestep of input for X and a timeseries for Y. Your input shape is (18830, 1, 8), when what you really want is (18830, 30, 8) so that the full 30 timesteps are fed into the LSTM. Otherwise the LSTM is only operating on one timestep and isn't really useful. To fix this, I changed the line in common.py from
X = X.reshape(X.shape[0], 1, X.shape[1])
to
X = windowfy(X, winsize)
Similarly, the output data should probably be only 1 value, from what I've gathered of your goals from the plotting function. There are certainly some situations where you want to predict a whole timeseries, but I don't know if that's what you want in this case. I changed Y_train to use fuels instead of fuels_w so that it only had to predict one step of the timeseries.
Training for 100 epochs might be way too much for this simple network architecture. In some cases when I ran it, it looked like there was some overfitting going on. Observing the decrease of loss in the network, it seems like maybe only 3-4 epochs are needed.
Here is the graph of predictions after 3 training epochs with the adjustments I mentioned. It's not a great prediction, but it looks like it's on the right track now at least. Good luck to you!
EDIT: Example predicting multiple output timesteps:
from sklearn import datasets, preprocessing
import numpy as np
from scipy import stats
from keras import models, layers
INPUT_WINDOW = 10
OUTPUT_WINDOW = 5 # Predict 5 steps of the output variable.
# Randomly generate some regression data (not true sequential data; samples are independent).
np.random.seed(11798)
X, y = datasets.make_regression(n_samples=1000, n_features=4, noise=.1)
# Rescale 0-1 and convert into windowed sequences.
X = preprocessing.MinMaxScaler().fit_transform(X)
y = preprocessing.MinMaxScaler().fit_transform(y.reshape(-1, 1))
X = np.array([X[i:i + INPUT_WINDOW] for i in range(len(X) - INPUT_WINDOW)])
y = np.array([y[i:i + OUTPUT_WINDOW] for i in range(INPUT_WINDOW - OUTPUT_WINDOW,
len(y) - OUTPUT_WINDOW)])
print(np.shape(X)) # (990, 10, 4) - Ten timesteps of four features
print(np.shape(y)) # (990, 5, 1) - Five timesteps of one features
# Construct a simple model predicting output sequences.
m = models.Sequential()
m.add(layers.LSTM(20, activation='relu', return_sequences=True, input_shape=(INPUT_WINDOW, 4)))
m.add(layers.LSTM(20, activation='relu'))
m.add(layers.RepeatVector(OUTPUT_WINDOW))
m.add(layers.LSTM(20, activation='relu', return_sequences=True))
m.add(layers.wrappers.TimeDistributed(layers.Dense(1, activation='sigmoid')))
print(m.summary())
m.compile(optimizer='adam', loss='mse')
m.fit(X[:800], y[:800], batch_size=10, epochs=60) # Train on first 800 sequences.
preds = m.predict(X[800:], batch_size=10) # Predict the remaining sequences.
print('Prediction:\n' + str(preds[0]))
print('Actual:\n' + str(y[800]))
# Correlation should be around r = .98, essentially perfect.
print('Correlation: ' + str(stats.pearsonr(y[800:].flatten(), preds.flatten())[0]))
I can't understand why dropout works like this in tensorflow. The blog of CS231n says that, "dropout is implemented by only keeping a neuron active with some probability p (a hyperparameter), or setting it to zero otherwise." Also you can see this from picture(Taken from the same site)
From tensorflow site, With probability keep_prob, outputs the input element scaled up by 1 / keep_prob, otherwise outputs 0.
Now, why the input element is scaled up by 1/keep_prob? Why not keep the input element as it is with probability and not scale it with 1/keep_prob?
This scaling enables the same network to be used for training (with keep_prob < 1.0) and evaluation (with keep_prob == 1.0). From the Dropout paper:
The idea is to use a single neural net at test time without dropout. The weights of this network are scaled-down versions of the trained weights. If a unit is retained with probability p during training, the outgoing weights of that unit are multiplied by p at test time as shown in Figure 2.
