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items [2, 4, 5, 6] items. insert (1, 7) print (items) items= [3, 4, 5, 6] items print (items) items [3, 4, 5, 6] items.reverse() items.reverse() print (items) items [2, 3, 4, 5, 6] print (items. index (4)) items.remove (4) print (items) 1

items [2, 4, 5, 6] items. insert (1, 7) print (items) items= [3, 4, 5, 6] items print (items) items [3, 4, 5, 6] items.reverse() items.reverse() print (items) items [2, 3, 4, 5, 6] print (items. index (4)) items.remove (4) print (items) 1

Database System Concepts
Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
Problem 1PE
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## Educational Content: Understanding Python List Operations ### Exercise: Analyze Code Fragments For each of the following code fragments (each box is a separate code fragment), determine what the program outputs or write "error" if it would crash or has a syntax error that prevents it from running. --- #### Code Fragment 1: ```python items = [2, 4, 5, 6] items.insert(1, 7) print(items) ``` - **Explanation:** - The list `items` initially contains `[2, 4, 5, 6]`. - `items.insert(1, 7)` inserts the number `7` at index `1`. - The list becomes `[2, 7, 4, 5, 6]`. - The `print(items)` statement outputs: **[2, 7, 4, 5, 6]**. --- #### Code Fragment 2: ```python items = [3, 4, 5, 6] items = items.reverse() print(items) items = [3, 4, 5, 6] items.reverse() print(items) ``` - **Explanation:** - The first part attempts to assign the result of `items.reverse()` to `items`. However, `reverse()` performs an in-place reversal and returns `None`. So `items` becomes `None`, causing the next line `print(items)` to output: **None**. - The second part correctly uses `reverse()` in place, so the list `[3, 4, 5, 6]` becomes `[6, 5, 4, 3]`. - The `print(items)` statement outputs: **[6, 5, 4, 3]**. --- #### Code Fragment 3: ```python items = [2, 3, 4, 5, 6] print(items.index(4)) items.remove(4) print(items) ``` - **Explanation:** - `print(items.index(4))` finds the index of `4`, which is `2`. It outputs: **2**. - `items.remove(4)` removes the first occurrence of `4` from the list. - The list becomes `[2, 3, 5, 6]`. - The `print(items)` statement
Transcribed Image Text:## Educational Content: Understanding Python List Operations ### Exercise: Analyze Code Fragments For each of the following code fragments (each box is a separate code fragment), determine what the program outputs or write "error" if it would crash or has a syntax error that prevents it from running. --- #### Code Fragment 1: ```python items = [2, 4, 5, 6] items.insert(1, 7) print(items) ``` - **Explanation:** - The list `items` initially contains `[2, 4, 5, 6]`. - `items.insert(1, 7)` inserts the number `7` at index `1`. - The list becomes `[2, 7, 4, 5, 6]`. - The `print(items)` statement outputs: **[2, 7, 4, 5, 6]**. --- #### Code Fragment 2: ```python items = [3, 4, 5, 6] items = items.reverse() print(items) items = [3, 4, 5, 6] items.reverse() print(items) ``` - **Explanation:** - The first part attempts to assign the result of `items.reverse()` to `items`. However, `reverse()` performs an in-place reversal and returns `None`. So `items` becomes `None`, causing the next line `print(items)` to output: **None**. - The second part correctly uses `reverse()` in place, so the list `[3, 4, 5, 6]` becomes `[6, 5, 4, 3]`. - The `print(items)` statement outputs: **[6, 5, 4, 3]**. --- #### Code Fragment 3: ```python items = [2, 3, 4, 5, 6] print(items.index(4)) items.remove(4) print(items) ``` - **Explanation:** - `print(items.index(4))` finds the index of `4`, which is `2`. It outputs: **2**. - `items.remove(4)` removes the first occurrence of `4` from the list. - The list becomes `[2, 3, 5, 6]`. - The `print(items)` statement
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