I have a Python-Code and want to rewrite it in Octave, but I meet so many problems during the converting. I found a solution for some of them and some of them still need your help. Now i would start with this part of the code :
INVOLUTE_FI = 0
INVOLUTE_FO = 1
INVOLUTE_OI = 2
INVOLUTE_OO = 3
def coords_inv(phi, geo, theta, inv):
"""
Coordinates of the involutes
Parameters
----------
phi : float
The involute angle
geo : struct
The structure with the geometry obtained from get_geo()
theta : float
The crank angle, between 0 and 2*pi
inv : int
The key for the involute to be considered
"""
rb = geo.rb
ro = rb*(pi - geo.phi_fi0 + geo.phi_oo0)
Theta = geo.phi_fie - theta - pi/2.0
if inv == INVOLUTE_FI:
x = rb*cos(phi)+rb*(phi-geo.phi_fi0)*sin(phi)
y = rb*sin(phi)-rb*(phi-geo.phi_fi0)*cos(phi)
elif inv == INVOLUTE_FO:
x = rb*cos(phi)+rb*(phi-geo.phi_fo0)*sin(phi)
y = rb*sin(phi)-rb*(phi-geo.phi_fo0)*cos(phi)
elif inv == INVOLUTE_OI:
x = -rb*cos(phi)-rb*(phi-geo.phi_oi0)*sin(phi)+ro*cos(Theta)
y = -rb*sin(phi)+rb*(phi-geo.phi_oi0)*cos(phi)+ro*sin(Theta)
elif inv == INVOLUTE_OO:
x = -rb*cos(phi)-rb*(phi-geo.phi_oo0)*sin(phi)+ro*cos(Theta)
y = -rb*sin(phi)+rb*(phi-geo.phi_oo0)*cos(phi)+ro*sin(Theta)
else:
raise ValueError('flag not valid')
return x,y
def CVcoords(CVkey, geo, theta, N = 1000):
"""
Return a tuple of numpy arrays for x,y coordinates for the lines which
determine the boundary of the control volume
Parameters
----------
CVkey : string
The key for the control volume for which the polygon is desired
geo : struct
The structure with the geometry obtained from get_geo()
theta : float
The crank angle, between 0 and 2*pi
N : int
How many elements to include in each entry in the polygon
Returns
-------
x : numpy array
X-coordinates of the outline of the control volume
y : numpy array
Y-coordinates of the outline of the control volume
"""
Nc1 = Nc(theta, geo, 1)
Nc2 = Nc(theta, geo, 2)
if CVkey == 'sa':
r = (2*pi*geo.rb-geo.t)/2.0
xee,yee = coords_inv(geo.phi_fie,geo,0.0,'fi')
xse,yse = coords_inv(geo.phi_foe-2*pi,geo,0.0,'fo')
xoie,yoie = coords_inv(geo.phi_oie,geo,theta,'oi')
xooe,yooe = coords_inv(geo.phi_ooe,geo,theta,'oo')
x0,y0 = (xee+xse)/2,(yee+yse)/2
beta = atan2(yee-y0,xee-x0)
t = np.linspace(beta,beta+pi,1000)
x,y = x0+r*np.cos(t),y0+r*np.sin(t)
return np.r_[x,xoie,xooe,x[0]],np.r_[y,yoie,yooe,y[0]]
https://docs.scipy.org/doc/numpy/reference/generated/numpy.r_.html I just don´t understand the last Output, and I am still confuse what´s mean _r here, and how can I write it by Octave?....I read what is written in the link, but it still not clear for me.
return np.r_[x,xoie,xooe,x[0]], np.r_[y,yoie,yooe,y[0]]
The function returns 2 values, both arrays created by np.r_.
np.r_[....] has indexing syntax, and ends up being translated into a function call to the np.r_ object. The result is just the concatenation of the arguments:
In [355]: np.r_[1, 3, 6:8, np.array([3,2,1])]
Out[355]: array([1, 3, 6, 7, 3, 2, 1])
With the [] notation it can accept slice like objects (6:8) though I don't see any of those here. I'd have to study the rest of the code to identify whether the other arguments are scalars (single values) or arrays.