Rather than adding ops to scale down the weights by keep_prob at test time, the TensorFlow implementation adds an op to scale up the weights by 1. / keep_prob at training time. The effect on performance is negligible, and the code is simpler (because we use the same graph and treat keep_prob as a tf.placeholder() that is fed a different value depending on whether we are training or evaluating the network).
Let's say the network had n neurons and we applied dropout rate 1/2
Training phase, we would be left with n/2 neurons. So if you were expecting output x with all the neurons, now you will get on x/2. So for every batch, the network weights are trained according to this x/2
Testing/Inference/Validation phase, we dont apply any dropout so the output is x. So, in this case, the output would be with x and not x/2, which would give you the incorrect result. So what you can do is scale it to x/2 during testing.
Rather than the above scaling specific to Testing phase. What Tensorflow's dropout layer does is that whether it is with dropout or without (Training or testing), it scales the output so that the sum is constant.
Here is a quick experiment to disperse any remaining confusion.
Statistically the weights of a NN-layer follow a distribution that is usually close to normal (but not necessarily), but even in the case when trying to sample a perfect normal distribution in practice, there are always computational errors.
Then consider the following experiment:
DIM = 1_000_000 # set our dims for weights and input
x = np.ones((DIM,1)) # our input vector
#x = np.random.rand(DIM,1)*2-1.0 # or could also be a more realistic normalized input
probs = [1.0, 0.7, 0.5, 0.3] # define dropout probs
W = np.random.normal(size=(DIM,1)) # sample normally distributed weights
print("W-mean = ", W.mean()) # note the mean is not perfect --> sampling error!
# DO THE DRILL
h = defaultdict(list)
for i in range(1000):
for p in probs:
M = np.random.rand(DIM,1)
M = (M < p).astype(int)
Wp = W * M
a = np.dot(Wp.T, x)
h[str(p)].append(a)
for k,v in h.items():
print("For drop-out prob %r the average linear activation is %r (unscaled) and %r (scaled)" % (k, np.mean(v), np.mean(v)/float(k)))
Sample output:
x-mean = 1.0
W-mean = -0.001003985674840264
For drop-out prob '1.0' the average linear activation is -1003.985674840258 (unscaled) and -1003.985674840258 (scaled)
For drop-out prob '0.7' the average linear activation is -700.6128015029908 (unscaled) and -1000.8754307185584 (scaled)
For drop-out prob '0.5' the average linear activation is -512.1602655283492 (unscaled) and -1024.3205310566984 (scaled)
For drop-out prob '0.3' the average linear activation is -303.21194422742315 (unscaled) and -1010.7064807580772 (scaled)
Notice that the unscaled activations diminish due to the statistically imperfect normal distribution.
Can you spot an obvious correlation between the W-mean and the average linear activation means?
If you keep reading in cs231n, the difference between dropout and inverted dropout is explained.
In a network with no dropout, the activations in layer L will be aL. The weights of next layer (L+1) will be learned in such a manner that it receives aL and produces output accordingly. But with a network containing dropout (with keep_prob = p), the weights of L+1 will be learned in such a manner that it receives p*aL and produces output accordingly. Why p*aL? Because the Expected value, E(aL), will be probability_of_keeping(aL)*aL + probability_of_not_keeping(aL)*0 which will be equal to p*aL + (1-p)*0 = p*aL. In the same network, during testing time there will be no dropout. Hence the layer L+1 will receive aL simply. But its weights were trained to expect p*aL as input. Therefore, during testing time you will have to multiply the activations with p. But instead of doing this, you can multiply the activations with 1/p during training only. This is called inverted dropout.
Since we want to leave the forward pass at test time untouched (and tweak our network just during training), tf.nn.dropout directly implements inverted dropout, scaling the values.