Say I had three variables (On a 2D-field):
NumSquares,
WindowSize (Both X and Y has this value),
Index
So if, for example, NumSquares were 8, WindowSize 256 and Index was at 64; How would the algorithm to get positional data out of these variables look?
I assume that by "positional data" you mean coordinates on the grid. And index means position of the tile in linearized array calculated with formula X + width * Y. Then:
X = index % width
Y = index / width
I'm not sure what language you are working in, so just to be sure: % is modulo, / is in this context integer division.
Related
f = #(x)(abs(x))
fplot(f, [-1, 1]
Freshly installed octave, with no configuration edited. It results in the following image, where it looks as if it is constant for a while around 0, looking more like a \_/ than a \/:
Why does it look so different from a usual plot of the absolute value near 0? How can this be fixed?
Since fplot is written in Octave it is relatively easy to read. Its location can be found using the which command. On my system this gives:
octave:1> which fplot
'fplot' is a function from the file /usr/share/octave/5.2.0/m/plot/draw/fplot.m
Examining fplot.m reveals that the function to be plotted, f(x), is evaluated at n equally spaced points between the given limits. The algorithm for determining n starts at line 192 and can be summarised as follows:
n is initially chosen to be 8 (unless specified differently by the user)
Construct a vector of arguments using a coarser grid of n/2 + 1 points:
x0 = linspace (limits(1), limits(2), n/2 + 1)'
(The linspace function will accept a non-integer value for the number of points, which it rounds down)
Calculate the corresponding values:
y0 = f(x0)
Construct a vector of arguments using a grid of n points:
x = linspace (limits(1), limits(2), n)'
Calculate the corresponding values:
y = f(x0)
Construct a vector of values corresponding to the members of x but calculated from x0 and y0 by linear interpolation using the function interp1():
yi = interp1 (x0, y0, x, "linear")
Calculate an error metric using the following formula:
err = 0.5 * max (abs ((yi - y) ./ (yi + y + eps))(:))
That is, err is proportional to the maximum difference between the calculated and linearly interpolated values.
If err is greater than tol (2e-3 unless specified by the user) then put n = 2*(n-1) and repeat. Otherwise plot(x,y).
Because abs(x) is essentially a pair of straight lines, if x0 contains zero then the linearly interpolated values will always exactly match their corresponding calculated values and err will be exactly zero, so the above algorithm will terminate at the end of the first iteration. If x doesn't contain zero then plot(x,y) will be called on a set of points that doesn't include the 'cusp' of the function and the strange behaviour will occur.
This will happen if the limits are equally spaced either side of zero and floor(n/2 + 1) is odd, which is the case for the default values (limits = [-5, 5], n = 8).
The behaviour can be avoided by choosing a combination of n and limits so that either of the following is the case:
a) the set of m = floor(n/2 + 1) equally spaced points doesn't include zero or
b) the set of n equally spaced points does include zero.
For example, limits equally spaced either side of zero and odd n will plot correctly . This will not work for n=5, though, because, strangely, if the user inputs n=5, fplot.m substitutes 8 for it (I'm not sure why it does this, I think it may be a mistake). So fplot(#abs, [-1, 1], 3) and fplot(#abs, [-1, 1], 7) will plot correctly but fplot(#abs, [-1, 1], 5) won't.
(n/2 + 1) is odd, and therefore x0 contains zero for symmetrical limits, only for every 2nd even n. This is why it plots correctly with n=6 because for that value n/2 + 1 = 4, so x0 doesn't contain zero. This is also the case for n=10, 14, 18 and so on.
Choosing slightly asymmetrical limits will also do the trick, try: fplot(#abs, [-1.1, 1.2])
The documentation says: "fplot works best with continuous functions. Functions with discontinuities are unlikely to plot well. This restriction may be removed in the future." so it is probably a bug/feature of the function itself that can't be fixed except by the developers. The ordinary plot() function works fine:
x = [-1 0 1];
y = abs(x);
plot(x, y);
The weird shape comes from the sampling rate, i.e. at how many points the function is evaluated. This is controlled by the parameter N of fplot The default call seems to accidentally skip x=0, and with fplot(#abs, [-1, 1], N=5) I get the same funny shape like you:
However, trying out different values of N can yield the correct shape, try e.g. fplot(#abs, [-1, 1], N=6):
Although in general I would suggest to use way higher numbers, like N=100.
I have a task: given a value N. I should generate a list of length L > 1 such that the sum of the squares of its elements is equal to N.
I wrote a code:
deltas = np.zeros(L)
deltas[0] = (np.random.uniform(-N, N))
i = 1
while i < L and np.sum(np.array(deltas)**2) < N**2:
deltas[i] = (np.random.uniform(-np.sqrt(N**2 - np.sum(np.array(deltas)**2)),\
np.sqrt(N**2 - np.sum(np.array(deltas)**2))))
i += 1
But this approach takes long time, if I generate such list many times. (I think because of loop).