My Octave is rusty (though I could experiment with the conversion).
t = np.lispace... # I think that exists in Octave, a 1000 values
x = x0+r*np.cos(t) # a derived array of 1000 values
xoie one of the values returned by coords_inv; may be scalar or array. x[0] the first value of x. So the r_ probably produces a 1d array made up of x, and the subsequent values.
Related
I'm new to Matlab and want to write a program that chooses the value of a parameter (P) to minimize the difference between two vectors, where each vector is a variable in a dataframe. The first vector (call it A) is a predetermined vector of 1s and 0s, and the second vector (call it B) has each of its entries determined as an indicator function that depends on the value of the parameter P and other variables in the dataframe. For instance, let C be a third variable in the dataset, so
A = [1, 0, 0, 1, 0]
B = [x, y, z, u, v]
where x = 1 if (C[1]+10)^0.5 - P > (C[1])^0.5 and otherwise x = 0, and similarly, y = 1 if (C[2]+10)^0.5 - P > (C[2])^0.5 and otherwise y = 0, and so on.
I'm not really sure where to start with the code, except that it might be useful to use the fminsearch command. Any suggestions?
Edit: I changed the above by raising to a power, which is closer to the actual example that I have. I'm also providing a complete example in response to a comment:
Let A be as above, and let C = [10, 1, 100, 1000, 1]. Then my goal with the Matlab code would be to choose a value of P to minimize the differences between the coordinates of the vectors A and B, where B[1] = 1 if (10+10)^0.5 - P > (10)^0.5 and otherwise B[1] = 0, and similarly B[2] = 1 if (1+10)^0.5 - P > (1)^0.5 and otherwise B[2] = 0, etc. So I want to choose P to maximize the likelihood that A[1] = B[1], A[2] = B[2], etc.
I have the following setup in Matlab, where ds is the name of my dataset:
ds.B = zeros(size(ds,1),1); % empty vector to fill
for i = 1:size(ds,1)
if ((ds.C(i) + 10)^(0.5) - P > (ds.C(i))^(0.5))
ds.B(i) = 1;
else
ds.B(i) = 0;
end
end
Now I want to choose the value of P to minimize the difference between A and B. How can I do this?
EDIT: I'm also wondering how to do this when the inequality is something like (C[i]+10)^0.5 - P*D[i] > (C[i])^0.5, where D is another variable in my dataset. Now P is a scalar being multiplied rather than just added. This seems more complicated since I can't solve for P exactly. How can I solve the problem in this case?
EDIT 1: It seems fminbnd() isn't optimal, likely due to the stairstep nature of the indicator function. I've updated to test the midpoints of all the regions between indicator function flips, plus endpoints.
EDIT 2: Updated to include dataset D as a coefficient of P.
If you can package your distance calculation up in a single function based on P, you can then search for its minimum.
arraySize = 1000;
ds.A = double(rand([arraySize,1]) > 0.5);
ds.C = rand(size(ds.A));
ds.D = rand(size(ds.A));
B = #(P)double((ds.C+10).^0.5 - P.*ds.D > ds.C.^0.5);
costFcn = #(P)sqrt(sum((ds.A-B(P)).^2));
% Solving the equation (C+10)^0.5 - P*D = C^0.5 for P, and sorting the results
BCrossingPoints = sort(((ds.C+10).^0.5-ds.C.^0.5)./ds.D);
% Taking the average of each crossing point with its neighbors
BMidpoints = (BCrossingPoints(1:end-1)+BCrossingPoints(2:end))/2;
% Appending endpoints onto the midpoints
PsToTest = [BCrossingPoints(1)-0.1; BMidpoints; BCrossingPoints(end)+0.1];
% Calculate the distance from A to B at each P to test
costResult = arrayfun(costFcn,PsToTest);
% Find the minimum cost
[~,lowestCostIndex] = min(costResult);
% Find the optimum P
optimumP = PsToTest(lowestCostIndex);
ds.B = B(optimumP);
semilogx(PsToTest,costResult)
xlabel('P')
ylabel('Distance from A to B')
1.- x is assumed positive real only, because with x<0 then complex values show up.
