将矩阵分为两部分 [英] Dividing a matrix into two parts

问题描述

我正在尝试对数据集进行分类.为此，我将使用数据集的第4列.如果数据集的第4列等于1，则该行将添加到名为Q1的新矩阵中.如果数据集的第4列等于2，则该行将添加到矩阵Q2中.

I am trying to classify my dataset. To do this, I will use the 4th column of my dataset. If the 4th column of the dataset is equal to 1, that row will added in new matrix called Q1. If the 4th column of the dataset is equal to 2, that row will be added to matrix Q2.

我的代码:

i = input('Enter a start row: ');
j = input('Enter a end row: ');
search = importfiledataset('search-queries-features.csv',i,j);
[n, p] = size(search);

if j>n
   disp('Please enter a smaller number!');
end

for s = i:j
    class_id = search(s,4);
    if class_id == 1 
       Q1 = search(s,1:4)        
    elseif class_id ==2  
       Q2 = search(s,1:4)
    end
end

这将计算Q1和Q2矩阵，但它们均为1x4，当给出新的Q1时，将删除旧的.如果条件为真，我需要添加新行并将其设置为2x4.我需要扩展我的Q1矩阵.

This calculates the Q1 and Q2 matrices, but they all are 1x4 and when it gives new Q1 the old one is deleted. I need to add new row and make it 2x4 if conditions are true. I need to expand my Q1 matrix.

简而言之，我正在尝试使用for循环和if语句将数据集分为两部分.

Briefly I am trying to divide my dataset into two parts using for loops and if statements.

数据集:

我需要如下结果:

Q1 = [30  64  1  1
      30  62  3  1
      30  65  0  1
      31  59  2  1
      31  65  4  1
      33  58 10  1
      33  60  0  1
      34  58 30  1
      34  60  1  1 
      34  61 10  1]

Q2 = [34 59 0 2
      34 66 9 2]

如何防止我的代码删除Q1和Q2的前几行并获取整个矩阵?

How can I prevent my code from deleting previous rows of Q1 and Q2 and obtain the entire matrices?

推荐答案

计算中的主要问题是每次循环迭代覆盖 Q1和Q2.最佳解决方案:摆脱循环并使用逻辑索引.

The main problem in your calculation is that you overwrite Q1 and Q2 each loop iteration. Best solution: get rid of the loops and use logical indexing.

您可以使用逻辑索引快速确定列等于1或2的位置:

You can use logical indexing to quickly determine where a column is equal to 1 or 2:

search = [
  30 64 1 1 
  30 62 3 1
  30 65 0 1
  31 59 2 1
  31 65 4 1
  33 58 10 1
  33 60 0 1
  34 59 0 2
  34 66 9 2
  34 58 30 1
  34 60 1 1 
  34 61 10 1
];
Q1 = search(search(:,4)==1,:)  % == compares each entry in the fourth column to 1

Q2 = search(search(:,4)==2,:)

Q1 =
    30    64     1     1
    30    62     3     1
    30    65     0     1
    31    59     2     1
    31    65     4     1
    33    58    10     1
    33    60     0     1
    34    58    30     1
    34    60     1     1
    34    61    10     1
Q2 =
    34    59     0     2
    34    66     9     2

---

警告:解决方法慢！

如果您不愿意使用循环，请确保不要覆盖变量.每次迭代都扩展它们(这非常非常慢):

---

Warning: Slow solution!

If you are hell bent on using loops, make sure to not overwrite your variables. Either extend them each iteration (which is very, very slow):

Q1=[];
Q2=[];

for ii = 1:size(search,1) % loop over all rows
   if search(ii,4)==1
       Q1 = [Q1;search(ii,:)];
   end
   if search(ii,4)==2
       Q2 = [Q2;search(ii,:)];
   end
end

MATLAB会在Q1和Q2下放置橙色的摆动，因为在原地生长数组不是一个好主意.另外，您可以预分配它们search的大小，然后去除多余的部分:

MATLAB will put orange wiggles beneath Q1 and Q2, because it's a bad idea to grow arrays in-place. Alternatively, you can preallocate them as large as search and strip off the excess:

Q1 = zeros(size(search)); % Initialise to be as large as search
Q2 = zeros(size(search));

Q1kk = 1; % Intialiase counters
Q2kk = 1;

for ii = 1:size(search,1) % loop over all rows
   if search(ii,4)==1
       Q1(Q1kk,:) = search(ii,:); % store
       Q1kk = Q1kk + 1; % Increase row counter
   end
   if search(ii,4)==2
       Q2(Q2kk,:) = search(ii,:);
       Q2kk = Q2kk + 1;
   end
end

Q1 = Q1(1:Q1kk-1,:); % strip off excess rows
Q2 = Q2(1:Q2kk-1,:);

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