Arithmetic Operation on Elements of a Vector satisfying a given condition using Logical Indexing

I have a problem statement where i have to change the elements of a vector depending on some condition

the vector comprises of random number between 4 to 12 i.e. 4,5,6,7,8,9,10,11,12

for eg. the given vector is A = [ 5,6,8,9,7,5,4,12,11,9,10]

now i have to add a scalar to the given vector , which can be either a -ive number or a +ive number

and any element of the resultant vector greater than 12 should be wrap around and any if the element is smaller than 4 then also it should be wrapped around.

for eg. if my input scalar is 3 , then the element 11 in input vector after addition operation will become 15 which is greater than 12

and should be wrapped around and changed to 5 , like shift in a circular manner.

in second case taking the input scalar as -4 to be added to the vector A , the element 6 will become 2 and should be wrapped around and changed to 11

I can solve this using for loop and if else statement , but i am interested in solving this via logical indexing.

any help on that will be greately appreciated.

[ANKUR WADHWA]

another point is for eg. i have a vector v where i want to replace all the negative number with 0

v = [ 5 4 3 -2 8 9 10 5 8 3 4 7 12 5]

i can use

v(v<0) = 0 (this works fine and return me a modified vector v of lenght 14)

also i can do the operation v = v-3 and this also return me the modified vector v of lenght 14 again

i am not able to understand why can i combine the above to operation in the same command like

v(v<0) = v-3;

this command gives me an error that

In an assignment A(:) = B, the number of elements in A and B must be the same.

Answer 1

[Ameer Hamza]

For first question

A = [5,6,8,9,7,5,4,12,11,9,10];
scalar = -4;
lb = 4;  % lower bound
ub = 12; % upper bound
A_shift = mod((A-lb)+scalar, ub-lb+1)+lb;

Result

>> A
A =
     5     6     8     9     7     5     4    12    11     9    10
>> A_shift
A_shift =
    10    11     4     5    12    10     9     8     7     5     6

For second question. Correct syntax is

v(v<0) = v(v<0)-3;

[ANKUR WADHWA]

ThanQ Ameer for your time and providing such a quick reply and the solution to the problems,

your solution works great :-)

I was also interested in using logical indexing to get the same results , below is a snippet of my code for the same , but it isn't giving the correct answer in all the cases , request your help on the same in understanding the bug in the below code and also why we use (upper_bound - lower_bound + 1)

% the bounds are 32 and 126 in this case
function coded = shift(a,b)
if abs(b) > 126
    b = rem(b,95); 
end
original = a + b;
original(original > 126) = original(original > 126) - 126 + 31;
original(original < 32) = original(original < 32) - 31 + 126;
coded = char(original)

[Ameer Hamza]

Try this

A = [5,6,8,9,7,5,4,12,11,9,10];
scalar = 4;
lb = 4;  % lower bound
ub = 12; % upper bound

A_ = A + scalar;
A_(A_ > ub) = A_(A_ > ub) - ub + lb + 1;
A_(A_ < lb) = A_(A_ < lb) - lb + ub + 1;

A_shift = mod((A-lb)+scalar, ub-lb+1)+lb;

12-4+1=9 is necessary because the mod will keep the value from 0 to 9-1. If you set it to 8 then you will only get digits from 0 to 7.

[ANKUR WADHWA]

Hi Ahmed , in the new solution you provided A_ is not used to compute the final variable A_shift

so the new solution is exactly equivalent to the previous solution you provided.

[Ameer Hamza]

Yes. I just added the last line as example. These lines

A_ = A + scalar;
A_(A_ > ub) = A_(A_ > ub) - ub + lb + 1;
A_(A_ < lb) = A_(A_ < lb) - lb + ub + 1;

shows the correct way to write the function shift().

[ANKUR WADHWA]

The code is giving incorrect results for coding and decoding when the scalar is greater than upper bound

or smaler than lower bound , for eg.

when i executed the above code with the below driver , the coded and decoded values are not same

and in some case they are wrong.

wrap     = shift('1234', 96)
back     = shift(wrap,  -96)

[Ameer Hamza]

Also wrap the shifting factor

wrap     = shift('1234', 96)
back     = shift(wrap, -96)
% the bounds are 32 and 126 in this case
function coded = shift(A, b)
    ub = 126;
    lb = 32;
    b = rem(b, ub-lb);
    A_ = A + b;
    A_(A_ > ub) = A_(A_ > ub) - ub + lb + 1;
    A_(A_ < lb) = A_(A_ < lb) - lb + ub + 1;
    coded = char(A_);
end