Note, that I don't want my list to consist of just one unique value. The distribution of values does not have to be uniform - I took uniform just for example.
Could you suggest any faster approach? May be there is special function in any lib?
If you didn't mind a few repeating 1s, you could do something like this:
def square_list(integer):
components = []
total = 0
remaining = integer
while total != integer:
component = int(remaining ** 0.5)
remaining -= component ** 2
components.append(component)
total = sum([x ** 2 for x in components])
return components
This code works by finding the taking the largest square, and then decreasing to the next largest square. It continues until the largest square is 1, which could at worse result in 3 1s in a list.
If you are looking for a more random distribution, it might make sense to randomly transform remaining as a separate variable before subtracting from it.
IE:
value = transformation(remaining)
component = int(value ** 0.5)
which should give you more "random" values.
Say you have an ordered array of values representing x coordinates.
[0,25,50,60,75,100]
You might notice that without the 60, the values would be evenly spaced (25). This would be indicative of a repeating pattern, something that I need to extract using this list (regardless of the length and the values of the list). In this particular example, the algorithm should find and remove the 60.
There are no time or space complexity requirements.
Both the values in the list and the ideal spacing (e.g 25) are unknown. So the algorithm must obtain this by looking at the values. In addition, the number of values, and where the outliers are in the array are not guaranteed. There may be more than one outlier. The algorithm should return a list with the outliers removed. Extra points if the algorithm uses a threshold for the spacing.
Edit: Here is an example image
Here there is one outlier on the x axis. (green-line) There are two on the y axis. The x-coordinates of the array represent the rho of the line on that axis.
arr = [0,25,50,60,75,100]
First construct the distances array
dist = np.array([arr[i+1] - arr[i] for (i, _) in enumerate(arr) if i < len(arr)-1])
print(dist)
>> [25 25 10 15 25]
Now I'm using np.where and np.percentile to cut the array in 3 part: the main , the upper values and the lower values. I arbitrary set them to 5%.
cond_sup = np.where(dist > np.percentile(dist, 95))
print(cond_sup)
>> (array([]),)
cond_inf = np.where(dist < np.percentile(dist, 5))
print(cond_inf)
>> (array([2]),)
You now got indexes where the value is different from the others.
So, dist[2] has a problem, which mean by construction the problem is between arr[2] and arr[2+1]
I don't know if you want to remove 1 or more numbers from this array. So I think the way to solve this problem will be like this:
array A[] = [0,25,50,60,75,100];
sort array (if needed).
create a new array B[] with value i-th: B[i] = A[i+1] - A[i]
find the value of B[] elements that appear most time. It's will be our distance.
find i such that A[i+1]-A[i] != distance
find k (k>i and k min) such that A[i+k]-A[i] == distance
so, we need remove A[i+1] => A[i+k-1]
I hope it is right.
I must solve the Euler Bernoulli differential beam equation which is:
w’’’’(x) = q(x)
and boundary conditions:
w(0) = w(l) = 0
and
w′′(0) = w′′(l) = 0
The beam is as shown on the picture below:
beam
The continious force q is 2N/mm.
I have to use shooting method and scipy.integrate.odeint() func.
I can't even manage to start as i do not understand how to write the differential equation as a system of equation
Can someone who understands solving of differential equations with boundary conditions in python please help!
Thanks :)
The shooting method
To solve the fourth order ODE BVP with scipy.integrate.odeint() using the shooting method you need to:
1.) Separate the 4th order ODE into 4 first order ODEs by substituting:
u = w
u1 = u' = w' # 1
u2 = u1' = w'' # 2
u3 = u2' = w''' # 3
u4 = u3' = w'''' = q # 4
2.) Create a function to carry out the derivation logic and connect that function to the integrate.odeint() like this:
function calc(u, x , q)
{
return [u[1], u[2], u[3] , q]
}
w = integrate.odeint(calc, [w(0), guess, w''(0), guess], xList, args=(q,))
Explanation:
We are sending the boundary value conditions to odeint() for x=0 ([w(0), w'(0) ,w''(0), w'''(0)]) which calls the function calc which returns the derivatives to be added to the current state of w. Note that we are guessing the initial boundary conditions for w'(0) and w'''(0) while entering the known w(0)=0 and w''(0)=0.
