I am working with a biological model of the distribution of microbial biomass (b1) on a 2D grid. From the biomass a protein (p1) is produced. The biomass diffuses over the grid, while the protein does not. Only if a certain amount of protein is produced (p > p_lim), the biomass is supposed to diffuse.
I try to implement this by using a dummy cell variable z multiplied with the diffusion coefficient and setting it from 0 to 1 only in cells where p > p_lim.
The condition works fine and when the critical amount of p is reached in a cell, z is set to 1, and diffusion happens. However, the diffusion still does not work with the rate I would like, because to calculate diffusion, the face variable, not the value of the cell itself is used. The faces of z are always a mean of the cell with z=1 and its neighboring cells with z=0. I I, however, would like the diffusion to work at its original rate even if the neighbouring cell is still at p < p_lim.
So, my question is: Can i somehow access a faceVariable and change it? For example, set a face to 1 if any neigboring cell has reached p1 > p_lim? I guess this is not a proper mathematical thing to do, but I couldn't think of another way to simulate this problem.
I will show a very reduced form of my model below. In any case, I thank you very much for your time!
##### produce mesh
nx= 5.
ny= nx
dx = 1.
dy = dx
L = nx*dx
mesh = Grid2D(nx=nx,ny=ny,dx=dx,dy=dy)
#parameters
h1 = 0.5 # production rate of p
Db = 10. # diffusion coeff of b
p_lim=0.1
# cell variables
z = CellVariable(name="z",mesh=mesh,value=0.)
b1 = CellVariable(name="b1",mesh=mesh,hasOld=True,value=0.)
p1= CellVariable(name="p1",mesh=mesh,hasOld=True,value=0.)
# equations
eqb1 = (TransientTerm(var=b1)== DiffusionTerm(var=b1,coeff=Db*z.arithmeticFaceValue)-ImplicitSourceTerm(var=b1,coeff=h1))
eqp1 = (TransientTerm(var=p1)==ImplicitSourceTerm(var=b1,coeff=h1))
# set b1 to 10. in the center of the grid
b1.setValue(10.,where=((x>2.)&(x<3.)&(y>2.)&(y<3.)))
vi=Viewer(vars=(b1,p1),FIPY_VIEWER="matplotlib")
eq = eqb1 & eqp1
from builtins import range
for t in range(10):
b1.updateOld()
p1.updateOld()
z.setValue(z + 0.1,where=((p1>=p_lim) & (z < 1.)))
eq.solve(dt=0.1)
vi.plot()
In addition to .arithmeticFaceValue, FiPy provides other interpolators between cell and face values, such as .harmonicFaceValue and .minmodFaceValue.
These properties are implemented using subclasses of _CellToFaceVariable, specifically _ArithmeticCellToFaceVariable, _HarmonicCellToFaceVariable, and _MinmodCellToFaceVariable.
You can also make a custom interpolator by subclassing _CellToFaceVariable. Two such examples are _LevelSetDiffusionVariable and ScharfetterGummelFaceVariable (neither is well documented, I'm afraid).
You need to override the _calc_() method to provide your custom calculation. This method takes three arguments:
alpha: an array of the ratio (0-1) of the distance from the face to the cell on one side, relative to the distance from distance from the cell on the other side to the cell on the first side
id1: an array of indices of the cells on one side of the face
id2: an array of indices of the cells on the other side of the face
Note: You can ignore any clause if inline.doInline: and look at the _calc_() method defined under the else: clause.
I am trying to make a graph of the brightness of a pixel vs the distance from center of that pixel. To do so I used for loops to check each pixel for these values. But when adding them to my array I find that I can't. One of the issues is I have to define the array size first so no values get placed in the right spot. I believe everything else to be working except adding values to the arrays.
I've tried various methods of concatenation to add the values of each pixel to the array. I didn't have any more solutions to try.
