为什么Python函数返回错误值？

Question

我的函数在调用时返回错误的值。首先，迭代返回正确的值，但是如果我调用一个新函数，则返回正确的值。为什么返回非空列表？

我在shell中运行了以下代码。最后一个调用应返回一个空列表，但它会返回之前调用中的返回值。

>>> from solvers.algorithms.dancing_links import solve
>>> solve({}, {})
[]
>>> solve(ctor, rtoc)
['B', 'D', 'F']
>>> solve(ctor, rtoc)
['B', 'D', 'F']
>>> solve({}, {})
['B', 'D', 'F']

这是我的python代码。

# Solves the Sudoku using dancing links algorithm using implementation from:
# https://www.cs.mcgill.ca/~aassaf9/python/algorithm_x.html


def solve(columns_to_rows, rows_to_columns, solution=[]):
    """
    This function solves the exact cover problem, i.e., given a matrix A in which each element is 0 or 1.
    Is there a subset of rows which have exactly one 1 in each column?
    The exact cover problem can also be stated as following,
    "Given a set X and a set of subsets of X called Y, is there a subset of Y such that all of its elements
    are disjoint and union of its elements is X."
    :param columns_to_rows: It is a dictionary where value of key `i` is the set of columns in which ith row is 1.
    :param rows_to_columns: It is a dictionary where value of key `i` is the set of rows in which ith column is 1.
    :param solution: Currently selected rows
    :return:
    """
    if not columns_to_rows:
        return list(solution)
    else:
        # Column with minimum 1's in it.
        selected_column = min(columns_to_rows, key=lambda col: columns_to_rows[col])
        # For each row which has a 1 in the `selected_column`.
        for selected_row in columns_to_rows[selected_column]:
            solution.append(selected_row)
            # Select and remove all columns which have 1's in the `selected_row`
            # Also remove all the rows which have 1's in the same column as the `selected_row`
            removed_columns = select_and_remove(columns_to_rows, rows_to_columns, selected_row)
            tmp = solve(columns_to_rows, rows_to_columns, solution)
            if tmp is not None:
                return tmp
            deselect_and_insert(columns_to_rows, rows_to_columns, selected_row, removed_columns)
            solution.pop()

编辑1：我添加了一条打印语句以打印“ solution”变量，然后发现了这一点。

>>> from solvers.algorithms.dancing_links import solve>>> solve({}, {})
[]
[]
>>> solve({}, {})
[]
[]
>>> solve(ctor, rtoc)
[]
['A']
['B']
['B', 'D']
['B', 'D', 'F']
['B', 'D', 'F']
>>> solve({}, {})
['B', 'D', 'F']
['B', 'D', 'F']
>>> 

因此，当我第一次调用该函数时，它返回空列表，解决方案也为空。然后，当我第二次使用实际参数调用该函数时，该函数将按预期工作并进入递归调用。但是第三次​​将解决方案变量设置为先前的值，但应为空。为什么会这样呢？以及如何解决？

Answer 1

不要将solution=[]传递给函数，它会将其保存为引用并可能影响您的结果，每次函数运行时都传递一个新列表

更多信息：here