Addition of derivatives to the current state of w occurs like this:
# the current w(x) value is the previous value plus the current change of w in dx.
w(x) = w(x-dx) + dw/dx
# others are calculated the same
dw(x)/dx = dw(x-dx)/dx + d^2w(x)/dx^2
# etc.
This is why we are returning values [u[1], u[2], u[3] , q] instead of [u[0], u[1], u[2] , u[3]] from the calc function, because u[1] is the first derivative so we add it to w, etc.
3.) Now we are able to set up our shooting method. We will be sending different initial boundary values for w'(0) and w'''(0) to odeint() and then check the end result of the returned w(x) profile to determine how close w(L) and w''(L) got to 0 (the known boundary conditions).
The program for the shooting method:
# a function to return the derivatives of w
def returnDerivatives(u, x, q):
return [u[1], u[2], u[3], q]
# a shooting funtion which takes in two variables and returns a w(x) profile for x=[0,L]
def shoot(u2, u4):
# the number of x points to calculate integration -> determines the size of dx
# bigger number means more x's -> better precision -> longer execution time
xSteps = 1001
# length of the beam
L= 1.0 # 1m
xSpace = np.linspace(0, L, xSteps)
q = 0.02 # constant [N/m]
# integrate and return the profile of w(x) and it's derivatives, from x=0 to x=L
return odeint(returnDerivatives, [ 0, u2, 0, u4] , xSpace, args=(q,))
# the tolerance for our results.
tolerance = 0.01
# how many numbers to consider for u2 and u4 (the guess boundary conditions)
u2_u4_maxNumbers = 1327 # bigger number, better precision, slower program
# you can also divide into separate variables like u2_maxNum and u4_maxNum
# these are already tested numbers (the best results are somewhere in here)
u2Numbers = np.linspace(-0.1, 0.1, u2_u4_maxNumbers)
# the same as above
u4Numbers = np.linspace(-0.5, 0.5, u2_u4_maxNumbers)
# result list for extracted values of each w(x) profile => [u2Best, u4Best, w(L), w''(L)]
# which will help us determine if the w(x) profile is inside tolerance
resultList = []
# result list for each U (or w(x) profile) => [w(x), w'(x), w''(x), w'''(x)]
resultW = []
# start generating numbers for u2 and u4 and send them to odeint()
for u2 in u2Numbers:
for u4 in u4Numbers:
U = []
U = shoot(u2,u4)
# get only the last row of the profile to determine if it passes tolerance check
result = U[len(U)-1]
# only check w(L) == 0 and w''(L) == 0, as those are the known boundary cond.
if (abs(result[0]) < tolerance) and (abs(result[2]) < tolerance):
# if the result passed the tolerance check, extract some values from the
# last row of the w(x) profile which we will need later for comaprisons
resultList.append([u2, u4, result[0], result[2]])
# add the w(x) profile to the list of profiles that passed the tolerance
# Note: the order of resultList is the same as the order of resultW
resultW.append(U)
# go through the resultList (list of extracted values from last row of each w(x) profile)
for i in range(len(resultList)):
x = resultList[i]
# both boundary conditions are 0 for both w(L) and w''(L) so we will simply add
# the two absolute values to determine how much the sum differs from 0
y = abs(x[2]) + abs(x[3])
# if we've just started set the least difference to the current
if i == 0:
minNum = y # remember the smallest difference to 0
index = 0 # remember index of best profile
elif y < minNum:
# current sum of absolute values is smaller
minNum = y
index = i
# print out the integral for w(x) over the beam
sum = 0
for i in resultW[index]:
sum = sum + i[0]
print("The integral of w(x) over the beam is:")
print(sum/1001) # sum/xSteps
This outputs:
The integral of w(x) over the beam is:
0.000135085272117
To print out the best profile for w(x) that we found:
print(resultW[index])
which outputs something like:
# w(x) w'(x) w''(x) w'''(x)
[[ 0.00000000e+00 7.54147813e-04 0.00000000e+00 -9.80392157e-03]
[ 7.54144825e-07 7.54142917e-04 -9.79392157e-06 -9.78392157e-03]
[ 1.50828005e-06 7.54128237e-04 -1.95678431e-05 -9.76392157e-03]
...,
[ -4.48774290e-05 -8.14851572e-04 1.75726275e-04 1.01560784e-02]
[ -4.56921910e-05 -8.14670764e-04 1.85892353e-04 1.01760784e-02]
[ -4.65067671e-05 -8.14479780e-04 1.96078431e-04 1.01960784e-02]]
To double check the results from above we will also solve the ODE using the numerical method.
The numerical method
To solve the problem using the numerical method we first need to solve the differential equations. We will get four constants which we need to find with the help of the boundary conditions. The boundary conditions will be used to form a system of equations to help find the necessary constants.
