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blocksize的块大小overlap
blocksize = 4
overlap = 2
结果将是包含4值的6大小的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实现此功能：
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，并且，细分。
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，认为它完成了上述操作，然而，它只会存储：
X[0] = 1, 2, 3, 4
x[1] = 3, 4, 5, 6
.....
有人可以解释我如何修改上述功能以允许overlap跳过？
此功能类似于：Rolling window
编辑：
我有以下矢量：
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2D向量），并且参数overlap重叠，因此在这种情况下，参数将为：size=4 overlap = 2，然后会创建以下2D矢量：
`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]`
基本上，已经创建了4个块，每个块都包含一个值，overlap
编辑2：
这是我需要去的地方：
overlap的值会与向量中的展示位置x的结果重叠：
block1 = [1, 3, 5, 7, 9, 11]
来自实际矢量块的注意事项：
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"
第2块
Overlap = 2;
value 2 - > Pushed back into block "2"
value 4 -> Pushed back into block "2"
value 6, 8, 10 etc..
所以每次，在这种情况下，矢量的位置被“重叠”跳过，它的值是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]]
答案 0 :(得分：1)
如果我理解正确，那你就非常接近了。您需要以下内容。我使用了int，因为坦率地说它比double = P
#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;
}
<强>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