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Neuron Layers SC4001 – Tutorial 3
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GD for Softmax layer Given training set ࠵?, ࠵? Set learning rate ࠵? Initialize ࠵? and ࠵? Iterate until convergence: ࠵? = ࠵?࠵? + ࠵? ࠵? ࠵? = ! ࠵? ∑ "#$ % ! ࠵? ࠵? ࠵? # ࠵? = − ࠵? − ࠵? ࠵? ࠵? ← ࠵? − ࠵?࠵? $ ࠵? ࠵? ࠵? ࠵? ← ࠵? − ࠵? ࠵? ࠵? ࠵? $ ࠵? &
Labels for classes: ࠵?࠵?࠵?࠵?࠵? ࠵? → 1, ࠵?࠵?࠵?࠵?࠵? ࠵? → 2, ࠵?࠵?࠵?࠵?࠵? ࠵? → 3 The data matrix and target vector: ࠵? = 0 4 −1 3 2 3 −2 2 0 2 1 2 −1 2 ⋮ ⋮ 4 −1 , ࠵? = 1 1 1 2 2 1 2 ⋮ 3 , ࠵? = 1 0 0 1 0 0 1 0 0 0 1 0 0 1 0 1 0 0 0 1 0 ⋮ ⋮ ⋮ 0 0 1
࠵? ! ࠵? " ࠵? # ࠵? ! ࠵? " +1 ࠵? ࠵? = ࠵? ࠵? ∑ "#$ % ࠵? ࠵? ࠵? = ࠵?(࠵? = ࠵?|࠵?) ࠵? = argmax " ࠵? ࠵? Learning rate ࠵? = 0.05. Initialize weights and biases: ࠵? = 0.88 0.08 −0.34 0.68 −0.39 −0.19 and ࠵? = 0.0 0.0 0.0
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Epoch 1: ࠵? = ࠵?࠵? + ࠵? = 0 4 −1 3 2 3 −2 2 ⋮ ⋮ 4 −1 0.88 0.08 −0.34 0.68 −0.39 −0.19 + 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 ⋮ ⋮ ⋮ 0.0 0.0 0.0 = 2.72 −1.54 −0.75 1.17 −1.23 −0.23 3.8 −1.0 −1.23 −0.39 −0.93 0.30 ⋮ ⋮ ⋮ 2.82 0.71 −1.16
࠵? = 2.72 −1.54 −0.75 1.17 −1.23 −0.23 3.8 −1.0 −1.23 −0.39 −0.93 0.30 ⋮ ⋮ ⋮ 2.82 0.71 −1.16 ࠵? ࠵? & = ࠵? ’ " ∑ "#$ % ࠵? ’ ࠵? ࠵? ࠵? == ࠵? (.*( ࠵? (.*( + ࠵? +$.,- + ࠵? +..*, ࠵? +$.,- ࠵? (.*( + ࠵? +$.,- + ࠵? +..*, ࠵? +..*, ࠵? (.*( + ࠵? +$.,- + ࠵? +..*, ࠵? $.$* ࠵? $.$* + ࠵? +$.(/ + ࠵? +..(/ ࠵? +$.(/ ࠵? $.$* + ࠵? +$.(/ + ࠵? +..(/ ࠵? +..(/ ࠵? $.$* + ࠵? +$.(/ + ࠵? +..(/ ⋮ ⋮ ⋮ ࠵? (.0( ࠵? (.0( + ࠵? ..*$ + ࠵? +$.$1 ࠵? ..*$ ࠵? (.0( + ࠵? ..*$ + ࠵? +$.$1 ࠵? +$.$1 ࠵? (.0( + ࠵? ..*$ + ࠵? +$.$1 = 0.96 0.01 0.03 0.75 0.07 0.18 ⋮ ⋮ ⋮ 0.88 0.11 0.02
࠵? = argmax ’ ࠵? ࠵? = argmax ’ 0.96 0.01 0.03 0.75 0.07 0.18 0.99 0.01 0.01 0.28 0.16 0.56 ⋮ ⋮ ⋮ 0.88 0.11 0.02 = 1 1 1 3 ⋮ 1 Classificationerror = ∑ ()* & 1 ࠵? ≠ ࠵? = 14 ࠵?࠵?࠵?࠵?࠵?࠵?࠵? = − X ()* & ࠵?࠵?࠵? ࠵? ࠵? (+ ’ = − log 0.96 − log 0.75 − log 0.99 − log 0.16 ⋯ − log 0.02 = 34.36 ࠵? = 1 1 1 2 ⋮ 3
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࠵? # ࠵? = − ࠵? − ࠵? ࠵? = − 1 0 0 1 0 0 ⋮ ⋮ ⋮ 0 0 1 − 0.96 0.01 0.03 0.75 0.07 0.18 ⋮ ⋮ ⋮ 0.88 0.11 0.02 = −0.04 0.01 0.03 −0.25 0.07 0.18 ⋮ ⋮ ⋮ 0.88 0.11 −0.98 ࠵? = ࠵? − ࠵? ࠵? $ ࠵? ࠵? ࠵? = 0.88 0.08 −0.34 0.68 −0.39 −0.19 − 0.05 0 −1 ⋯ 4 4 3 ⋯ −1 −0.04 0.01 0.03 −0.25 0.07 0.18 ⋮ ⋮ ⋮ 0.88 0.11 −0.98 = 0.28 −0.54 0.89 0.54 −0.12 −0.31 ࠵? = ࠵? − ࠵? ࠵? ࠵? ࠵? $ ࠵? & = 0.0 0.0 0.0 + 0.05 −0.04 0.01 0.03 −0.25 0.07 0.18 ⋮ ⋮ ⋮ 0.88 0.11 −0.98 $ 1 1 1 ⋮ 1 = −0.32 0.27 0.06
