integers error in matlab
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i keep getting this error in matlab
this equation is in a for loop
Array indices must be positive integers or logical values.
n= (( 4.6 / 2 ) / 0.1 ) + 1
for i = 1 : n
k( i ) = (4 * x ( i )*( x ( i ) / 2 ))-( R_2b * x ( i ));
1 个评论
madhan ravi
2019-3-13
post the full code , you seem to be riddling the readers
回答（2 个）
The 3 lines of code you presented are fine. My guess is that you're executing the 3rd line manually, perhaps from the command window, before 'i' is defined. When 'i' is not defined, matlab uses 'i' to represent the imaginary number 0 + 1i. This cannot be used as an index value. To test that I'm correct, define i=1 and then run the 3rd line of code manually. If my hunch is correct, the error doesn't happen when the code is executed naturally from the script/function.
Here's a reproduction of the error
clear all
x = [1 2 3 4 5];
x(i)
%<error>
% Array indices must be positive integers or logical values.
i
%ans =
% 0 + 1i
3 个评论
madhan ravi
2019-3-13
True but OP has defined i right?
madhan ravi
2019-3-13
Yes seems plausible.
Steven Lord
2019-3-13
n= (( 4.6 / 2 ) / 0.1 ) + 1
In theory, n is an integer value, specifically 24.
In practice, because neither 4.6 nor 0.1 are exactly representable as double precision numbers, it isn't. It's very, very close but it's not exactly an integer value.
>> n-24
ans =
-3.5527136788005e-15
You can round n so it's exactly an integer value and use that in your for loop.
>> rn = round(n);
>> rn-24
ans =
0
Alternately if you're trying to compute the number of elements the x variable should have, don't. Ask the variable how many elements it has.
n = numel(x)
Depending on whether or not what you've shown is the entire loop body, there may be ways to vectorize the loop and eliminate it entirely.
2 个评论
Super good point! However, that's not what's causing the error "Array indices must be positive integers or logical values." It merely only computes 23 of the 24 iterations.
n= (( 4.6 / 2 ) / 0.1 ) + 1
x= 1:24;
R_2b = 1;
for i = 1 : n
k( i ) = (4 * x ( i )*( x ( i ) / 2 ))-( R_2b * x ( i ));
end
length(k)
%ans = 23 (should be 24 if n were an integer)
Steven Lord
2019-3-13
You're right, I was too focused on the floating-point issue. Because n is just smaller than 24 the expression 1:n is the same as 1:23.
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2019-3-13
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