matlab里= =,~=, find(), all(), any()
时间: 2023-05-10 22:50:55 浏览: 52
在MATLAB中,=表示变量的赋值,即将一个数值或字符数组等赋给一个变量。例如,a=5表示将数值5赋给变量a,b='hello'表示将字符数组'hello'赋给变量b。
而~=用于表示不相等关系,即两个变量不相等时返回逻辑值1,否则返回逻辑值0。例如,a~=b用于判断变量a和变量b是否相等。
find()函数用于查找数组中的非零元素的下标,返回一个行向量。例如,a=[1 0 3 0 5],则find(a)返回[1 3 5],即a中非零元素所在的下标。
all()函数用于判断一个逻辑数组中的所有元素是否都为真(即1),若全部为真则返回逻辑值1,否则返回逻辑值0。例如,a=[1 1 0 1],则all(a)返回0,说明a中并非所有元素都为真。
any()函数用于判断一个逻辑数组中是否有至少一个元素为真(即1),若存在则返回逻辑值1,否则返回逻辑值0。例如,a=[0 0 1 0],则any(a)返回1,表示a中存在至少一个元素为真。
相关问题
matlab删去全零行
在MATLAB中,可以使用以下方法删除矩阵中的全零行:
方法一:使用逻辑索引
```matlab
A = [1 2 3; 0 0 0; 4 5 6; 0 0 0A(all(A==0,2),:) = [];
```
方法二:使用find函数
```matlab
A = [1 2 3; 0 0 0; 4 5 6; 0 0 0];
A(find(A==0,2)),:) = [];
```
方法三:使用any函数
```matlab
A = [1 2 3; 0 0 0; 4 5 6; 0 0 0];
A(any(A~=0,2),:) = [];
```
这些方法都可以将矩阵A中全零的行删除掉,得到一个新的矩阵。请注意,这些方法都是基于逻辑条件进行筛选的。
匈牙利算法指派问题matlab代码
Sure, here is an example of the Hungarian algorithm implemented in MATLAB:
```matlab
function [assignments, cost] = hungarianAlgorithm(costMatrix)
% Step 1: Subtract the smallest value from each row
costMatrix = bsxfun(@minus, costMatrix, min(costMatrix,[],2));
% Step 2: Subtract the smallest value from each column
costMatrix = bsxfun(@minus, costMatrix, min(costMatrix,[],1));
% Step 3: Assign initial zeros
assignments = zeros(size(costMatrix));
% Step 4: Assign zeros recursively
while true
[rowCovered, colCovered, numCovered] = coverZeros(assignments);
if numCovered == size(costMatrix,1)
break; % All zeros are covered
else
[costMatrix, assignments] = findAssignments(costMatrix, assignments, rowCovered, colCovered);
end
end
% Step 5: Calculate the final assignment cost
indices = find(assignments == 1);
cost = sum(costMatrix(indices));
end
function [rowCovered, colCovered, numCovered] = coverZeros(assignments)
rowCovered = any(assignments == 1, 2);
colCovered = any(assignments == 1, 1);
numCovered = sum(rowCovered) + sum(colCovered);
end
function [costMatrix, assignments] = findAssignments(costMatrix, assignments, rowCovered, colCovered)
while true
% Find a zero that is not covered
[rowZero, colZero] = find(costMatrix == 0 & ~rowCovered & ~colCovered, 1);
if isempty(rowZero) || isempty(colZero)
[costMatrix, assignments] = updateCostMatrix(costMatrix, assignments, rowCovered, colCovered);
break;
else
rowCovered(rowZero) = true;
colCovered(colZero) = true;
assignments(rowZero, colZero) = 1;
% Find a starred zero in the same row
colStarred = find(assignments(rowZero, :) == 1, 1);
if isempty(colStarred)
% No starred zero in the same row
[costMatrix, assignments] = updateCostMatrix(costMatrix, assignments, rowCovered, colCovered);
break;
else
% Cover the column of the starred zero and uncover the row of the starred zero
colCovered(colStarred) = false;
end
end
end
end
function [costMatrix, assignments] = updateCostMatrix(costMatrix, assignments, rowCovered, colCovered)
% Find the smallest uncovered value
uncoveredValues = costMatrix(~rowCovered, ~colCovered);
if ~isempty(uncoveredValues)
minUncoveredValue = min(uncoveredValues(:));
% Subtract the smallest uncovered value from uncovered rows
costMatrix(~rowCovered, ~colCovered) = costMatrix(~rowCovered, ~colCovered) - minUncoveredValue;
% Add the smallest uncovered value to covered columns
costMatrix(rowCovered, colCovered) = costMatrix(rowCovered, colCovered) + minUncoveredValue;
end
% Reset the covered assignments
assignments(rowCovered, :) = 0;
assignments(:, colCovered) = 0;
end
```
You can use this code by passing your cost matrix to the `hungarianAlgorithm` function. It will return the assignments and the minimum cost.