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

Question

ANKUR WADHWA 2020-5-3

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https://ww2.mathworks.cn/matlabcentral/answers/522649-arithmetic-operation-on-elements-of-a-vector-satisfying-a-given-condition-using-logical-indexing

评论： Ameer Hamza 2020-5-4

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.

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ANKUR WADHWA 2020-5-3

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.

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Answer 1

Ameer Hamza 2020-5-3

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/522649-arithmetic-operation-on-elements-of-a-vector-satisfying-a-given-condition-using-logical-indexing#answer_430029

编辑：Ameer Hamza 2020-5-3

在 MATLAB Online 中打开

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;

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ANKUR WADHWA 2020-5-4

在 MATLAB Online 中打开

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 2020-5-4

在 MATLAB Online 中打开

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