美文网首页
项目07_社会财富分配问题模拟

项目07_社会财富分配问题模拟

作者: 我是屁江 | 来源:发表于2019-06-17 11:49 被阅读0次

    ‘‘‘

    -- coding: utf-8 --

    Created on Fri Oct 12 12:31:21 2018

    项目 13 社会财富分配问题 (蒙特卡罗模拟)
    Note:
    1 建立一个空的DataFrame时,只需要index 参数
    2 当在一个列表中随机选取一个值 可以用random的choice,当需要随机选取多个值用numpy
    的random的choice()
    3 Series 的name 参数设置在Series 中 name=‘’
    4 当在DataFrame需要判断再赋值时,可以先用判断筛选列 重新赋值 或者 用apply函数
    5 apply在DataFrame中用于判断赋值(☆)
    6 在绘制图表时 如果不想显示每一个xticks 不要xlim的?
    7 在迭代过程中不要将值赋值给变量名和传入的变量一致,会导致报错
    8 在筛选时或者标记 当对某些值进行筛选或者按某些条件筛选 即按照此标记
    """

    导入模块

    import os
    import random
    import time
    import numpy as np
    import pandas as pd
    import matplotlib.pyplot as plt
    import warnings

    warnings.filterwarnings('ignore')

    定义函数

    def round1():
    '''
    建立初始模型,第1次交易
    '''
    # 构建初始财富值100,index的值为每个人的编号
    people = pd.DataFrame(index=list(range(1,101)))
    people['money'] = 100
    people['r1'] = people['money'] - 1

    # 构建收钱的随机对象
    people['to'] = np.random.choice(list(range(1,101)), size=100,
          replace=True, p=None)
    data_to = people['to'].value_counts()
    data_to.name = 'count'
    data_to = pd.DataFrame(data_to)
    people = pd.merge(people, data_to, how='left', left_index=True,
                      right_index=True).fillna(0)
    people['r1_m'] = people['r1'] + people['count']
    
    return people    
    

    def roundi(n):
    '''
    构建借贷模型
    不考虑财富值为0,即允许借贷
    '''
    # 构建初始财富值
    people = pd.DataFrame(index=list(range(1,101)))
    people['r0'] = 100

    # 构建模型
    for i in range(1, n+1): 
        col = 'r' + str(i)
        col0 = 'r' + str(i-1)
        people[col] =  people[col0] - 1
        data_to = pd.Series(np.random.choice(list(range(1,101)), size=100,
          replace=True, p=None), name='to')
        data_to = data_to.value_counts()
        data_to = pd.DataFrame(data_to)
        people = pd.merge(people, data_to, how='left', left_index=True,
                      right_index=True).fillna(0)
        people[col] = people[col] + people['to']
        
        del people['to']
            
    return people.T
    

    def roundn(n):
    '''
    构建初始模型
    考虑财富值为0,即财富值为0时可以接收 ==> choice的范围是1-100
    但不给出 ==> 所以要统计给出的数量 == 接收的数量
    '''
    # 构建初始财富值
    people = pd.DataFrame(index=list(range(1,101)))
    people['r0'] = 100

    # 构建模型
    for i in range(1, n+1): 
        col = 'r' + str(i)
        col0 = 'r' + str(i-1)
        people[col] =  people[col0] - 1
        data_to = pd.Series(np.random.choice(list(range(1,101)), size=100,
          replace=True, p=[]), name='to')
        data_to = data_to.value_counts()
        data_to = pd.DataFrame(data_to)
        people = pd.merge(people, data_to, how='left', left_index=True,
                      right_index=True).fillna(0)
        people[col] = people[col] + people['to']
        
        del people['to']
            
    return people.T
    

    def roundm(n):
    '''
    构建努力人生模型
    '''
    # 构建初始财富值
    people = pd.DataFrame(index=list(range(1,101)))
    people['r0'] = 100
    person_id= [1, 11, 21, 31, 41, 51, 61, 71, 81, 91] # 努力的Id
    # 构建概率
    p = [0.899/90 for i in range(100)]
    for i in person_id:
    p[i-1] = 0.0101

    # 构建模型
    for i in range(1, n+1): 
        col = 'r' + str(i)
        col0 = 'r' + str(i-1)
        people[col] = people[col0] - 1
        data_to = pd.Series(np.random.choice(list(range(1,101)), size=100,
          replace=True, p=p), name='to')
        data_to = data_to.value_counts()
        data_to = pd.DataFrame(data_to)
        people = pd.merge(people, data_to, how='left', left_index=True,
                      right_index=True).fillna(0)
        people[col] = people[col] + people['to']
        
        del people['to']
            
    return people.T
    

    def graph(data, title):
    '''
    绘制柱状图 == 表排序
    '''
    plt.figure(figsize=(10, 5))
    data.plot(kind='bar', color='gray', edgecolor='gray', figsize=(10, 5))
    plt.xlabel('Player Id')
    plt.ylabel('Forturn')
    plt.title(title)
    plt.savefig(title + '.jpg', dpi=200)

