How can I do matrix indexing

I want to select 3 elements from the matrix C, one element from each column, the selecting will be regarding to the maximum value in in each column in matrix f.

C=

41.0875 66.4048 31.5374

84.3670 25.9733 73.1052

68.8608 99.4798 75.5581

the vector needed like this [ 68.8608 66.4048 73.1052 ]

f=

1.0027 0.9766 1.2106

0.4001 0.5136 1.8174

1.5451 0.9488 0.8571

and because this values and positions are subject to change randomly

any help will be very thankful

[KSSV]

REad about max

Answer 1

[Stephen23]

A simple and reliable solution using linear indexing:

>> f = [1,22,333;4,33,111;1,4,33]
f =
     1    22   333
     4    33   111
     1     4    33
>> C = [41.0875,66.4048,31.5374;84.3670,25.9733,73.1052;68.8608,99.4798,75.5581]
C =
   41.087   66.405   31.537
   84.367   25.973   73.105
   68.861   99.480   75.558
>> [vec,idx] = max(f,[],1);     % value and row index of max in each column.
>> S = size(f);                 % size of input array. 
>> idx = idx + S(1)*(0:S(2)-1); % convert row index into linear index.
>> out = C(idx)                 % use linear index to get elements of C.
out =
   84.367   25.973   31.537
>> vec
vec =
     4    33   333

[Ashraf Sherif]

yeah you are completly right Mr. Stephen Cobeldick and it is work well, thank you a lot,

[Ashraf Sherif]

How to add a row vector when my M is 4 dimensional

F=random('exp',1,3,3,4);
C=F*100
M=bsxfun(@times,  C./cumsum(C,3) , reshape(1:4,1,1,[])   );
[vec,idx] = max(M,[],1);      % value and row index of max in each column.
 S = size(M);                 % size of input array.
idx = idx + S(1)*(0:S(2)-1); % convert row index into linear index !

 C(idx)                % use linear index to get elements of C.

Answer 2

[madhan ravi]

C(any(f(:)==max(f),2)).'

[Torsten]

I get

[84.3670 99.4798 31.5374]

not

[68.8608 66.4048 73.1052]

[madhan ravi]

>> f=[  1.0027    0.9766    1.2106
        0.4001    0.5136    1.8174
        1.5451    0.9488    0.8571 ]
    
C=[41.0875   66.4048   31.5374
   84.3670   25.9733   73.1052
   68.8608   99.4798   75.5581]

C(any(f(:)==max(f),2)).'

f =

    1.0027    0.9766    1.2106
    0.4001    0.5136    1.8174
    1.5451    0.9488    0.8571


C =

   41.0875   66.4048   31.5374
   84.3670   25.9733   73.1052
   68.8608   99.4798   75.5581


ans =

   68.8608   66.4048   73.1052

>> 

[Torsten]

Sorry, my mistake.

[madhan ravi]

No problem ;-)

[Ashraf Sherif]

thank you madhan ravi yes exactly thats what I want, but this message appears

Error using == Matrix dimensions must agree.

[madhan ravi]

C(any(bsxfun(@eq,f(:),max(f)),2)).'

[Ashraf Sherif]

I think it related to my MATLAB version I'm using MATLAB R2013a

[madhan ravi]

Yes , sorry I have no experience with older versions , didn't my latter comment work?

[Ashraf Sherif]

Yes It's work I tried it on MATLAB cloud by my phone. thank you very much my friend.

[Ashraf Sherif]

yes this one work well with my MATLAB version

C(any(bsxfun(@eq,f(:),max(f)),2)).'

Thanks a lot

[madhan ravi]

No problem.

[Stephen23]

Note that this is very fragile code, and it can easily fail when values repeat in matrix f:

>> f = [1,22,333;4,33,111;1,4,33]
f =
     1    22   333
     4    33   111
     1     4    33
>> C = [41.0875,66.4048,31.5374;84.3670,25.9733,73.1052;68.8608,99.4798,75.5581]
C =
   41.087   66.405   31.537
   84.367   25.973   73.105
   68.861   99.480   75.558
>> C(any(f(:)==max(f),2)).'
ans =
   84.367   25.973   99.480   31.537   75.558  % <- why five output values?
 

See my answer for code that actually delivers what the question asked for: " I want to select 3 elements from the matrix C, one element from each column..."

[madhan ravi]

Agree with Stephen , @Ashraf accept Stephen's answer.

Answer 3

[Ashraf Sherif]

Yes It's work I tried it on MATLAB cloud by my phone. thank you very much my friend.

Answer 4

[Ashraf Sherif]

Mr. Stephen Cobeldick and Mr. madhan ravi I appreciate you help, this was sloved my problem when I have a simple matrix 3 by 3, but now am facing an other problem that when I want to calculate new index matrix. here is my code with some explaining for what am looking for

clear all
close all
clc
for u=[2:30]% the size of matrix 
x1=[];x2=[];x3=[];x4=[];x5=[];x6=[];x6=[];x7=[];x8=[];x9=[];% this is number of elements in may matrix when it was 3 by 3 
for i=1:4 % number of itteration (time slot in my case)
    f=random('exp',1,u,20);  % size of matrix will change from 2 by 20 up to 30 by 20
     Pt=1;                          % Transmit power 1 watt
    B=50*10^3;               % Bandwidth = 50 KHz * 10^3 Hz
    d=5;                           % Distance 5 meter 
    lamda= -3;                 % Used in Urban area
    No=10^-8;                  % AWGN noise  
    SNR=Pt*d^lamda*f/(B*No);  % Signal-to-Noise Ratio 
                                                                   %%  Capacity  when Max-SNR and Proportional Fairness  %%
    C=(B/3*log2(1+SNR))/1000;   % matrix of my data
% calculting the instantaneous data rate / average data rate for each user 
%1- Instantaneous values of elements
x1=[x1;C(1,1)]; x2=[x2;C(1,2)];x3=[x3;C(1,3)];x4=[x4;C(2,1)];x5=[x5;C(2,2)];x6=[x6;C(2,3)];x7=[x7;C(3,1)];x8=[x8;C(3,2)];x9=[x9;C(3,3)];
% 2- Average of each elements during for example 4 itterations
M1=C(1,1)/mean(x1);M2=C(1,2)/mean(x2);M3=C(1,3)/mean(x3);M4=C(2,1)/mean(x4);M5=C(2,2)/mean(x5);M6=C(2,3)/mean(x6);M7=C(3,1)/mean(x7);M8=C(3,2)/mean(x8);M9=C(3,3)/mean(x9);
% this is my index matrix
PFindex=[M1,M2,M3;M4,M5,M6;M7,M8,M9];
maxPF=[max([M1,M4,M7]), max([M2,M5,M8]), max([M3,M6,M9])];
    MaxSNR=sum(max(C))
    
    % Capacity with Proportional Fairness
[vec,idx] = max(PFindex,[],1);     % value and row index of max in each column.
 S = size(PFindex);                 % size of input array. 
idx = idx + S(1)*(0:S(2)-1); % convert row index into linear index.
 PF = sum(C(idx))                 % use linear index to get elements of C.
end
end

fainally I'm collecting Values of MaxSNRs and PF, thank you any way.

[Stephen23]

Replace all of those numbered variables, use indexing instead.

Numbered variables are a sign that you are doing something wrong.

[Ashraf Sherif]

Thank you Stephen that what am going to do