Finding difference of array using alternative indexes

Hello everyone,

I hope you're doing, I've simple question I've array (1, 24), now I want to findout the difference and divide, like [element1-element2 element2- element3 element3-element4...............element24-element23], what I did as follows

for i=1:24
a=1x24
%first case
a1(i)=a(i)-a(i+1)./i-(i+1), 
% Second case
 a1(i)=a(i)-a(i-1)./i-(i-1)
end

However, it is clear the index causing error (first case: Index exceeds the number of array elements (24). second case: Array indices must be positive integers or logical values.)

, could you please help me out in this situation.

[shane watson]

@Adam Danz, can you help here?

[Stephen23]

https://de.mathworks.com/help/matlab/ref/diff.html

[shane watson]

Thanks @Stephen Cobeldickfor link, however, the problem is one dimension is always reduced, how to compensate that? and this is mentioned in the question

[Adam Danz]

Consider a 1x4 vector x,

x = [2 5 3 6]

The difference you describe would be

d = [2-5, 5-3, 3-6]

or

d = [-3 2 -3]

which is a 1x3 vector. So your loop must either be

for i = 2:numel(x)

or

for i = 1:numel(x)-1

[shane watson]

@Adam Danz I'm sorry for the late response, however, the issue was "reduction of index and value" so in my case I need 1 x 24 matix at the end while it is reduces to 23, and your given last two steps are good but the it creates the same problem.

[Adam Danz]

I'm trying to show you that the loss of one value when numerically differentiating is not a problem - it's exactly the expected behavior. Carefully look at my previous comment again to understand why you're losing a value.

I'll add an answer to suggest an alternative.

Answer 1

[Adam Danz]

The loss of 1 value when differentiating with diff(x,1) is the expected behavior. This function computes the difference between adjacent values in a matrix [a b c d] and there is n-1 comparisons.

Perhaps you're looking for the numeric gradient.

Comparison between gradient and diff

y = exp([1:.1:3]);
d = diff(y); 
g = gradient(y); 

size(y)

ans = 1×2

1 21

size(d)

ans = 1×2

1 20

size(g)

ans = 1×2

1 21

x = 1:numel(y)

x = 1×21

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21

hold on
plot(x(2:end),d,'b-','DisplayName','diff')
plot(x,g,'r-','DisplayName','gradient')
legend

[shane watson]

@Adam Danz, Thanks for the detail answer, I understood what you mentioned about loss of 1 value when differentiating, however as you showed the comparision of diff and gradient almost same, it is difficult to digest, espically in case when i'm computing the load demand of households and any single value changes it is problem for me. For example I used your example with sightly different values of "y" then the graph seems pretty different incase of diff and gradient.

ni=[6 7 8 9 7 5 5 6 7 8];
d = diff(ni); 
g = gradient(ni); 
x = 1:numel(ni);
hold on
plot(x(2:end),d,'b-','DisplayName','diff')
plot(x,g,'r-','DisplayName','gradient')
legend

[Adam Danz]

I need to know more about your goal. In your original question, the indexing error was caused by the loss of 1 value due to differentiating using diff(). Maybe you don't need to differentiate. Maybe you need to differentiate using a different method. Or maybe what you're doing is fine and you need to understand the output.

Answer 2

[KSSV]

n = length(a) ; 
iwant = (a(2:n)-a(1:n-1))./((2:n)-(1:n-1)) ;

Also have a look on geadient.

[shane watson]

Hello @KSSV, Thanks for the answer but the final dimension of iwant is 1x23 and not 24,

secondly can we do it with loop?

[KSSV]

Use gradient. You will get 1*24 dimension.

[KSSV]

iwant = gradient(a) ; 

[shane watson]

@KSSVIt's not showing the right results unfortunately. :(