Since no comment is made in the question it seems reasonable to assume x real and x>0 only.
As requested, P 'the parameter' a scalar, P only has 2 significant states >0 or <0, let's see how is this:
2.- The following lines generate kind-of random A and C.
Then a sweep of p is carried out and distances d1 and d2 are calculated.
d1 is euclidean distance and d2 is the absolute of the difference between A and and B converting both from binary to decimal:
N=10
% A=[1 0 0 1 0]
A=randi([0 1],1,N);
% C=[10 1 1e2 1e3 1]
C=randi([0 1e3],1,N)
p=[-1e4:1:1e4]; % parameter to optimize
B=zeros(1,numel(A));
d1=zeros(1,numel(p)); % euclidean distance
d2=zeros(1,numel(p)); % difference distance
for k1=1:1:numel(p)
B=(C+10).^.5-p(k1)>C.^.5;
d1(k1)=(sum((B-A).^2))^.5;
d2(k1)=abs(sum(A.*2.^[numel(A)-1:-1:0])-sum(B.*2.^[numel(A)-1:-1:0]));
end
figure;
plot(p,d1)
grid on
xlabel('p');title('d1')
figure
plot(p,d2)
grid on
xlabel('p');title('d2')
The only degree of freedom to optimise seems to be the sign of P regardless of |P| value.
3.- f(p,x) has either no root, or just one root, depending upon p
The threshold funtion is
if f(x)>0 then B(k)==1 else B(k)==0
this is
f(p,x)=(x+10)^.5-p-x^.5
Now
(x+10).^.5-p>x.^.5 is same as (x+10).^.5-x.^.5>p
There's a range of p that keeps f(p,x)=0 without any (real) root.
For the particular case p=0 then (x+10).^.5 and x.^.5 do not intersect (until Inf reached = there's no intersection)
figure;plot(x,(x+10).^.5,x,x.^.5);grid on
[![enter image description here][3]][3]
y2=diff((x+10).^.5-x.^.5)
figure;plot(x(2:end),y2);
grid on;xlabel('x')
title('y2=diff((x+10).^.5-x.^.5)')
[![enter image description here][3]][3]
% 005
This means the condition f(x)>0 is always true holding all bits of B=1. With B=1 then d(A,B) turns into d(A,1), a constant.
However, for a certain value of p then there's one root and f(x)>0 is always false keeping all bits of B=0.
In this case d(A,B) the cost function turns into d(A,0) and this is A itself.
4.- P as a vector
The optimization gains in degrees of freedom if instead of P scalar, P is considered as vector.
For a given x there's a value of p that switches B(k) from 0 to 1.
Any value of p below such threshold keeps B(k)=0.
Equivalently, inverting f(x) :
g(p)=(10-p^2)^2/(4*p^2)>x
Values of x below this threshold bring B closer to A because for each element of B it's flipped to the element value of A.
