Question regarding a condition difference.

Hi, I'm new on this forum, hope I will be doing good.

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Here's my question: I have 3 matrix, for this example let's say 3 vectors

A, B and C (is a scalar)

I want to do the following operation :

for each element in vector A, if the element > C then A-B, else the element =0.

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The only way I know is the do a loop for echa element and check the condition. I don't like this option since A and B are going tho be 25200 rowa and 15 column. I assume it's going to take tome much time this way.

Any would be appreciate.

regards

Gimpy

[Gimpy]

I want to do the following operation :

for each element in vector A, if the element > C then A-B, else the element =0. note : else A-B=0 if A<C (it's only a precision of my question)

[Albert Yam]

What have you tried?

[Azzi Abdelmalek]

what about A=c? your conditions are > and <

Answer 1

[Azzi Abdelmalek]

here an example

 clear

 a1=[88;57;42;100];
a2=[98;87;32;80];
A=[a1 a2];
b1=[78;5;32;106];
b2=[88;87;35;88];
B=[b1 b2];
c=[ones(4,1)*80.5 ones(4,1)*89.27];
result=A-B; result=A-B;result(arrayfun(@(x) x<0,A-c))=0

% the result is result =

    10    10
     0     0
     0     0
    -6     0

[Gimpy]

Looks pretty good but I will have to tried it on my computer when I get home. You're very good Sir. What about if I want to find the Scalar "C" thats going to maximise:

A, B and C(is a scalar)

For each element in vector A, if the element > C then A-B, else the element =0.

Do I have to do sophisticated optimisation to do that? If you now what method Iwould be great. I smart anwser like previously would be more efficient.

Thanls again for your help.

[Gimpy]

To be more precise : if A>=C then A-B, else A-B=0 (result wanted in a mtrix for each rows).

Then I want to find wich "C" is going to maximise the previous condition.

Thanks again

Regards

Gimpy

[Azzi Abdelmalek]

if i have understood your question that will be

 c= min(A)

Answer 2

[Gimpy]

first let's get back to my initial question. Here's what I wan't to do exactly:

A=rand(10,2)*100;

B=rand(10,2)*100;

c1=ones(10,1)*80.5;

c2=ones(10,1)*89.27;

c=[c1 c2];

result=A-B;

result(find(A-c>=0))

the only problem I have with this is that the variable"result" is an "nx1", I would like to have the results as nx2 (maybe with 0 or something). In this case the variable "results" would be a 10x2 (i guess with zeros where the condition (A-c>=0 is not met.

[Gimpy]

I mean all the results of the test (A-c>=0 came in a vector instead of a matrix with their respective column.

[Azzi Abdelmalek]

instead

 result(find(A-c>=0))

use

 result(arrayfun(@(x) x<0,A-c))=0

[Gimpy]

not exactly(it don't seems to be working some conditions are respected but not report in"R"). Here's what I'm looking exactly:

a1=[88;57;42;100];

a2=[98;87;32;80];

A=[a1 a2];

b1=[78;5;32;106];

b2=[88;87;35;88];

B=[b1 b2];

c=[ones(4,1)*80.5 ones(4,1)*89.27];

result=A-B; result=A-B;result(find(A-c>=0))

ans =

    10
    -6
    10

*** the result is good but I want it this way:

10 10

0 0

0 0

-6 0

[Azzi Abdelmalek]

i have already updated a code

clear
a1=[88;57;42;100];
a2=[98;87;32;80];
A=[a1 a2];
b1=[78;5;32;106];
b2=[88;87;35;88];
B=[b1 b2];
c=[ones(4,1)*80.5 ones(4,1)*89.27];
result=A-B; result=A-B;result(arrayfun(@(x) x<0,A-c))=0

% the result is result =

    10    10
     0     0
     0     0
    -6     0

Answer 3

[Gimpy]

Yes that's exactly what I'm looking for.

The seceond step(and the last one)I want to do the sum of the variable"results" and get the maximum.In this case the result would be in the form of: 1x2

So clearly I want to find the "c" that maximise "result2":

result2=sum(result)

(in the previous example the "c" were 80.5 and 89.27

So the point is when we do

result=A-B;

result(~(A-c>=0)) = 0

Some "A-B" yield a negative results so I want to find the c that

optimize (thus skip the A-B that are negative.

Let's set the "c" as c1 and c2

the final optization is going to be in the following dimesnion:

c=[ones(4,1)*c1 ones(4,1)*c2];

Thank you in advance

Gimpy

[Azzi Abdelmalek]

it's not obvious

[Gimpy]

ok here's my example again: A=[45;37;32;50]

B=[17;100;200;10]

c=??(the value we look for)

The condition to test: A=[45;37;32;50]

B=[17;100;200;10]

c=? (the value we are looking for)

the condition to test: A >= C then A-B, else A-B=0 in my example the C we look for is something between 37.01 and 50 because.

In excel for example to be more specific: cells A2= 45,A3=37,A4=32,A5=50

B2=17, B3=100, B4=200, B5=10

C2:C5=37.01 (in this case i know the results by trial and error).

The condition to test is:

=IF(A2>=C2,(A2-B2),0)= 28 =IF(A3>=C3,(A3-B3),0)=0 =IF(A4>=C4,(A4-B4),0)=0 =IF(A5>=C5,(A5-B5),0)=40

Total: 68

Any value between 37.01 and 50 would give the save result. A way to find the lower and uper bound would be appreciate of only one...

Thanks again for your patience and help.

Answer 4

[Azzi Abdelmalek]

if you give the interval, we can use a while loop

 A=[45;37;32;50]
 B=[17;100;200;10]
 c=37.01; som=[];
 while c<50
   result=A-B;
   result=A-B;result(arrayfun(@(x) x<0,A-c))=0
   som=[som ;[sum(result) c]];
   c=c+0.1
 end
   somax=max(som(:,1))
   c_result=som(find(som==somax),2)
   lower_c=min(c_result)
   upper_c=max(c_result)

[Gimpy]

We don't know the interval. It s the result we are looking for. Again the equation to maximize is:

For each element in "A"(let's call them ai) if this ai is > to c then

A-B, else 0

Need to find the C that maximize result2

result=A-B;

result(~(A-c>=0)) = 0

result2 =sum(result)

Answer 5

[Gimpy]

Here’s my problem with more explanation again:

Matrix A,B,C :22 500 x18

Matrix c : each colum are scalar

Exemple :[50;50;50;50 72.25;72.25;72.25;72.25]

… Here’s what I want to do :

For each elements in A, lets call them ai, if ai >ci then

ai - bi

else

ai - bi =0

Here’s what I do in matlab(for simplification I do a 4x1)

a1=[88;57;42;100];

a2=[98;87;32;80];

A=[a1 a2];

b1=[78;5;32;106];

b2=[88;87;35;88];

B=[b1 b2];

c=[ones(4,1)*80.5 ones(4,1)*89.27];

result=A-B;

result(~(A-c>=0)) = 0

result2=sum(result)

****Now here’s the point : I want to find the matrix c that’s going to maximise :

result2=sum(result).

The results should be an interval

In the revious example I’am expecting MATLAB to give something like:

resultat2= 68

C_lower= 37

C_upper=45

Or at least one one of the 2 results (37 or 45).

I found this result by calculating result2=sum(result) by trial and error until

I get the maximum value(result2=68).

Please keep I mind that I want to work with 22500x18