Help me find the index value for first minimum value in a matrix.

I have a matrix

A=[ 34 67 78 88 4 15 23 67 78 56 3 2 24 67]

First minimum value in this matrix is 4. (this is first drop). Index of 4 is 5.

After 4 many minimum value is there (i.e) 2.

I don't want to consider the other minimum value 2.

I want index of first minimum value (i.e first drop) as a output...

output =( first minimum value 4 and its index is 5)

someone please suggest me a code to solve this..

Answer 1

[Walter Roberson]

A=[ 34 67 78 88 4 15 23 67 78 56 3 2 24 67]

A = 1×14

34 67 78 88 4 15 23 67 78 56 3 2 24 67

find(diff(A) < 0, 1)

ans = 4

Answer 2

[Bruno Luong]

[~,idx] = min(A)

[Birdman]

Why directly answering to the question? Why not navigating in a direction where this simple line can be easily learned by the OP?

[Divya Ganesan]

Thankyou for your response @Bruno Luong... the code you suggested is considering the whole matrix and gives the index of minimum value... But i want first minimum value (i.e first drop) index to be printed...Please suggest me

[Divya Ganesan]

@Birdman I couldnot get you sir..what is OP

[Walter Roberson]

OP stands for "Original Poster"

[Bruno Luong]

Not sure your definition of "first drop", but try this

d=diff(A);
find(d(1:end-1)<=0 & d(2:end)>=0,1,'first')+1

Answer 3

[Riley]

A=[ 34 67 78 88 4 15 23 67 78 56 3 2 24 67];
flag=0;
for i=1:length(A)-1
    if flag==0&&A(i+1)<A(i)
        flag=1;
        continue;
    end
    if flag==1&&A(i)<A(i+1)
        first_min=A(i);
        idx=i;
        break;
    else
        flag=0;
    end
end
first_min
idx

Hi, OP means original poster which is you here. The code above can find the first minimum value and its indice, only if 4 is the first drop. As you can see, if the number ahead of 4 is smaller than 88, this code fails. This is not the perfect solution. Hope you can find a better one.