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Вивчайте Extracting and Transforming Data | Working with Structured Data Formats
Working with Strings and Data Formats

bookExtracting and Transforming Data

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# Suppose you have CSV data loaded as a list of rows, where each row is a list of strings. rows = [ ["id", "name", "age", "city"], ["1", "Alice", "30", "New York"], ["2", "Bob", "25", "Los Angeles"], ["3", "Charlie", "35", "Chicago"] ] # To extract the "name" column (index 1) from all rows except the header: name_column = [row[1] for row in rows[1:]] print(name_column) # Output: ['Alice', 'Bob', 'Charlie']
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When working with structured data such as JSON, you often deal with a list of dictionaries, where each dictionary represents an object with key-value pairs. To extract values for a given key from all dictionaries in the list, use a list comprehension. For instance, if you have a list of dictionaries representing people and want to extract all ages, you can use [person["age"] for person in people]. This approach gives you a new list containing only the values associated with the specified key from each dictionary.

1. Which of the following is the best way to access a value for a specific key in a dictionary?

2. Which approaches can be used to extract a column from a list of lists?

question mark

Which of the following is the best way to access a value for a specific key in a dictionary?

Select the correct answer

question mark

Which approaches can be used to extract a column from a list of lists?

Select the correct answer

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Can you show an example using a list of dictionaries instead of a list of lists?

How can I extract multiple columns at once from the CSV data?

What if some rows are missing the "name" value?

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bookExtracting and Transforming Data

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1234567891011
# Suppose you have CSV data loaded as a list of rows, where each row is a list of strings. rows = [ ["id", "name", "age", "city"], ["1", "Alice", "30", "New York"], ["2", "Bob", "25", "Los Angeles"], ["3", "Charlie", "35", "Chicago"] ] # To extract the "name" column (index 1) from all rows except the header: name_column = [row[1] for row in rows[1:]] print(name_column) # Output: ['Alice', 'Bob', 'Charlie']
copy

When working with structured data such as JSON, you often deal with a list of dictionaries, where each dictionary represents an object with key-value pairs. To extract values for a given key from all dictionaries in the list, use a list comprehension. For instance, if you have a list of dictionaries representing people and want to extract all ages, you can use [person["age"] for person in people]. This approach gives you a new list containing only the values associated with the specified key from each dictionary.

1. Which of the following is the best way to access a value for a specific key in a dictionary?

2. Which approaches can be used to extract a column from a list of lists?

question mark

Which of the following is the best way to access a value for a specific key in a dictionary?

Select the correct answer

question mark

Which approaches can be used to extract a column from a list of lists?

Select the correct answer

Все було зрозуміло?

Як ми можемо покращити це?

Дякуємо за ваш відгук!

Секція 3. Розділ 3
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