如何自动生成1-81的9个分块list并按需打印指定list

Here's a straightforward Python implementation to automatically create 9 lists, each containing 9 consecutive numbers from 1 to 81. You can access any list using its index (starting from 0) just like print(number_lists[0]).

# Initialize an empty list to store all sublists
number_lists = []

# Loop to generate each of the 9 sublists
for idx in range(9):
    # Calculate the starting number for the current sublist
    start_num = idx * 9 + 1
    # Create the sublist with 9 consecutive numbers
    current_list = list(range(start_num, start_num + 9))
    # Add the sublist to our main list
    number_lists.append(current_list)

# Example usage
print(number_lists[0])  # Output: [1, 2, 3, 4, 5, 6, 7, 8, 9]
print(number_lists[1])  # Output: [10, 11, 12, 13, 14, 15, 16, 17, 18]
print(number_lists[8])  # Output: [73, 74, 75, 76, 77, 78, 79, 80, 81]

Key Details: #

• The main list number_lists acts as a container for all 9 sublists, no manual creation of individual lists needed.
• For each iteration in the loop:
  • start_num calculates the first number of each sublist (1, 10, 19... up to 73).
  • range(start_num, start_num +9) generates exactly 9 consecutive integers, converted to a list.
• Access any sublist using 0-based indexing: number_lists[2] gives the third list, number_lists[5] gives the sixth, etc.

内容的提问来源于stack exchange，提问作者steven-14