Alternative for linspace?

function [xvals, yvals]=A5fourier(L,nmax,numpoints)

%xvals=linspace(-L,L,numpoints);

for z=-0.1:0.001:0.1
    xvals=z;
end 

yvals=zeros(1,numpoints);
for i=1:numpoints
    for j=1:2:nmax
        yvals(i)=(4/pi)*(yvals(i)+(1/j)*sin(j*pi*xvals(i)/L));
        j=j+1;
    end
    i=i+1;
end

L in this function is 0.1, and numpoints is 200. I've been challenged to use a loop rather than linspace to create the same 200 equally spaced points. My current solution ruins my yvals part of the function returning an error "Index exceeds the number of array elements". How can I remedy linspace while mainting the rest of my function?

Answer 1

[DGM]

Why not just use the colon operator?

r = [-4 4];
n = 10;

x1 = linspace(r(1),r(2),n)

x1 = 1×10

-4.0000 -3.1111 -2.2222 -1.3333 -0.4444 0.4444 1.3333 2.2222 3.1111 4.0000

x2 = r(1):range(r)/(n-1):r(2)

x2 = 1×10

-4.0000 -3.1111 -2.2222 -1.3333 -0.4444 0.4444 1.3333 2.2222 3.1111 4.0000

This can definitely be simplified. The iterator in a for-loop doesn't need to be incremented.

yvals=zeros(1,numpoints);
for i=1:numpoints
    for j=1:2:nmax
        yvals(i)=(4/pi)*(yvals(i)+(1/j)*sin(j*pi*xvals(i)/L));
        % j=j+1; % this doesn't do anything
    end
    % i=i+1; % this doesn't do anything
end

[VBBV]

@Sarah Gomez, if your question involves, to use the for loops /other loops to create linearly spaced points, then this could be one option to do so, however, there are several redundant lines of code in your function whcih @DGM pointed out. Further, ihe simplest altrenative would be to use the colon : operator which also was mentioned by @DGM

[xvals yvals] = A5fourier(0.1,200,200)

xvals = 1×201

-0.1000 -0.0990 -0.0980 -0.0970 -0.0960 -0.0950 -0.0940 -0.0930 -0.0920 -0.0910 -0.0900 -0.0890 -0.0880 -0.0870 -0.0860 -0.0850 -0.0840 -0.0830 -0.0820 -0.0810 -0.0800 -0.0790 -0.0780 -0.0770 -0.0760 -0.0750 -0.0740 -0.0730 -0.0720 -0.0710

yvals = 1×200

1.0e+10 * -0.0000 -0.4436 -0.8361 -1.1547 -1.4027 -1.5940 -1.7426 -1.8596 -1.9533 -2.0295 -2.0925 -2.1452 -2.1898 -2.2280 -2.2609 -2.2896 -2.3148 -2.3370 -2.3567 -2.3743 -2.3901 -2.4043 -2.4171 -2.4286 -2.4392 -2.4487 -2.4575 -2.4654 -2.4727 -2.4794

linspace(-0.1,0.1,200)

ans = 1×200

-0.1000 -0.0990 -0.0980 -0.0970 -0.0960 -0.0950 -0.0940 -0.0930 -0.0920 -0.0910 -0.0899 -0.0889 -0.0879 -0.0869 -0.0859 -0.0849 -0.0839 -0.0829 -0.0819 -0.0809 -0.0799 -0.0789 -0.0779 -0.0769 -0.0759 -0.0749 -0.0739 -0.0729 -0.0719 -0.0709

function [xvals, yvals]=A5fourier(L,nmax,numpoints)

%xvals=linspace(-L,L,numpoints);
% alternative to linspace using for loops !!! 
k = 1;
for z=-0.1:0.001:0.1
    xvals(k)=z;
    k = k+1;
end 

yvals=zeros(1,numpoints);
for i=1:numpoints
    for j=1:2:nmax
        yvals(i)=(4/pi)*(yvals(i)+(1/j)*sin(j*pi*xvals(i)/L));        
    end    
end
end

Answer 2

[Pre]

The simplest alternative to linspace that outputs an identical array to what you are seeking...

example

step = 4/10 # for 10 elements between -2,2
    x = []
    for i in range(11):
        x.append(round(-2 + i * step, 1)) # this removes any floating point error that occurs and rounds to 1 decimal place

more general solution

        def my_linspace(start, end, size):
        step = (end - start) / size - 1
        array = []
        for i in range(size):
            array.append( round ( start + i * step, 1)) # here i have specified 1 degree of accuracy for neatness
            return array
            my_linspace(-2, 2, 10)  # yields 10 evenly spaced values between -2 and 2 exactly like linspace

note that because I used 'round' the values outputted by the function will have exactly one degree of accuracy, whereas, outputs of linsapce are floats with undefined degrees of accuracy.

hope this helps someone :D

sorry about formatting, first post.

[Walter Roberson]

This solution appears to be written in python, not in MATLAB.