Using python3 and reportlab I'm trying to generate two columns of barcodes:
from reportlab.graphics.barcode import code39
from reportlab.lib.pagesizes import A4
from reportlab.lib.units import mm
from reportlab.pdfgen import canvas
c = canvas.Canvas("barcode_example.pdf", pagesize=A4)
code_list = [
'E100', 'RA100',
'E101', 'RA101',
'E102', 'RA102',
'E103', 'RA103',
'E104', 'RA104',
'E105', 'RA105',
'E106', 'RA106',
'E107', 'RA107',
'E108', 'RA108',
'E109', 'RA109']
x = 1 * mm
y = 278 * mm
x1 = 6.4 * mm
for i, code in enumerate(code_list):
barcode = code39.Extended39(code, barWidth=0.6*mm, barHeight=15*mm)
if i % 2 == 0:
barcode.drawOn(c, x, y)
x1 = x + 6.4 * mm
c.drawString(x1, y- 5 * mm, code)
else:
x1 = x + 100 * mm
barcode.drawOn(c, x1, y)
y = y - 5 * mm
c.drawString(x1 + 6.4 * mm, y, code)
y = y - 25 * mm
# if int(y) == 0:
# x = x + 50 * mm
# y = 285 * mm
c.showPage()
c.save()
It generates a file and seems to be just fine but after scanning them in with a barcode scanner they read as follows:
E100F RA100
E101G RA101$
E102H RA102/
E103I RA103+
E104J RA104%
E105K RA1050
E106L RA1061
E107M RA1072
E108N RA1083
E109O RA1094
RA100 has a space after it.
It is clearly going in order of ASCII precedence I just don't know why. Is there a setting I'm suppose to enable/disable?
Also, it's definitely not the barcode scanner, I reset it to read code39 and tested it on other non-reportlab generated barcodes and they worked just fine.
Thank you for your help!
Also, I modified the code from here: https://stackoverflow.com/questions/2179269/python-barcode-generation-library
Edit
When I tested with code128, it worked perfectly. So it appears to be specific to code39
EDIT 2: And I'm an idiot. My barcode scanner doesn't support the checksum option. Disabling the checksum in reportlab fixed my issue. I'm going to leave this just in case someone else comes across this problem.
Disable checksum via:
barcode = code39.Extended39(code, barWidth=0.6*mm, barHeight=15*mm, checksum=0)
checksum=False would probably work just fine, but after looking at their source they use 1 and 0 for true/false and figured it would be safer to adhere to their practices.
Related
How to convert float to a specific format in hexadecimal:
1 bit for sign, 15 bit for the integer value, and the rest 16 bits for values after the decimal point.
Example output should be ffff587a for -0.6543861, fff31a35 for -12.897631, 006bde10 for 107.8674316, 003bd030 for 59.8132324
I have written a program that can do the unsigned conversion, I am stuck at the signed part. Could anyone guide me on how I can achieve this in a very compact way?
def convert(num):
binary2 = ""
Int = int(num)
fract = num - Int
binary = '{:16b}'.format(Int & 0b1111111111111111)
for i in range (16):
fract *= 2
fract_bit = int(fract)
if fract_bit == 1:
fract -= fract_bit
binary2 += '1'
else:
binary2 += '0'
return int(binary + binary2, 2)
value = 107.867431640625
x = convert(value)
hex(x)
output: 0x6bde10
This is simply the Q16.16 fixed-point format. To convert a floating-point number to this format, simply multiply it by 216 (in Python, 1<<16 or 65536) and convert the product to an integer:
y = int(x * (1<<16))
To show its 32-bit two’s complement representation, add 232 if it is negative and then convert it to hexadecimal:
y = hex(y + (1<<32 if y < 0 else 0))
For example, the following prints “0xfff31a35”:
#!/usr/bin/python
x=-12.897631
y = int(x * (1<<16))
y = hex(y + (1<<32 if y < 0 else 0))
print(y)
This conversion truncates. If you want rounding, you can add .5 inside the int or you can add additional code for other types of rounding. You may also want to add code to handle overflows.
I need some help with a quiete simple problem in FiPy. My goal is to simulate a fluid flowing through a concrete block while phase change.
