我正在使用暴力策略来寻找解决问题的答案。其思想是以10为增量在3个不同的范围内循环,例如嵌套循环,以显著减少组合的数量
然后取第一个函数中给出的三个解,为函数重新定义循环范围参数,从而得到更精确的解
这是我的代码:
# first broad looping function to zone down to the relevant area of my solution
def t1_solve(cgoal):
max_value = None
nc_f = None
c_f = None
cd_f = None
for i, j, k in [(i,j,k) for i in range(nc_rev.idxmax(),int((cgoal*100)+200),5) for
j in range(c_rev.idxmax(),int((cgoal*100)+200),5) for k in range(cd_rev.idxmax(),
int((cgoal*100)+200),5)]:
if (t1rev(i,j,k) > max_value and t1c(i,j,k) > cgoal):
#storing the optimal value result, and my three solution in nc_f, c_f, cd_f
max_value = t1rev(i,j,k)
nc_f = i
c_f = j
cd_f = k
print max_value
print nc_f, c_f, cd_f
return nc_f
return c_f
return cd_f
# second reduced looping problem to fine-tune my answer
def t1_finetune():
# run the broad looping function
t1_solve(3.61)
# this is where I have trouble with passing my solutions stored in the
# previous function's nc_f, c_f, cd_f
#ERROR OCCURS HERE!!!!!
if nc_f - 20 > 0:
nc_lowerbound = nc_f - 20
else:
nc_lowerbound = 1
if nc_f + 20 < 1499:
nc_upperbound = nc_f + 20
else:
nc_upperbound = 1499
if c_f - 20 > 0:
c_lowerbound = c_f - 20
else:
c_lowerbound = 1
if c_f + 20 < 1499:
c_upperbound = c_f + 20
else:
c_upperbound = 1499
if cd_f - 20 > 0:
cd_lowerbound = cd_f - 20
else:
cd_lowerbound = 1
if cd_f + 20 < 1499:
cd_upperbound = cd_f + 20
else:
cd_upperbound = 1499
for i, j, k in [(i,j,k) for i in range(nc_lowerbound, nc_upperbound) for
j in range(c_lowerbound, c_upperbound) for k in range(cd_lowerbound, cd_upperbound)]:
if (t1rev(i,j,k) > max_value and t1c(i,j,k) > cgoal):
max_value = t1rev(i,j,k)
nc_f = i
c_f = j
cd_f = k
print max_value
print nc_f, c_f, cd_f
return nc_f, c_f, cd_f
t=time.time()
t1_finetune()
print time.time() - t
我收到的错误消息是:
UnboundLocalError: local variable 'nc_f' referenced before assignment
本质上,我只需要将nc_f、c_f和cd_f从我的t1_solve()传递到我的t1_finetune()。单独运行t1_solve()可以正常工作,当在t1_finetune()中调用它时,它仍然可以正常工作,直到它继续运行到代码的其余部分,在那里我对错误发生进行了注释
我希望这是明确的,请让我知道,如果有什么我可以澄清
提前谢谢
首先,您的
t1_solve
函数有三个return
语句,而不是一个。一旦到达第一个,函数就结束了,其他函数就永远不会发生。所以,你需要这个:接下来,当您调用
t1_solve
并将这些值返回给您时,您只需忽略结果。你需要把它们存放在某个地方。例如:为了直观地理解,这里要做的是返回三个值而不是一个,并将这三个返回值分配给三个变量
如果您想知道实际发生了什么:第一个
nc_f, c_f, cd_f
创建一个3元素元组并返回该元组。然后,后面的nc_f, c_f, cd_f =
使用iterable赋值解包。请参阅教程第Tuples and Sequences节,以了解详细的介绍相关问题 更多 >
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