返回所有行，或只返回最后一行 [英] Returning all rows, or just the last one

问题描述

我正在写一堆（约十几个）算法，按照以下模式迭代处理一个向量：

I'm writing up a bunch (~ a dozen) algorithms that iteratively process a vector in the following pattern:

ArrayXd prev;
ArrayXd curr;

prev = some_initial_vector();
for (i = 0; i < N; ++i) {
    // do lots of stuff, many lines to compute curr
    // use lots of algorithm specific variables initialised above
    ...
    prev = curr;
}
return curr;

我想有一种方法来返回 curr 以及 ArrayXXd 中的行。

I would like to have a way of returning the entire history of the values of curr as well, as rows in an ArrayXXd.

这是通过写两个类，它们显示一个 curr 句柄，一个作为 ArrayXd& 另一个作为 Block< ArrayXXd，1，-1> ，但失败为无法重新分配阻止 。

I have tried solving this by writing two classes that exhibit a curr handle, one as an ArrayXd & the other as a Block<ArrayXXd, 1, -1>, but that failed as it's not possible to reassign Blocks.

是解决这个问题的好办法吗？也许我可以在 std :: vector中存储 Block s或 ArrayXd ，然后将其转换为 ArrayXXd 。

What is a good way of solving this problem? Maybe I could store the Blocks or the ArrayXds themselves in a std::vector and then convert that to an ArrayXXd at the end.

修改：添加了示例输入和输出

Added sample input, output

struct NotAccumulator {
    typedef ArrayXd return_type;
    ArrayXd curr;
    ArrayXd prev;
    void record () {}
    return_type result() {
        return prev;
    }
};

struct RowAccumulator {
    typedef ArrayXXd return_type;
    ArrayXd curr;
    ArrayXd prev;
    RowAccumulator(const uint N) {
        history.reserve(N);
    }

    void record () {
        history.push_back(curr);
    }
    return_type result () {
        uint rows = history.size();
        uint cols = history[0].size();
        ArrayXXd result_matrix (rows, cols);
        for(uint i = 0; i < rows; ++i) {
            result_matrix.row(i) = Map<ArrayXd> (history[i].data(), cols);
        }
        return result_matrix;
    }

private:
    std::vector<ArrayXd> history;
};

template <typename Accumulator>
typename Accumulator::return_type add_one(const ArrayXd & start, const uint how_many_times, Accumulator & u) {
    u.prev = start;
    for (uint i = 0; i < how_many_times; ++i) {
        u.curr = 1 + u.prev;
        u.record();
        u.prev = u.curr;
    }
    return u.result();
}

ArrayXd start (3);
start << 1, 0, -1;
NotAccumulator notAccumulator;
RowAccumulator rowAccumulator (5);
cout << add_one(start, 5, notAccumulator) << endl;
// outputs 6 5 4
cout << add_one(start, 5, rowAccumulator) << endl;
// outputs 2 1 0\n 3 2 1\n ... 6 5 4

推荐答案

如果你想尽可能避免数据复制，可以分配 N -row ArrayXXd ，并避免使用 return 。

If you want to avoid the data copy as much as possible, you could allocate the N-row ArrayXXd in advance and avoid using return.

c $ c> prev 和 curr 也是不必要的。以下代码还假设预先已知迭代次数 N 。

Buffers like prev and curr are unnecessary too. The following code also assumes the number of iterations N is known in advance.

void GenerateHistory(int N, ArrayXXd* result) {
    result.row(0) = gen_row0_without_prev();
    for(int curr=1; curr<N; curr++) {
        result.row(curr) = gen_row_with_prev_row(result.row(curr-1));
    }
}

您可以使用

int N = 100;
int m = 20;
ArrayXXd history(N, m);
GenerateHistory(N, &history);

EDIT

在旁边。

EDIT

Let's put the copying issue aside. Here's an updated code that should fulfill the requirement according to your code sample.

ArrayXXd add_one(const ArrayXd& start, int num_iterations, bool acc) {
    if (acc) {
        ArrayXXd result = start;
        for(int i=0; i<num_iterations; i++) {
            result = result + 1;
        }
        return result;
    } else {
        ArrayXXd result(num_iterations, start.cols());
        result.row(0)=start+1;
        for(int i=1; i<num_iterations; i++) {
            result.row(i)=result.row(i-1);
        }
        return result;
    }
}

ArrayXd start (3);
start << 1, 0, -1;
bool acc = true;
cout << add_one(start, 5, !acc) << endl;
// outputs 6 5 4
cout << add_one(start, 5, acc) << endl;
// outputs 2 1 0\n 3 2 1\n ... 6 5 4

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