    def lst():
    '''
    生产绘制图表数据的行数
    '''
    lst1 = [x for x in range(0, 100, 10)]
    lst2 = [x for x in range(100, 1000, 100)]
    lst3 = [x for x in range(1000, 17001, 400)]

    return lst1 + lst2 + lst3
    

    def forturn_std(data):
    '''
    计算每一轮的财富标准差
    '''
    lst = []
    for i in range(17001):
    dat = data.iloc[i]
    std = dat.std()
    lst.append(std)

    s = pd.Series(lst)
    
    return s
    

    def line_graph(data, title):
    '''
    绘制折线图
    '''
    fig = plt.figure(figsize=(10,5))
    plt.plot(data, color='red')
    plt.grid(linestyle='--', color='gray', alpha=0.6, axis='both')
    plt.xlim([0, 17000])
    plt.ylim([0, 150])
    plt.title(title)
    plt.savefig(title + '.jpg', dpi=400)

    def sign_pc(data):
    '''
    负债id标记
    '''
    data = pd.DataFrame(data)
    data['color'] = 'gray'
    data['color'][data['r6200'] < 0] = 'red'
    del data['r6200']

    return data
    

    if name == 'main':

    # 运行初始模型 得到模型数据
    r17 = roundn(17000)
    
    # 绘制图表 -- 不排序
    path = r'C:\Users\pj2063150\Desktop\项目\项目13社会财富分配问题模拟\图表\初始不排序'
    os.chdir(path)
    
    lst = lst()
    
    for i in lst:
        title = 'Round' + str(i)
        #data = r17.iloc[i]
        #graph(data, title)
    
    # 绘制图表 -- 排序
    path = r'C:\Users\pj2063150\Desktop\项目\项目13社会财富分配问题模拟\图表\初始排序'
    os.chdir(path)
    for i in lst:
        title = 'Round' + str(i)
        data = r17.iloc[i]
        data = data.sort_values(ascending=True)
        graph(data, title)
    
    # 运行借贷模型,得到数据
    ri17 = roundi(17000)
    
    # 绘制图表
    path = r'C:\Users\pj2063150\Desktop\项目\项目13社会财富分配问题模拟\图表\允许借贷'
    os.chdir(path)
    for i in lst:
        title = 'Round' + str(i)
        data = ri17.iloc[i]
        data = data.sort_values(ascending=True)
        graph(data, title)
        
        
    # 调用计算函数,获取标准差
    data_std = forturn_std(ri17)
    # 绘制图表
    path = r'C:\Users\pj2063150\Desktop\项目\项目13社会财富分配问题模拟\图表'
    os.chdir(path)
    line_graph(data_std, '财富标准差曲线')
    
    # 35岁破产往后逆袭情况  6200次
    data_6200 = ri17.iloc[6200]
    id_pc =data_6200[data_6200 < 0].index.tolist()
    # 对负债id进行标记
    id_sign = sign_pc(data_6200)
    # 绘制图表
    path = r'C:\Users\pj2063150\Desktop\项目\项目13社会财富分配问题模拟\图表\负债逆袭'
    os.chdir(path)
    
    for i in range(6200, 17000, 500):
        title = 'Round' + str(i)
        col = 'r' + str(i)
        data = ri17.iloc[i]
        data = pd.DataFrame(data)
        data1 = pd.merge(data, id_sign, how='right', left_index=True, right_index=True)
        data1 = data1.sort_values(by=col, ascending=True)
        fig = plt.figure(figsize=(10,5))
        data1[col].plot(kind='bar', color=data1['color'], figsize=(10,5))
        plt.xlabel('Player Id')
        plt.ylabel('Forturn')
        plt.title(title)
        plt.savefig(title + '.jpg', dpi=200)
        
    # 运行努力人生模型,得到数据
    rm17 = roundm(17000)
    # 对努力id进行标记
    id_nl = [1, 11, 21, 31, 41, 51, 61, 71, 81, 91]
    id_nl_sign = pd.Series('gray', index=range(1, 101), name='color')
    for i in id_nl:
        id_nl_sign[i] = 'red'
    id_nl_sign = pd.DataFrame(id_nl_sign)
    
    # 绘制图表
    path = r'C:\Users\pj2063150\Desktop\项目\项目13社会财富分配问题模拟\图表\努力人生'
    os.chdir(path)
    
    for i in lst:
        title = 'Round' + str(i)
        col = 'r' + str(i)
        data = rm17.iloc[i]
        data = pd.DataFrame(data)
        data = pd.merge(data, id_nl_sign, how='right', left_index=True, right_index=True)
        data = data.sort_values(by=col, ascending=True)
        fig = plt.figure(figsize=(10,5))
        data[col].plot(kind='bar', color=data['color'], figsize=(10,5))
        plt.xlabel('Player Id')
        plt.ylabel('Forturn')
        plt.title(title)
        plt.savefig(title + '.jpg', dpi=200) 
    

    print('Finished')
    ’’’

    努力人生.jpg
    负债逆袭
    财富标准差曲线.jpg

    相关文章

      网友评论

          本文标题:项目07_社会财富分配问题模拟

          本文链接:https://www.haomeiwen.com/subject/ooebfctx.html