I have this function in matlab-
function [c,arr2]=dist1(i,c,arr1,arr2,A,mx,point)
for j=i+1:mx
if arr1(i,j)==1 & A(j)~=0
x1=point(i,1);
y1=point(i,2);
x2=point(j,1);
y2=point(j,2);
d=((((x1-x2).^2)+((y1-y2).^2)).^(0.5));
if d< 0.5
arr2(c)=i;
c=c+1;
[c,arr2]=dist1(j,c,arr1,arr2,A,mx,point);
end
end
end
end
When I call this function this function I get following error-
Integers can only be raised to positive integral powers.
Error in dist1 (line 9)
d=((((x1-x2).^2)+((y1-y2).^2)).^(0.5));
This works fine if I remove power of 0.5 in the calculation of d.Why am I getting this error,there seems to be nothing wrong in this statement.Also I checked the values of x1,x2,y1,y2 in the preceding lines and they are
x1=208 y1=171 x2=207 y2=162
The error is pretty explicit:
Integers can only be raised to positive integral powers.
Your x1, x2, y1, y2 variables seem to be of an integer data type (such as uint8, int32, ...). They need to be double (or single) to perform that operation. So, try
d = double((((x1-x2).^2)+((y1-y2).^2)).^(0.5))^0.5;
Note also that, since x1, x2, y1, y2 are scalars, you could remove the dots:
d = double((((x1-x2)^2)+((y1-y2)^2))^0.5)^0.5;
Related
I'm trying to build a function in Matlab which generates a Taylor series around 0 for sine and the user can insert x (the value for which the sine is approximated) and a maximum error. Thus I want the function to check the maximum error and from this maximum it generates the amount of elements in the Taylor series.
I get the following error:
Error using factorial (line 20)
N must be an array of real non-negative integers.
Error in maxError (line 19)
y(x) = y(x) + (-1)^(j) * x^(2j+1)/factorial(2j+1)
Below my code.
function [n] = maxError(x,e);
%Computes number of iterations needed for a given absolute error.
n=1;
while abs(x)^(n+1)/factorial(n+1) >= e
n = n+1;
end
if mod(n,2) == 0
n=n+1;
end
y=#(x) x;
j=1;
while j<n
y(x) = y(x) + (-1)^(j) * x^(2j+1)/factorial(2j+1)
j=j+1;
end
return
I think I get the error because the factorial function can only take up integers, but the way I see it, I am feeding it an integer. Since j=1; and then gets larger by one per iteration, I don't see how Matlab can perceive this as something else than a integer.
Any help is appreciated.
You are using j as an indexing variable, which is also the complex number in Matlab, and your are forgetting a * multiply.
You can use j as a variable (not recommended!) but when you are putting a number in front of it, Matlab will stil interpret is as the complex number, and not as the variable.
Adding the multiplication symbol will solve the issue, but using i and j as variables will give you these hard to debug errors. If you had used a, the error would have been easier to understand:
>> a=10;
>> 2a+1
2a+1
↑
Error: Invalid expression. Check for missing multiplication operator, missing or
unbalanced delimiters, or other syntax error. To construct matrices, use brackets
instead of parentheses.
Why does the error message of this code return: "Subscript indices must either be real positive integers or logicals.", when I am using ceil for every subscript?
A=1:1:100;
B=1:1:100;
C=1;
D=1:1:100;
E=2;
F=1:1:100;
G=1:1:100;
H=0.1:0.1:10;
fun_1=#(t)integral(#(ti)G(ceil(ti)).*H(ceil(t-ti)),0.1,t-1);
fun_2=#(t)integral(#(ti)G(ceil(ti)).*B(ceil(ti)).*(C.*D(t).^E)./F(t).*...
exp(-integral(#(x)(C.*D(ceil(x)).^E)./F(ceil(x)),ti,5)-K.*(t-ti)),0.1,t-
1,'ArrayValued',true);
I=500;
J=1000;
K=2;
fun_3=#(t)I*integral(#(ti)min(fun_2(ceil(ti)),J).*exp(-(K+I).*(t-ti)),0.1,t-
1);
t=1:1:5;
figure(1)
fplot(fun_1,t);
figure(2)
fplot(fun_2,t);
figure(3)
fplot(fun_3,t);
fplot see documentation Called as fplot(f,xinterval) evaluates your function handle f over the interval xinterval. IT will evaluate f at automatically determined steps along that given interval.
