由于内存限制，将向量循环转换为 for 循环

Question

我有一段 Octave / Matlab 代码，我从 Andy 和团队那里得到了很多帮助。我现在运行遇到的问题是没有足够的内存来发出持续时间更长的信号。

我的解决方案是：

1) 将向量循环转换为for循环。 （这里有问题）
2) 让 for 循环将循环的每一段导出为 wav 文件，而不是执行矢量代码所做的附加它。 （这里有问题）
3) 使用 sox 连接每个 wave 文件段。

大多数在线示例都是从 for 循环到矢量化循环，而不是相反，有什么想法吗？我也愿意接受其他建议来解决我的记忆问题。注意：我在 rasberry pi 2 上使用 1 gig ram，它工作正常，速度非常快，我只是想获得持续时间更长的信号，并导出每个信号段应该允许这样做。

我正在使用与 Matlab 兼容的 Octave。

请参阅下面的工作矢量化代码：它基于在这里找到的 Paul Nasca 拉伸算法 http://www.paulnasca.com/algorithms-created-by-me#TOC-PaulStretch-extreme-sound-stretching-algorithm

urlwrite('http://www.onewithall.net/rttmpfiles/3sec8000.wav','3sec8000.wav'); 
inputfn='3sec8000.wav' %change this to test another file
[d, fs, bps]=wavread(inputfn);
inputlen=rows  (d)/fs;

printf ("Original duration of file in seconds = %.2f s\n", rows (d)/fs);

dur=60; %duration / length you want the file to be in seconds
stretch = dur/rows(d)*fs; %how much I need to stretch the file to get it to be the duration I want
windowsize = round (0.25 * fs);

step = round ((windowsize/2)/stretch);

## original window
fwin = @(x) (1-x.^2).^1.25;
win = fwin (linspace (-1, 1, windowsize));

#win = hanning (windowsize)';

## build index
ind = (bsxfun (@plus, 1:windowsize, (0:step:(rows(d)-windowsize))'))';
cols_ind = columns(ind);

## Only use left channel
left_seg = d(:,1)(ind);
clear d ind;

## Apply window
left_seg = bsxfun (@times, left_seg, win');

## FFT
fft_left_seg = fft (left_seg);
clear left_seg

#keyboard

## overwrite phases with random phases
fft_rand_phase_left = fft_left_seg.*exp(i*2*pi*rand(size(fft_left_seg)));
clear fft_left_seg;

ifft_left = ifft (fft_rand_phase_left);
clear fft_rand_phase_left;

## window again
ifft_left = bsxfun (@times, real(ifft_left), win');

## restore the windowed segments with half windowsize shift
restore_step = floor(windowsize/2);
ind2 = (bsxfun (@plus, 1:windowsize, (0:restore_step:(restore_step*(cols_ind-1)))'))';
left_stretched = sparse (ind2(:), repmat(1:columns (ind2), rows(ind2), 1)(:), real(ifft_left(:)), ind2(end, end), cols_ind);
clear ind2 ifft_left win;

left_stretched = full (sum (left_stretched, 2));

## normalize
left_stretched = 0.8 * left_stretched./max(left_stretched);
printf ("converted %s =%.2f(s) file to stretched.wav = %.2f(s)\n", inputfn, inputlen, rows (left_stretched)/fs);
wavwrite (left_stretched, fs, bps, "streched.wav");

我尝试通过在关键点放置 display('line') 来追踪问题。它看起来像行

left_stretched = sparse (ind2(:), repmat(1:columns (ind2), rows(ind2), 1)(:), real(ifft_left(:)), ind2(end, end), cols_ind);

只有当我运行 out ram 时，上面的行似乎才有问题。 它说错误下标索引必须是正整数或逻辑数。请注意，只有当我运行试图通过设置 dur=60 来使用长持续时间时内存不足时才会发生这种情况* 1800。如果我设置 dur=60*10 一切正常。

Answer 1

你还记得我吗？我是您发布的初始代码的作者。下面的代码作为 for 循环。我已经用 800s 的输出 len 测试了它。

## based on http://hypermammut.sourceforge.net/paulstretch/
## https://github.com/paulnasca/paulstretch_python/blob/master/paulstretch_steps.png

more off
inputfn = "original.wav"
[d, fs, bps] = wavread (inputfn);
inputlen=rows  (d)/fs;

printf ("Original  duration of file in seconds = %.2f s\n", rows (d)/fs);

target_duration = 60; # in seconds
stretch = target_duration/inputlen;
# 1/4 s window len
windowsize = round (0.25 * fs);

# stepwidth between windows
step = round ((windowsize/2)/stretch);
numsteps = floor((rows(d)-windowsize)/step);

## restore the windowed segments with half windowsize shift
restore_step = floor (windowsize / 2);

## stetched duration
stretched_len = (numsteps*restore_step+windowsize)/fs;
printf ("Stretched duration of file in seconds = %.2f s\n", stretched_len);
stretched = zeros (numsteps*restore_step+windowsize, 1);
if (!exist ("out", "dir"))
  mkdir ("out");
endif

## Matrix which holds the freq of the maximum amplitude and the max. amplitude
chunks_stats = zeros (numsteps, 2);

## original window
fwin = @(x) (1-x.^2).^1.25;
win = fwin (linspace (-1, 1, windowsize));

## loop over all windows
for k = 1:numsteps
  if (! mod(k, 50))
    printf ("Calculating chunk %i of %i...\n", k, numsteps);
    fflush (stdout);
  endif

  ## Only use left channel
  s_ind = (k - 1) * step + 1;
  e_ind = s_ind + windowsize - 1;
  tmp = win' .* d(s_ind:e_ind, 1);

  ## FFT, overwrite phases with random phases and IFFT
  tmp = fft(tmp);
  [m, ind] = max (abs(tmp(1:numel(tmp)/2)));
  # Frequency in Hz
  chunks_stats(k, 1) = (ind-1)/windowsize*fs;
  # max Amplitude
  chunks_stats(k, 2) = m;
  printf ("Freq =  %.2f Hz, max = %.2f\n", chunks_stats(k, :));
  tmp = ifft (tmp .* exp(i*2*pi*rand(size(tmp))));

  ## window again
  tmp = win' .* real (tmp);
  fn = sprintf ("out/out_%04i.wav", k);
  wavwrite (tmp, fs, bps, fn);
  s_ind = (k - 1) * restore_step + 1;
  e_ind = s_ind + windowsize - 1;
  stretched (s_ind:e_ind) += tmp;
endfor

## normalize
stretched = 0.8 * stretched./max(stretched);
wavwrite (stretched, fs, bps, "stretched.wav");

如果你想编写多个 wav 以稍后将它们连接起来，这会有点困难，因为重叠 windows。但我认为这段代码运行在 BeagleBoneBlack 上没问题。

编辑：添加了保存块以分隔文件，并将每个块的此信号的最大幅度和频率收集到 chunk_stats。

由于内存限制，将向量循环转换为 for 循环

converting a vector loop to a for loop because of memory limitations

matlab

octave

sox