如何自动实现嵌套循环以生成mⁿ行的全唯一组合表格？

Manually writing nested loops works for fixed values of n, but it's tedious to adjust the code every time n changes. Instead, we can leverage Python's built-in tools or a custom iterative approach to create a fully flexible solution. Here's how to do it:

Using itertools.product (Simplest Approach) #

Python's itertools.product is designed specifically to generate Cartesian products—exactly what we need for all possible state combinations. It eliminates the need for any manual loop nesting.

Solution Code #

import itertools

def generate_state_combinations(m, n):
    # Generate all n-length tuples of values from 1 to m
    all_combinations = itertools.product(range(1, m + 1), repeat=n)
    # Print each combination as a space-separated string
    for combo in all_combinations:
        print(' '.join(map(str, combo)))

# Test with your examples
print("=== m=2, n=3 ===")
generate_state_combinations(2, 3)

print("\n=== m=2, n=4 ===")
generate_state_combinations(2, 4)

How It Works #

• itertools.product(range(1, m+1), repeat=n) creates every possible tuple of length n where each element is between 1 and m (matching your original output's 1-based numbering).
• We convert each tuple to a space-separated string for printing, just like your manual loop examples.
• If you need to store the combinations instead of printing them, convert the product object to a list: list(itertools.product(...)).

Verified Output #

For m=2, n=3, the output will match your example exactly:

1 1 1
1 1 2
1 2 1
1 2 2
2 1 1
2 1 2
2 2 1
2 2 2

Bonus: No External Libraries (Custom Iterative Approach) #

If you can't use itertools, you can build combinations incrementally without recursion:

def generate_state_combinations(m, n):
    # Start with all 1-length combinations
    combinations = [[num] for num in range(1, m + 1)]
    
    # Build up to n-length combinations
    for _ in range(n - 1):
        new_combos = []
        for combo in combinations:
            for num in range(1, m + 1):
                new_combos.append(combo + [num])
        combinations = new_combos
    
    # Print each combination
    for combo in combinations:
        print(' '.join(map(str, combo)))

This works by starting with single-element lists, then repeatedly appending each possible value to existing combinations until we reach the desired length n.

内容的提问来源于stack exchange，提问作者sohrab haghayegh