From the docs:
xinterval — Interval for x [–5 5] (default) | two-element vector of
form [xmin xmax]
You seem to be trying to specify exactly where you want your functions evaluated
t=1:1:5;
...
fplot(fun_1,t);
But it doesn't work that way. What is happening is that fplot is evaluating the function from 1 to 2 (the first 2 elements of t). So for example it might feed values of t = 1, 1.05, 1.1,... ,2 into your fun_# functions.
You can tell this because you first function which does work actually plots over the x-range of 1 to 2.
The reason you are getting a subscript indices error is because in fun_2 you have this ...(C.*D(t).^E)./F(t).*... Since fplot is feeding in values for t which are spaced between 1 and 2 (ex. 1.1) that is not a valid index.
If you really just want the values of your functions at t = 1:1:5 The you probably do not want to use fplot and just want evaluate the functions at those times and plot it.
y = feval(fun_1,t);
plot(t,y)
EDIT: The above code doesn't work
You will need to do something like the code below. This is because the 2nd & 3rd trems to the intergral function need to be scalar (1x1). If you feed them an array for t then they crash. So evaluate at each t not all at once.
figure(1)
y_1 = arrayfun(fun_1,t);
plot(t,y_1);
figure(2)
y_2 = arrayfun(fun_2,t);
plot(t,y_2);
figure(3)
y_3 = arrayfun(fun_3,t);
plot(t,y_3);
Note: the Third function still errors ... and I'm not 100% sure why. I didn't really look at it.
Matlab (2015a) is behaving weirdly: a number of builtin functions are not responding as expected. For instance, typing
ttest([1 2], [1 2])
results in
Error using size
Dimension argument must be a positive integer scalar within indexing range.
Error in nanstd (line 59)
tile(dim) = size(x,dim);
Error in ttest (line 132)
sdpop = nanstd(x,[],dim);
If I do a which for each of these functions:
which size
which nanstd
which ttest
I get, respetively:
built-in (C:\Program Files\MATLAB\R2015a\toolbox\matlab\elmat\size)
C:\Program Files\MATLAB\R2015a\toolbox\stats\eml\nanstd.m
C:\Program Files\MATLAB\R2015a\toolbox\stats\stats\ttest.m
Each of these files looks fine, except that size.m has each one of its rows commented out.
What could be the problem here?
Perhaps related to your problem:
ttest for R2013a makes the following call:
sdpop = nanstd(x,[],dim);
The helpfile for R2013a version of nanstd states:
Y = nanstd(X,FLAG,DIM) takes the standard deviation along dimension DIM of X.
On the other hand, nanstd in the 2005 nansuite package downloaded off Mathworks file exchange states:
FORMAT: Y = nanstd(X,DIM,FLAG)
Notice how DIM and FLAG are reversed!
If I call R2013a's ttest such that it makes a call to the old, 2005 nansuite function nanstd, Matlab generates an error similar to yours:
Error using size
Dimension argument must be a positive integer scalar within indexing range.
Error in nanmean (line 46)
count = size(x,dim) - sum(nans,dim);
Error in nanstd (line 54)
avg = nanmean(x,dim);
Error in ttest (line 132)
sdpop = nanstd(x,[],dim);
If [] is passed as DIM instead of FLAG, then nanstd's call to size(x, DIM) triggers an error because [] is not a positive integer scalar. If something like this is the cause, the broader question is, what's going on with your Matlab install or setup or downloads or whatever such that you're calling archaic code? Or why is that archaic code even around? I don't know at what point in Matlab's release history that nanstd(x, FLAG, DIM) became supported (instead of simply nanstd(x, DIM))?
Archive: below is my old answer which misdiagnosed your problem
Both of your sample vectors x and y are the same (i.e. [1,2]). The estimated variance of the difference is 0, and all your stats are going to blow up with NaN.
