Why deep learning handles complex patterns in ML Python - Test Your Understanding

Practice - 5 Tasks

1fill in blank

easy

Complete the code to create a deep learning model with multiple layers.

ML Python

model = Sequential()
model.add(Dense(64, activation=[1], input_shape=(100,)))

Drag options to blanks, or click blank then click option'

Asigmoid

Brelu

Clinear

Dsoftmax

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'linear' activation which does not add non-linearity.

The 'relu' activation helps the model learn complex patterns by adding non-linearity.

2fill in blank

medium

Complete the code to add a second hidden layer to the model.

ML Python

model.add(Dense(32, activation=[1]))

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Arelu

Bsoftmax

Csigmoid

Dlinear

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'softmax' activation in hidden layers.

Adding another 'relu' activated layer helps the model capture more complex features.

3fill in blank

hard

Fix the error in the code to compile the model with the correct loss function for classification.

ML Python

model.compile(optimizer='adam', loss=[1], metrics=['accuracy'])

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Amean_absolute_error

Bhinge

Ccategorical_crossentropy

Dmse

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'mse' which is for regression tasks.

For multi-class classification, 'categorical_crossentropy' is the correct loss function.

4fill in blank

hard

Fill both blanks to create a dictionary comprehension that maps each word to its length if the length is greater than 3.

ML Python

{word: [1] for word in words if len(word) [2] 3}

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Alen(word)

Bword

C>

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using the word itself as value instead of its length.

Using '<' instead of '>' in the condition.

This comprehension creates a dictionary with words as keys and their lengths as values, only for words longer than 3 characters.

5fill in blank

hard

Fill all three blanks to create a dictionary comprehension that maps uppercase words to their counts if count is greater than zero.

ML Python

result = [1]: [2] for word, count in word_counts.items() if count [3] 0}

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Aword.upper()

Bcount

C>

Dword

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'word' instead of 'word.upper()' for keys.

Using '<' instead of '>' in the condition.

This comprehension creates a dictionary with uppercase words as keys and their counts as values, only for counts greater than zero.