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At convergence: ࠵? = −0.15 −3.41 4.18 5.27 −1.02 −4.15 and ࠵? = −7.82 5.81 2.02 Entropy = 0.58 Classification Error = 0
࠵? = −1 2 0 4 −1 3 0 2 3 0 −2 −1 4 1 1 2 2 −1 2 3 2 1 −2 0 −3 −1 1 0 −1 1 4 −1 −3 1 −2 2 ࠵? ࠵? = 1.0 0.0 0.0 0.88 0.12 0.0 1.0 0.0 0.0 0.0 1.0 0.0 0.26 0.74 0.0 0.89 0.1 0.0 0.01 0.99 0.0 0.0 1.0 0.0 0.0 1.0 0.0 0.0 0.0 1.0 0.0 0.0 1.0 0.0 1.0 0.0 0.0 0.02 0.98 0.0 0.0 1.0 0.0 1.0 0.0 0.0 1.0 0.0 0.0 0.0 1.0 0.0 0.0 1.0 , ࠵? = 1 1 1 2 2 1 2 2 2 3 3 2 3 3 2 2 3 3 Probabilities that input patterns belong to target classes are given in RED . At convergence:
At convergence: ࠵? = −0.15 −3.41 4.18 5.27 −1.02 −4.15 and ࠵? = −7.82 5.81 2.02 Synaptic inputs at the softmax layer for an input ࠵? = ࠵? $ , ࠵? ( : Neuron of class A, ࠵? $ = ࠵? $ 2 ࠵? + ࠵? $ = −0.15࠵? $ + 5.27࠵? ( − 7.82 Neuron of class B, ࠵? ( = ࠵? ( 2 ࠵? + ࠵? ( = −3.41࠵? $ − 1.02࠵? ( + 5.81 Neuron of class C, ࠵? / = ࠵? / 2 ࠵? + ࠵? / = 4.18࠵? $ − 4.15࠵? ( + 2.02 Decision boundaries: Between class A and class B is given when ࠵? $ = ࠵? ( −0.15࠵? $ + 5.27࠵? ( − 7.82 = −3.41࠵? $ − 1.02࠵? ( + 5.81 3.25࠵? $ + 6.29࠵? ( − 13.63 = 0 Similarly, between class B and class C ࠵? ( = ࠵? / : −7.59࠵? $ + 3.13࠵? ( + 3.79 = 0 between class A and class C ࠵? / = ࠵? $ : 4.33࠵? $ − 9.42࠵? ( + 9.84 = 0
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class A and class B class A and class C class B and class C
Decision boundaries leant by the softmax layer
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from sklearn.datasets i mport load_linnerud X, y = load_linnerud (return_X_y=True) X y Parameters of the data targets inputs
# preprocess input and output data from sklearn import preprocessing # normal Gaussian scaling for inputs X_scaler = preprocessing.StandardScaler(). fit (X_train) X_scaled = X_scaler. transform (X_train) [ 9.06666667 161.6 77.8 ] [ 28.46222222 3795.17333333 2861.62666667] ࠵? 3 = ࠵? − ࠵? ࠵? Normalizing input variables such that ࠵? , ~࠵? 0, 1 . print (X_scaler.mean_) print (X_scaler.var_)
Sclaing output variables such that ࠵? 3 ~[0, 1]. # linear scaling up to [0,1] for outputs y_scaler = preprocessing.MinMaxScaler(). fit (y_train) y_scaled = y_scaler. transform (y_train) [0.00917431 0.06666667 0.03571429] [-1.26605505 -2.06666667 -1.64285714] prin t(y_scaler.scale_) # print (y_scaler.min_) ࠵? , = * - ()* .- (+, ࠵? + .- (+, - ()* .- (+, ࠵?࠵?࠵?࠵?࠵? ࠵?࠵?࠵?
