How to make simpler this at 'if condition'?

There is 'a' vector [6x9].

And I want to define 'ahat' depends on 'ii' value.

For example, if ii=3, 'ahat' use a(1,1),a(2,1),a(3,1),a(4,1),a(5,1),a(6,1)] this value.

But could I make simpler this?

if (ii<5)
ahat=[a(1,1),a(2,1),a(3,1),a(4,1),a(5,1),a(6,1)];
elseif ((4<ii)&&(ii<9))
ahat=[a(1,2),a(2,2),a(3,2),a(4,2),a(5,2),a(6,2)]; 
elseif ((8<ii)&&(ii<13))
ahat=[a(1,3),a(2,3),a(3,3),a(4,3),a(5,3),a(6,3)]; 
elseif ((12<ii)&&(ii<17))
ahat=[a(1,4),a(2,4),a(3,4),a(4,4),a(5,4),a(6,4)]; 
elseif ((16<ii)&&(ii<21))
ahat=[a(1,5),a(2,5),a(3,5),a(4,5),a(5,5),a(6,5)]; 
elseif ((20<ii)&&(ii<25))
ahat=[a(1,6),a(2,6),a(3,6),a(4,6),a(5,6),a(6,6)]; 
elseif ((24<ii)&&(ii<29))
ahat=[a(1,7),a(2,7),a(3,7),a(4,7),a(5,7),a(6,7)]; 
elseif ((28<ii)&&(ii<33))
ahat=[a(1,8),a(2,8),a(3,8),a(4,8),a(5,8),a(6,8)]; 
else ((32<ii)&&(ii<37))
ahat=[a(1,9),a(2,9),a(3,9),a(4,9),a(5,9),a(6,9)];
end
a = ahat;

Answer 1

[Star Strider]

I am not certain this is more efficient than your code, however it takes advantage of the repeating patterns to make it simpler:

a = randi(9, 6, 9);                                                 % Create ‘a’

C1 = 4 : 4 : 32;
ahatfcn = @(k) [a(1,k),a(2,k),a(3,k),a(4,k),a(5,k),a(6,k)];         % Create Anonymous Function
for ii = 1 : 40                                                     % Loop Over ‘ii’
    if ii < 5
        ahat = ahatfcn(1);
    else
        for k1 = 1:numel(C1)
            if ((C1(k1)<ii) && (ii<(C1(k1)+5)))
                ahat = ahatfcn(k1+1);
            end
        end
    end
end
a = ahat

It runs without error. I defer to you to determine if it produces the result you want.

Answer 2

[Steven Lord]

Make some sample data.

a = reshape(randperm(6*9), [6 9])
ii = 15;

Define the cutoffs.

vec = [0 4 8 12 16 20 24 28 32 36];
% or
vec = 0:4:36;

Determine which bin your selected ii falls into. For this sample ii, it falls into bin #4 (between 12 and 16.)

d = discretize(ii, vec)

Create ahat.

ahat = a(1:6, d);

This will even work with a row vector ii.

ii = [15 21 28];
d = discretize(ii, vec)
ahat2 = a(1:6, d)

[CHOI HYUNDUK]

Thank you a lot!

Answer 3

[madhan ravi]

a = rand(6,9)  % some data to test
ii = 6 % according to our expectation the result should be the second column

if ii < 5
ahat = a(:,1) ;
elseif ( 4 < ii ) && ( ii < 9 )
ahat = a(:,2) ;
elseif ( 8 < ii ) && ( ii < 13 )
ahat = a(:,3) ;
elseif ( 12 < ii ) && ( ii < 17 )
ahat = a(:,4) ; 
elseif ( 16 < ii ) && ( ii < 21 )
ahat = a(:,5) ; 
elseif ( 20 < ii ) && ( ii < 25 )
ahat = a(:,6) ; 
elseif ( 24 < ii ) && ( ii < 29 )
ahat = a(:,7) ; 
elseif ( 28 < ii ) && ( ii < 33 )
ahat = a(:,8) ; 
else ( 32 < ii ) && ( ii < 37 )
ahat = a(:,9) ;
end
a = ahat.' 

Gives:

a =

  Columns 1 through 7

    0.1672    0.9516    0.5479    0.6663    0.9991    0.1904    0.6448
    0.1062    0.9203    0.9427    0.5391    0.1711    0.3689    0.3763
    0.3724    0.0527    0.4177    0.6981    0.0326    0.4607    0.1909
    0.1981    0.7379    0.9831    0.6665    0.5612    0.9816    0.4283
    0.4897    0.2691    0.3015    0.1781    0.8819    0.1564    0.4820
    0.3395    0.4228    0.7011    0.1280    0.6692    0.8555    0.1206

  Columns 8 through 9

    0.5895    0.6171
    0.2262    0.2653
    0.3846    0.8244
    0.5830    0.9827
    0.2518    0.7302
    0.2904    0.3439

ii =

     6

a =

    0.9516    0.9203    0.0527    0.7379    0.2691    0.4228

[CHOI HYUNDUK]

Thank you a lot!