胜率与薪资,球队是否有竞争性优势

2017-08-13  本文已影响0人  kerwin0905

讨论的问题:综合考虑胜率与薪资,OKA球队相比其他的球队是否有竞争性优势。
数据来源:http://seanlahman.com/files/database/lahman-csv_2014-02-14.zip
1、从网络上下载需要的CSV文档,这里采用request,stringIO,zipfile进行数据提取:

def getZIP(zipFileName):
    #以字节的方式请求
    r = requests.get(zipFileName).content
    #创建内存文件
    s = StringIO.StringIO(r)
    zf = zipfile.ZipFile(s,'r')
    return zf
url = 'http://seanlahman.com/files/database/lahman-csv_2014-02-14.zip'
zf = getZIP(url)

数据展示如下:

['SchoolsPlayers.csv', 'SeriesPost.csv', 'Teams.csv',       'TeamsFranchises.csv', 'TeamsHalf.csv', 'AllstarFull.csv', 'Appearances.csv', 'AwardsManagers.csv', 'AwardsPlayers.csv', 'AwardsShareManagers.csv', 'AwardsSharePlayers.csv', 'Batting.csv', 'BattingPost.csv', 'Fielding.csv', 'FieldingOF.csv', 'FieldingPost.csv', 'HallOfFame.csv', 'Managers.csv', 'ManagersHalf.csv', 'Master.csv', 'Pitching.csv', 'PitchingPost.csv', 'readme2013.txt', 'Salaries.csv', 'Schools.csv']

这里把需要的salaries和teams这两个CSV文件读取出来:

salaries = pd.read_csv(zf.open(tablenames[tablenames.index('Salaries.csv')]))
print salaries.head()
teams = pd.read_csv(zf.open(tablenames[tablenames.index('Teams.csv')]))
#这里只需要这几列
teams = teams[['yearID', 'teamID', 'W']]
print teams.head()
     yearID teamID lgID   playerID   salary
0    1985    BAL   AL    murraed02  1472819
1    1985    BAL   AL     lynnfr01  1090000
2    1985    BAL   AL    ripkeca01   800000
3    1985    BAL   AL     lacyle01   725000
4    1985    BAL   AL    flanami01   641667
   yearID teamID   W
0    1871    PH1  21
1    1871    CH1  19
2    1871    BS1  20
3    1871    WS3  15
4    1871    NY2  16

接下来计算各个队每年的总工资,并把两个列表合并起来,W代表胜场:

#一般情况下,聚合数据都需要唯一的分组键组成的索引,但也可以通过向groupby传入as_index=False以禁用该功能
totleSalaries = salaries.groupby(['yearID','teamID'],as_index=False).sum()
print totleSalaries.head()
#how="inner"指当左右两个对象存在不重合的键时,inner 代表交集;outer 代表并集;on指的是用于连接的列索引名称,如果没有指定且其他参数也未指定则以两个DataFrame的列名交集做为连接键
joined = pd.merge(totleSalaries, teams, how="inner", on=['yearID', 'teamID'])
print joined.head()
 yearID    teamID    salary
0    1985    ATL   14807000
1    1985    BAL   11560712
2    1985    BOS   10897560
3    1985    CAL   14427894
4    1985    CHA    9846178
   yearID teamID    salary   W
0    1985    ATL  14807000  66
1    1985    BAL  11560712  83
2    1985    BOS  10897560  81
3    1985    CAL  14427894  90
4    1985    CHA   9846178  85

接下来画出各个球队每年总的薪水和获胜次数的关系图,并标记处OKA这只球队:

teamName ='OAK'
years = np.arange(2000,2004)
for year in years:
    df = joined[joined['yearID'] == year]
    print df
    #画出薪资和胜场的散点图
    plt.scatter(df['salary'] / 1e6,df['W'])
    plt.title(str(year)+'年'+'胜场与薪资')
    plt.xlabel('总薪水(百万)')
    plt.ylabel('胜场')
    plt.xlim(0, 180)
    plt.ylim(30, 130)
    plt.grid()
    #标记出OKA球队
    plt.annotate(teamName,
                 xy=(df['salary'][df['teamID'] == teamName] / 1e6, df['W'][df['teamID'] == teamName]),
                 xytext=(-20, 20), textcoords='offset points', ha='right', va='bottom',
                 bbox=dict(boxstyle='round,pad=0.5', fc='yellow', alpha=0.5),
                 arrowprops=dict(arrowstyle='->', facecolor='black', connectionstyle='arc3,rad=0'))
    plt.show() 
image.png image.png image.png image.png

可以看出OKA的在2000年到2004年间付出总薪水较少的情况下获得了比较好的胜利场数,接下来用回归分析证明这一点,并看看更长时间内的数据怎么样,算出各支球队的残差,就能知道是否如上述推论:

teamName = 'OAK'
years = np.arange(1999, 2014)
def Residual(year):
    residData = pd.DataFrame()
    df = joined[joined['yearID'] == year]
    #原始数据横坐标
    x_list = df['salary'].values / 1e6
    #纵坐标
    y_list = df['W'].values
    #最小二乘估计
    A = np.array([x_list, np.ones(len(x_list))])#构造系数矩阵
    y = y_list
    w = np.linalg.lstsq(A.T,y)[0] #求出斜率以及纵截距,w[0]斜率w[1]纵截距
    yhat = (w[0]*x_list+w[1]) # 回归线
    residData['teamID'] = df['teamID']
    residData[year] = y - yhat
    residData.index = residData['teamID']
    residData = residData.drop(residData.columns[0], axis=1)
    #print residData
    return residData
#将dataframe放入数组
Residuals = [Residual(year) for year in years]
#按照队名合并
Residual_df = reduce(lambda  left,right:pd.merge(left,right,how='outer',left_index=True, right_index=True),Residuals)
print Residual_df
Residual_df = Residual_df.T
Residual_df.plot(title = '各支球队的残差图', figsize = (15, 8),
               color=map(lambda x: 'blue' if x==teamName else 'gray',Residual_df.columns))
plt.xlabel('年')
plt.ylabel('残差')
plt.show()

这里主要在于如何将多个将dataframe拆分成多个小的dataframe并重新按照不重合的主键名合并。

image.png

如图可以看出,在2000年到2003年间,OKA球队偏移回归线较远,且残差为正,说明其能在付出较少薪水的情况下获得较好的成绩,特别是在2002与2003年,偏移最远,此时球队的性价比在联盟中应该是最高的。但在2004年后,残差往负的方向走,并持续多年,说明此时球队成绩不太好,但在2010年后有复苏的趋势。

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