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Given a training dataset ࠵?, ࠵? Set learning parameter ࠵? Initialize ࠵? and ࠵? Repeat until convergence: ࠵? = ࠵?࠵? + ࠵? ࠵? = ࠵? ࠵? ࠵? # ࠵? = − ࠵? − ࠵? b ࠵? , ࠵? ࠵? ← ࠵? − ࠵?࠵? $ ࠵? ࠵? ࠵? ࠵? ← ࠵? − ࠵? ࠵? ࠵? ࠵? $ ࠵? & GD for a perceptron layer with direct gradients ࠵? ! ࠵? " ࠵? # +1 ࠵? ! ࠵? # ࠵? "
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Implementing Perceptron Layer using pytorch libraries : from torch import nn # torch nn API provides methods for build custom neural networks nn.Module - Base class for all neural network modules. - Your models should also subclass this class. nn.Linear - applies a linear transformation ࠵? = ࠵? 2 ࠵? + ࠵? nn.Sigmoidal - implements a perceptron layer nn.Sequential A sequential container. Modules will be added to it in the order they are passed in the constructor. The value a Sequential provides over manually calling a sequence of modules is that it allows treating the whole container as a single module Implementing MSE loss and SGD optim izer loss_fn = nn.MSELoss () optimizer = torch.optim.SGD (model.parameters(), lr=lr)
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# Perceptron layer class class PerceptronLaye r(nn.Module): def __init__(self, no_inputs, no_outputs): super().__init__() self.perceptron_layer = nn. Sequential ( nn. Linear (no_inputs, no_outputs), nn. Sigmoid () ) def forward(self, x): logits = self.perceptron_layer(x) return logits # create an instance of the layer no_inputs, no_outputs = 3, 3 model = PerceptronLayer (no_inputs, no_outputs) GD for a perceptron layer with PyTorch nn.Module class
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loss_fn = torch.nn. MSELoss () optimizer = torch.optim. SGD (model.parameters(), lr=.001) no_epochs, lr = 50000, 0.001 for epoch in range(no_epochs): # Compute prediction and loss pred = model(torch.tensor(X_scaled, dtype=torch.float)) loss = loss_fn(pred, torch.tensor(y_scaled, dtype=torch.float)) optimizer.zero_grad() loss.backward() optimizer.step() GD for a perceptron layer with PyTorch autograd
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Direct gradients Autograd Time for weight update = 0.074ms Number of epochs = 10,000 Time for weight update = 0.114ms Number of epochs = 50,000
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# scaling testing inputs X_scaled = X_scaler.transform(X_test) y_pred = model (X_scaled) # scaling predicted outputs y_scaled = y_scaler. inverse_transform (y_pred.detach().numpy()) # computing mean square error from sklearn.metrics import mean_squared_error rms = mean_squared_error (y_scaled, y_test, squared=True, multioutput='raw_values') # computing R 2 from sklearn.metrics import r2_score r2 = r2_score (y_scaled, y_test, multioutput='raw_values')
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Direct gradients Autograd MSE = [407.06558087, 1.21179368, 22.40310961] R 2 = [-15.16134924, 0.53783585 , -11.22642492] MSE = [356.95227178, 1.61016719, 21.11763412] R 2 = [ -3.18467906, 0.06741378 , -11.90990362] Exercise can best predict the waist!
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R 2 : coefficient of determination Let ‘࠵? & be the predicted value of data point ࠵? & and the number of data points be ࠵?: The residual sum of square error ࠵?࠵? 456 = ∑ $ 7 ‘࠵? & − ࠵? & ( If mean of data, b࠵? = $ 7 ∑ &#$ 7 ࠵? & . The total sum of squares (proportional to the variance), ࠵?࠵? 898 = ∑ $ 7 b࠵? − ࠵? & ( ࠵? ( = 1 − ࠵?࠵? 456 ࠵?࠵? 898 In the best case, the modelled value exactly match the observed value ࠵?࠵? 456 =0, then ࠵? ( = 1. The baseline model which predicts b࠵? , then ࠵? ( = 0. The models that have worse prediction than baseline models have negative ࠵? ( .
