Numerical derivative in matlab
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Hello,
I'm using Matlab 2012 and would need the numerical derivative of a function. What is the Malab command for that ?
Thank you in advance
1 个评论
Leah
2013-9-26
I like your user name
采纳的回答
Azzi Abdelmalek
2013-9-26
If you have a series of data
t=[0 1 2 3 4 5] % Vector time
y=[1 2 4 8 12 13] % Your signal
% The derivative can be approximated :
diff(y)./diff(t)
%You can also use symbolic derivative
syms t y
y=cos(t)
dy=diff(y)
更多回答（2 个）
Rohit Garud
2020-8-16
编辑：Rohit Garud
2020-8-16
You can use the gradient() function in newer versions of matlab
dy_by_dt = gradient(y(:)) ./ gradient(t(:))
11 个评论
Daphne PARLIARI
2020-12-29
Indeed this works like a charm.
Do you have any idea how to convert dy_by_dt to percentage change?
Walter Roberson
2020-12-29
percentage_change = dy_by_dt ./ y(:) * 100;
Daphne PARLIARI
2020-12-29
I have the attached data, which are daily values of temperature and deaths. I want to calculate percent increase in mortality per 1oC increase above a certain heat threshold. Eg. if the threshold is 30oC, there is a 2.1% increase in mortality per 1oC increase in temperature (31oC, 32oC, ...).
Currently I have written this code although I am not sure it works the way it should:
idx = Daily_Data.Daily_T(:) > 30;
Daily_Data_new = Daily_Data(idx,:);
dy_by_dt = gradient(Daily_Data_new.Daily_Deaths(:)) ./ gradient(Daily_Data_new.Daily_T(:));
percentage_change = dy_by_dt ./ Daily_Data_new.Daily_Deaths(:) * 100;
I would appreciate any help...
I don't think that is right at all.
Daily_Data = readtable('Daily_Data_MoT.xlsx');
idx = Daily_Data.Daily_T(:) > 30;
Daily_Data_new = Daily_Data(idx,:);
[G, Temperature] = discretize(Daily_Data_new.Daily_T, 30:37);
d_count = accumarray(G, Daily_Data_new.Daily_Deaths, [length(Temperature), 1]);
d_mean = accumarray(G, Daily_Data_new.Daily_Deaths, [length(Temperature), 1], @mean);
bar(Temperature, d_count); xlabel('temperature (C)'); ylabel('Total deaths')
bar(Temperature, d_mean); xlabel('temperature (C)'); ylabel('average deaths')
You could do some kind of statistical analysis to show that the mean deaths increases with temperature 30 to 32, but it decreases after that up to 34, and then increases again at 36 but not to as much as 32.
Walter Roberson
2020-12-30
You could look at gradient(d_mean) but I do not think you will find it enlightening.
Daphne PARLIARI
2021-1-4
编辑：Daphne PARLIARI
2021-1-4
Wow, the graphs were great, thank you!
Could you please explain what does every term in
d_count = accumarray(G, Daily_Data_new.Daily_Deaths, [length(Temperature), 1]);
do and what is the difference with
d_mean = accumarray(G, Daily_Data_new.Daily_Deaths, [length(Temperature), 1], @mean) ?
In particular, I don't quite get what G is doing.
Walter Roberson
2021-1-4
编辑：Walter Roberson
2021-1-5
G is a grouping number, similar in spirit to using findgroups(), except that findgroups() treats every unique value as a different bin but you want to treat ranges as different bins.
[G, Temperature] = discretize(Daily_Data_new.Daily_T, 30:37);
is pretty much the same as
G = floor(Daily_Data_new.Daily_T - 30) + 1;
in that it maps a degree at a time, [30, 31) -> 1, [31,32) -> 2, [32,33) -> 3 and so on. It is the same thing exactly as histogram bin number doing a histogram of temperatures above 30 one degree per bin.
d_count = accumarray(G, Daily_Data_new.Daily_Deaths, [length(Temperature), 1]);
That code is the same as
d_count = zeros(length(Temperature),1);
DD = Daily_Data_new.Daily_Deaths;
for K = 1 : length(G)
g = G(K);
d_count(g) = d_count(g) + DD(K);
end
That is, it uses the G entries as index numbers into the array, and accumulates all of the Daily_Death values corresponding to the same index. Another way of looking at it would be
d_count = zeros(length(Temperature),1);
for g = 1 : 7
d_count(g) = sum(DD(G == g));
end
and
d_mean = accumarray(G, Daily_Data_new.Daily_Deaths, [length(Temperature), 1], @mean)
would correspond to
d_mean = zeros(length(Temperature),1);
for g = 1 : 7
d_mean(g) = mean(DD(G == g));
end
and in turn you could rewrite the code to:
Temperature = 30:37;
nt = length(Temperature)-1;
d_count = zeros(nt,1);
d_mean = zeros(nt,1);
DD = Daily_Data_new.Daily_Deaths;
DT = Daily_Data_new.Daily_T;
for g = 1 : nt
d = DD(DT >= Temperature(g) & DT < Temperature(g+1));
d_count(g) = sum(d);
d_mean(g) = mean(d);
end
Daphne PARLIARI
2021-1-5
Thanks for the detailed answer, that made all clear! Let me pose another question, if you have the time of course.
I want to calculate percent change with reference to a constant temperature, e.g. 30oC, and add that percentage on top of each column. When I try
x = b.XData;
y = b.YData;
reference_temp = y(1); % 30oC in plot
percent_change = 100 * (y - reference_temp) / reference_temp;
for k = 1:length(percent_change)
text(x(k+1),y(k+1),sprintf('%0.1f%%',percent_change(k)),...
'HorizontalAlignment','center',...
'VerticalAlignment','bottom')
end
I am getting the Error: Index exceeds the number of array elements (8).
Walter Roberson
2021-1-5
Is b the handle of the bar() ?
length(percent_change) is going to be the same as length(y) . You loop k 1 to length percent_change which is the same as loop k 1 to length y. You access y(k+1) but when k becomes length(y) then you are accessing y(length(y)+1) which is after the end of y.
Daphne PARLIARI
2021-1-6
Oh I got confused... What would be the solution to that?
Walter Roberson
2021-1-6
for k = 2:length(percent_change)
text(x(k),y(k),sprintf('%0.1f%%',percent_change(k)),...
'HorizontalAlignment','center',...
'VerticalAlignment','bottom')
end
Tamas Kis
2021-4-14
1 个投票
Function that can differentiate a set of points given as vectors (such as how you would use cumtrapz to integrate), or differentiate a function defined using a function handle. It can differentiate at any specified point(s) or over an entire domain. Also is of slightly higher accuracy than using diff(y)./diff(t) because it uses central approximation at the interior nodes.
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