folder3 = 'C:\Users\slenka\Desktop\Image_Analysis\Subtracted';
cd('C:\Users\slenka\Desktop\Image_Analysis\Subtracted');
subtractedFiles = [dir(fullfile(folder3,'*.TIF')); dir(fullfile(folder3,'*.PNG')); dir(fullfile(folder3,'*.BMP')); dir(fullfile(folder3,'*.jpg'))];
numberOfSubImages= length(subtractedFiles);
for b = 1 : numberOfSubImages
subFileName=fullfile(folder3, subtractedFiles(b).name);
chartImage=imread(subFileName);
[chartY, chartX, chartNumberOfColorChannels] = size(chartImage);
ccY= chartY/2;
ccX= chartX/2;
c=[ccX,ccY];
distanceArray=zeros(1,chartX);
intensityArray=zeros(1,chartY);
f=1;
g=1;
for y=1:chartY
for x=1:chartX
D = sqrt((y - c(1)) .^ 2 + (x - c(2)) .^ 2);
grayScale= impixel(chartImage, x, y);
distanceArray(f)=[D];
intensityArray(g)=[grayScale];
f=f+1;
g=g+1;
end
end
xAxis=distanceArray;
yAxis=intensityArray;
plot(xAxis,yAxis);
end
I'm expecting 2 arrays one full of the data values for the light intensity of each pixel in the image, and another for that pixels distance from the center of the image. I am wanting to plot these two arrays as the y and x axis respectively. At the moment the actual results is an entirely empty array full of zeros.
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)),'.--')
I am trying to generate a graph that should look similar to:
My arrays are:
Array4:[Nan;Nan;.......;20;21;22;23;24;..........60]
Array3:[[Nan;Nan;.......;20;21;22;23;24;..........60]
Array2:[0;1;2;3;4;5;6;Nan;Nan;Nan;Nan;17;18;.....60]
Array1:[0;1;2;3;4;5;6;Nan;Nan;Nan;Nan;17;18;.....60]
I cannot find the right way to group my arrays in order to plot them in the way shown on the above graph.
I tried using the following function explained in: http://uk.mathworks.com/help/matlab/ref/barh.html
barh(1:numel(x),y,'hist')
where y=[Array1,Array2;Array3,Array4] and x={'1m';'2m';'3m';......'60m'}
but it does not work.
Why Your Current Approach Isn't Working
Your intuition makes sense to me, but the barh function you are using doesn't work the way you think it does. Specifically, you are interpreting the meaning of the x and y inputs to that function incorrectly. Those are inputs are constant values, not entire axes. The first y input refers to the end-point of the bar that stretches horizontally from x = 0 and the first x input refers to location on the y-axis of the horizontal bar. To illustrate what I mean, I've provided the below horizontal bar graph:
You can find this same picture in the official documentation of the MATLAB barh function. The code used to generate this bar graph is also given in the documentation, shown below:
x = 1900:10:2000;
y = [57,91,105,123,131,150,...
170,203,226.5,249,281.4];
figure;
barh(x, y);
The individual elements of the x array, rather confusingly, show up on the y-axis as the starting locations of each bar. The corresponding elements of the y array are the lengths of each bar. This is the reason that the arrays must be the same length, and this illustrates that they are not specifications of the x and y axes as one might intuitively believe.
An Approach To Solve Your Problem
First things first, the easiest approach is to do this manually with the plot function and a set of lines that represent floating bars. Consult the official documentation for the plot function if you'd like to plot the lines with some sort of color coordination in mind - the code I present (modified version of this answer on StackOverflow) just switches the color of the floating bars between red and blue. I tried to comment the code so that the purpose of each variable is clear. The code I present below matches the floating bar graph that you want to be plotted, if you are alright with replacing thick floating bars with 2D lines floating on a plot.
I used the data that you gave in your question to specify the floating horizontal bars that this script would output - a screenshot is shown below the code. Array1 & Array2:[0;1;2;3;4;5;6;Nan;Nan;Nan;Nan;17;18;.....60], these arrays go from 0 to 6 (length = 6) and 17 to 60 (length = 60 - 17 = 43). Because there is a "discontinuity" of sorts from 7 to 16, I have to define two floating bars for each array. Hence, the first four values in my length array are [6, 6, 43, 43]. Where the first 6 and the first 43 correspond to Array1 and the second 6 and the second 43 correspond to Array2. Recognizing this "discontinuity", the starting point of the first floating bar for Array1 and Array2 is x = 0 and the starting point of the second floating bar for Array1 and Array2 is x = 7. Putting that all together, you arrive at the x-coordinates for the first four points in the floating_bars array, [0 0; 0 1.5; 17 0; 17 1.5]. The y-coordinates in this array only serve to distinguish Array1, Array2, and so on from each other.
