我正在尝试生成矩阵 A == [针对范围(0,n)中的i,j的f(i,j)] .矩阵是对称(f(i,j)== f(j,i)),对角线元素为零(f(i,i)== 0).
I am trying to generate a matrix A == [f(i,j) for i,j in range(0,n)]. The matrix is symmetric (f(i,j) == f(j,i)) and the diagonal elements are zero (f(i,i) == 0).
问题:是否有可能产生一个列表理解,该列表生成会生成此矩阵,但只为 i<调用函数 f(i,j).j ?
Question: Is it possible to produce a list comprehension that generates this matrix but that only calls the function f(i,j) for i < j?
不是我的问题:以其他方式生成对称矩阵.
Not my question: Generate the symmetric matrix in some other way.
可能的解决方案:通过辅助功能 g 调用 f ,该辅助功能保存了 f 的值存储在附加存储中或返回存储的值.
Possible solution: Make the call to f through an auxiliary function g that saves the value of f in an additional storage or returns the stored value.
是否有可能解决避免额外存储的问题?我不确定是否有其他障碍意味着是否需要从自身引用列表理解(我读到的内容在Python中不存在),希望我还缺少其他技巧.
Would it be possible to solve avoiding the additional storage? I am not sure if this additional handicap implies that a reference to the list comprehension from itself (which I read doesn't exist in Python) is necessary, hopefully I am missing some other trick.
推荐答案如果您希望函数仅在i<时执行.j,您可以使用lru_cache(@lru_cache)将函数的结果保存在缓存中,并在i> j时直接使用它而无需重新计算它.
If you want the function to execute only when i < j, you can use lru_cache (@lru_cache) to save the results of the function in cache, and use it directly without re-calculating it when i > j.
from typing import Any from functools import lru_cache @lru_cache() def f(i: int, j: int) -> Any: return i * (-j) n = 5 m = [f(i, j) if i < j else f(j, i) if i > j else 0 for i in range(n) for j in range(n)] for i in range(0, n * n, n): print(m[i: i + n])结果
[0, 0, 0, 0, 0] [0, 0, -2, -3, -4] [0, -2, 0, -6, -8] [0, -3, -6, 0, -12] [0, -4, -8, -12, 0]更多推荐
Python:使用列表推导生成对称数组
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