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+1 ࠵? $ ࠵? " 1 2 3 ࠵? # ࠵? ! Sepal length Sepal width Petal length Petal width Setosa Versicolour Virginica Four features and three labels 90 data points for training and 60 data points for testing
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# datasets import from sklearn import datasets from sklearn.model_selection import train_test_split iris = datasets.load_iris () iris.data -= np.mean (iris.data, axis=0) # mean correct data # dataset split for train and test x_train, x_test, y_train, y_test = train_test_split (iris.data, iris.target, test_size=0.2, random_state=2)
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class SoftmaxLayer (nn.Module): def __init__(self, no_inputs, no_outputs): super().__init__() self.softmax_layer = nn. Sequential ( nn. Linear (no_inputs, no_outputs), # applies a linear transformation ࠵? = ࠵? 2 ࠵? + ࠵? nn. Softmax (dim=1) # implements softmax; sum up across rows to 1.0 ) def forward(self, x): logits = self.softmax_layer(x) return logits from torch import nn
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Mini-batch gradient descent Batch size = 16, learning factor = 0.1 Torch Dataset and Dataloader utils provide easy function for mini-batch learning from torch.utils.data import Dataset from torch.utils.data import DataLoader We will be creating a subclass of Dataset class and using Dataloaders to implement mini-batch gradient descent
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# create a Dataset class # A custom Dataset class must implement three functions: __init__ , __len__ , and __getitem__ . class MyDataset (Dataset): def __init__ (self, X, y): self.X =torch.tensor(X, dtype=torch.float) self.y =torch.tensor(y) def __len__( self): return len(self.y) def __getitem__ (self,idx): return self.X[idx], self.y[idx] # create Dataset objects for train and test data train_data = MyDataset (x_train, y_train) test_data = MyDataset (x_test, y_test) # create DataLoader objects for train and test data train_dataloader = DataLoader (train_data, batch_size=batch_size, shuffle= True ) test_dataloader = DataLoader (test_data, batch_size=batch_size, shuffle= True )
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def train_loop (dataloader, model, loss_fn, optimizer): size = len(dataloader.dataset) num_batches = len(dataloader) train_loss, correct = 0, 0 for batch, (X, y) in enumerate(dataloader): # Compute prediction and loss pred = model(X) loss = loss_fn(pred, y) optimizer.zero_grad(). #initialize gradient calculations loss.backward() # compute gradients optimizer.step() # execute one step of SGD train_loss += loss.item() correct += (pred.argmax(1) == y).type(torch.float).sum().item() train_loss /= size correct /= size return train_loss, correct
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def test_loop (dataloader, model, loss_fn): size = len(dataloader.dataset) num_batches = len(dataloader) test_loss, correct = 0, 0 with torch.no_grad(): for X, y in dataloader: pred = model(X) test_loss += loss_fn(pred, y).item() correct += (pred.argmax(dim=1) == y).type(torch.float).sum().item() test_loss /= size correct /= size return test_loss, correct
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train_loss_, train_acc_, test_loss_, test_acc_ = [], [], [], [] for epoch in range(no_epochs): train_loss, train_acc = train_loop(train_dataloader, model, loss_fn, optimizer) test_loss, test_acc = test_loop(test_dataloader, model, loss_fn) train_loss_.append(train_loss), train_acc_.append(train_acc) test_loss_.append(test_loss), test_acc_.append(test_acc)
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Learning curves at batch-size = 16
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weight = [[-0.875575 1.9825934 -4.016873 -1.6209128 ] [ 0.70730793 -1.0039392 -0.2982792 -2.5409048 ] [-0.3355393 -0.79086286 3.4620962 3.9241226 ]] bias = [-0.6561739 3.990793 -2.9169736] train_loss = 0.039542 train_acc = 0.966667 test_loss = 0.040103, test_acc = 0.950000 At convergence (1000 epochs):
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Accuracies and time-to-update weights against batch-size Elbow point
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E Managerial Accountin x B 3-1 Problem Set: Mod X * CengageNOWv2 |On x b My Questions | bartlel x * Cengage Learning Cengage Learning + A v2.cengagenow.com/ilrn/takeAssignment/takeAssignmentMain.do?invoker=&takeAssignmentSessionLocator=&inprogress=false Update : Problem Set: Module 6 O eBook I Show Me How 1. EX.07.02.ALGO Income Statements under Absorption Costing and Variable Costing Gallatin County Motors Inc. assembles and sells snowmobile engines. The company began operations on July 1 and operated at 100% of capacity during 2. TMM.07,01 the first month. The following data summarize the results for July: 3. EX.07.12.ALGO Sales (18,500 units) $2,405,000 Production costs (24,000 units): 4. EX.07.05.ALGO Direct materials $1,154,400 5. TMM.07.05 Direct labor 554,400 Variable factory overhead 276,000 Fixed factory overhead 184,800 2,169,600 Selling and administrative expenses: Variable selling and administrative expenses $336,300 Fixed selling and administrative expenses 130,200 466,500…
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What is the product? ole 1. NaOEt Br 2. H3O+, heat Br O a. O b. HO OH Ос ora ○ d. w HO
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