基于AMPLPY的Python LCP求解器
正如@denfromufa所指出的那样,PATH
求解器有一个AMPL
接口。他建议使用开源的Pyomo
来处理AMPL
。然而,Pyomo
被证明是慢且繁琐的。我最终用cython编写了自己的PATH
求解器接口,并希望在某个时候发布它,但目前它没有输入检查,是快速而粗糙的,我不想花时间在上面。
目前,我可以分享一个使用AMPL
的Python扩展的答案。它不像直接接口到PATH
那么快:对于每个要解决的LCP
,它创建一个(临时)模型文件,运行AMPL
并收集输出。它有点快速而粗糙,但我觉得至少应该报告自从提出这个问题以来过去几个月的一些结果。
import os
os.environ['PATH_LICENSE_STRING'] = "3413119131&Courtesy&&&USR&54784&12_1_2016&1000&PATH&GEN&31_12_2017&0_0_0&5000&0_0"
from amplpy import AMPL, Environment, dataframe
import numpy as np
from scipy import sparse
from tempfile import mkstemp
import os
import sys
import contextlib
class DummyFile(object):
def write(self, x): pass
@contextlib.contextmanager
def nostdout():
save_stdout = sys.stdout
sys.stdout = DummyFile()
yield
sys.stdout = save_stdout
class modFile:
content = """
set Rn;
param B {Rn,Rn} default 0;
param q {Rn} default 0;
var x {j in Rn};
s.t. f {i in Rn}:
0 <= x[i]
complements
sum {j in Rn} B[i,j]*x[j]
>= -q[i];
"""
def __init__(self):
self.fd = None
self.temp_path = None
def __enter__(self):
fd, temp_path = mkstemp()
file = open(temp_path, 'r+')
file.write(self.content)
file.close()
self.fd = fd
self.temp_path = temp_path
return self
def __exit__(self, exc_type, exc_val, exc_tb):
os.close(self.fd)
os.remove(self.temp_path)
def solveLCP(B, q, x=None, env=None, binaryDirectory=None, pathOptions={'logfile':'logpath.tmp' }, verbose=False):
if binaryDirectory is not None:
env = Environment(binaryDirectory='/home/foo/amplide.linux64/')
if verbose:
pathOptions['output'] = 'yes'
ampl = AMPL(environment=env)
with modFile() as mod:
ampl.read(mod.temp_path)
n = len(q)
dfQ = dataframe.DataFrame('Rn', 'c')
for i in np.arange(n):
dfQ.addRow(int(i)+1, np.float(q[i]))
dfB = dataframe.DataFrame(('RnRow', 'RnCol'), 'val')
if sparse.issparse(B):
if not isinstance(B, sparse.lil_matrix):
B = B.tolil()
dfB.setValues({
(i+1, j+1): B.data[i][jPos]
for i, row in enumerate(B.rows)
for jPos, j in enumerate(row)
})
else:
r = np.arange(n) + 1
Rrow, Rcol = np.meshgrid(r, r, indexing='ij')
dfB.setColumn('RnRow', list(Rrow.reshape((-1), order='C').astype(float)))
dfB.setColumn('RnCol', list(Rcol.reshape((-1), order='C').astype(float)))
dfB.setColumn('val', list(B.reshape((-1), order='C').astype(float)))
ampl.getSet('Rn').setValues([int(x) for x in np.arange(n, dtype=int)+1])
if x is not None:
dfX = dataframe.DataFrame('Rn', 'x')
for i in np.arange(n):
dfX.addRow(int(i)+1, np.float(x[i]))
ampl.getVariable('x').setValues(dfX)
ampl.getParameter('q').setValues(dfQ)
ampl.getParameter('B').setValues(dfB)
ampl.setOption('solver', 'pathampl')
pathOptions = ['{}={}'.format(key, val) for key, val in pathOptions.items()]
ampl.setOption('path_options', ' '.join(pathOptions))
if verbose:
ampl.solve()
else:
with nostdout():
ampl.solve()
if False:
bD = ampl.getParameter('B').getValues().toDict()
qD = ampl.getParameter('q').getValues().toDict()
xD = ampl.getVariable('x').getValues().toDict()
BB = ampl.getParameter('B').getValues().toPandas().values.reshape((n, n,), order='C')
qq = ampl.getParameter('q').getValues().toPandas().values[:, 0]
xx = ampl.getVariable('x').getValues().toPandas().values[:, 0]
ineq2 = BB.dot(xx) + qq
print((xx * ineq2).min(), (xx * ineq2).max() )
return ampl.getVariable('x').getValues().toPandas().values[:, 0]
if __name__ == '__main__':
n = 4
B = np.array([[0, 0, -1, -1], [0, 0, 1, -2], [1, -1, 2, -2], [1, 2, -2, 4]])
q = np.array([2, 2, -2, -6])
BSparse = sparse.lil_matrix(B)
env = Environment(binaryDirectory='/home/foo/amplide.linux64/')
print(solveLCP(B, q, env=env))
print(solveLCP(BSparse, q, env=env))
from numpy import random
n = 1000
B = np.diag((random.randn(n)))
q = np.ones((n,))
print(solveLCP(B, q, env=env))
min(rv - (u + Av), v - s) = 0
时,您具体指的是什么:(a) min_v (rv - u - Av).dot(v-s)(如最小化二次型); (b) 查找v,使得(min_v (rv - u - Av).dot(v-s)) == 0; 或者(c/d)上述两者之一但是按元素运算(不是内积.dot,而是按元素最小值,因此0是一个由零向量组成的向量?所有这些都有不同的答案; 有些相当好(有些不太好)。 - muskrat