But first of all I tried to do a simple 1D simulation assumed a fluid massflow and a constant wall temperature without any phase change.
from fipy import *
from fipy.meshes import CylindricalGrid2D, Grid1D
import matplotlib.pyplot as plt
import numpy as np
#%%
L = 1.5 #length transfer surface
bS = 0.75 #wide
AV = L * bS #transfer surface
tS0 = 350. #tWall
rhoWF = 880. #density fluid
mWF = 0.036 #mass flow
u = 5e-4 #Fluid speed
hWF = mWF / AV / rhoWF / u #height "fluid block"
nx = 50
VWF = hWF * L * bS/nx #fluid volumen
lambdaWF = 0.6 # thermal conductivity
alpha = 500. #heat transfer coefficient
tWF0 = 371.
mesh = Grid1D(dx=L/nx, nx=nx)
tWF = CellVariable(name="Fluid",
mesh=mesh,
value= tWF0,
hasOld=True)
tS = CellVariable(name="storage",
mesh=mesh,
value=tS0,
hasOld=True)
sourceWF=CellVariable(name="source Fluid", #Variable der Konvektion
mesh=mesh,
value=0.)
cvrho = CellVariable(name = 'cprho',#Fluid
mesh = mesh,
value = rhoWF * 4215.2,
hasOld = True)
tWF.constrain(tWF0, mesh.facesLeft()) #constant inlet temperature
t = 6*3600. #time
timeStepDuration = 1e2
#outflow boundary condition
outlet = mesh.facesRight
ConvCoeff = FaceVariable(mesh,value=u,rank=1)
exteriorCoeff = FaceVariable(mesh,value=0.,rank=1)
exteriorCoeff.setValue(value=ConvCoeff, where=outlet)
ConvCoeff.setValue(0., where=outlet)
residual1 = 1.
elapsedTime = 0.
tWFall = np.zeros(nx)[None,:]
while elapsedTime < t:
tWF.updateOld()
it = 0 #iterations
while residual1> 1e-2:
sourceWF.value = - AV / nx * alpha*(tWF - tS)/ cvrho / VWF #this will be a variable convection source
eq1 = HybridConvectionTerm(coeff=ConvCoeff) + TransientTerm(coeff=1.) == \
+ sourceWF\
- ImplicitSourceTerm(exteriorCoeff.divergence) \
#+ DiffusionTerm(coeff= lambdaWF / cvrho) #not necessary(?)
residual1 = eq1.sweep(dt = timeStepDuration, var = tWF)
print('res1: ' + str(residual1) )
it += 1
if it > 10:
raise ValueError (r'MaxIter reached')
elapsedTime += timeStepDuration ; print('t= ' + str(round(elapsedTime,2)))
residual1 = 1.
tWFall = np.r_[tWFall, tWF.value[None,:]] #value collection
#%% outlet fluid temperature and storage temperature
plt.plot(np.linspace(0,t/3600.,int(t/timeStepDuration)), tWFall[1:,-1], label=r'$\vartheta_{WF}$')
plt.legend()
I would expect a constant fluid outlet temperature because of the constant wall temperature and constant fluid inlet temperature. I have not defined the wall temperature as a boundary condition because some day I would like to analyse heat conduction and variable temperature gradients too. Running my mwe you can see that the fluid temperature at the outlet declines.
Could someone please help at this case?
Thanks in advance!
I changed the script around and it seem to give a constant temperature of 371.0. See this link.
The sourceWF term has been removed. I was unsure what this was for, but I think it would take time for the wall temperature to adjust to this.
The equation declaration has been moved outside the loop. This is the correct way to use FiPy, but shouldn't impact the results in this case.
I'm try to run an ocean temperature model for 25 years using the explicit method (parabolic differential equation).
If I run for a year a = 3600 or five years a = 18000 it works fine.
However, when I run it for 25 years a = 90000 it crashes.
a is the amount of time steps used. And a year is considered to be 360 days. The time step is 4320 seconds, delta_t = 4320..
Here is my code:
program task
!declare the variables
implicit none
! initial conditions
real,parameter :: initial_temp = 4.
! vertical resolution (delta_z) [m], vertical diffusion coefficient (av) [m^2/s], time step delta_t [s]
real,parameter :: delta_z = 2., av = 2.0E-04, delta_t = 4320.
! gamma
real,parameter :: y = (av * delta_t) / (delta_z**2)
! horizontal resolution (time) total points
integer,parameter :: a = 18000
!declaring vertical resolution
integer,parameter :: k = 101
! declaring pi
real, parameter :: pi = 4.0*atan(1.0)
! t = time [s], temp_a = temperature at upper boundary [°C]
real,dimension(0:a) :: t
real,dimension(0:a) :: temp_a
real,dimension(0:a,0:k) :: temp
integer :: i
integer :: n
integer :: j
t(0) = 0
do i = 1,a
t(i) = t(i-1) + delta_t
end do
! temperature of upper boundary
temp_a = 12. + 6. * sin((2. * t * pi) / 31104000.)
temp(:,0) = temp_a(:)
temp(0,1:k) = 4.
