通常，您将创建一个变量字典（
x在这种情况下）和一个模型变量（
mod在这种情况下）。要创建目标，您需要
sum对变量乘以一些标量，然后将该结果添加到中
mod。你通过再次计算的变量的线性组合，使用构建体的限制
>=，
<=或者
==，和并称约束
mod。最后，您将使用
mod.solve()以获得解决方案。
import pulp# Create variables and modelx = pulp.LpVariable.dicts("x", df.index, lowBound=0)mod = pulp.LpProblem("Budget", pulp.LpMaximize)# Objective functionobjvals = {idx: (1.0/(df['30-day Cost'][idx]/df['Trials'][idx]))*(df['Success'][idx]/float(df['Trials'][idx])) for idx in df.index}mod += sum([x[idx]*objvals[idx] for idx in df.index])# Lower and upper bounds:for idx in df.index: mod += x[idx] >= df['Cost Min'][idx] mod += x[idx] <= df['Cost Max'][idx]# Budget summod += sum([x[idx] for idx in df.index]) == 5000.0# Solve modelmod.solve()# Output solutionfor idx in df.index: print idx, x[idx].value()# 0 2570.0# 1 1350.0# 2 1080.0print 'Objective', pulp.value(mod.objective)# Objective 1798.70495012数据：
import numpy as npimport pandas as pdidx = [0, 1, 2]d = {'channel': pd.Series(['Channel1', 'Channel2', 'Channel3'], index=idx), '30-day Cost': pd.Series([1765.21, 2700., 2160.], index=idx), 'Trials': pd.Series([9865, 1500, 1200], index=idx), 'Success': pd.Series([812, 900, 333], index=idx), 'Cost Min': pd.Series([882.61, 1350.00, 1080.00], index=idx), 'Cost Max': pd.Series([2647.82, 4050.00, 3240.00], index=idx)}df = pd.Dataframe(d)df# 30-day Cost Cost Max Cost Min Success Trials channel# 0 1765.21 2647.82 882.61 812 9865 Channel1# 1 2700.00 4050.00 1350.00 900 1500 Channel2# 2 2160.00 3240.00 1080.00 333 1200 Channel3
