I have a code which perfectly draws a sine wave graph
y = mgl_create_data_size(100,1,0);
mgl_data_modify(y,"0.4*y+0.1+sin(6*pi*x)",0);
gr = mgl_create_graph(WINDOW_WIDTH, WINDOW_HEIGHT);
mgl_plot(gr,y,". ","");
mgl_box(gr);
Now that the variable y is a MathGL data type HMDT
How I will get numeric values from it
I see there is function which mgl_data_get_value(y,i,j,0);
but it only accepts long data type and it always returns NaN
I found the answer that, I am trying to assume j as incremental value.
But, base on trial and error based, when I set j as 0 I am getting all the 100 graph values which applied based on the MathGL formula 0.4*y+0.1+sin(6*pi*x).
So the assumption of the function mgl_create_data_size is
If the data is set using 100,1,0 which is nx,ny,nz
nx = 100 - Total number of X axis values
ny = 1 - is only one Y value for each X value
nz = 0 - is assumed as 0, there is no z axis
the function which can get each graph value is
mgl_data_get_value(y,i,0,0)
Example of 10 values for the same formula
var y1 = mgl_create_data_size(10,1,0)
mgl_data_modify(y1,"0.4*y+0.1+sin(6*pi*x)",0)
for(i in 0..10-1) {
println(mgl_data_get_value(y1,i,0,0))
}
The output is
0.1
0.9660254037844387
-0.7660254037844384
0.09999999999999976
0.9660254037844391
-0.7660254037844387
0.09999999999999952
0.9660254037844401
-0.7660254037844377
0.09999999999999927
The graph in Excel for the above values is
This is very help full to use MathGL in applied mathematics in data generation as one of the light weight libraries
Related
This is probably a trivial question, but I want to select a portion of a complex array in order to plot it in Matlab. My MWE is
n = 100;
t = linspace(-1,1,n);
x = rand(n,1)+1j*rand(n,1);
plot(t(45):t(55),real(x(45):x(55)),'.--')
plot(t(45):t(55),imag(x(45):x(55)),'.--')
I get an error
Error using plot
Vectors must be the same length.
because the real(x(45):x(55)) bit returns an empty matrix: Empty matrix: 1-by-0. What is the easiest way to fix this problem without creating new vectors for the real and imaginary x?
It was just a simple mistake. You were doing t(45):t(55), but t is generated by rand, so t(45) would be, say, 0.1, and t(55), 0.2, so 0.1:0.2 is only 0.1. See the problem?
Then when you did it for x, the range was different and thus the error.
What you want is t(45:55), to specify the vector positions from 45 to 55.
This is what you want:
n = 100;
t = linspace(-1,1,n);
x = rand(n,1)+1j*rand(n,1);
plot(t(45:55),real(x(45:55)),'.--')
plot(t(45:55),imag(x(45:55)),'.--')
The following problem is the extension of This problem
I have written the following codes :
load y; P = y;k = length(P);
% the following matrix is used to sum each 'n' elements in a row:
`n = 2; %For sum of n elements in a row
summer = diag(ones(k,1));
for d = 1:n-1
summer = summer + diag(ones(k-d,1),-d);
end
X = -2.*log(P(:).')*summer;`
The value of X I am getting is 'NAN' values for all given P datasets array(of size 200x1) but when I am testing this with say 10 values of P then it is working very fine and not giving me any 'NAN' values.Can anyone help me why I am not getting values for large datasets while for small number it is working...??
There are probably -Inf, Inf, or NaN values in your P vector to begin with. Based on the arithmetic operations being done this seems to be the only possible source of NaN values resulting in X
i am trying to learn matlab.
I am trying to make a program that draw these imaginary numbers: ("," = decimal number)
and determine what of the 500 numbers that is closest the real axis.
And i need a little guidance.
What do i have to do to solve this task?
I was thinking about making a loop where all the "values" get stored in a array:
[code]
n= 1
while n < 500
value=1+0.1^n;
disp(value)
n=n+1[/code]
(seems like value is printing wrong values? and how to store in a array?)
And then somehow determine what number that is nearest the real axis and then display the value.
would be really grateful if someone could help me.
thanks in advance.
MATLAB creates imaginary numbers by appending an i or j term with the number. For example, if you wanted to create an imaginary number such that the real component was 1 and the imaginary component was 1, you would simply do:
>> A = 1 + i
A =
1.0000 + 1.0000i
You can see that there is a distinct real component as well as an imaginary component and is stored in A. Similarly, if you want to make the imaginary component have anything other than 1, you would need to add a constant in front of the i (or j). Something like:
>> A = 3 + 6i
A =
3.0000 + 6.0000i
Therefore, for your task, you simply need to create a vector of n between 1 to 500, input this into the above equation, then plot the resulting imaginary numbers. In this case, you would plot the real component on the x axis and the imaginary component on the y axis. Something like:
>> n = 1 : 500;
>> A = (1 + 0.1i).^n;
>> plot(real(A), imag(A));
real and imag are functions in MATLAB that access the real and imaginary components of complex numbers stored in arrays, matrices or single values. As noted by knedlsepp, you can simply plot the array itself as plot can handle complex-valued arrays:
>> plot(A);
Nice picture btw! Be mindful of the . operator appended with the ^ operator. The . means an element-wise operation. This means that we wish to apply the power operation for each value of n from 1 to 500 with 1 + 0.1i as the base. The result would be a 500 element array with the resulting calculations. If we did ^ by itself, we would be expecting to perform a matrix power operation, when this is not the case.
