Python 解线性不等式
我想解一个不等式组，它似乎是你所需要的超集。这应该很容易处理线性不等式 更新 我又在考虑这个问题，我想我会尝试看看没有Python 解线性不等式,python,numpy,matplotlib,inequalities,Python,Numpy,Matplotlib,Inequalities,我想解一个不等式组，它似乎是你所需要的超集。这应该很容易处理线性不等式 更新 我又在考虑这个问题，我想我会尝试看看没有fillplots可以做什么，只要使用标准库，比如scipy和numpy 在这样一个不等式系统中，每个方程定义了一个半空间。该系统是所有这些半空间的交集，是一个凸集 查找该集合的顶点（例如，绘制顶点）称为。幸运的是，有强大的算法来处理凸包，计算n维的半空间交点（以及做许多其他奇妙的事情）。示例实现是 更幸运的是，我们可以通过scipy.spacial直接访问该库的各个方面，特别是
fillplotsscipynumpyscipy.spacial半空间相交所需的内点
InfnanAx有一个优秀的库pypoman，它解决了顶点枚举问题，可以帮助您解决问题，但不幸的是它只输出集合的顶点，不是视觉化。顶点可能无序，如果没有其他操作，可视化将不正确。为了克服这个问题，您可以使用这个站点的算法（Graham scan或algo Jarvis）
下面是一个示例代码：
import pypoman
import cdd
import matplotlib.pyplot as plt
def grahamscan(A):
def rotate(A,B,C):
return (B[0]-A[0])*(C[1]-B[1])-(B[1]-A[1])*(C[0]-B[0])
n = len(A)
if len(A) == 0:
return A
P = np.arange(n)
for i in range(1,n):
if A[P[i]][0]<A[P[0]][0]:
P[i], P[0] = P[0], P[i]
for i in range(2,n):
j = i
while j>1 and (rotate(A[P[0]],A[P[j-1]],A[P[j]])<0):
P[j], P[j-1] = P[j-1], P[j]
j -= 1
S = [P[0],P[1]]
for i in range(2,n):
while rotate(A[S[-2]],A[S[-1]],A[P[i]])<0:
del S[-1]
S.append(P[i])
return S
def compute_poly_vertices(A, b):
b = b.reshape((b.shape[0], 1))
mat = cdd.Matrix(np.hstack([b, -A]), number_type='float')
mat.rep_type = cdd.RepType.INEQUALITY
P = cdd.Polyhedron(mat)
g = P.get_generators()
V = np.array(g)
vertices = []
for i in range(V.shape[0]):
if V[i, 0] != 1: continue
if i not in g.lin_set:
vertices.append(V[i, 1:])
return vertices
A = np.array([[-1, 1],
[0, 1],
[0.5, 1],
[1.5, 1],
[-1, 0],
[0, -1]])
b = np.array([1, 2, 3, 6, 0, 0])
vertices = np.array(compute_poly_vertices(A, b))
print(vertices)
vertices = np.array(vertices[grahamscan(vertices)])
x, y = vertices[:, 0], vertices[:, 1]
fig=plt.figure(figsize=(15,15))
ax = fig.add_subplot(111, title="Solution")
ax.fill(x, y, linestyle = '-', linewidth = 1, color='gray', alpha=0.5)
ax.scatter(x, y, s=10, color='black', alpha=1)
我对我最初的答案做了很多扩展。。。看看，嗯。。。不完全是我想要的，因为在片场上仍然有很多线要建，但已经有了进展！）谢谢你说的“台词”是什么意思？您可以只绘制凸包，而这只是红色多边形，没有与不等式对应的线。换句话说，如果您不想可视化这些线，可以删除plot\u不等式的最后两行。我把它们标出来是为了演示，我会想办法的。你的回答真的帮了我大忙，谢谢！！！）我修改了代码：只需使用plot\u凸集（）。你在哪里使用了皮波曼图书馆？你的函数lineqs（-A，-b）也给出了正确的结果，我从库本身获取了compute_poly_顶点函数，并对其进行了一些修改。你可以在GitHub上查找，在那里很容易找到。我希望你的问题解决了
import matplotlib.pyplot as plt
import numpy as np
from scipy.spatial import HalfspaceIntersection, ConvexHull
from scipy.optimize import linprog
def feasible_point(A, b):
# finds the center of the largest sphere fitting in the convex hull
norm_vector = np.linalg.norm(A, axis=1)
A_ = np.hstack((A, norm_vector[:, None]))
b_ = b[:, None]
c = np.zeros((A.shape[1] + 1,))
c[-1] = -1
res = linprog(c, A_ub=A_, b_ub=b[:, None], bounds=(None, None))
return res.x[:-1]
def hs_intersection(A, b):
interior_point = feasible_point(A, b)
halfspaces = np.hstack((A, -b[:, None]))
hs = HalfspaceIntersection(halfspaces, interior_point)
return hs
def plt_halfspace(a, b, bbox, ax):
if a[1] == 0:
ax.axvline(b / a[0])
else:
x = np.linspace(bbox[0][0], bbox[0][1], 100)
ax.plot(x, (b - a[0]*x) / a[1])
def add_bbox(A, b, xrange, yrange):
