Finding the maximum of rows

Say that we have the following matrix:
I=[3 4; 5 3; 6 3; 7 4];
If we want to find the maximum value in each row, we can do the following:
m=max(I,[],2);
For m, how do we read this? How is the statement interpreted? What should we do if we want to find the maxim of the columns?
Thanks.

2 Commenti

HI how would you take the max of every other row?
There are probably multiple ways to find the max of every other row, but the simplest I can think of is to just use matrix indexing:
i=1; % Changes the starting row. Can be either 1 or 2.
m=max(I(i:2:end,:),[],2);
Of course, this can be generalized if you want to take every nth row:
i=1; % Changes the starting row. Positive integer between 1 and n.
n=4; % Changes the increment. Positive integer.
m=max(I(i:n:end,:),[],2);

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Risposte (2)

Wayne King
Wayne King il 1 Gen 2014
Modificato: Wayne King il 1 Gen 2014
[m,idx] = max(I,[],2);
m gives you the maximum value in each row and idx gives you the column in which it occurs. I used this in my other response to you about fcm()
To find the maximum of the columns, just operate along the rows
[m,idx] = max(I,[],1);

3 Commenti

med-sweng
med-sweng il 1 Gen 2014
Thanks for your reply. What does [] represent here?
Are you reading the documentation?? The syntax max(x,y) takes the maximum of x or y, so if you just called:
max(I,1)
that would not give you what you want. You want 1 or 2 to represent the dimension along which you want the maximum and that has to be the third input argument.
max(A) means the maximum of A column-wise
max(A,B) means the maximum element-by-element of A(I,J) vs B(I,J)
max(A,[],k) means the maximum of A along the k'th dimension. You need the [] as a place-holder so that MATLAB does not confuse this with max(A,B)

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Ariunbolor Purvee
Ariunbolor Purvee il 6 Ago 2020
Modificato: Ariunbolor Purvee il 6 Ago 2020

0 voti

clc;clear
absDiff=[1 2 3; 4 5 6; 7 8 9; 9 10 12];
MaxMinAveRow= MaxMinAveOfRow(absDiff);
display(MaxMinAveRow);
function MaxMinAveRow= MaxMinAveOfRow(absDiff)
n=length(absDiff);
maxRow=zeros(n,1);
minRow=zeros(n,1);
aveRow=zeros(n,1);
sum=0;
for i=1:n
maxRow(i)=sum+max(absDiff(i,:));% max of row
minRow(i)=sum+min(absDiff(i,:));% min of row
aveRow(i)=sum+mean(absDiff(i,:));% average of row
end
MaxMinAveRow=[ maxRow,minRow, aveRow ];
end
Result on Command Window
MaxMinAveRow =
3.0000 1.0000 2.0000
6.0000 4.0000 5.0000
9.0000 7.0000 8.0000
12.0000 9.0000 10.3333

Tag

Richiesto:

il 1 Gen 2014

Commentato:

il 17 Mag 2022

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