Python二维列表复制答疑：如何正确复制4×4网格列表

Great question—since you're working with a 2D list (your 4×4 grid), the shallow copy methods that work for 1D lists (like list(l1) or l1[:]) won't fully break the reference association. Here's why, plus the reliable ways to create a true independent copy:

Why Shallow Copies Fail for 2D Lists #

A 2D list is a list of lists. When you do a shallow copy (e.g., copied_grid = original_grid[:]), you're creating a new outer list—but each row inside that new list is still a reference to the rows in the original grid. Modifying any element inside a row of the original grid will also change the copied grid, because they share the same inner list references.

Reliable Methods to Copy Your 4×4 Grid #

1. List Comprehension (Most Pythonic for 2D Grids with Immutable Elements)

Since your grid contains numbers (immutable values), you can copy each row individually with a slice or list():

# Original 4×4 grid
original_grid = [
    [1, 2, 3, 4],
    [5, 6, 7, 8],
    [9, 10, 11, 12],
    [13, 14, 15, 16]
]

# Copy using list comprehension + row slicing
copied_grid = [row[:] for row in original_grid]

# OR: Use list() to copy each row
copied_grid = [list(row) for row in original_grid]

This creates a new outer list, and a new copy of each inner row—so there's no shared reference between original_grid and copied_grid.

2. copy.deepcopy() (Universal Solution for Any Nested Depth)

If you ever need to copy a grid with deeper nesting (e.g., 3D lists) or mutable elements inside rows, use deepcopy from the copy module. It recursively copies all nested objects:

import copy

original_grid = [
    [1, 2, 3, 4],
    [5, 6, 7, 8],
    [9, 10, 11, 12],
    [13, 14, 15, 16]
]

copied_grid = copy.deepcopy(original_grid)

This is the most foolproof method, as it works regardless of how deeply your list is nested.

Verify the Copy Works #

To confirm the copies are independent, modify the original grid and check the copied one:

# Change an element in the original grid
original_grid[0][0] = 99

print(original_grid[0][0])  # Output: 99
print(copied_grid[0][0])    # Output: 1 (no change—success!)

内容的提问来源于stack exchange，提问作者V. Raman