A = np.vstack((A, [
[-1, 0],
[ 1, 0],
[ 0, -1],
[ 0, 1],
]))
b = np.hstack((b, [-xrange[0], xrange[1], -yrange[0], yrange[1]]))
return A, b
def solve_convex_set(A, b, bbox, ax=None):
A_, b_ = add_bbox(A, b, *bbox)
interior_point = feasible_point(A_, b_)
hs = hs_intersection(A_, b_)
points = hs.intersections
hull = ConvexHull(points)
return points[hull.vertices], interior_point, hs
def plot_convex_set(A, b, bbox, ax=None):
# solve and plot just the convex set (no lines for the inequations)
points, interior_point, hs = solve_convex_set(A, b, bbox, ax=ax)
if ax is None:
_, ax = plt.subplots()
ax.set_aspect('equal')
ax.set_xlim(bbox[0])
ax.set_ylim(bbox[1])
ax.fill(points[:, 0], points[:, 1], 'r')
return points, interior_point, hs
def plot_inequalities(A, b, bbox, ax=None):
# solve and plot the convex set,
# the inequation lines, and
# the interior point that was used for the halfspace intersections
points, interior_point, hs = plot_convex_set(A, b, bbox, ax=ax)
ax.plot(*interior_point, 'o')
for a_k, b_k in zip(A, b):
plt_halfspace(a_k, b_k, bbox, ax)
return points, interior_point, hs
plt.rcParams['figure.figsize'] = (6, 3)
A = np.array([[-1, 1],
[0, 1],
[0.5, 1],
[1.5, 1],
[-1, 0],
[0, -1]])
b = np.array([1, 2, 3, 6, 0, 0])
bbox = [(-1, 5), (-1, 4)]
fig, ax = plt.subplots(ncols=2)
plot_convex_set(A, b, bbox, ax=ax[0])
plot_inequalities(A, b, bbox, ax=ax[1]);
A = np.array([
[-1, 1],
[0, 1],
[-1, 0],
[0, -1],
])
b = np.array([1, 2, 0, 0])
fig, ax = plt.subplots(ncols=2)
plot_convex_set(A, b, bbox, ax=ax[0])
plot_inequalities(A, b, bbox, ax=ax[1]);
import pypoman
import cdd
import matplotlib.pyplot as plt
def grahamscan(A):
def rotate(A,B,C):
return (B[0]-A[0])*(C[1]-B[1])-(B[1]-A[1])*(C[0]-B[0])
n = len(A)
if len(A) == 0:
return A
P = np.arange(n)
for i in range(1,n):
if A[P[i]][0]<A[P[0]][0]:
P[i], P[0] = P[0], P[i]
for i in range(2,n):
j = i
while j>1 and (rotate(A[P[0]],A[P[j-1]],A[P[j]])<0):
P[j], P[j-1] = P[j-1], P[j]
j -= 1
S = [P[0],P[1]]
for i in range(2,n):
while rotate(A[S[-2]],A[S[-1]],A[P[i]])<0:
del S[-1]
S.append(P[i])
return S
def compute_poly_vertices(A, b):
b = b.reshape((b.shape[0], 1))
mat = cdd.Matrix(np.hstack([b, -A]), number_type='float')
mat.rep_type = cdd.RepType.INEQUALITY
P = cdd.Polyhedron(mat)
g = P.get_generators()
V = np.array(g)
vertices = []
for i in range(V.shape[0]):
if V[i, 0] != 1: continue
if i not in g.lin_set:
vertices.append(V[i, 1:])
return vertices
A = np.array([[-1, 1],
[0, 1],
[0.5, 1],
[1.5, 1],
[-1, 0],
[0, -1]])
b = np.array([1, 2, 3, 6, 0, 0])
vertices = np.array(compute_poly_vertices(A, b))
print(vertices)
vertices = np.array(vertices[grahamscan(vertices)])
x, y = vertices[:, 0], vertices[:, 1]
fig=plt.figure(figsize=(15,15))
ax = fig.add_subplot(111, title="Solution")
ax.fill(x, y, linestyle = '-', linewidth = 1, color='gray', alpha=0.5)
ax.scatter(x, y, s=10, color='black', alpha=1)
from intvalpy import lineqs
import numpy as np
A = np.array([[-1, 1],
[0, 1],
[0.5, 1],
[1.5, 1],
[-1, 0],
[0, -1]])
b = np.array([1, 2, 3, 6, 0, 0])
lineqs(-A, -b)