Finding difference of array using alternative indexes
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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.
6 个评论
shane watson
2021-6-4
shane watson
2021-6-4
编辑：shane watson
2021-6-4
Adam Danz
2021-6-4
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
2021-6-7
Adam Danz
2021-6-7
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.
采纳的回答
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.
Comparison between gradient and diff
y = exp([1:.1:3]);
d = diff(y);
g = gradient(y);
size(y)
size(d)
size(g)
x = 1:numel(y)
hold on
plot(x(2:end),d,'b-','DisplayName','diff')
plot(x,g,'r-','DisplayName','gradient')
legend
2 个评论
shane watson
2021-6-9
@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
2021-6-9
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.
更多回答（1 个）
n = length(a) ;
iwant = (a(2:n)-a(1:n-1))./((2:n)-(1:n-1)) ;
Also have a look on geadient.
4 个评论
shane watson
2021-6-4
KSSV
2021-6-4
Use gradient. You will get 1*24 dimension.
KSSV
2021-6-4
iwant = gradient(a) ;
shane watson
2021-6-4
编辑：shane watson
2021-6-4
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