跨步窗户

Striding windows

本文关键字：窗户      更新时间：2023-10-16

假设我有一个向量：

x = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14]

我需要做的是将此向量拆分为具有blocksize的块大小，并带有overlap

blocksize = 4

overlap = 2

结果将是一个大小为4包含6值的 2D 矢量。

x[0] = [1, 3, 5, 7, 9, 11]

x[1] = [ 2 4 6 8 10 12]

....

我尝试使用以下功能来实现这一点：

std::vector<std::vector<double> > stride_windows(std::vector<double> &data, std::size_t 
NFFT, std::size_t overlap)
{
std::vector<std::vector<double> > blocks(NFFT); 
for(unsigned i=0; (i < data.size()); i++)
{
blocks[i].resize(NFFT+overlap);
for(unsigned j=0; (j < blocks[i].size()); j++)
{
std::cout << data[i*overlap+j] << std::endl;
}
}
}

这是错误的，而且，段。

std::vector<std::vector<double> > frame(std::vector<double> &signal, int N, int M)
{
unsigned int n = signal.size();
unsigned int num_blocks = n / N;

unsigned int maxblockstart = n - N;
unsigned int lastblockstart = maxblockstart - (maxblockstart % M);
unsigned int numbblocks = (lastblockstart)/M + 1;
std::vector<std::vector<double> > blocked(numbblocks);
for(unsigned i=0; (i < numbblocks); i++)
{
blocked[i].resize(N);
for(int j=0; (j < N); j++)
{
blocked[i][j] = signal[i*M+j];
}
}
return blocked;
}

我写了这个函数，认为它做到了上述操作，但是，它只会存储：

X[0] = 1, 2, 3, 4

x[1] = 3, 4, 5, 6

.....

谁能解释一下我将如何修改上述功能以允许overlap进行跳过？

此函数类似于以下内容： 滚动窗口

编辑：

我有以下向量：

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14

我想将此向量拆分为子块(从而创建一个 2D 向量)，参数重叠overlap因此在这种情况下，参数将是：size=4overlap=2，这将创建以下 2D 向量：

`block0 = [ 1  3  5  7  9 11]
block1 = [ 2  4  6  8 10 12]
block2 = [ 3  5  7  9 11 13]
block3 = [ 4  6  8 10 12 14]`

因此，基本上已经创建了 4 个块，每个块包含一个值，其中元素被overlap跳过

编辑2：

这是我需要到达的地方：

overlap的值将在向量内的放置方面与x的结果重叠：

block1 = [1, 3, 5, 7, 9, 11]

实际向量块的注意事项：

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14
Value: 1 -> This is pushed into block "1"
Value 2 -> This is not pushed into block "1" (overlap is skip 2 places in the vector)
Value 3 -> This is pushed into block "1" 
value 4 -> This is not pushed into block "1" (overlap is skip to places in the vector)
value 5 -> This is pushed into block "1"
value 6 -> "This is not pushed into block "1" (overlap is skip 2 places in the vector)
value 7 -> "This value is pushed into block "1"
value 8 -> "This is not pushed into block "1" (overlap is skip 2 places in the vector)"
value 9 -> "This value is pushed into block "1"
value 10 -> This value is not pushed into block "1" (overlap is skip 2 places in the 
vector)
value 11 -> This value is pushed into block "1"

区块 2

Overlap = 2; 
value 2 - > Pushed back into block "2" 
value 4 -> Pushed back into  block "2"
value 6, 8, 10 etc..

因此，每次向量中的位置都会被"重叠"跳过，在这种情况下，它是 2.

这是预期的输出：

[[ 1  3  5  7  9 11]
[ 2  4  6  8 10 12]
[ 3  5  7  9 11 13]
[ 4  6  8 10 12 14]]

如果我理解正确的话，你们已经很接近了。您需要类似以下内容的内容。我使用int因为坦率地说，它比double=P 更容易输入

#include <iostream>
#include <algorithm>
#include <vector>
#include <limits>
#include <iterator>
std::vector<std::vector<int>>
split(const std::vector<int>& data, size_t blocksize, size_t overlap)
{
// compute maximum block size
std::vector<std::vector<int>> res;
size_t minlen = (data.size() - blocksize)/overlap + 1;
auto start = data.begin();
for (size_t i=0; i<blocksize; ++i)
{
res.emplace_back(std::vector<int>());
std::vector<int>& block = res.back();
auto it = start++;
for (size_t j=0; j<minlen; ++j)
{
block.push_back(*it);
std::advance(it,overlap);
}
}
return res;
}
int main()
{
std::vector<int> data { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14 };
for (size_t i=2; i<6; ++i)
{
for (size_t j=2; j<6; ++j)
{
std::vector<std::vector<int>> blocks = split(data, i, j);
std::cout << "Blocksize = " << i << ", Overlap = " << j << std::endl;
for (auto const& obj : blocks)
{
std::copy(obj.begin(), obj.end(), std::ostream_iterator<int>(std::cout, " "));
std::cout << std::endl;
}
std::cout << std::endl;
}
}
return 0;
}

输出

Blocksize = 2, Overlap = 2
1 3 5 7 9 11 13 
2 4 6 8 10 12 14 
Blocksize = 2, Overlap = 3
1 4 7 10 13 
2 5 8 11 14 
Blocksize = 2, Overlap = 4
1 5 9 13 
2 6 10 14 
Blocksize = 2, Overlap = 5
1 6 11 
2 7 12 
Blocksize = 3, Overlap = 2
1 3 5 7 9 11 
2 4 6 8 10 12 
3 5 7 9 11 13 
Blocksize = 3, Overlap = 3
1 4 7 10 
2 5 8 11 
3 6 9 12 
Blocksize = 3, Overlap = 4
1 5 9 
2 6 10 
3 7 11 
Blocksize = 3, Overlap = 5
1 6 11 
2 7 12 
3 8 13 
Blocksize = 4, Overlap = 2
1 3 5 7 9 11 
2 4 6 8 10 12 
3 5 7 9 11 13 
4 6 8 10 12 14 
Blocksize = 4, Overlap = 3
1 4 7 10 
2 5 8 11 
3 6 9 12 
4 7 10 13 
Blocksize = 4, Overlap = 4
1 5 9 
2 6 10 
3 7 11 
4 8 12 
Blocksize = 4, Overlap = 5
1 6 11 
2 7 12 
3 8 13 
4 9 14 
Blocksize = 5, Overlap = 2
1 3 5 7 9 
2 4 6 8 10 
3 5 7 9 11 
4 6 8 10 12 
5 7 9 11 13 
Blocksize = 5, Overlap = 3
1 4 7 10 
2 5 8 11 
3 6 9 12 
4 7 10 13 
5 8 11 14 
Blocksize = 5, Overlap = 4
1 5 9 
2 6 10 
3 7 11 
4 8 12 
5 9 13 
Blocksize = 5, Overlap = 5
1 6 
2 7 
3 8 
4 9 
5 10
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