GEKKO - 如何修复 Python Gekko Max Equation 错误 - 元素数
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【中文标题】GEKKO - 如何修复 Python Gekko Max Equation 错误 - 元素数【英文标题】：GEKKO - How to fix Python Gekko Max Equation error - number of elements 【发布时间】：2021-11-20 21:45:22 【问题描述】：我使用 Gekko 优化函数开发了一个脚本。下面的脚本针对许多元素运行。我测试了 20 和 47 个细胞（shapefile 数据集）的优化算法，脚本实现了解决方案。但是，当我测试更大的数据集时，例如包含 160 个元素，会显示以下错误消息：
“APM 模型错误：字符串 > 15000 个字符
考虑将线分解成多个方程”
我阅读了一些解决此问题的建议。我尝试使用 m.sum，但问题仍然存在。
拜托，你能帮我解决这个问题吗？
请在我们转移链接下方找到下载包含 47 个单元格和 160 个单元格的数据集。
https://wetransfer.com/downloads/64cc631237adacc926c67f56124b327a20210928212223/d8a2d7
谢谢
亚历山大。
import geopandas as gpd
import time
import csv
from gekko import GEKKO
import numpy as np
import math
import pandas as pd
m = GEKKO()
A = -0.00000536
B = -0.0000291
E = 0.4040771
r = 0.085
input_path = 'D:/Alexandre/shapes/Threats/Prototype/BHO50k/Velhas_BHO50k1summ4_47cells.shp'
output_folder = 'D:/Alexandre/shapes/Threats/Prototype/Small_area/resultados'
input_layer = gpd.read_file(input_path)
input_layer = input_layer[
['cocursodag', 'cobacia', 'nuareacont', 'nudistbact', 'D0c', 'Ki0', 'Kj0', 'nuareamont', 'deltai', 'It',
'cost_op_BR', 'Ii_ub', 'Itj', 'cj', 'deltaj2']]
input_layer = input_layer.astype('cobacia': 'string', 'cocursodag': 'string')
count_input_feat = input_layer.shape[0]
row=count_input_feat
col=10
input_cobacia =
ubi =
numareacont =
Ki0 =
Kj0 =
X =
deltai2 =
ai =
aj =
D0 =
Itj =
It =
deltaj =
for row1 in input_layer.iterrows():
i = row1[0]
input_cobacia[i] = row1[1]['cobacia']
Ki0[i] = row1[1]['Ki0']+0.001
Kj0[i] = row1[1]['Kj0']
X[i] = row1[1]['nuareamont']
deltai2[i] = row1[1]['deltai']
ai[i] = 5423304*(pow(X[i],-0.1406852))
aj[i] = row1[1]['cj']*100 + row1[1]['cost_op_BR']*100
ubi[i] = row1[1]['Ii_ub']
numareacont[i] = row1[1]['nuareacont']
D0[i] = row1[1]['D0c']
It[i] = row1[1]['It']
Itj[i] = row1[1]['Itj']
if Itj[i]<1:
deltaj[i] = row1[1]['deltaj2'] * 0.0001
elif Itj[i]<2:
deltaj[i] = row1[1]['deltaj2'] * 0.0001
else:
deltaj[i] = row1[1]['deltaj2'] * 0.0001
Ii = m.Array(m.Var, (row, col))
Ij = m.Array(m.Var, (row, col))
for i in range(row):
for j in range(col):
if It[i] == 0:
Ii[i, j].value = 0
Ii[i, j].lower = 0
Ii[i, j].upper = 5
Ij[i,j].value = 0
Ij[i,j].lower = 0
Ij[i,j].upper = numareacont[i]*0.05*Itj[i]/3.704545
elif It[i] <= 2:
Ii[i, j].value = 0
Ii[i, j].lower = 0
Ii[i, j].upper = 10
Ij[i, j].value = 0
Ij[i, j].lower = 0
Ij[i, j].upper = numareacont[i]*0.05*Itj[i]/3.704545
elif It[i] <= 2.5:
Ii[i, j].value = 0
Ii[i, j].lower = 0
Ii[i, j].upper = 15
Ij[i, j].value = 0
Ij[i, j].lower = 0
Ij[i, j].upper = numareacont[i]*0.05*Itj[i]/3.704545
elif It[i] <= 3:
Ii[i, j].value = 0
Ii[i, j].lower = 0
Ii[i, j].upper = 15
Ij[i, j].value = 0
Ij[i, j].lower = 0
Ij[i, j].upper = numareacont[i]*0.05*Itj[i]/3.704545
else:
Ii[i,j].value = 0
Ii[i,j].lower = 0
Ii[i,j].upper = 20
Ij[i,j].value = 0
Ij[i,j].lower = 0
Ij[i,j].upper = numareacont[i]*0.05*Itj[i]/3.704545
Ki = m.Array(m.Var, (row, col))
Kj = m.Array(m.Var, (row, col))
indicator = m.Array(m.Var, (row, col))
p = 2
numerator = m.Array(m.Var, (row, col))
denominator = m.Array(m.Var, (row, col))
for row2 in input_layer.iterrows():
input_cobacia2 = row2[1]['cobacia']
input_cocursodag = row2[1]['cocursodag']
input_distance = row2[1]['nudistbact']
numerator = 0
denominator = 0
exp = f"cobacia > 'input_cobacia2' and cocursodag.str.startswith('input_cocursodag')"
for j in range(col):
for target_feat in input_layer.query(exp).iterrows():
i=target_feat[0]
target_green_area = Ij[i,j]
target_distance = target_feat[1]['nudistbact']
