Python 向矩阵添加纸浆变量值
我一直把“nan”作为矩阵中的值。我已经试过了。value,只是x1本身。如何获取值并将其插入矩阵? 为什么呢Python 向矩阵添加纸浆变量值,python,matrix,pulp,Python,Matrix,Pulp,我一直把“nan”作为矩阵中的值。我已经试过了。value,只是x1本身。如何获取值并将其插入矩阵? 为什么呢 def square(n,m): my_lp_problem = pulp.LpProblem("My LP Problem", pulp.LpMinimize) x1 = pulp.LpVariable('x1', lowBound=0, cat='Integer') x2 = pulp.LpVariable('x2', lowBound=0, cat='In
def square(n,m):
my_lp_problem = pulp.LpProblem("My LP Problem", pulp.LpMinimize)
x1 = pulp.LpVariable('x1', lowBound=0, cat='Integer')
x2 = pulp.LpVariable('x2', lowBound=0, cat='Integer')
x3 = pulp.LpVariable('x3', lowBound=0, cat='Integer')
x4 = pulp.LpVariable('x4', lowBound=0, cat='Integer')
x5 = pulp.LpVariable('x5', lowBound=0, cat='Integer')
x6 = pulp.LpVariable('x6', lowBound=0, cat='Integer')
x7 = pulp.LpVariable('x7', lowBound=0, cat='Integer')
x8 = pulp.LpVariable('x8', lowBound=0, cat='Integer')
# Objective function
Sum = x1+x2+x3+x4+x5+x6+x7+x8
my_lp_problem += Sum, "Z"
my_lp_problem += 1+x1-x2+x3-x4+2*x5-2*x6+2*x7-2*x8 == n
my_lp_problem += 1-(2*x1+2*x2-2*x3-2*x4+x5+x6-x7-x8) == m
my_lp_problem.solve()
pulp.LpStatus[my_lp_problem.status]
for variable in my_lp_problem.variables():
print( "{} = {}".format(variable.name, variable.varValue))
return
for n in range(1,3):
for m in range(1,3):
square(n,m)
Matrix[n-1,m-1]=x1.varValue
要定义x1,此行需要在
平方(n,m)
函数中。矩阵在示例代码中未初始化。此外,x1
未在分配给矩阵
元素的范围内定义。你能给我一个答案吗?我忘了包括在内。但我发现了问题所在。
Matrix[n-1,m-1]=x1.varValue