! Vertical resolution
do j = 1,a
do n = 1,k
temp(j,n) = temp(j-1,n) + (y * (temp(j-1,n+1) - (2. * temp(j-1,n)) + temp(j-1,n-1)))
end do
temp(:,101) = temp(:,100)
end do
print *, temp(:,:)
end program task
The variable a is on line 11 (integer,parameter :: a = 18000)
As said, a = 18000 works, a = 90000 doesn't.
At 90000 get I get:
Program received signal SIGSEGV: Segmentation fault - invalid memory reference.
Backtrace for this error:
RUN FAILED (exit value 1, total time: 15s)
I'm using a fortran on windows 8.1, NetBeans and Cygwin (which has gfortran built in).
I'm not sure if this problem is caused through bad compiler or anything else.
Does anybody have any ideas to this? It would help me a lot!
Regards
Take a look at the following lines from your code:
integer,parameter :: k = 101
real,dimension(0:a,0:k) :: temp
integer :: n
do n = 1,k
temp(j,n) = temp(j-1,n) + (y * (temp(j-1,n+1) - (2. * temp(j-1,n)) + temp(j-1,n-1)))
end do
Your array temp has bounds of 0:101, you loop n from 1 to 101 where in iteration n=101 you access temp(j-1,102), which is out of bounds.
This means you are writing to whatever memory lies beyond temp and while this makes your program always incorrect, it is only causing a crash sometimes which depends on various other things. Increasing a triggers this because column major ordering of your array means k changes contiguously and is strided by a, and as a increases your out of bounds access of the second dimension is further in memory beyond temp changing what is getting overwritten by your invalid access.
After your loop you set temp(:,101) = temp(:,100) meaning there is no need to calculate temp(:,101) in the above loop, so you can change its loop bounds from
do n = 1,k
to
do n = 1, k-1
which will fix the out of bounds access on temp.
Will somebody please explain why the initial conditions are properly taken care of in the following openmodelica model compiled and simulated in OMEdit v1.9.1 beta2 in Windows, but if line 5 is commentd and 6 uncommented (x,y) is initialized to (0.5,0)?
Thank you.
class Pendulum "Planar Pendulum"
constant Real PI = 3.141592653589793;
parameter Real m = 1,g = 9.81,L = 0.5;
Real F "Force of the Rod";
output Real x(start=L*sin(PI/4)) ,y(start=-0.35355);
//output Real x(start = L * sin(PI / 4)), y(start=-L*sin(PI/4));
output Real vx,vy;
equation
m * der(vx) = -x / L * F;
m * der(vy) = (-y / L * F) - m * g;
der(x) = vx;
der(y) = vy;
x ^ 2 + y ^ 2 = L ^ 2;
end Pendulum;
The short answer is that initial values are treated merely as hints, you have to add the fixed=true attribute to force them as in:
output Real x(start=L*cos(PI/4),fixed=true);
If initialized variables are constrained, the fixed attribute should not be used on all initialized variables but on a 'proper' subset, in this case on just one.
The long answer can be found here
This is a geometry question.
I have a line between two points A and B and want separate it into k equal parts. I need the coordinates of the points that partition the line between A and B.
Any help is highly appreciated.
Thanks a lot!
You just need a weighted average of A and B.
C(t) = A * (1-t) + B * t
or, in 2-D
Cx = Ax * (1-t) + Bx * t
Cy = Ay * (1-t) + By * t
When t=0, you get A.
When t=1, you get B.
When t=.25, you a point 25% of the way from A to B
etc
So, to divide the line into k equal parts, make a loop and find C, for t=0/k, t=1/k, t=2/k, ... , t=k/k
for(int i=0;i<38;i++)
{
Points[i].x = m_Pos.x * (1 - (i/38.0)) + m_To.x * (i / 38.0);
Points[i].y = m_Pos.y * (1 - (i/38.0)) + m_To.y * (i / 38.0);
if(i == 0 || i == 37 || i == 19) dbg_msg("CLight","(%d)\nPos(%f,%f)\nTo(%f,%f)\nPoint(%f,%f)",i,m_Pos.x,m_Pos.y,m_To.x,m_To.y,Points[i].x,Points[i].y);
}
prints:
[4c7cba40][CLight]: (0)
Pos(3376.000000,1808.000000)
To(3400.851563,1726.714111)
Point(3376.000000,1808.000000)
[4c7cba40][CLight]: (19)
Pos(3376.000000,1808.000000)
To(3400.851563,1726.714111)
Point(3388.425781,1767.357056)
[4c7cba40][CLight]: (37)
Pos(3376.000000,1808.000000)
To(3400.851563,1726.714111)
Point(3400.851563,1726.714111)
which looks fine but then my program doesn't work :D.
but your method works so thanks