First of all the specific error message you are getting:
TypeError: __init__() got multiple values for argument 'lowBound'
In python when calling a function you can pass arguments either by 'position' - which means the order in which you pass the arguments tells the function what each of them is - or by naming them. If you look up the documentation for the pulp.LpVariable method you'll see the second position argument is 'lowbound' which you then also pass as a named argument - hence the error message.
I think you might also be slighly misunderstanding how a dataframe works. It is not like excel where you set a 'formula' in a column and it stays updated to that formula as other elements on that row change. You can assign values to columns but if the input data change - the cell would only be updated if that bit of code was run again.
In terms of solving your problem - I'm not convinced I've understood what you're trying to do but I've understood the following.
- We want to select values of
df['speed'] to maximise the sum-product of heading and choices columns
- The value of the choices column depends on the
ratio of speed to heading (as per the given 5 ranges)
Heading column is fixed
By inspection the optimum will be achieved by setting all of the speeds so that the ratios are in the [0.2 - 0.3] range, and where they fall in that range doesn't matter. Code to do this in PuLP within pandas dataframes below. It relised on using binary variables to keep track of which range the ratios fall in.
The syntax is a little awkward though - I'd recommend doing the optimisation completely outside of dataframes and just loading results in at the end - using the LpVariable.dicts method to create arrays of variables instead.
from pandas import DataFrame
import numpy as np
from pulp import *
headings = [184.0, 153.0, 140.0, 122.0, 119.0]
df = DataFrame (headings, columns=['heading'])
df['speed'] = 50
max_speed = 500.0
max_ratio = max_speed / np.min(headings)
df['ratio'] = df.speed/df.heading
conditions_lb = [0, 0.1, 0.2, 0.3, 0.4]
conditions_ub = [0.1, 0.2, 0.3, 0.4, max_speed / np.min(headings)]
choices = [3, 1, 8, 5, 2]
n_range = len(choices)
n_rows = len(df)
# Create primary ratio variables - one for each variable:
df['speed_vars'] = [LpVariable("speed_"+str(j)) for j in range(n_rows)]
# Create auxilary variables - binaries to control
# which bit of range each speed is in
df['aux_vars'] = [[LpVariable("aux_"+str(i)+"_"+str(j), cat='Binary')
for i in range(n_range)]
for j in range(n_rows)]
# Declare problem
prob = LpProblem("max_pd_column",LpMaximize)
# Define objective function
prob += lpSum([df['aux_vars'][j][i]*choices[i]*headings[j] for i in range(n_range)
for j in range(n_rows)])
# Constrain only one range to be selected for each row
for j in range(n_rows):
prob += lpSum([df['aux_vars'][j][i] for i in range(n_range)]) == 1
# Constrain the value of the speed by the ratio range selected
for j in range(n_rows):
for i in range(n_range):
prob += df['speed_vars'][j]*(1.0/df['heading'][j]) <= \
conditions_ub[i] + (1-df['aux_vars'][j][i])*max_ratio
prob += df['speed_vars'][j]*(1.0/df['heading'][j]) >= \
conditions_lb[i]*df['aux_vars'][j][i]
# Solve problem and print results
prob.solve()
# Dislay the optimums of each var in problem
for v in prob.variables ():
print (v.name, "=", v.varValue)
# Set values in dataframe and print:
df['speed_opt'] = [df['speed_vars'][j].varValue for j in range(n_rows)]
df['ratio_opt'] = df.speed_opt/df.heading
print(df)
The last bit of which prints out:
heading speed_vars b spd_opt rat_opt
0 184.0 speed_0 [b_0_0, b_1_0, b_2_0, b_3_0, b_4_0] 36.8 0.2
1 153.0 speed_1 [b_0_1, b_1_1, b_2_1, b_3_1, b_4_1] 30.6 0.2
2 140.0 speed_2 [b_0_2, b_1_2, b_2_2, b_3_2, b_4_2] 28.0 0.2
3 122.0 speed_3 [b_0_3, b_1_3, b_2_3, b_3_3, b_4_3] 24.4 0.2
4 119.0 speed_4 [b_0_4, b_1_4, b_2_4, b_3_4, b_4_4] 23.8 0.2
