在matlab中连接向量

你好

我需要构造向量xx和yy。问题是，在每个周期，过去的xx和yy都会被删除，但我想要的是相反的结果。我希望他们保留过去的信息，并在每个周期中成长。我能做什么？我也知道我应该预先分配xx和yy

谢谢你的帮助。多谢各位

**这是一个可能的解决办法。这里的问题是我需要预先分配。但如果我这样做，xx和yy保留了零，并继续增长，而“零在里面”，这是错误的

:

---

您需要在

xx

和

yy

中明确说明要放置数据的位置。从您的代码中很难理解您想要它的位置，但它应该在下面几行中找到一些东西

distances = sqrt((x - max(x)).^2 + (y - max(y)).^2);
[peaks, iPeaks] = findpeaks(distances);%to find out where the curve turns around
xx=[];
yy=[];
for i = 1 : length(iPeaks)-1
        iPeaks1 = iPeaks(i);
        iPeaks2 = iPeaks(i+1)-1;%analyse of consecutive pair of peaks
         %skip small noise peaks
       if length(iPeaks1:iPeaks2)>=5 
          xx = [xx; x(iPeaks1:iPeaks2)];%''concatenate''(connect)
          yy = [yy; y(iPeaks1:iPeaks2)];
       end
end

对于

yy

也一样<代码>开始你必须自己计算

另一方面，如果您想将新的

x

和

y

向量部分附加到

xx

和

yy

的末尾，您可以这样做

start = some_index_in_xx;
v = x(iPeaks1:iPeaks2);
xx(start:start+numel(v)-1)=v;

您必须在循环之前初始化

xx

和

yy

：

xx = [xx x(iPeaks1:iPeaks2)];
yy = [yy y(iPeaks1:iPeaks2)];

---

使用串联可以轻松地增长向量：

xx = [];
yy = [];

在你的情况下，你会这样做

x = [1, 2, 3];
y = [4, 5, 6, 7];
z = [x, y]; %# z will be [1, 2, 3, 4, 5, 6, 7]

独立于

iPeaksNoise

的形状（即，无论是行还是 列向量）您可以使用

%# Start with an empty vector.
xx = [];
for i = 1 : length(iPeaksNoise)-1
    [...]
    xx = [xx, x(iPeaks1:iPeaks2)];
end

预分配意味着创建一个零向量，该向量具有最终输出的长度。你会这样做吗

for [...]
    t = x(iPeaks1:iPeaks2);
    xx = [xx; t(:)];
end

然而，在你的情况下，你有额外的困难，不知道最终的大小 因为您不知道

length（iPeaks1:iPeaks2）>=5的频率
在循环内完成

编辑：

预分配为
xx
的代码可能如下所示：

final_size = what_so_ever;
xx = zeros(final_size, 1);
start_index = 1;
for [...]
    t = x(iPeaks1:iPeaks2);
    xx(start_index : start_index + length(t) - 1) = t;
    start_index = start_index + length(t);
end

在这里，我假设在每次迭代中要添加到xx和yy的元素的数量不是每次都相同。然后，如果你想预先分配空间，你需要知道你要加多少。因此，如果有一个额外的循环，代码将变得更加复杂。一个版本可能是

distances = sqrt((x - max(x)).^2 + (y - max(y)).^2);
[peaks, iPeaks] = findpeaks(distances);

%# Compute the distance between two peaks.
peakDistance = diff(iPeaks);
%# Filter peaks which are of length 5 or less.
peakMask = peakDistance > 5;
finalSize = sum(peakDistance(peakMask));
%# Copy the values in a new vector.
xx = zeros(1, finalSize);
idx = 1;
for count = 1 : length(peakDistance)
  if peakMask(count)
    pD = peakDistance(count);
    xx(idx : idx + pD - 1) = x(iPeaks(count) : iPeaks(count) + pD - 1);
    idx = idx + pD;
  end
end

