Simple Question (I think) about finding zeros(and other number) in an array

I have the following array.

R = [ 1 2 0 1 0 2]

if values in R = 0 I want to replace it with 3, so I did

If R == 0
R = 3
end

So it should end up

R = [ 1 2 3 1 3 2]

However as you probably know, this does not work. How do I make this work? Thanks! Same problem with the following:

P = [ 1 2 3 4 5]
Q = [ 5 4 3 2 5]

I want to find positions where P == Q so I tried

if P == Q
Q = 6
end

ie all positions where P = Q change to 6 so it should end up with

Q = [ 5 4 6 2 6]

Answer 1

[madhan ravi]

TRY THIS:

    R = [ 1 2 0 1 0 2]
    R(R==0)=3

    P = [ 1 2 3 4 5]
    Q = [ 5 4 3 2 5]
    Q(Q==P)=6

[madhan ravi]

If it works please do accept my answer.

[Manne Plok]

Hi, what if I wanted to change two things whenever Q == P, eg I wanted to change Q values to 6 and increase count by 1?

[madhan ravi]

Can you give an example and your required output ?

[madhan ravi]

Then you should be using loop.

for i=1:length(Q)

if Q(i)==P(i)

Q(i) = 6+1

else

continue

end end

[Image Analyst]

Count? Count of what?

[madhan ravi]

But give an example so that we could meet your requirements.

[Manne Plok]

I have two other arrays C and D.

C = [ 3 4 5 6 7 8]
R = [ 1 2 0 1 0 2]
D = [ 8 7 6 5 4 3]
if R == 0
R = 3
C = C - 1
D = 1
end

So whenever R = 0, that same position in C should lower by 1 too. Also that same position in D should be 1.

Intended output:

R = [1 2 3 1 3 2]
C = [3 4 4 6 6 8]
D = [8 7 1 5 1 3]

[madhan ravi]

C = [ 3 4 5 6 7 8]

r = [ 1 2 0 1 0 2]

D = [ 8 7 6 5 4 3]

for i=1:numel(r)

if r(i) == 0

r(i) = 3

C(i) = C(i) - 1

D(i) = 1

else continue

end

end

[Manne Plok]

Yes this works! I was trying to avoid for loops (for speed) but I guess you cant avoid them forever :(

[madhan ravi]

C = [ 3 4 5 6 7 8]

R = [ 1 2 0 1 0 2]

D = [ 8 7 6 5 4 3]

D(R==0)=1

C(R==0)=C(R==0)-1

R(R==0)=3

Without loop

[madhan ravi]

Give a vote if you find the latter without loop useful.