问题

考虑以下代码:

对于任何5维索引，我都需要获取相关系数，而无需了解或计算 i.例如，给定[2,4,8,16,32]，我需要获得3.0而无需计算i.

For any 5 dimensional index I need to obtain the associated coefficient, without knowing or calculating i. For instance, given [2,4,8,16,32] I need to obtain 3.0 without computing i.

是否存在一个合理的解决方案，也许使用稀疏矩阵，它对于n_indices适用于100的数量级(尽管n_coeffs仍为1000的数量级)?

Is there a reasonable solution, perhaps using sparse matrices, that would work for n_indices in the order of 100 (though n_coeffs still in the order of 1000)?

不良解决方案

一种解决方案是定义一个5维数组，如

One solution would be to define a 5 dimensional array as in

然后，要获取与[2,4,8,16,32]相关的系数，请调用

then, to get the coefficient associated with [2,4,8,16,32], call

但是，除了非常浪费内存外，考虑到数组的7个维的限制，该解决方案也不允许将n_indices设置为大于7的数字.

However, besides being very wasteful of memory, this solution would not allow n_indices to be set to a number higher than 7 given the limit of 7 dimensions to an array.

OBS:这个问题是这个问题的衍生产品.我试图更精确地问这个问题，第一次尝试失败了，这一努力极大地受益于@Rodrigo_Rodrigues的回答.

OBS: This question is a spin-off of this one. I have tried to ask the question more precisely having failed in the first attempt, an effort that greatly benefited from the answer of @Rodrigo_Rodrigues.

实际代码

以防万一，这里是我要解决的实际问题的代码.它是一种用于逼近函数的自适应稀疏网格方法.主要目标是尽可能快地进行插值:

In case it helps here is the code for the actual problem I am trying to solve. It is an adaptive sparse grid method for approximating a function. The main goal is to make the interpolation at the and as fast as possible:

对于任何5维索引，我需要获取相关的 系数，不知道或不计算 i.例如，给定 [2,4,8,16,32]我需要在不计算i的情况下获得3.0.

For any 5 dimensional index I need to obtain the associated coefficient, without knowing or calculating i. For instance, given [2,4,8,16,32] I need to obtain 3.0 without computing i.

这就是您的使用方式:

我必须承认，我不愿意发布此代码，因为即使回答了您的问题，我还是老实认为这是您真正想要/需要的，但您没有提供足够的信息让我知道您真正想要/需要什么.

I must confess I was reluctant to post this code because, even though this answers your question, I honestly think this is not what you really want/need, but you dind't provide enough information for me to know what you really do want/need.

编辑(在OP中提供实际代码之后):

Edit (After actual code from OP):

如果我正确解密了您的代码(并考虑了您在上一个问题中所说的内容)，则说明:

If I decrypted your code correctly (and considering what you said in your previous question), you are declaring:

(其中L_max = 9，因此size(coeff_grid) = 21233664 =〜160MB)，然后使用以下命令填充它:

(where L_max = 9, so size(coeff_grid) = 21233664 =~160MB) and then populating it with:

(其中count大约为1000，即其元素的0.005％)，因此您可以通过数组符号用其4个索引获取值.

(where count is of the order of 1000, i.e. 0.005% of its elements), so this way you can fetch the values by its 4 indices with the array notation.

请不要这样做.在这种情况下，您也不需要稀疏矩阵.您提出的新方法要好得多:将索引存储在较小数组的每一行中，然后通过这些索引在其自己的数组中的对应位置来提取系数数组.这样会更快(避免大分配)，并且内存效率更高.

Please, don't do that. You don't need a sparse matrix in this case either. The new approach you proposed is much better: storing the indices in each row of an smaller array, and fetching on the array of coefficients by the corresponding location of those indices in its own array. This is way faster (avoiding the large allocation) and much more memory-efficient.

PS:坚持使用Fortran 90是强制性的吗?它是标准的非常旧的版本，可能是您使用的编译器实现了较新的版本.您可以使用内部move_alloc(对于较少的数组副本)，内部模块iso_fortran_env的种类常量(用于可移植性)，[], >, <, <=,...表示法(用于可读性)来大大提高代码质量.

PS: Is it mandatory for you to stick to Fortran 90? Its a very old version of the standard and chances are that the compiler you're using implements a more recent version. You could improve the quality of your code a lot with the intrinsic move_alloc (for less array copies), the kind constants from the intrinsic module iso_fortran_env (for portability), the [], >, <, <=,... notation (for readability)...