For example:
w’’’’(x) = q(x);
means that we have this:
d^4(w(x))/dx^4 = q(x)
Since q(x) is constant after integrating we have:
d^3(w(x))/dx^3 = q(x)*x + C
After integrating again:
d^2(w(x))/dx^2 = q(x)*0.5*x^2 + C*x + D
After another integration:
dw(x)/dx = q(x)/6*x^3 + C*0.5*x^2 + D*x + E
And finally the last integration yields:
w(x) = q(x)/24*x^4 + C/6*x^3 + D*0.5*x^2 + E*x + F
Then we take a look at the boundary conditions (now we have expressions from above for w''(x) and w(x)) with which we make a system of equations to solve the constants.
w''(0) => 0 = q(x)*0.5*0^2 + C*0 + D
w''(L) => 0 = q(x)*0.5*L^2 + C*L + D
This gives us the constants:
D = 0 # from the first equation
C = - 0.01 * L # from the second (after inserting D=0)
After repeating the same for w(0)=0 and w(L)=0 we obtain:
F = 0 # from first
E = 0.01/12.0 * L^3 # from second
Now, after we have solved the equation and found all of the integration constants we can make the program for the numerical method.
The program for the numerical method
We will make a FOR loop to go through the entire beam for every dx at a time and sum up (integrate) w(x).
L = 1.0 # in meters
step = 1001.0 # how many steps to take (dx)
q = 0.02 # constant [N/m]
integralOfW = 0.0; # instead of w(0) enter the boundary condition value for w(0)
result = []
for i in range(int(L*step)):
x= i/step
w = (q/24.0*pow(x,4) - 0.02/12.0*pow(x,3) + 0.01/12*pow(L,3)*x)/step # current w fragment
# add up fragments of w for integral calculation
integralOfW += w
# add current value of w(x) to result list for plotting
result.append(w*step);
print("The integral of w(x) over the beam is:")
print(integralOfW)
which outputs:
The integral of w(x) over the beam is:
0.00016666652805511192
Now to compare the two methods
Result comparison between the shooting method and the numerical method
The integral of w(x) over the beam:
Shooting method -> 0.000135085272117
Numerical method -> 0.00016666652805511192
That's a pretty good match, now lets see check the plots:
From the plots it's even more obvious that we have a good match and that the results of the shooting method are correct.
To get even better results for the shooting method increase xSteps and u2_u4_maxNumbers to bigger numbers and you can also narrow down the u2Numbers and u4Numbers to the same set size but a smaller interval (around the best results from previous program runs). Keep in mind that setting xSteps and u2_u4_maxNumbers too high will cause your program to run for a very long time.
You need to transform the ODE into a first order system, setting u0=w one possible and usually used system is
u0'=u1,
u1'=u2,
u2'=u3,
u3'=q(x)
This can be implemented as
def ODEfunc(u,x): return [ u[1], u[2], u[3], q(x) ]
Then make a function that shoots with experimental initial conditions and returns the components of the second boundary condition
def shoot(u01, u03): return odeint(ODEfunc, [0, u01, 0, u03], [0, l])[-1,[0,2]]
Now you have a function of two variables with two components and you need to solve this 2x2 system with the usual methods. As the system is linear, the shooting function is linear as well and you only need to find the coefficients and solve the resulting linear system.
I'm trying to calculate histogram for an image. I'm using the following formula to calculate the bin
%bin = red*(N^2) + green*(N^1) + blue;
I have to implement the following Matlab functions.
[row, col, noChannels] = size(rgbImage);
hsvImage = rgb2hsv(rgbImage); % Ranges from 0 to 1.
H = zeros(4,4,4);
for col = 1 : columns
for row = 1 : rows
hBin = floor(hsvImage(row, column, 1) * 15);
sBin = floor(hsvImage(row, column, 2) * 4);
vBin = floor(hsvImage(row, column, 3) * 4);
F(hBin, sBin, vBin) = hBin, sBin, vBin + 1;
end
end
When I run the code I get the following error message "Subscript indices must either be real positive integers or logical."
As I am new to Matlab and Image processing, I'm not sure if the problem is with implementing the algorithm or a syntax error.
There are 3 problems with your code. (Four if you count that you changed from H to F your accumulator vector, but I'll assume that's a typo.)
First one, your variable bin can be zero at any moment if the values of a giving pixel are low. And F(0) is not a valid index for a vector or matrix. This is why you are getting that error.
You can solve easily by doing F(bin+1) and keep in mind that your F vector will have your values shifted one position over.
Second error, you are assigning the value bin + 1 to your accumulator vector F, which is not what you want, you want to add 1 every time a pixel in that range is found, what you should do is F(bin+1) = F(bin+1) + 1;. This way the values of F will be increasing all the time.
Third error is simpler, you forgot to implement your bin = red*(N^2) + green*(N^1) + blue; equation