Do the stats step by step, and it will be clear what's going on.
x = [1; 2]; % Data you used in the example.
y = [1; 2]; % Data you used in the example.
z = x - y; % Your call to ttest tests whether this vector is different from zero at a statistically significant level.
Now we do all the stats on z
r.n = length(z);
r.mu = mean(z);
r.standard_error = sqrt(var(z,1) / (r.n-1)); % For your data, this will be zero since z is constant!
r.t = r.mu ./ r.standard_error; % For your data, this will be inf because dividing by zero!
r.df = r.n - 1;
r.pvals(r.t >= 0) = 2 * (1 - tcdf(r.t(r.t>=0), r.df)); % For your data, tcdf returns NaN and this all fails...
r.pvals(r.t < 0) = 2 * tcdf(r.t(r.t<0), r.df);
etc...
This should match a call to
[h, p, ci, stats] = ttest(x-y);
I am stuck on trying to write the output of a for loop into a vector. The issue is that once it goes through the equation, it stops and spits out this error: "Subscript indices must either be real positive
integers or logicals."
I've looked into what this error is and I've checked my code. There are no negative integers being output, nor any zeros. The for-loop, function and the fzero command are all good, if I comment out MW4p(y4,:) = MW4 I get all of my answers on the command window. I just can't put them into an array.
for y4 = linspace(1.4,1.67,100)
x0 = 12; %Starting Point
fun = #(MW4) (y4.*MW1.*T3)./(y1.*MW4.*T2) - ...
(((1+((y4+1)./(y4-1)).*p5p2).*((2.*y4)./(y4-1))) ./ ((1+...
((y1+1)./(y1-1)).*p5p2).*((2.*y1)./(y1-1)))) .* (((2./(y1-1))...
./ (2./(y4-1))).^2);
MW4 = fzero(fun,x0)
MW4p(y4,:) = MW4
end
where
y1 = 1.67;
MW1 = 39.55;
T3 = 250; %k
T2 = 700; %k
p5p2 = 2.307; %Determined from T5 desired
MW4p = ones(1,100);
In earlier attempts, I've tried MW4p(y4) = MW4 which had this error: "Attempted to access MW4p(1.4); index must be a
positive integer or logical." So I added a colon. Putting this command outside of the for-loop simply had it access the last value of the loop and return the same error.
The error message is telling you exactly what you need to know. y4 is not an integer. When using vectors, matrices, cells, or arrays you can't index with a non-integer number.
vectors, matrices and arrays are discrete structures that can only be indexed with discrete values, though obviously they can be used to store non-discrete values.
Unless I'm missing something, your choices are:
Through whatever data structure: store y4 and its corresponding MW4 value as a tuple; for example, you might have an array of structs or cells where one element is y4 and another element is the array returned from MW4
Use something like this answer, using a hash map. I haven't looked into this closely, but my guess is you can use a float as the key.
I want to evaluate the simple example of integral
a = max(solve(x^3 - 2*x^2 + x ==0 , x));
fun = #(x) exp(-x.^2).*log(x).^2;
q = integral(fun,0,a)
and the error is
Error using integral (line 85)
A and B must be floating-point scalars.
Any tips? The lower limit of my integral must be a function, not a number.
The Matlab command solve returns symbolic result. integral accepts only numeric input. Use double to convert symbolic to numeric. As your code is written now, already max should throw an error due to symbolic input. The following works.
syms x;
a = max(double(solve(x^3 - 2*x^2 + x)));
fun = #(x) exp(-x.^2).*log(x).^2;
q = integral(fun,0,a)
Output: 1.9331.
the lower limit of my integral must be a function, not a number
integral is a numeric integration routine; the limits of integration must be numeric.
Check values of a by mouse over in breakpoint or removing the ; from the end of the line so it prints a. Based on the error, a is not a scalar float. You might need another max() or double() statement to transform the vector to a single value.
Solve Help : http://www.mathworks.com/help/symbolic/solve.html
Integral Help : http://www.mathworks.com/help/ref/integral.html