Therefore, it's convenient to consider P as a vector, not a ascalar, and :
For all, or as many (as possible) elements of C to meet c(k)<(10-p^2)^2/(4*p^2) in order to get C=A or
minimize d(A,C)
5.- roots of f(p,x)
syms t positive
p=[-1000:.1:1000];
zp=NaN*ones(1,numel(p));
sol=zeros(1,numel(p));
for k1=1:1:numel(p)
p(k1)
eq1=(t+10)^.5-p(k1)-t^.5-p(k1)==0;
s1=solve(eq1,t);
if ~isempty(s1)
zp(k1)=s1;
end
end
nzp=~isnan(zp);
zp(nzp)
returns
=
620.0100 151.2900 64.5344 34.2225 20.2500 12.7211
8.2451 5.4056 3.5260 2.2500 1.3753 0.7803
0.3882 0.1488 0.0278
I need to splice two vectors based on a condition that also takes a vector as an argument. Example:
vec_cond = -5:5; % The exact values are calculated differently
vec1 = 0:10;
vec2 = 5:15;
I need a resulting vector to be comprised from values out of both vectors based on a condition from the third vector. Let's assume this is the condition: vec_cond >= 0
Then if this is true, I want vec_result to have values from vec1 on appropriate indexes, and if not, take values from vec2 on appropriate indexes:
vec_result = vec1 if (vec_cond >=0) else vec2
This is portion of my MATLAB script (original comments were Czech) where I would need to use that:
%% Draw output current and voltage characteristics
R = 100:5:2*10^3; % Load rezistor [ohm]
U_2 = R .* (I * 10^(-3)); % Load voltage [V]
U_1stab = U_LM + U_x + U_2; % Min. required input voltage
% for stabilization [V]
U_delta = U_1 - U_1stab; % Difference between actual and
% min. req. input voltage [V]
U_2norm = U_1 - U_LM - U_x % Calculating output load
% voltage based on params [V]
I_z = U_2norm ./ R .* 10^3; % Load current param based[mA]
I_r1 = I * I_z.^0; % Stabilizator current [mA]
So the condition would be U_delta >= 0.
I tried to use a ternary operator, which I found here:
I_graph = (U_delta >= 0) : (#() I) : (#() I_z); % Current splice [mA]
U_graph = (U_delta >= 0) : (#() U_2) : (#() U_2norm); % Voltage splice [V]
That means that for I_graph, if the condition is met, take a constant value I and vectorize it, otherwise take values from I_z vector. For U_graph, if the condition is met, take values from U_2 vector, otherwise take constant value of U_2norm and vectorize it.
But it didn't work, this is what it tells me:
Operator ':' is not supported for operands of type 'function_handle'.
Error in vypocet1 (line 52)
I_graph = (U_delta >= 0) : (#() I) : (#() I_z); % Current splice [mA]
I guess that I might want to use for loop, but I'm not sure how it will help me and how can I actually construct the necessary vector using a for loop.
Given:
vec_cond = -5:5;
vec1 = 0:10;
vec2 = 5:15;
You can set:
out = vec2;
I = vec_cond >= 0;
out(I) = vec1(I);
This uses logical indexing, which is indexing with a logical array.
By the way, the ternary operator you found is an exercise to overload the : operator for a specific class to do something that it normally doesn’t do. Note how you use the colon when creating vec_cond. This is what the colon operator does normally.
I'm trying to plot the Yukawa Potential in Matlab and I want to have my program go through user inputs for the values alpha (called alph in my program) and l. The values I need to use are 0.1, 0.2 and 0.3 for alpha with values of 0, 1 and 2 of l for each value of alpha. I know I could set up a loop for this but it doesn't have to be pretty and I want to test the values one at a time. Anyway I keep getting an error after I input the values for alpha, the error I keep getting is in my function, saying that I don't have enough input arguments. The output should be the T matrix, the Hamiltonian matrix and a plot of the first 10 eigenfunctions.
I've tried going in and simply defining alpha as the numbers I want to look at and the program works fine with displaying the output I'm looking for. I just want to be able to change the values for alpha without having to change the program itself. I haven't had any problems with the l inputs.
r = linspace(0.05,19.95,1999)
n = 1999
dr = 0.05
a = full(gallery("tridiag",n,1,-2,1))
T = -0.5*a/(dr^2)
l = input('Input a value for l.')
alph = input('Input a value for alpha.')
v = arrayfun(#(r) yuk_pot(r,l),r);
V = diag(v)
H = T + V
[O,D] = eig(H);
plot(r,O(:,1),r,O(:,2),r,O(:,3),r,O(:,4),r,O(:,5),r,O(:,6),r,O(:,7),r,O(:,8),r,O(:,9),r,O(:,10))
function v = yuk_pot(r,alph,l)
v = (-exp(-alph*r)/r) + 0.5*(l*(l+1)/(r^2));
end
your function function v = yuk_pot(r,alph,l) has 3 input arguments.
you call it with 2 arguments (r and l)
v = arrayfun(#(r) yuk_pot(r,l),r);
what about the second alph argument?
