Question

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我正在努力 1）将数组拆分成多个部分 2）将每个部分导出为单独的波形文件 3）重新导入wav文件并将它们连接在一起以确保拆分的数组数据没有改变。

我可以在测试时完成所有这些步骤错误我希望它应该像2.232e-15那样几乎没有错误，但我得到了意想不到的大数字错误。
MAE = 0.046232 MXE = 0.14522 RMSE = 0.064035

如何解决此问题，以便错误率降低？我认为数组被分割成部分并且正在复制单元格数据确切但看起来可能不是这样，我该如何解决这个问题？

以下代码：

%split_file
%create sine wave signal
clear all, clc
tic
fs = 44100;                   % Sampling frequency
t=linspace(0,1,fs);
freq=340;
ya = sin(2*pi*freq*t); %+ 1*sin(2*pi*250*t); 
[size_r,size_c]=size(ya');

jj=[];
kk=0;
wavefilesplit=[];

%need to delete diretory and recreate it to clean out files
fileprepathStr='/home/rat/Documents/octave/pre/'; %
rmdir(fileprepathStr,'s');
fprintf('\n-1- deleting %s directory %2.4f sec',fileprepathStr,toc);
mkdir(fileprepathStr);
fprintf('\n-1- creating %s directory %2.4f sec',fileprepathStr,toc);


jj=1;
for ii=1:fs/4:size_r,  %build array of desired ranges or fs/2
    jj(end+1,:)=ii-1;  %minus 1 to get correct array index in cell
end;

[size_rjj,size_cjj]=size(jj); %used to get size of jj array
jj(end+1,:)=size_r-(size_rjj-2); %adds the end of the sound file to the end of the jj array minus the amount of files joined

jj(2,:)=[]; %deletes second cell with zero and shifts the cells up

for ii=1:1:size_rjj-1,kk=kk+1;

    wavefilesplit=ya(jj(kk):jj(kk+1));
    wavefn=strcat('wavefn_',num2str(kk,'%04d')); %build filename dynamiclly with 4 leading zeros
    wavwrite([wavefilesplit],fs,16,strcat('/home/rat/Documents/octave/pre/',wavefn,'.wav')); 
    fprintf('\n-1- wavwrite split %s.wav %3.0f of %3.0f %6.3fsec  %6.3fmins\n',wavefn,kk,size_rjj-1,toc,toc/60);
end; 
fprintf('\n-2- Elapsed time in seconds after wavwrite split %6.3fsec  %6.3fmins\n',toc,toc/60);

%rejoin to check if arrays are the same
y2=[]; %
yb2=[];
filepathprocStr='/home/rat/Documents/octave/pre/';
files2=strcat(filepathprocStr,'*.wav');
files2=dir(files2);
[rwsz_files2,clsz_files2]=size(files2); %used to get ro and col size
for i=1:numel(files2) 
    [yb2, fs2, nbits] = wavread(strcat(filepathprocStr,files2(i).name));
    yb2=yb2';
    y2=[y2;yb2];    %Append files2
    fprintf('\n %4.0f of %4.0f joined %s',i,rwsz_files2,files2(i).name) 
end;
wavwrite([y2],fs2,16,'/home/rat/Documents/octave/pre/All_joined.2wav')

fprintf(' \n Done!!!\n');
ya=ya';
dy   = abs(ya-y2);           % absolute error
MAE  = mean(dy)               % 7.2292e-015   mean-absolute-error
MXE  = max(dy)                 % 3.4195e-014   maximum-absolute-error
RMSE = sqrt(mean(dy.^2))  % 9.5049e-015   root-mean-sqare-error

Answer 1

我决定使用reshape命令并使用零填充，因为它会大大减少代码并且应该更快。

clear all, clc
ya=1:64;
fs=9;

padlen = mod(-length(ya), fs); %creates number of zeros needed to get correct array reshaped
ya_reshaped = reshape([ya zeros(1,padlen)], fs, []); % used to pad zeros on 1 col x rows
[size_r,size_c]=size(ya_reshaped)

wavefilesplit=[];
wavefilesplit2=[];
for ii=1:size_c
    wavefilesplit=ya_reshaped(:,ii) %this line can be used to export data/audio to file
    %could use if else statment to strip zeros off end if inserted here 
    wavefilesplit2=[wavefilesplit2; ya_reshaped(:,ii)] %will append to end all in one col to error check
end;
wavefilesplit2(end-(padlen-1):end)=[] %will erase zeros at the end of array wavefilesplit2(end-(padlen-1):end,1)=2 will ad 2's

在matlab中拆分数组

1 个答案: