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Problems re-implementing the fipy mesh20x20 example from my own IDE

I am currently using fipy but am still relatively new to the nuiances associated with the package. While I have been able to regenerate the desired heatmap from the examples folder in for the mesh20x20 diffusion example using the command line, I have struggled to replicate it within a Spyder IDE. I am using python version 3.8 . It is simple enough to generate it using the "examples" folder from the command line the command line image generated, however, when I attempt to "re-program" it I end up with iterations of the following. the following result. I am hoping to be able to regenerate the smooth color transition from the examples folder, as opposed to the discrete dichromatic option that I have been limited to at present. I believe there is some issues with the viewer in some capacity I believe some related issues may have cropped up in the past for others, potentially as it relates to colorbar reformatting, though I have not yet been capable of effectively implementing these workarounds to generate the desired imagery. datamin and datamax in Viewer() did not work
I would be greatly indebted for any assitance the community could provide.
from fipy.terms.transientTerm import TransientTerm
from fipy.terms.implicitDiffusionTerm import ImplicitDiffusionTerm
from fipy.terms.explicitDiffusionTerm import ExplicitDiffusionTerm
from fipy.meshes.nonUniformGrid2D import NonUniformGrid2D
from fipy.variables.cellVariable import CellVariable
from fipy.viewers.matplotlibViewer.matplotlib2DViewer import Matplotlib2DViewer
####
#Global Inputs
D=1
steps=10
#Dimensional Inputs
nx=20
dx=1
ny=20
dy=1
L=dx*nx
#Temporal Inputs
#nt=20
#dt=1
#cell variable initial values
value=0
#construct mesh from dimensional pts
mesh=NonUniformGrid2D(nx=nx, dx=dx, ny=ny, dy=dy)
#construct term variables phi with name, mesh design
phi=CellVariable(name="solutionvariable", mesh=mesh, value=0)
#construct boundary conditions
#dirichlet ---> we can an automatic application of neumann to top right and bottom left
valueTopLeft=0
valueBottomRight=1
#assign boundary conditions to a face or cell
X, Y=mesh.faceCenters
facesTopLeft=((mesh.facesLeft & (Y > L/2 )) | (mesh.facesTop &( X < L/2)))
facesBottomRight=((mesh.facesRight & (Y < L/2)) | (mesh.facesBottom & (X > L/2)))
#constrain variables
phi.constrain(valueTopLeft, facesTopLeft)
phi.constrain(valueBottomRight, facesBottomRight)
#equation construction
eq=TransientTerm()==ExplicitDiffusionTerm(coeff=D)
#equation solving and either viewing and/or extraction
timestepduration=0.9 *(dx**2)/(2*D)
for step in range(steps):
eq.solve(var=phi, dt=timestepduration)
print(phi[step])
viewer=Matplotlib2DViewer(vars=phi, datamin=0, datamax=1)
viewer.axes.set_title("Solutionvbl(Step %d)" % (step+1,))

Figured it out I think. I was using ExplicitDiffusion and the example utilizes ImplicitDiffusion. When I tried this all I got back was a blank monochromatic image (and returned zeros for my phi[step] at the end. I am happy to report that once a "kickstart" value is provided in the value section for cellVariable (I used 0.001), and utilized in conjunction with ImplicitDiffusion, and the timestepduration is increased from its limit of 0.9x**2/2D to the utilized 9x**2/2D used in the example documentation it more or less adheres to the image generated when run from the command line. Grateful to have this sorted. Hope this provides assistance to anyone else who might run into a similar problem.

Can python code results be written in Latex?

I am really new in programming, sorry for the awful way of asking.
So for a class of kids I'm helping I am trying to make a program in python which must assign a random integer to 2 variables "A" & "B", once that is done, we must check if the ratio A/B yields an integer.
If that is the case, then we must have python print "A/B=~" so we must print this as a question without displaying the answer.
I achieved this by printing the variables as text once I checked the previously stated condition. So far everything is fine. I did this with a loop 5 times and got 5 different questions. I made it in such a way that changing I couple numbers I can make as many questions as I want.
Just to give an example I got:
14/7=
56/8=
35/5=
7/1=
81/3=
So the python part was basically done.
What I am unable to do and would appreciate if anyone could help me is the next part.
I have to take this results and and be able to make a pdf, if possible with latex, with the caveat that i don't like the idea of manually typing said results since for all the kids I have to do this around 180 times (30 times per kid). Is there a way to do this, since typing all manually in latex would take forever.

Yes, you can generate a latex by Python automatically with pylatex package. Here is a full example about the pylatex: PyLatex full examples
I write a small demo if I understand you correctly. These demo will create a latex named "test.tex" in the current directory, and an equation "a/b=0" is printed.
from pylatex import Document, Section, Subsection, Command,Package, Alignat
doc = Document(default_filepath='basic.tex', documentclass='article')
with doc.create(Subsection('Alignat math environment')):
with doc.create(Alignat(numbering=False, escape=False)) as agn:
agn.append(r'\frac{a}{b} &= 0 \\')
# trying to generate the tex and pdf, and do not clean the tex file after generating the pdf
doc.generate_pdf("test", clean_tex=False)
If you have tools like latexmk which can convert tex to pdf installed, then a pdf name test.pdf will also created. Otherwise you can use your latex editor open the test.tex file.

