如何使用预定义列表分割列表生成子列表或字典

Got it, let's work through this problem step by step. Since every title in your titles list shows up exactly once in x, we can build clean, straightforward solutions for both the sublist and dictionary outputs you're targeting.

Solution 1: Generate Sublist Output (new_list) #

We'll iterate through x, keep track of the current group of elements, and whenever we hit a title from titles, we wrap up the current group (if it exists) and start a new one with the title. Here's the code:

titles = ['a','c','g','final']
x = ['a',3,4,2,'c','b','b1',2,'g','final', 5, 6, 'end']

new_list = []
current_group = []

for item in x:
    if item in titles:
        # If we have an ongoing group, add it to the result before starting fresh
        if current_group:
            new_list.append(current_group)
        # Start a new group with the current title
        current_group = [item]
    else:
        # Add non-title items to the current group
        current_group.append(item)
# Don't forget to add the last group after the loop ends
if current_group:
    new_list.append(current_group)

print(new_list)
# Output: [['a', 3, 4, 2], ['c', 'b', 'b1', 2], ['g'], ['final', 5, 6, 'end']]

Breakdown: #

• We start with empty containers for the final result (new_list) and the group we're building (current_group).
• For each item in x:
  • If it's a title, we check if current_group has content from the previous section. If yes, we add it to new_list, then reset current_group to start with the current title.
  • If it's not a title, we just append it to the ongoing current_group.
• After the loop finishes, we add the last remaining current_group to new_list (since there's no title left to trigger the append inside the loop).

Solution 2: Generate Dictionary Output (my_dict) #

For the dictionary version, we'll track which title we're collecting elements for, then map each title to its following elements. Here's how to implement it:

titles = ['a','c','g','final']
x = ['a',3,4,2,'c','b','b1',2,'g','final', 5, 6, 'end']

my_dict = {}
current_title = None

for item in x:
    if item in titles:
        # Save the previous title's elements to the dict if we were tracking one
        if current_title is not None:
            my_dict[current_title] = current_elements
        # Update to the new title and start an empty elements list
        current_title = item
        current_elements = []
    else:
        # Add non-title items to the current title's element list
        current_elements.append(item)
# Save the last title and its elements to the dict
if current_title is not None:
    my_dict[current_title] = current_elements

print(my_dict)
# Output: {'a': [3, 4, 2], 'c': ['b', 'b1', 2], 'g': [], 'final': [5, 6, 'end']}

Breakdown: #

• We initialize an empty dictionary (my_dict), plus variables to track the current title we're working with (current_title) and its associated elements (current_elements).
• For each item in x:
  • If it's a title, we first save the previous title's elements to the dictionary (if there was one), then set current_title to this new title and reset current_elements to an empty list.
  • If it's not a title, we append it to current_elements.
• After the loop, we save the last title and its elements to the dictionary.

Edge Cases Handled: #

• Titles with no following elements (like 'g' in your example) will map to an empty list, which matches your expected output.
• The last title can have any number of trailing elements, which will all be captured in its value list.

内容的提问来源于stack exchange，提问作者hadesfv