Question

我正在尝试更改以下代码，以便第一个矩阵成为第二个矩阵：

function BellTri = matrix(n)
BellTri = zeros(n);
BellTri(1,1) = 1;
for i = 2:n
  BellTri(i,1) = BellTri(i-1,i-1);
  for j = 2:i
    BellTri(i,j) = BellTri(i - 1,j-1) + BellTri(i,j-1);
  end
end
BellTri

第一个矩阵（当n = 7时）

 1     0     0     0     0     0     0
 1     2     0     0     0     0     0
 2     3     5     0     0     0     0
 5     7    10    15     0     0     0
15    20    27    37    52     0     0
52    67    87   114   151   203     0
203   255   322   409   523   674   877

第二个矩阵

1     1     2     5     15    52    877
1     3     10    37    151   674     0
2     7     27    114   523    0      0 
5    20     87    409    0     0      0
15   67     322    0     0     0      0
52   255    0      0     0     0      0
203   0     0      0     0     0      0

Answer 1

一种方法：

out = zeros(size(A));
out(logical(fliplr(triu(ones(size(A,1)))))) = A(logical(tril(ones(size(A,1)))));

注意：正如Divakar指出的那样，第一行应该有一个拼写错误。这种方法给出了纠正的方法。

<强>结果：

A = [1     0     0     0     0     0     0;
     1     2     0     0     0     0     0;
     2     3     5     0     0     0     0;
     5     7    10    15     0     0     0;
    15    20    27    37    52     0     0;
    52    67    87   114   151   203     0;
    203   255   322   409   523   674   877];

>> out

out =

   1     2     5    15    52   203   877
   1     3    10    37   151   674     0
   2     7    27   114   523     0     0
   5    20    87   409     0     0     0
  15    67   322     0     0     0     0
  52   255     0     0     0     0     0
 203     0     0     0     0     0     0

Answer 2

选项是使用#include <iostream> #include <string> #include <fstream> #include <deque> using namespace std; string findGroupKey(const string &line) { size_t first = line.find('_'); if (first == string::npos) first = 0; size_t last = line.find_last_of('.'); size_t len = (last == string::npos ? string::npos : last - first + 1); // The key includes the start and end chars themselves in order to // distinguish lines like "x_test." and "xtest" return line.substr(first,len); } int main() { // *** Var defs // Read the input file as stream ifstream inStream("test.txt"); // line by line placing each just read line into inLine string inLine; // Place each inLine into its one group deque<deque<string> *> linesGrouped; // according to the grouping key deque<string> keys; // *** Read the input file and group the lines in memory collections while (getline(inStream, inLine)) { string groupKey = findGroupKey(inLine); // Find the key in our keys-met-so-far collection int keyIndex = -1; for (int i = 0, keyCount = (int)keys.size(); i < keyCount; i++) if (keys.at(i) == groupKey) { keyIndex = i; break; }; if (keyIndex == -1) { // If haven't found the key so far, add it to our key index collection keys.push_back(groupKey); // and add a new group collection deque<string> *newGroup = new deque<string>(); newGroup->push_back(inLine); linesGrouped.push_back(newGroup); } else { // Otherwise just add the line into its respective group linesGrouped.at(keyIndex)->push_back(inLine); } } // *** Write the groups into the output file ofstream outStream("output.txt"); for (int i = 0, groupCount = (int)linesGrouped.size(); i < groupCount; i++) { for (int j = 0, lineCount = (int)linesGrouped.at(i)->size(); j < lineCount; j++) outStream << linesGrouped.at(i)->at(j) << endl; // Add a delimiter line (uncomment if you need one) //if (i < groupCount - 1) // outStream << "-------------------" << endl; } return 0; }循环置换列。

circshift

输入：function [BellTri, Second] = matrix(n) BellTri = zeros(n); BellTri(1,1) = 1; for i = 2:n BellTri(i,1) = BellTri(i-1,i-1); for j = 2:i BellTri(i,j) = BellTri(i - 1,j-1) + BellTri(i,j-1); end end Second = BellTri; for i = 1:n Second(:, i) = circshift(Second(:,i), 1-i); end for i = n-1:-1:2 Second(1, i) = Second(1, i-1); end end

输出：

[BellTri, Second] = matrix(7)

改变矩阵的指数

2 个答案: