How to find vertices of a polytope in python? - python

I want to find the vertices given the following:
`A = np.array([
[-1,2/3,0,0,0,0,0,0,0],
[1,-2/3,0,0,0,0,0,0,0],
[-1,0,0,2/3,0,0,0,0,0],
[-1,0,2/3,0,0,0,0,0,0],
[1,0,-2/3,0,0,0,0,0,0],
[-1,0,0,0,2/3,0,0,0,0],
[-1,0,0,0,2/3,0,0,0,0],
[1,0,0,-2/3,0,0,0,0,0],
[-1,2/3,0,0,0,0,0,0,0]])`
`b = np.array([-1/3, 1/3, -2/3, 1/3, -1/3, 0, -2/3, 2/3, -1/3])` .
I tried to compute the vertices by using the
vertices = pypoman.compute_polytope_vertices(A,b).
However, I get the following error:
raise Exception("Polyhedron is not a polytope")
Exception: Polyhedron is not a polytope".
Did anyone have a problem like this?

Use pycddlib:
# -*- coding: utf-8 -*-
import numpy as np
import cdd as pcdd
import fractions as frac
A = np.array([
[-1,frac.Fraction(2,3),0,0,0,0,0,0,0],
[1,-frac.Fraction(2,3),0,0,0,0,0,0,0],
[-1,0,0,frac.Fraction(2,3),0,0,0,0,0],
[-1,0,frac.Fraction(2,3),0,0,0,0,0,0],
[1,0,-frac.Fraction(2,3),0,0,0,0,0,0],
[-1,0,0,0,frac.Fraction(2,3),0,0,0,0],
[-1,0,0,0,frac.Fraction(2,3),0,0,0,0],
[1,0,0,-frac.Fraction(2,3),0,0,0,0,0],
[-1,frac.Fraction(2,3),0,0,0,0,0,0,0]])
b = np.array([
[-frac.Fraction(1,3)],
[frac.Fraction(1,3)],
[-frac.Fraction(2,3)],
[frac.Fraction(1,3)],
[-frac.Fraction(1,3)],
[0],
[-frac.Fraction(2,3)],
[frac.Fraction(2,3)],
[-frac.Fraction(1,3)]
])
M = np.hstack( (b, -A) )
mat = pcdd.Matrix(M, linear=False, number_type="fraction")
mat.rep_type = pcdd.RepType.INEQUALITY
poly = pcdd.Polyhedron(mat)
ext = poly.get_generators()
print(ext)
Giving:
1 2/3 1/2 3/2 0 0 0 0 0 0
0 1 3/2 3/2 3/2 0 0 0 0 0
0 2/3 1 1 1 1 0 0 0 0
0 0 0 0 0 0 1 0 0 0
0 0 0 0 0 0 0 1 0 0
0 0 0 0 0 0 0 0 1 0
0 0 0 0 0 0 0 0 0 1
Indeed, this is not a polytope. A "1" in the first column means it is an extreme point, and here there is only one.

Related

Counting repeated sequences in transition table

I'm using the following function to generate a transition table:
import numpy as np
import pandas as pd
def make_table(allSeq):
n = max([ max(s) for s in allSeq ]) + 1
arr = np.zeros((n,n), dtype=int)
for seq in allSeq:
ind = (seq[1:], seq[:-1])
arr[ind] += 1
return pd.DataFrame(arr).rename_axis(index='Next', columns='Current')
However, my result is incorrect:
list1 = [1,2,3,4,5,4,5,4,5]
list2 = [4,5,4,5]
make_table([list1, list2])
Current 0 1 2 3 4 5
Next
0 0 0 0 0 0 0
1 0 0 0 0 0 0
2 0 1 0 0 0 0
3 0 0 1 0 0 0
4 0 0 0 1 0 2
5 0 0 0 0 2 0
For example, the transition 4->5 should be counted 5 times, but it's only counted once per sequence (2). I know the issue is the arr[ind] += 1 line, but I just can't figure it out! Do I nest another loop, or is there a slick way to add the total number of instances at once? Thanks!
Figured it out! Switched to the following:
def make_table(allSeq):
n = max([ max(s) for s in allSeq ]) + 1
arr = np.zeros((n,n), dtype=int)
for seq in allSeq:
for i,j in zip(seq[1:],seq[:-1]):
ind = (i,j)
arr[ind] += 1
return pd.DataFrame(arr).rename_axis(index='Next', columns='Current')
Another loop seems like the easiest solution, with a bit of a twist of using zip:
import numpy as np
import pandas as pd
def make_table(allSeq):
n = max([ max(s) for s in allSeq ]) + 1
arr = np.zeros((n,n), dtype=int)
for seq in allSeq:
ind = zip(seq[1:], seq[:-1])
for i in ind:
arr[i] += 1
return pd.DataFrame(arr).rename_axis(index='Next', columns='Current')
list1 = [1,2,3,4,5,4,5,4,5]
list2 = [4,5,4,5]
make_table([list1, list2])
returns
Next 0 1 2 3 4 5
------ --- --- --- --- --- ---
0 0 0 0 0 0 0
1 0 0 0 0 0 0
2 0 1 0 0 0 0
3 0 0 1 0 0 0
4 0 0 0 1 0 3
5 0 0 0 0 5 0

Complex structure of int32_t and char array in python ctypes

I am trying to create a structure to use in a C library provided (DLL),
How the following structure (given in the documentation) can be defined?
#define A 10
#define B 20
typedef struct
{
int32_t size;
int32_t num;
char buf1[A][B];
char buf2[A][B];
char buf3[A][B];
} INSTRUCT;
My attempt to define it in python using ctypes was like so:
from ctypes import*
char_buff1 = ((c_char * 10) * 20)
char_buff2 = ((c_char * 10) * 20)
char_buff3 = ((c_char * 10) * 20)
class INSTRUCT(Structure):
_fields_=[("size",c_int32),("num",c_int32),("buf1",char_buff1),("buf2",char_buff2),("buf3",char_buff3)]
Can int32_t be replaced with c_int_32 in ctypes?
Is it correct way to define the structure?
Then I tried to feed the pointer of the structure to the DLL function and check what it returns as follows:
dlllib = CDLL("some.dll")
somefunction = dlllib.some_function
somefunction.argtypes = [POINTER(INSTRUCT)]
INSTRUCT().size
INSTRUCT().num
print(np.ctypeslib.as_array(INSTRUCT().buf1))
However, I can only the return is 0 and unmodified by the function -- equal to the one defined before the C function call.
I am not sure at which stage the problem occurs, however, there are no errors, the code executes normally.
Unfortunately, I don't have the C code available, only the input parameters for the function.
Best regards
The array definition is wrong. In ctypes, the array indices need to be reversed to index the array the way C does. For example, the equivalent of char buf[x][y] in Python with ctypes is buf = (c_char * y * x)(). Note that the bounds are reversed. Otherwise, your definition was correct.
Note that using c_char will return text characters for array values. If you want integers, use c_int8. I'll use the latter below.
Example:
from ctypes import *
import numpy as np
A,B = 10,20
ARRAY = c_int8 * B * A # build as B,A
class INSTRUCT(Structure):
_fields_=[("size",c_int32),
("num",c_int32),
("buf1",ARRAY),
("buf2",ARRAY),
("buf3",ARRAY)]
i = INSTRUCT()
i.buf1[9][19] = 1 # access indices as A,B
print(np.ctypeslib.as_array(i.buf1))
[[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1]] # 1 in correct location
Your example of accessing used INSTRUCT() which creates a new, zeroed object each time. Create a single instance and pass it to a function like so:
dlllib = CDLL("some.dll")
somefunction = dlllib.some_function
somefunction.argtypes = [POINTER(INSTRUCT)]
i = INSTRUCT() # make an instance
somefunction(byref(i)) # byref() passes address of a ctypes object.

How to split a list using two nested conditions

Basically I have list of 0s and 1s. Each value in the list represents a data sample from an hour. Thus, if there are 24 0s and 1s in the list that means there are 24 hours, or a single day. I want to capture the first time the data cycles from 0s to 1s back to 0s in a span of 24 hours (or vice versa from 1s to 0s back to 1s).
signal = [1,1,1,1,1,0,0,0,0,0,1,1,1,1,1,0,0,0,1,1,1,1,1,0,0,0,0,0,0,0,1]
expected output:
# D
signal = [1,1,1,1,1,0,0,0,0,0,1,1,1,1,1,0,0,0,1,1,1,1,1,0,0,0,0,0,0,0,1,1,0,0,0]
output = [0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0]
# ^ cycle.1:day.1 |dayline ^cycle.1:day.2
In the output list, when there is 1 that means 1 cycle is completed at that position of the signal list and at rest of the position there are 0. There should only 1 cycle in a days that's why only 1 is there.
I don't how to split this list according to that so can someone please help?
It seams to me like what you are trying to do is split your data first into blocks of 24, and then to find either the first rising edge, or the first falling edge depending on the first hour in that block.
Below I have tried to distill my understanding of what you are trying to accomplish into the following function. It takes in a numpy.array containing zeros and ones, as in your example. It checks to see what the first hour in the day is, and decides what type of edge to look for.
it detects an edge by using np.diff. This gives us an array containing -1's, 0's, and 1's. We then look for the first index of either a -1 falling edge, or 1 rising edge. The function returns that index, or if no edges were found it returns the index of the last element, or nothing.
For more info see the docs for descriptions on numpy features used here np.diff, np.array.nonzero, np.array_split
import numpy as np
def get_cycle_index(day):
'''
returns the first index of a cycle defined by nipun vats
if no cycle is found returns nothing
'''
first_hour = day[0]
if first_hour == 0:
edgetype = -1
else:
edgetype = 1
edges = np.diff(np.r_[day, day[-1]])
if (edges == edgetype).any():
return (edges == edgetype).nonzero()[0][0]
elif (day.sum() == day.size) or day.sum() == 0:
return
else:
return day.size - 1
Below is an example of how you might use this function in your case.
import numpy as np
_data = [1,1,1,1,1,0,0,0,0,0,1,1,1,1,1,0,0,0,1,1,1,1,1,0,0,0,0,0,0,0,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]
#_data = np.random.randint(0,2,280, dtype='int')
data = np.array(_data, 'int')
#split the data into a set of 'day' blocks
blocks = np.array_split(data, np.arange(24,data.size, 24))
_output = []
for i, day in enumerate(blocks):
print(f'day {i}')
buffer = np.zeros(day.size, dtype='int')
print('\tsignal:', *day, sep = ' ')
cycle_index = get_cycle_index(day)
if cycle_index:
buffer[cycle_index] = 1
print('\toutput:', *buffer, sep=' ')
_output.append(buffer)
output = np.concatenate(_output)
print('\nfinal output:\n', *output, sep=' ')
this yeilds the following output:
day 0
signal: 1 1 1 1 1 0 0 0 0 0 1 1 1 1 1 0 0 0 1 1 1 1 1 0
output: 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0
day 1
signal: 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
output: 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
day 2
signal: 0 0 0 0 0 0
output: 0 0 0 0 0 0
final output:
0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

