python - 如何在不保存迭代的最后一个值的情况下替换 Python 嵌套列表中的值

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我正在尝试在列表的第 i 个列表中添加/替换 i 元素，但是，仅保存最后一次迭代。了解为什么会发生这种情况以及这将是什么解决方法会很棒。我对看起来像这样的输出感兴趣: [[[0], [], [], []], [[1], [], [], []], [[2], [], [], []], [[3], [], [], []]] ，但我得到了这个: [[[3], [], [], []], [[3], [], [], []], [[3], [], [], []], [[3], [], [], []]]

matrix = [ [ [] * 4] * 4] * 4
print('pre_matrix=',matrix)
for m in range(len(a) + 1):
    matrix[m][0] = [m]
    print('m={}, matrix[m]={}'.format(m, matrix[m]))
print('post_matrix=', matrix)

# This is what the response looks like:
# pre_matrix= [[[], [], [], []], 
# [[], [], [], []], 
# [[], [], [], []], 
# [[], [], [], []]]
#
# m=0, matrix[m]=[[0], [], [], []]
# m=1, matrix[m]=[[1], [], [], []]
# m=2, matrix[m]=[[2], [], [], []]
# m=3, matrix[m]=[[3], [], [], []]
#
# post_matrix= [[[3], [], [], []], 
# [[3], [], [], []], 
# [[3], [], [], []], 
# [[3], [], [], []]]

最佳答案

当您执行以下操作时:

matrix = [ [ [] * 4] * 4] * 4

您实际上并不是在每次乘法时创建新列表。相反，您正在创建对同一基础列表的许多引用。当我们简化示例时可以看出这一点:

matrix = [[0, 0, 0]] * 3 # <- Wrong!
matrix[0][0] = 1
print(matrix)
# [[1, 0, 0], [1, 0, 0], [1, 0, 0]]

要解决此问题，请每次使用列表推导式正确创建一个新列表:

matrix = [[[] for _ in range(4)] for _ in range(4)]

在您的代码中(用 len(a) + 1 替换 4 ):

matrix = [[[] for _ in range(4)] for _ in range(4)]
print('pre_matrix=',matrix)
for m in range(4):
    matrix[m][0] = [m]
    print('m={}, matrix[m]={}'.format(m, matrix[m]))
print('post_matrix=', matrix)

输出:

pre_matrix= [[[], [], [], []], [[], [], [], []], [[], [], [], []], [[], [], [], []]]
m=0, matrix[m]=[[0], [], [], []]
m=1, matrix[m]=[[1], [], [], []]
m=2, matrix[m]=[[2], [], [], []]
m=3, matrix[m]=[[3], [], [], []]
post_matrix= [[[0], [], [], []], [[1], [], [], []], [[2], [], [], []], [[3], [], [], []]]

关于python - 如何在不保存迭代的最后一个值的情况下替换 Python 嵌套列表中的值，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/56876991/

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Answer 1

当您执行以下操作时:

matrix = [ [ [] * 4] * 4] * 4

您实际上并不是在每次乘法时创建新列表。相反，您正在创建对同一基础列表的许多引用。当我们简化示例时可以看出这一点:

matrix = [[0, 0, 0]] * 3 # <- Wrong!
matrix[0][0] = 1
print(matrix)
# [[1, 0, 0], [1, 0, 0], [1, 0, 0]]

要解决此问题，请每次使用列表推导式正确创建一个新列表:

matrix = [[[] for _ in range(4)] for _ in range(4)]

在您的代码中(用 len(a) + 1 替换 4 ):

matrix = [[[] for _ in range(4)] for _ in range(4)]
print('pre_matrix=',matrix)
for m in range(4):
    matrix[m][0] = [m]
    print('m={}, matrix[m]={}'.format(m, matrix[m]))
print('post_matrix=', matrix)

输出:

pre_matrix= [[[], [], [], []], [[], [], [], []], [[], [], [], []], [[], [], [], []]]
m=0, matrix[m]=[[0], [], [], []]
m=1, matrix[m]=[[1], [], [], []]
m=2, matrix[m]=[[2], [], [], []]
m=3, matrix[m]=[[3], [], [], []]
post_matrix= [[[0], [], [], []], [[1], [], [], []], [[2], [], [], []], [[3], [], [], []]]