我正在做项目欧拉问题15 。 我使用了正确的算法，但它似乎没有用。 这是我的代码：
f = [[0] * 21] * 21 # init the list for i in range(21): f[0][i] = 1 f[i][0] = 1 for i in range(21): for j in range(21): f[i][j] = f[i-1][j] + f[i][j-1] print f[20][20]当我完成列表的初始化时，我将其打印出来。 我期望它像[[1, 1, 1...], [1, 0, 0...]...] ，但它变成[[1, 1, 1...], [1, 1, 1...]...]我无法理解为什么。
我曾经使用类C语言，我认为Python中的列表有点像C中的数组，所以我以相同的方式使用它们。
I am doing Project Euler Problem 15. I used the right algorithm, but it didn't seem to work. Here's my code:
f = [[0] * 21] * 21 # init the list for i in range(21): f[0][i] = 1 f[i][0] = 1 for i in range(21): for j in range(21): f[i][j] = f[i-1][j] + f[i][j-1] print f[20][20]when I finished initializing the list, I printed it. I expected it like [[1, 1, 1...], [1, 0, 0...]...], but it turned [[1, 1, 1...], [1, 1, 1...]...] and I can't figure why.
I used to use C-like language and I thought the list in Python is kind of like the array in C, so I used them in the same way.
最满意答案
乘以列表时，您不是创建单独的列表，而是创建对同一列表的多个引用 。
相反，这样做：
f = [[0 for _ in range(21)] for _ in range(21)]使用id()函数时可以看到区别：
>>> f = [[0]*21]*21 >>> for nested in f[:3]: ... print id(nested) ... 4523317152 4523317152 4523317152 >>> f = [[0 for _ in range(21)] for _ in range(21)] >>> for nested in f[:3]: ... print id(nested) ... 4523317512 4523317440 4523317656单独的对象具有单独的内存ID值，而您的列表对每个id()调用具有相同的结果，显示它们都是相同的列表 。
When multiplying a list, you are not creating separate lists but rather create multiple references to the same list.
Instead, do this:
f = [[0 for _ in range(21)] for _ in range(21)]You can see the difference when using the id() function:
>>> f = [[0]*21]*21 >>> for nested in f[:3]: ... print id(nested) ... 4523317152 4523317152 4523317152 >>> f = [[0 for _ in range(21)] for _ in range(21)] >>> for nested in f[:3]: ... print id(nested) ... 4523317512 4523317440 4523317656Separate objects have separate memory id values, while your lists have the same result for each id() call, showing they are all the same list.
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