- k is i+1 in all cases, so omit k and use i+1.
- Pre-allocate f1 by zeros() to avoid n iterative growing.
- This is equivalent:
Turning a sequence of operations into a function
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I have a sequence of operations that attempts to calculate, whether there is an increase or a decrease in the next step, in comparison with the previous element of a matrix f:
f = [5 6 8 1 2 10 1 0 9 4 6 4]
f1 = [0]
k = 2
for i = 1:length(f)-1
if f(i+1) > f(i)
f1(k) = 1
else
f1(k) = 0
end
k = k+1
end
The code gives correct results.
How can I generalize the code into a function?
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回答(1 个)
Jan
2021-11-18
编辑:Jan
2021-11-18
Start with simplifying the code:
% Version 1:
if condition
a(k) = 1;
else
a(k) = 0;
end
% Version 2:
a(k) = double(condition);
If a is preallocated as double, the casting by double() can be omitted.
f = [5 6 8 1 2 10 1 0 9 4 6 4];
f1 = zeros(size(f));
for i = 1:length(f)-1
f1(i+1) = f(i+1) > f(i);
end
To convert this into a function:
f = [5 6 8 1 2 10 1 0 9 4 6 4];
f1 = compareWithNext(f)
function f1 = compareWithNext(f)
f1 = zeros(size(f));
for i = 1:length(f)-1
f1(i+1) = f(i+1) > f(i);
end
end
Shorter versions:
f1 = [0, f(2:end) > f(1:end-1)]
% Or even simpler:
f1 = [0, diff(f) > 0];
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