I am trying to use interpn (in python using Scipy) to replicate results from Matlab using interp3. However, I am struggling to structure my arguments. I tried the following line:
f = interpn(blur_maps, fx, fy, pyr_level)
Where blur maps is a 600 x 800 x 7 representing a grayscale image at seven levels of blur,
fx and fy are indices of the seven maps. Both fx and fy are 2d arrays. pyr_level is a 2d array that contains values from 1 to 7 representing the blur map to be interpolated.
My question is since I incorrectly arranged the arguments, how can I arrange them in a way that works? I tried to look up examples but I didn't see anything similar. Here is an example of the data I am trying to interpolate:
import numpy as np
import cv2, math
from scipy.interpolate import interpn
levels = 7
img_path = '/Users/alimahdi/Desktop/i4.jpg'
img = cv2.cvtColor(cv2.imread(img_path), cv2.COLOR_BGR2GRAY)
row, col = img.shape
x_range = np.arange(0, col)
y_range = np.arange(0, row)
fx, fy = np.meshgrid(x_range, y_range)
e = np.exp(np.sqrt(fx ** 2 + fy ** 2))
pyr_level = 7 * (e - np.min(e)) / (np.max(e) - np.min(e))
blur_maps = np.zeros((row, col, levels))
blur_maps[:, :, 0] = img
for i in range(levels - 1):
img = cv2.pyrDown(img)
r, c = img.shape
tmp = img
for j in range(int(math.log(row / r, 2))):
tmp = cv2.pyrUp(tmp)
blur_maps[:, :, i + 1] = tmp
pixelGrid = [np.arange(x) for x in blur_maps.shape]
interpPoints = np.array([fx.flatten(), fy.flatten(), pyr_level.flatten()])
interpValues = interpn(pixelGrid, blur_maps, interpPoints.T)
finalValues = np.reshape(interpValues, fx.shape)
I am now getting the following error: ValueError: One of the requested xi is out of bounds in dimension 0 I do know that the problem is in interpPoints but I am not sure how to fix it. Any suggestions?
The documentation for scipy.interpolate.interpn states that the first argument is a grid of the data you are interpolating over (which is just the integers of the pixel numbers), second argument is data (blur_maps) and third arguments is the interpolation points in the form (npoints, ndims). So you would have to do something like:
import scipy.interpolate
pixelGrid = [np.arange(x) for x in blur_maps.shape] # create grid of pixel numbers as per the docs
interpPoints = np.array([fx.flatten(), fy.flatten(), pyr_level.flatten()])
# interpolate
interpValues = scipy.interpolate.interpn(pixelGrid, blur_maps, interpPoints.T)
# now reshape the output array to get in the original format you wanted
finalValues = np.reshape(interpValues, fx.shape)
I want to use the y value corresponding to the given x value from the table (my current table has 1000 values with 10-4 decimal points so I use :
load question_table.mat
eta_p = %assign a value
F12_p=find( (eta <eta_p+0.01) & (eta > eta_p-0.01), 1, 'first' )
what is missing ?
Here is how I have created the table, run this program.
i = 1;
etaspan = -500:0.001:500;
y = zeros(length(etaspan),1);
f = #(x,eta) (x.^(1/2))./(1+exp(x-eta));
for eta = etaspan
g = #(x) f(x,eta);
y(i) = integral(g,0,500);
i = i + 1;
end
f=y
eta=etaspan
save question_table.mat eta f
Just have MATLAB do the interpolation for you:
y_p = interp1(eta, y, eta_p);
interp1 uses linear interpolation by default, but can instead use higher order interpolation methods. Even with linear, your table seems much denser than necessary.