投资组合和回溯的动态规划背包问题

Question

我正在开展一个 python 项目，该项目利用动态规划的背包问题来根据可以投资的金额找到最佳投资。到目前为止，我能够按名称提出最佳投资，但我在格式化和获取其余信息以及实施回溯 table 方面遇到了麻烦。这是我到目前为止的内容和输出：

import pandas as pd
from itertools import chain

def investmentFilename(file):
    df = pd.read_csv(file)
    frame = pd.DataFrame(df)
    frame = frame.drop(0) # dropping the United States
    # print(frame)
    return frame

def loadInvestments(frame):
    portfolio = []
    state = frame['RegionName'].tolist()
    avg = frame['Zhvi'].tolist()
    dfAvg = pd.DataFrame(avg)
    # print(dfAvg)
    tenyr = (frame['10Year'].tolist())
    tenyr = pd.DataFrame(tenyr)
    roi = tenyr.multiply(dfAvg, axis='columns', level=None, fill_value=None)
    # print(roi)
    # roi = pd.DataFrame(roi)
    ROI = roi.values.tolist()
    ROI = list(chain.from_iterable(ROI))
    print("InvestmentName InvestmentCost EstimatedReturnOnInvestment")
    for i in range(len(state)):
        portfolio.append([state[i], int(avg[i]), int(ROI[i])])
        print(state[i], '\t', avg[i], '\t', ROI[i])

    # print(portfolio)
    # portfolio = list(chain.from_iterable(portfolio))
    # print(portfolio)

    # printing list data
    # print("Investment Name:",state)
    # print("Investment Cost:", avg)
    # print("Estimated Return on Investment:", ROI)
    return portfolio

def optimizeInvestments(invstmt, money):
    """ knapsack problem """
    n = len(invstmt)
    val = []
    name = []
    roi = []

    for i in invstmt:
        name.append(i[0])
        val.append(i[1])
        roi.append(i[2])

    K = [[0 for x in range(money + 1)] for x in range(n + 1)]
    I = [[0 for x in range(money + 1)] for x in range(n + 1)]

    for i in range(n + 1):
        for w in range(money + 1):
            if i == 0 or w == 0:
                K[i][w] = 0
                I[i][w] = ""

            elif roi[i - 1] <= w:

                if (val[i - 1] + K[i - 1][w - roi[i - 1]] > K[i - 1][w]):
                    K[i][w] = val[i - 1] + K[i - 1][w - roi[i - 1]]
                    I[i][w] = name[i - 1] + I[i - 1][w - roi[i - 1]]
                else:
                    K[i][w] = K[i - 1][w]
                    I[i][w] = I[i - 1][w]

            else:
                K[i][w] = K[i - 1][w]
                I[i][w] = I[i - 1][w]

    return (I[n][money])

dataFrame = investmentFilename('zhvi-short.csv')
items = loadInvestments(dataFrame)
print(items)
money = 15000 # change the amount of money you want to invest here
# items = [["A", 60, 120], ["B", 100, 20],["C", 120, 30]] # test
# print(items)
val = []
roi = []
for i in items:
    val.append(i[1])
    roi.append(i[2])
print(optimizeInvestments(items, money))

这给出了输出：佛罗里达纽约

我希望用“and”或逗号分隔。然后我还希望输出每个名称的特定投资回报率。

我还需要为最佳投资实施回溯 table。我知道如何在理论上实现回溯 tables 但我不确定如何实现背包问题。

Answer 1

这就是我最终想出的办法来解决这个问题。

def optimizeInvestments(invstmt, money):
""" knapsack problem """
n = len(invstmt)
val = []
name = []
roi = []
traceback = [[0 for i in range(n)] for i in range(n)]

for i in invstmt:
    name.append(i[0])
    val.append(i[-1])
    roi.append(i[1])

K = [[0 for x in range(money + 1)] for x in range(n + 1)]
I = [[0 for x in range(money + 1)] for x in range(n + 1)]

for i in range(n + 1):
    for w in range(money + 1):
        if i == 0 or w == 0:
            K[i][w] = 0
            I[i][w] = ""

        elif roi[i - 1] <= w:

            if (val[i - 1] + K[i - 1][w - roi[i - 1]] > K[i - 1][w]):
                K[i][w] = val[i - 1] + K[i - 1][w - roi[i - 1]]
                if len(I[i - 1][w - roi[i - 1]]) > 0:
                    I[i][w] = name[i - 1] + " & " + I[i - 1][w - roi[i - 1]]
                else:
                    I[i][w] = name[i - 1]

            else:
                K[i][w] = K[i - 1][w]
                I[i][w] = I[i - 1][w]

        else:
            K[i][w] = K[i - 1][w]
            I[i][w] = I[i - 1][w]

portfolio = 'With $'+ str(money) + ", invest in " + str(I[n][money]) + " for a ROI of $" + str(K[n][money])
return portfolio

投资组合和回溯的动态规划背包问题

Knapsack problem with dynamic programming for investment portfolio and traceback

python

algorithm

knapsack-problem

dynamic-programming

pandas