如何从一个更大的矩阵中提取一个2x2子矩阵 [英] How to extract a 2x2 submatrix from a bigger matrix

问题描述

我是一个很基本的用户，不知道太多关于在C中使用命令，所以请原谅......我不能使用非常复杂的codeS。我在stdio.h和库文件ctype.h一些知识，但多数民众赞成它。
我有一个txt文件矩阵，我要加载的矩阵根据我的行数和列数的输入

I am a very basic user and do not know much about commands used in C, so please bear with me...I cant use very complicated codes. I have some knowledge in the stdio.h and ctype.h library, but thats about it. I have a matrix in a txt file and I want to load the matrix based on my input of number of rows and columns

例如，我在文件中的5×5矩阵。我想提取特定的2×2子矩阵，我该怎么办呢？

For example, I have a 5 by 5 matrix in the file. I want to extract a specific 2 by 2 submatrix, how can I do that ?

我创建使用嵌套循环：

FILE *sample
sample=fopen("randomfile.txt","r"); 
for(i=0;i<rows;i++){
  for(j=0;j<cols;j++){
     fscanf(sample,"%f",&matrix[i][j]);
   }
 fscanf(sample,"\n",&matrix[i][j]);
}
fclose(sample);

可悲的是code不起作用。如果我有这样的矩阵：

Sadly the code does not work .. If I have this matrix :

5.00 4.00 5.00 6.00 
5.00 4.00 3.00 25.00 
5.00 3.00 4.00 23.00 
5.00 2.00 352.00 6.00

和输入3行和3列，我得到：

And inputting 3 for row and 3 for column, I get :

5.00 4.00 5.00
6.00 5.00 4.00
3.00 25.00 5.00

这不仅心不是一个2×2子矩阵，但即使我想第3行第3列，其无法正常打印出来....

Not only this isnt a 2 by 2 submatrix, but even if I wanted the first 3 rows and first 3 columns, its not printing it correctly....

我需要3行和第3栏开始，然后坐2×2子矩阵！

I need to start at row 3 and col 3, then take the 2 by 2 submatrix !

我应该已经结束了：

4.00 23.00 
352.00 6.00

我听说，我可以使用与fgets和sscanf做到这一点。这里是我的审判code：

I heard that I can use fgets and sscanf to accomplish this. Here is my trial code :

fgets(garbage,1,fin);
sscanf(garbage,"\n");

但是，这并不工作之一：（

But this doesnt work either :(

我在做什么错了？

请帮忙。
谢谢！

Please help. Thanks !

推荐答案

好了，你想读大小的子矩阵的 N X M 的开始，在位置 X 的是的大小的大矩阵的 p X 问：的。你需要两样东西：

OK, so you want to read a submatrix of size n x m, starting at positions x, y in the big matrix of size p x q. You need two things:

1. （验证的 X + N 的＆LT; = P 和是 + M ＆LT; =的问：的）


2. 跳到你想要阅读的矩阵的第一要素。这需要先跳过第一的是的 - 1行


3. 跳过的 X 的 - 从下一行1个元素，然后读的 N 的元素融入到您的子矩阵。重复的 M 的时间。

1. (verify that x + n <= p and y + m <= q)
2. skip to the first element of the matrix you want to read. This requires first skipping the first y - 1 rows
3. skip x - 1 elements from the next row, then read n elements into your submatrix. Repeat m times.

您目前执行开始从基体的第一个元件读取，然后读取元件邻接到基质中。一个更新版本：

Your current implementation starts reading from the very first element of the matrix, then reads elements contiguously into the submatrix. An updated version:

FILE *sample = fopen("randomfile.txt", "r");
// skip the first y-1 rows
for (i = 0; i < y - 1; i++) {
  fscanf(sample, "%*[^\n]\n", &matrix[i][j]);
}
for (i = 0; i < m; i++) {
  // skip the first x-1 numbers
  for (j = 0; j < x - 1; j++) {
     fscanf(sample, "%*f");
  }
  // read n numbers
  for (j = 0; j < n; j++) {
     fscanf(sample, "%f", &matrix[i][j]);
  }
  if (x + n < p) {
    // consume the rest of the line
    fscanf(sample, "%*[^\n]\n");
  }
}
fclose(sample);

更新：来读取一个数组的子矩阵，而不是更简单，只是需要更多的计算。主旨在于，大小的矩阵的 P X 问：的可以存储在大小的连续排列的 P X 问：这样矩阵[I，J]可以从阵列读取[我*（J-1）+ J]（约 - 有可能会关闭接一个错误，我从来不敢肯定这是列，这是行，但希望你的想法： - ）

Update: to read the submatrix from an array instead is even simpler, just requires a bit more calculation. The gist is, a matrix of size p x q can be stored in a contiguous array of size p x q such that matrix[i,j] can be read from array[i*(j-1)+j] (approximately - there may be off-by-one errors and I am never sure which is the column and which is the row, but hopefully you get the idea :-)

所以，code会是这样

So the code would be something like

for (i = 0; i < m; i++) {
  for (j = 0; j < n; j++) {
     submatrix[i][j] = array[(y + i) * p + x + j];
  }
}

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