Note: When generating a list, it is better to use
[0 for _ in range(n)]the method instead of[0] * nthe method for the following reasons.
Suppose we want to create a n = 2list of length a, there are two general methods: a = [0] * nand a = [0 for _ in range(n)], there is no difference between the two methods when it is one-dimensional. But if we want to create a list, each element in the list is m = 3a list of length The results of the two methods are different:
n, m = 2, 3
a = [[0] * n] * m # 用第一种方法生成的
# a = [[0, 0], [0, 0], [0, 0]]
b = [[0 for _ in range(n)] for _ in range(m)] # 用第二种方法生成的
# b = [[0, 0], [0, 0], [0, 0]]
a[0][0] = 1 # 令 a 列表的第一个元素为 1
# a = [[1, 0], [1, 0], [1, 0]]
b[0][0] = 1
# b = [[1, 0], [0, 0], [0, 0]]
Our expected result is like bthis, that is, the first element of the first list is equal to 1, but the list asets the first element of each list to 1.
The reason for this is that the first generation method is =generated in a way similar to , that is, assuming that both a, bare lists, we make a = b that if one of them changes, the other will change accordingly, for example :
a = [0, 0, 0]
b = a
b[0] = 1
# a = [1, 0, 0]
# b = [1, 0, 0]
To avoid this situation, it should be written like this b = list(a).
Going back to the original question, if you use the first method a = [[0] * n] * m, then the addresses stored in the lists ain mare the same, then if you change one of the lists, the other lists will change accordingly, and the above situation will appear.
c = [[0] * n for _ in range(m)] # 没问题
d = [[0 for _ in range(n)]] * m # 有问题
Therefore, it will be better to use all the lists to be generated in the future [0 for _ in range(n)].