但是，这是matlab。你不应该这么做。下面是一些代码，它们使用逻辑标志执行相同的操作，只需一次删除所有小峰值：

x=(5*sin(0:(pi/100):2*pi)+rand(201,1)'-0.5)';
y=(5*cos(0:(pi/100):2*pi)+rand(201,1)'-0.5)';
%%
distances = sqrt((x - max(x)).^2 + (y - max(y)).^2);
[peaks, iPeaks] = findpeaks(distances);%to find out where the curve turns around

counterx = 0;
for i = 1 : length(iPeaks)-1
    iPeaks1 = iPeaks(i);
    iPeaks2 = iPeaks(i+1)-1;%analyse of consecutive pair of peaks
    %skip small noise peaks
    if length(iPeaks1:iPeaks2)>=5
        counterx =counterx+ iPeaks2-iPeaks1+1;
    end
end
xx=zeros(counterx,1);
yy=xx;
counterx=0;
clc
for i = 1 : length(iPeaks)-1
    iPeaks1 = iPeaks(i);
    iPeaks2 = iPeaks(i+1)-1;%analyse of consecutive pair of peaks
    %skip small noise peaks
    if length(iPeaks1:iPeaks2)>=5
        %         xx = [xx; x(iPeaks1:iPeaks2)];%''concatenate''(connect)
        %         yy = [yy; y(iPeaks1:iPeaks2)];
        counterx=counterx+1;
        xx(counterx:counterx+iPeaks2-iPeaks1) = x(iPeaks1:iPeaks2);
        yy(counterx:counterx+iPeaks2-iPeaks1) = y(iPeaks1:iPeaks2);
        counterx=counterx+iPeaks2-iPeaks1;
    end
end
disp([xx yy]')

%查找与下一个ipeak相距5的所有ipeak
peakdiffs=find（（iPeaks（2:end）-iPeaks（1:end-1）-1）>=5）；
%快速创建与x相同大小的逻辑阵列
对不起，你席的意思是“开始=某事索引xxx”？我想从一开始就填写xx，那么1？我改变了问题并绘制了所有代码。也许你更容易帮助我。非常感谢您。如果您预先分配了xx，那么它有一个初始大小（例如
xx=0（11000）
。你的向量
v
是，比方说，10个元素。你想把它放在
xx
的哪里？从哪个索引开始？你能看到我在问题空间中写的解决方案吗？你知道我能做些什么来解决它吗？非常感谢。我想你把事情搞混了。你没有预先分配
xx
和
yy
。您将它们初始化为空。然后添加到它们的是您的数据，而不是一些不需要的零。好吧，是的，这就是问题所在。您能看看我在“问题空间”中编写的新代码吗。您知道我可以做些什么来解决问题吗？非常感谢。@user1675777:我添加了一个基于您的ed的预分配示例信息技术
x=(5*sin(0:(pi/100):2*pi)+rand(201,1)'-0.5)';
y=(5*cos(0:(pi/100):2*pi)+rand(201,1)'-0.5)';
%%
distances = sqrt((x - max(x)).^2 + (y - max(y)).^2);
[peaks, iPeaks] = findpeaks(distances);%to find out where the curve turns around

counterx = 0;
for i = 1 : length(iPeaks)-1
    iPeaks1 = iPeaks(i);
    iPeaks2 = iPeaks(i+1)-1;%analyse of consecutive pair of peaks
    %skip small noise peaks
    if length(iPeaks1:iPeaks2)>=5
        counterx =counterx+ iPeaks2-iPeaks1+1;
    end
end
xx=zeros(counterx,1);
yy=xx;
counterx=0;
clc
for i = 1 : length(iPeaks)-1
    iPeaks1 = iPeaks(i);
    iPeaks2 = iPeaks(i+1)-1;%analyse of consecutive pair of peaks
    %skip small noise peaks
    if length(iPeaks1:iPeaks2)>=5
        %         xx = [xx; x(iPeaks1:iPeaks2)];%''concatenate''(connect)
        %         yy = [yy; y(iPeaks1:iPeaks2)];
        counterx=counterx+1;
        xx(counterx:counterx+iPeaks2-iPeaks1) = x(iPeaks1:iPeaks2);
        yy(counterx:counterx+iPeaks2-iPeaks1) = y(iPeaks1:iPeaks2);
        counterx=counterx+iPeaks2-iPeaks1;
    end
end
disp([xx yy]')

% find all the iPeaks that are 5 apart from the next one
peakdiffs = find((iPeaks(2:end)-iPeaks(1:end-1)-1)>=5);
% make a logical array quickly with same size as x
removeflag = x<=inf;
for i=1:numel(peakdiffs)
    % for each peak that is bigger than your threshold, those parts of the
    % logical array will be made false.
    removeflag(iPeaks(peakdiffs(i)):iPeaks(peakdiffs(i)+1)-1)=false;
end
%remove all indices in x,y that are still false in removeflag and store in xx
%and yy resp
xx= x; xx(removeflag)=[];
yy= x; yy(removeflag)=[];
disp([xx yy]')