Code:
floating_bars=[0 0; 0 1.5; 17 0; 17 1.5; 20 6; 20 7.5]; % Each row is the [x,y] coordinate pair of the starting point for the floating bar
L=[6, 6, 43, 43, 40, 40]; % Length of each consecutive bar
thickness = 0.75;
figure;
for i=1:size(floating_bars,1)
curr_thickness = 0;
% It is aesthetically pleasing to have thicker bars, this makes the plot look for like the grouped horizontal bar graph that you want
while (curr_thickness < thickness)
% Each bar group has two bars; set the first to be red, the second to be blue (i.e., even index means red bar, odd index means blue bar)
if mod(i, 2)
plot([floating_bars(i,1), floating_bars(i,1)+L(i)], [floating_bars(i,2) + curr_thickness, floating_bars(i,2) + curr_thickness], 'r')
else
plot([floating_bars(i,1), floating_bars(i,1)+L(i)], [floating_bars(i,2) + curr_thickness, floating_bars(i,2) + curr_thickness], 'b')
end
curr_thickness = curr_thickness + 0.05;
hold on % Make sure that plotting the current floating bar does not overwrite previous float bars that have already been plotted
end
end
ylim([ -10 30]) % Set the y-axis limits so that you can see more clearly the floating bars that would have rested right on the x-axis (y = 0)
Output:
How Do I Do This With the barh Function?
The short answer is that you'd have to modify the function manually. Someone has already done this with one of the bar graph plotting functions provided by MATLAB, bar3. The logic implemented in this modified bar3 function can be re-applied for your purposes if you read their barNew.m function and tweak it a bit. If you'd like a pointer as to where to start, I'd suggest looking at how they specify z-axis minimum and maximums for their floating bars on the plot, and apply that same logic to specify x-axis minimum and maximums for your floating bars in your 2D case.
I hope this helps, happy coding! :)
I explain here my approach to generate these type of graphs. Not sure if it is the best but it works and there is no need to do anything manually. I came up with this solution based on the following Vladislav Martin's explained fact: "The y-coordinates in this array only serve to distinguish Array1, Array2, and so on from each other".
My original arrays are:
Array4=[Nan....;20;21;22;23;24;..........60]
Array3=[Nan....;20;21;22;23;24;..........60]
Array2=[0;1;2;3;4;5;6;Nan;Nan;Nan;Nan;17;18;.....60]
Array1=[0;1;2;3;4;5;6;Nan;Nan;Nan;Nan;17;18;.....60]
x={'0m';'1m';'2m';'3m';'4m';....'60m'}
The values contained in these arrays make reference to the x-axis on the graph. In order to make the things more simple and to avoid having to code a function to determine the length for each discontinuity in the arrays, I replace these values for y-axis position values. Basically I give to Array1 y-axis position values of 0 and to Array2 0+0.02=0.02. To Array3 I give y-axis position values of 0.5 and to Array4 0.5+0.02=0.52. In this way, Array2 will be plotted on the graph closer to Array1 which will form the first group and Array4 closer to Array3 which will form the second group.
Datatable=table(Array1,Array2,Array3,Array4);
cont1=0;
cont2=0.02;
for col=1:2:size(Datatable,2)
col2=col+1;
for row=1:size(Datatable,1)
if isnan(Datatable{row,col})==0 % For first array in the group: If the value is not nan, I replace it for the corresponnding cont1 value
Datatable{row,col}=cont1;
end
if isnan(Datatable{row,col2})==0 % For second array in the group: If the value is not nan, I replace it for the corresponnding cont2 value
Datatable{row,col2}=cont2;
end
end
cont1=cont1+0.5;
cont2=cont2+0.5;
end
The result of the above code will be a table like the following:
And now I plot the Arrays using 2D floating lines:
figure
for array=1:2:size(Datatable,2)
FirstPair=cell2mat(table2cell(Datatable(:,array)));
SecondPair=cell2mat(table2cell(Datatable(:,array+1)));
hold on
plot(1:numel(x),FirstPair,'r','Linewidth',6)
plot(1:numel(x),SecondPair,'b','Linewidth',6)
hold off
end
set(gca,'xticklabel',x)
And this will generate the following graph:
I have a large 2D-array (31100 x 4) built up like this:
Intensity | X | Y | Z
out of this array I would like to create a 3D-array to visualize it as a 3D-image.
I would somehow have to transform the 2D into a 3D-array. Since there will be plenty of points where there is no intensity value available all of these points should be zero values.
Side information:
min(x) = -152.0120;
max(x) = 161.4350;
min(y) = -256.2560;
max (y) = -52.3801;
min(z) = -428.5920;
max (z) = -152.4210;
Because of the larger number of decimal places I might have to round up. Yet, this would possibly lead to several intensities being assigned to the same coordinate combination. Logically, I would have to add up the intensities in these cases.
This makes everything really complicated.
I would be happy if someone could teach me how to transform a 2D into a 3D array with zero-values for spots without intensity values.
Thanks for reading.