i have a matrix size 378x5, how do i find the index column of the maximum value of each row? Thanks 댓글 수: 1 Rik2019년 5월 30일 (I'm on mobile so I can't check) The second output of max can do this. You may need to use a loop. ...
Sparse matrices have a different storage convention from full matrices in MATLAB. The parameterpais still an array of double-precision numbers or logical values, but this array contains only nonzero data elements. There are three additional parameters:nzmax,ir, andjc. Use themwSizeandmwIndextypes...
Open in MATLAB Online Ran in: Recreating the issue: x2 = (1:1:8); y2 = (1:2:16)'; Q32(x2,y2) functionQ32(x2,y2) %aim of function to have two vectors x, y and each has a size of m. Order the vectors according to value (max 1st, lowest last) %Make a (m x m)...
is a built-in function) prohibit the use of built-in functions. This information is in the title of the original question, not the body. [Yes, technically the question doesn't state this was a homework assignment, but I've been on Answers and the MATLAB newsgroup...
% Copy each example into a patch on the display array curr_ex = 1; for j = 1:display_rows for i = 1:display_cols if curr_ex > m, break; end % Copy the patch % Get the max value of the patch max_val = max(abs(X(curr_ex, :))); ...
Ran in: See the full error message: Indexexceeds the number of array elements. Index must not exceed 1. Errorin LE_RF_p1 (line 4) X= [x(4) x(7) x(10); Apparentlyxis a scalar, not a vector, meaning it only has a single value. There error is because your code...
MATLAB 编程风格指南 Richard Johnson 著 Genial 译 MATLAB 编程风格指南编程风格指南 Richard Johnson Version 1.5,Oct. 2002 版权; Datatool 所有 翻译:Ge
value of a: 12 value of a: 13 value of a: 14 value of a: 15 value of a: 16 value of a: 17 value of a: 18 value of a: 19 value of a: 20 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 在MATLAB 中 while循环的语法是: ...
value,the order of the elements are preserved in the sorted result and the indexes of equal elements will be ascending in any index matrix.Example:If X = [3,7,5; 0,4,2] then sort(X,1) is [0,4,2;3,7,5] and sort(X,2) is [3,5,7;0,2,4];See also SORTROWS,MIN,MAX,...
for index=1:2 if index==1 dosomething(index); ... end % End if end % End for 条件式 避免复杂的条件表达式,改用临时逻辑变量。 isValid = (v >= lowerLimit) & (v <= upperLimit); isNew = ismember(v, valueArray); 避免使用条件表达式if 0。