The values that you want to analyze for each value of n being applied to the equation in your post are stored in A. We then plot the real and imaginary components on the graph. Now if you want to find which numbers are closest to the real axis, you simply need to find the smallest absolute imaginary component of the numbers stored in A, then search for all of those numbers that share this number.
>> min_dist = min(abs(imag(A)));
>> vals = A(abs(imag(A)) == min_dist)
vals =
1.3681 - 0.0056i
This means that the value of 1.3681 - 0.0056i is the closest to the real axis.
I have a large data set with two arrays, say x and y. The arrays have over 1 million data points in size. Is there a simple way to do a scatter plot of only 2000 of these points but have it be representative of the entire set?
I'm thinking along the lines of creating another array r ; r = max(x)*rand(2000,1) to get a random sample of the x array. Is there a way to then find where a value in r is equal to, or close to a value in x ? They wouldn't have to be in the same indexed location but just throughout the whole matrix. We could then plot the y values associated with those found x values against r
I'm just not sure how to code this. Is there a better way than doing this?
I'm not sure how representative this procedure will be of your data, because it depends on what your data looks like, but you can certainly code up something like that. The easiest way to find the closest value is to take the min of the abs of the difference between your test vector and your desired value.
r = max(x)*rand(2000,1);
for i = 1:length(r)
[~,z(i)] = min(abs(x-r(i)));
end
plot(x(z),y(z),'.')
Note that the [~,z(i)] in the min line means we want to store the index of the minimum value in vector z.
You might also try something like a moving average, see this video: http://blogs.mathworks.com/videos/2012/04/17/using-convolution-to-smooth-data-with-a-moving-average-in-matlab/
Or you can plot every n points, something like (I haven't tested this, so no guarantees):
n = 1000;
plot(x(1:n:end),y(1:n:end))
Or, if you know the number of points you want (again, untested):
npoints = 2000;
interval = round(length(x)/npoints);
plot(x(1:interval:end),y(1:interval:end))
Perhaps the easiest way is to use round function and convert things to integers, then they can be compared. For example, if you want to find points that are within 0.1 of the values of r, multiply the values by 10 first, then round:
r = max(x) * round(2000,1);
rr = round(r / 0.1);
xx = round(x / 0.1);
inRR = ismember(xx, rr)
plot(x(inRR), y(inRR));
By dividing by 0.1, any values that have the same integer value are within 0.1 of each other.
ismember returns a 1 for each value of xx if that value is in rr, otherwise a 0. These can be used to select entries to plot.
Here is the graph I currently have
:
The Dotted Blue line represented the y value that corresponds to the x value I am looking for. I am trying to find the x values of the line's intersections with the blue curve(Upper).Since the interesections do not fall on a point that has already been defined, we need to interpolate a point that falls onto the Upper plot.
Here is the information I have:
LineValue - The y value of the intersection and the value of the dotted line( y = LineValue)
Frequency - an array containing the x value coordinates seen on this plot. The interpolated values of Frequency that corresponds to LineValue are what we are looking for
Upper/Lower - arrays containing the y value info for this graph
This solution is an improvement on Amro's answer. Instead of using fzero you can simply calculate the intersection of the line by looking for transition in the first-difference of the series created by a logical comparison to LineValue. So, using Amro's sample data:
>> x = linspace(-100,100,100);
>> y = 1-2.*exp(-0.5*x.^2./20)./(2*pi) + randn(size(x))*0.002;
>> LineValue = 0.8;
Find the starting indices of those segments of consecutive points that exceed LineValue:
>> idx = find(diff(y >= LineValue))
idx =
48 52
You can then calculate the x positions of the intersection points using weighted averages (i.e. linear interpolation):
>> x2 = x(idx) + (LineValue - y(idx)) .* (x(idx+1) - x(idx)) ./ (y(idx+1) - y(idx))
x2 =
-4.24568579887939 4.28720287203057
Plot these up to verify the results:
>> figure;
>> plot(x, y, 'b.-', x2, LineValue, 'go', [x(1) x(end)], LineValue*[1 1], 'k:');
The advantages of this approach are:
The determination of the intersection points is vectorized so will work regardless of the number of intersection points.
Determining the intersection points arithmetically is presumably faster than using fzero.
Example solution using FZERO:
%# data resembling your curve
x = linspace(-100,100,100);
f = #(x) 1-2.*exp(-0.5*x.^2./20)./(2*pi) + randn(size(x))*0.002;
VALUE = 0.8;
%# solve f(x)=VALUE
z1 = fzero(#(x)f(x)-VALUE, -10); %# find solution near x=-10
z2 = fzero(#(x)f(x)-VALUE, 10); %# find solution near x=+10
%# plot
plot(x,f(x),'b.-'), hold on
plot(z1, VALUE, 'go', z2, VALUE, 'go')
line(xlim(), [VALUE VALUE], 'Color',[0.4 0.4 0.4], 'LineStyle',':')
hold off
Are the step sizes in your data series the same?
Is the governing equation assumed to be cubic, sinuisoidal, etc..?
doc interpl
Find the zero crossings