Corner Refine Method, CORNER_REFINE_SUBPIX, from AcUro with Python and OpenCV

Hi There
I want to increase the accuracy of the marker detection from aruco.detectMarkers. So, I want to use Corner Refine Method with CORNER_REFINE_SUBPIX, but I do not understand how it is implemented in python.
Sample code:
frame = cv.imread("test.png")
gray = cv.cvtColor(frame, cv.COLOR_BGR2GRAY)
para = aruco.DetectorParameters_create()
det_corners, ids, rejected = aruco.detectMarkers(gray,dictionary,parameters=para)
aruco.drawDetectedMarkers(frame,det_corners,ids)
Things I have tried:
para.cornerRefinementMethod()
para.cornerRefinementMethod(aruco.CORNER_REFINE_SUBPIX)
para.cornerRefinementMethod.CORNER_REFINE_SUBPIX
para = aruco.DetectorParameters_create(aruco.CORNER_REFINE_SUBPIX)
para = aruco.DetectorParameters_create(para.cornerRefinementMethod(aruco.CORNER_REFINE_SUBPIX))
They did not work, I’m pretty new to python ArUco so I hope that there is a simple and obvious solution.
I would also Like to implement enclosed markers like in the Documentation(Page 4). Do you happen to know if there is a way to generate these enclosed markers in python?

Concerning the first part of your question, you were pretty close: I assume your trouble is in switching and tweaking the "para" options. If so, you only need to set the corresponding values in the parameters object like
para.cornerRefinementMethod = aruco.CORNER_REFINE_SUBPIX
Note that "aruco.CORNER_REFINE_SUBPIX" is simply an integer. You can verify this by typing type(aruco.CORNER_REFINE_SUBPIX) in the console. Thus assigning values to the "para" object works like mentioned above.
You might also want to tweak the para.cornerRefinementWinSize which seems to be implemented in units of code pixels, not actual image pixel units.
Concerning the second part, you might have to write a function, that adds the boxes at the corner points, which you can get using the detectMarker function. Note that the corner points are always ordered clockwise, thus you can easily assign the correct offset values (like "up & left", "up & right" etc.).

para.cornerRefinementMethod = 1
may work.

Python - print image bit data to a ESC/POS printer with python

I've been looking for an example of how to format and print bmp's to my receipt printer (so I can add logos) for a long time, so I doubt this is a duplicate post considering others were for java or other script languages. Usually I'm pretty good at understanding instructions, but all I seem to find is the same old instructions I can never fully understand.
I am using python 2.7 and I have a function pI(x) which uses win32print to send data to the printer, where x is the data in string format using "\x??" for hex data like formatting text. It seems to work well.
The programmer manual that came with my printer says (for downloading bit image, GS *) for syntax:
Hex 1D 2A x y d1...dk
and:
d=1 for printing the corresponding dot and d=0 for not printing the corresponding dot.
Here is my questions about these instructions:
Does this mean that all x, y, d1...dk is in hex (or "\x??")? I think so.
What is x and y representing? I read a while ago on a site (maybe this one) that x+y*255 = image width, and I assume that is using the order of operations. Is this correct?
The instructions on my generic printer also state that x and y are both supposed to be between 1 and 48, totaling no more than 1500, unlike some manuals which say x is supposed to be between 0 and 3 and y being between 1 and 128 (I think) which said x+y*255=width, totaling about 2000. it also says k=x*y*8 which I think that the example would be 8*8*8=512*"\x01", so where does the third 8 come from and how do I code that in the string??? Does x=width and y=height?... then how do I get an image width of the maximum 384 dots?
Does this mean that I have to enter "\x00" or "\x01" for each dot, so one instance (a small black block of 8x8) of GS* would be 64*"\x01"?
Do I have to GS * each group of 8 dots tall or line of 8 dots, or will that overwrite the previously programmed data?
I'd like to later include in my program a means of easily creating logos using a tkinter canvas widget and saving it to a text file for future printing using pI(), so I really need to know how to directly 'download' image data to the printer and using a third party module probably won't work since I want to continue using my pI() function. Yes, it's ambitious and I'm probably doing it the hard way. But I'm afraid if I start incorporating too much new stuff I'm not familiar with, I'll get too confused.
Basically, what string should I send to pI() to download an image of a solid 8x8-dot black box with a 2x wide white line down the center on the printer?
Here's an example of what I would like the printer to print so I can see a working code string

Abaqus Python getByBoundingBox: 6.10 vs 6.13+

I have written a Python Script for ABAQUS to create several parts with many partitions. To get a structered mesh I have to select several edges. Now there is one edge I apparently cannot select in ABAQUS 6.10 & 6.11. Oddly, everything is fine with ABAQUS 6.13+.
p = mdb.models[name_model].parts[name_part_1]
e = p.edges
pickedEdges = e.getByBoundingBox(((cos(alpha_rad)*ri)-delta_p),((sin(alpha_rad)*ri)-delta_p),0.0,
((cos(alpha_rad)*d_core/2)+delta_p),((sin(alpha_rad)*d_core/2)+delta_p),0.0)
p.seedEdgeByBias(biasMethod=SINGLE, end2Edges=pickedEdges, ratio=bias_f, number=elem_num_rad, constraint=FINER)
Here, 'ri' is used to describe a radius, 'delta_p' (=0.001) is used to get a boundingbox slightly bigger than the original edge.
I also tried to use a bigger boundingbox by increasing delta_p but nothing works.
Any ideas? Thank you in advance! :)
for a sketch:
click me
the described bounding box is box E and I try to get the orange line

Its not clear from your post why the method isn't working.
you could determine a point on your edge and use the findAt method instead of getByBoundingBox.
By setting delta_p to a very large number, you should select every edge in your model? Its not clear what you mean by "not working"