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

Yuna Luzi

我正在尝试在列表的第i个列表上添加/替换一个i元素，但是，仅保存了最后一次迭代。很高兴了解为什么会发生这种情况以及解决方法。我对看起来像这样的输出感兴趣：[[[0], [], [], []], [[1], [], [], []], [[2], [], [], []], [[3], [], [], []]]但我却得到了：[[[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], [], [], []]]

iz_

当您执行以下操作时：

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

实际上，您并不是每次乘法都创建新列表。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到：

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，请使用列表推导每次正确创建一个新列表：

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

在您的代码中（替换len(a) + 1为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], [], [], []]]

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