Converting this operation from matlab to python

I have this line in some matlab script that Im trying to convert to python. So, m=20, and n=20. The dimensions of I_true equals [400,1].
I want to convert following Matlab code:
A=zeros((2*m*n),(2*m*n)+2);
A(1:m*n,(2*m*n)+1)=-I_true(:);
Am I converting it right?
Converted code in Python:
for i in range(0,m*n):
for j in range((2*m*n)+1):
A[i][j] = I_true[i]
Let's look at a small example, with n = 2, m = 2:
In Octave (and presumably Matlab):
octave:50> m = 2; n = 2;
octave:51> I_true = [1;2;3;4];
octave:52> A = zeros((2*m*n),(2*m*n)+2);
octave:53> A(1:m*n,(2*m*n)+1)=-I_true(:)
A =
0 0 0 0 0 0 0 0 -1 0
0 0 0 0 0 0 0 0 -2 0
0 0 0 0 0 0 0 0 -3 0
0 0 0 0 0 0 0 0 -4 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
The equivalent in Python (with n = 20, m = 20) would be
import numpy as np
n, m = 20, 20
I_true = np.arange(1, n*m+1) # just as an example
A = np.zeros((2*m*n, 2*(n*m+1)), dtype=I.dtype)
A[:m*n, 2*m*n] = -I_true
The reason why the last line uses A[:m*n, 2*m*n] and not A[1:m*n, (2*m*n)+1] is
because Python uses 0-based indexing whereas Matlab uses 1-based indexing.
Check this so question as well.
You can define a matrix with 2*m*n rows and 2*m*n+2 columns in python like this:
m = 20
n = 20
a = [[0 for i in range(2*m*n)] for j in range((2*m*n)+2)]
Now you have your matrix you can assign values to its elements using different ways. One example would be using for loops to assign values from another matrix with same size:
for i in range(2*m*n):
for j in range((2*m*n)+2):
a[i][j] = I_true[i][j]
I hope it helps.

rotate an nxnxn matrix in python

I have a binary array of size 64x64x64, where a volume of 40x40x40 is set to "1" and rest is "0". I have been trying to rotate this cube about its center around z-axis using skimage.transform.rotate and also Opencv as:
def rotateImage(image, angle):
row, col = image.shape
center = tuple(np.array([row, col]) / 2)
rot_mat = cv2.getRotationMatrix2D(center, angle, 1.0)
new_image = cv2.warpAffine(image, rot_mat, (col, row))
return new_image
In the case of openCV, I tried, 2D rotation of each idividual slices in a cube (Cube[:,:,n=1,2,3...p]).
After rotating, total sum of the values in the array changes. This may be caused by interpolation during rotation. How can I rotate 3D array of this kind without adding anything to the array?
Ok so I understand now what you are asking. The closest I can come up with is scipy.ndimage. But there is a way interface with imagej from python if which might be easier. But here is what I did with scipy.ndimage:
from scipy.ndimage import interpolation
angle = 25 #angle should be in degrees
Rotatedim = interpolation.rotate(yourimage, angle, reshape = False,output = np.int32, order = 5,prefilter = False)
This worked for some angles to preserve the some and not others, perhaps by playing around more with the parameters you might be able to get your desired outcome.
One option is to convert into sparse, and transform the coordinates using a matrix rotation. Then transform back into dense. In 2 dimensions, this looks like:
import numpy as np
import scipy.sparse
import math
N = 10
space = np.zeros((N, N), dtype=np.int8)
space[3:7, 3:7].fill(1)
print(space)
print(np.sum(space))
space_coo = scipy.sparse.coo_matrix(space)
Coords = np.array(space_coo.nonzero()) - 3
theta = 30 * 3.1416 / 180
R = np.array([[math.cos(theta), math.sin(theta)], [-math.sin(theta), math.cos(theta)]])
space2_coords = R.dot(Coords)
space2_coords = np.round(space2_coords)
space2_coords += 3
space2_sparse = scipy.sparse.coo_matrix(([1] * space2_coords.shape[1], (space2_coords[0], space2_coords[1])), shape=(N, N))
space2 = space2_sparse.todense()
print(space2)
print(np.sum(space2))
Output:
[[0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0]
[0 0 0 1 1 1 1 0 0 0]
[0 0 0 1 1 1 1 0 0 0]
[0 0 0 1 1 1 1 0 0 0]
[0 0 0 1 1 1 1 0 0 0]
[0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0]]
16
[[0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0]
[0 0 0 1 0 0 0 0 0 0]
[0 0 1 1 1 1 0 0 0 0]
[0 0 1 1 1 1 1 0 0 0]
[0 1 1 0 1 1 0 0 0 0]
[0 0 0 1 1 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0]]
16
The advantage is that you'll get exactly as many 1 values before and after the transform. The downsides is that you might get 'holes', as above, and/or duplicate coordinates, giving values of '2' in the final dense matrix.

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