distance = float(target_distance) - float(input_distance)
idw = 1 / (distance + 1) ** p
numerator += target_green_area * idw
denominator += idw
i=row2[0]
sum = Ij[i,j]
if denominator:
indicator[i,j] = numerator / denominator + sum
else:
indicator[i,j] = sum
D0F = m.Array(m.Var, (row, col))
for i in range(row):
def constraintD0(x):
return x - 0.2
for j in range(col):
if j == 0:
m.fix(Ki[i,j],val = Ki0[i])
Ki[i,j].lower = 0
Ki[i,j].upper = 500000
m.fix(Kj[i,j], val = Kj0[i])
Kj[i,j].lower = 0
Kj[i,j].upper = 100000
m.Equation(D0F[i, j] == A * Ki[i, j] + B * Kj[i, j] + E)
D0[i] = D0F[i, j]
else:
D0F[i,j].lower = 0
D0F[i, j].upper = 1
Ki[i,j].lower = 0
Ki[i,j].upper = 500000
Kj[i, j].lower = 0
Kj[i, j].upper = 100000
m.Equation(Ki[i,j] - Ki[i,j-1] == Ii[i,j] - deltai2[i] * Ki[i,j-1])
m.Equation(Kj[i,j] - Kj[i,j-1] == Ij[i,j] + deltaj[i] * Kj[i,j-1]+indicator[i,j])
m.Equation(D0F[i,j] == A*Ki[i,j] + B*Kj[i,j] + E)
m.Equation(D0F[i,j]<=D0[i])
dep = 1 / (1+r)
z1 = m.sum([m.sum([pow(dep, j)*(ai[i]*Ii[i,j]+aj[i]*Ij[i,j]) for i in range(row)]) for j in range(col)])
# Objective
m.Obj(z1)
m.options.IMODE = 3
m.options.SOLVER = 3
m.options.DIAGLEVEL = 1
m.options.REDUCE=3
try:
m.solve() # solve
# Outputs
output_Ki = pd.DataFrame(columns=['cobacia'] + list(range(col)))
output_Kj = pd.DataFrame(columns=['cobacia'] + list(range(col)))
output_Ii = pd.DataFrame(columns=['cobacia'] + list(range(col)))
output_Ij = pd.DataFrame(columns=['cobacia'] + list(range(col)))
output_D0 = pd.DataFrame(columns=['cobacia'] + list(range(col)))
output_ai = pd.DataFrame(columns=['cobacia'] + list(range(col)))
output_aj = pd.DataFrame(columns=['cobacia'] + list(range(col)))
for i in range(row):
for j in range(col):
print(Ki)
output_Ii.loc[i, 'cobacia'] = input_cobacia[i]
output_Ii.loc[i, j] = Ii[i,j].value[0]
output_Ij.loc[i, 'cobacia'] = input_cobacia[i]
output_Ij.loc[i, j] = Ij[i,j].value[0]
output_Ki.loc[i, 'cobacia'] = input_cobacia[i]
output_Ki.loc[i, j] = Ki[i,j].value[0]
output_Kj.loc[i, 'cobacia'] = input_cobacia[i]
output_Kj.loc[i, j] = Kj[i,j].value[0]
output_D0.loc[i, 'cobacia'] = input_cobacia[i]
output_D0.loc[i, j] = D0F[i, j].value[0]
output_ai.loc[i, 'cobacia'] = input_cobacia[i]
output_ai.loc[i, j] = ai[i]
output_aj.loc[i, 'cobacia'] = input_cobacia[i]
output_aj.loc[i, j] = aj[i]
df_outputIi = pd.DataFrame(output_Ii)
df_outputIj = pd.DataFrame(output_Ij)
df_outputKi = pd.DataFrame(output_Ki)
df_outputKj = pd.DataFrame(output_Kj)
df_outputD0 = pd.DataFrame(output_D0)
df_outputai = pd.DataFrame(output_ai)
df_outputaj = pd.DataFrame(output_aj)
with pd.ExcelWriter('output.xlsx') as writer:
df_outputIi.to_excel(writer, sheet_name="resultado Ii")
df_outputIj.to_excel(writer, sheet_name="resultado Ij")
df_outputKi.to_excel(writer, sheet_name="resultado Ki")
df_outputKj.to_excel(writer, sheet_name="resultado Kj")
df_outputD0.to_excel(writer, sheet_name="resultado D0")
df_outputai.to_excel(writer, sheet_name="ai")
df_outputaj.to_excel(writer, sheet_name="aj")
except:
print('Not successful')
from gekko.apm import get_file
print(m._server)
print(m._model_name)
f = get_file(m._server,m._model_name,'infeasibilities.txt')
f = f.decode().replace('\r','')
with open('infeasibilities.txt', 'w') as fl:
fl.write(str(f))
for i in range(row):
for j in range(col):
print(Ki[i,j].value)
print(Kj[i,j].value)
print(D0F[i,j].value)```
【问题讨论】：
【参考方案1】：通过使用m.open_folder() 打开运行文件夹，查看m.path 中的APMonitor 模型文件gk0_model.apm。从较小的问题开始，随着问题的扩大，观察方程的大小。此限制（每个方程m.Intermediate() 的中间体可用于更有效地表达模型。 Additional information on Intermediates 可在文档中找到。
【讨论】：
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