If else 语句返回 TypeError：'int' 类型的对象没有 len() - 不知道为什么

Question

所以，我正在尝试解决一个优化问题。我想弄清楚的是，当我运行代码时，我的函数调用“to_fp_Cx”会引发错误，我不明白为什么。

回溯一直指向我定义的函数。我通过使用不同的值调用这些函数来独立测试这些函数，并且按预期工作。所以，我不确定发生了什么。

---------------------------------------------------------------------------
TypeError                                 Traceback (most recent call last)
<ipython-input-41-3f717a4f07e1> in <module>
     37 # intermediate variables with explicit equations
     38 
---> 39 if(fload_c3_1 < 0.1):
     40     alt_fload_c3_1 = m.Intermediate(0)
     41 else:

~\AppData\Local\Continuum\anaconda3\lib\site-packages\gekko\gk_operators.py in __len__(self)
     23         return self.name
     24     def __len__(self):
---> 25         return len(self.value)
     26     def __getitem__(self,key):
     27         return self.value[key]

~\AppData\Local\Continuum\anaconda3\lib\site-packages\gekko\gk_operators.py in __len__(self)
    142 
    143     def __len__(self):
--> 144         return len(self.value)
    145 
    146     def __getitem__(self,key):

TypeError: object of type 'int' has no len()

顾名思义，我是一个蟒蛇菜鸟，我被迷惑了。任何帮助，将不胜感激。谢谢

import numpy as np

# import gekko, pip install if needed
from gekko import GEKKO


# Compressor Performance curves
# Fraction capacity to Fractional power conversion

# Compressor C3
def to_fp_c3(fc):

    a = 5.16102738
    b = -16.25992208
    c = 18.52731113
    d = -8.859480201
    e = 2.096698885
    f = 0.334319989

    if (fc < 0.1):
        fp = 0.0
    else:
        fp = (a*fc**5)+(b*fc**4)+(c*fc**3)+(d*fc**2)+(e*fc**1)+(f*fc**0)

    return fp

...

### Optimization Model ####

# create new model
m = GEKKO(remote = False)

# Solver option - 1: APOPT, 2: BPOPT, 3: IPOPT 0:Benchmark all available
m.options.SOLVER = 3

# declare model parameters
maxcap_c3_1 = m.Param(value = 900)
maxcap_c3_2 = m.Param(value = 900)


load = m.Param(value = 1500)


## Model variables

# load distribution
fload_c3_1 = m.Var(value=0.50,lb=0.0,ub=1.0, integer = False)
fload_c3_2 = m.Var(value=0.50,lb=0.0,ub=1.0, integer = False)

# declare variables and initial guesses
#totalpowerdraw = m.Var()

# intermediate variables with explicit equations

if(fload_c3_1 < 0.1):
    alt_fload_c3_1 = m.Intermediate(0)
else:
    alt_fload_c3_1 = m.Intermediate(fload_c3_1)


if(fload_c3_2 < 0.1):
    alt_fload_c3_2 = m.Intermediate(0)
else:
    alt_fload_c3_2 = m.Intermediate(fload_c3_2)



assignedload_c3_1 = m.Intermediate(alt_fload_c3_1 * maxcap_c3_1)
assignedload_c3_2 = m.Intermediate(alt_fload_c3_2 * maxcap_c3_2)

powerdraw_c3_1 = m.Intermediate(to_fp_c3(alt_fload_c3_1) * maxcap_c3_1)
powerdraw_c3_2 = m.Intermediate(to_fp_c3(alt_fload_c3_2) * maxcap_c3_2)

totalpowerdraw = m.Intermediate(powerdraw_c3_1 + powerdraw_c3_2)


# implicit equations
m.Equation(load == assignedload_c3_1 + assignedload_c3_2 )


# minimize weight1
m.Obj(totalpowerdraw)

# solve optimization
m.solve()  # remote=False for local solve

print ('')
print ('--- Results of the Optimization Problem ---')
print (alt_fload_c3_1.value, powerdraw_c3_1.value)
print (alt_fload_c3_1.value, powerdraw_c3_2.value)

Answer 1

尝试使用 Gekko 中的 m.if3()（或 m.if2()）函数，根据 Gekko 变量进行条件语句切换。在Question about the conditional statement ('m.if3') in the GEKKO中有更多关于条件语句的信息

# use gekko if3 (or if2)
alt_fload_c3_1 = m.if3(fload_c3_1-0.1,0,fload_c3_1)
alt_fload_c3_2 = m.if3(fload_c3_2-0.1,0,fload_c3_2)

这是您的程序的一个版本，它提供了一个成功的解决方案。

import numpy as np
from gekko import GEKKO

# Compressor Performance curves
# Fraction capacity to Fractional power conversion
# Compressor C3
def to_fp_c3(fc):
    a = 5.16102738
    b = -16.25992208
    c = 18.52731113
    d = -8.859480201
    e = 2.096698885
    f = 0.334319989
    fp = m.if3(fc-0.1,0,(a*fc**5)+(b*fc**4)+(c*fc**3)\
               +(d*fc**2)+(e*fc**1)+(f*fc**0))
    return fp

### Optimization Model ####
# create new model
m = GEKKO(remote = False)

# declare model parameters
maxcap_c3_1 = m.Param(value = 900)
maxcap_c3_2 = m.Param(value = 900)
load = m.Param(value = 1500)

## Model variables

# load distribution
fload_c3_1 = m.Var(value=0.50,lb=0.0,ub=1.0, integer = False)
fload_c3_2 = m.Var(value=0.50,lb=0.0,ub=1.0, integer = False)

# use gekko if3 (or if2)
alt_fload_c3_1 = m.if3(fload_c3_1-0.1,0,fload_c3_1)
alt_fload_c3_2 = m.if3(fload_c3_2-0.1,0,fload_c3_2)
assignedload_c3_1 = m.Intermediate(alt_fload_c3_1 * maxcap_c3_1)
assignedload_c3_2 = m.Intermediate(alt_fload_c3_2 * maxcap_c3_2)
powerdraw_c3_1 = m.Intermediate(to_fp_c3(alt_fload_c3_1) * maxcap_c3_1)
powerdraw_c3_2 = m.Intermediate(to_fp_c3(alt_fload_c3_2) * maxcap_c3_2)
totalpowerdraw = m.Intermediate(powerdraw_c3_1 + powerdraw_c3_2)

# implicit equations
m.Equation(load == assignedload_c3_1 + assignedload_c3_2 )

# minimize weight1
m.Obj(totalpowerdraw)

# solve optimization
m.solve()  # remote=False for local solve

print ('')
print ('--- Results of the Optimization Problem ---')
print (alt_fload_c3_1.value, powerdraw_c3_1.value)
print (alt_fload_c3_1.value, powerdraw_c3_2.value)

解决办法：

 ---------------------------------------------------
 Solver         :  APOPT (v1.0)
 Solution time  :  0.0313 sec
 Objective      :  1576.7914326000025
 Successful solution
 ---------------------------------------------------

--- Results of the Optimization Problem ---
[0.66761123885] [677.4476587]
[0.66761123885] [899.3437739]

Answer 2

你必须使用这个

如果（fload_c3_1.value