Python groupby滚动agg组合测试版自定义函数
感谢阅读并提前给出答案 贝塔系数是衡量投资组合系统风险的指标。计算方法是将投资组合收益与基准/市场的协方差除以市场方差。我想对许多投资组合进行滚动计算 我有一个df如下Python groupby滚动agg组合测试版自定义函数,python,pandas,group-by,finance,Python,Pandas,Group By,Finance,感谢阅读并提前给出答案 贝塔系数是衡量投资组合系统风险的指标。计算方法是将投资组合收益与基准/市场的协方差除以市场方差。我想对许多投资组合进行滚动计算 我有一个df如下 PERIOD,PORT1,PORT2,BM 201504,-0.004,-0.001,-0.013 201505,0.017,0.019,0.022 201506,-0.027,-0.037,-0.039 201507,0.026,0.033,0.017 201508,-0.045,-0.054,-0.081 201509,-0
PERIOD,PORT1,PORT2,BM
201504,-0.004,-0.001,-0.013
201505,0.017,0.019,0.022
201506,-0.027,-0.037,-0.039
201507,0.026,0.033,0.017
201508,-0.045,-0.054,-0.081
201509,-0.033,-0.026,-0.032
201510,0.053,0.07,0.09
201511,0.03,0.032,0.038
201512,-0.05,-0.034,-0.044
201601,-0.016,-0.043,-0.057
201602,-0.007,-0.007,-0.011
201603,0.014,0.014,0.026
201604,0.003,0.001,0.01
201605,0.046,0.038,0.031
除了更多的列,如port1和port2
我想创建一个带有滚动beta和BM列的数据集
我创建了一个类似的滚动关联数据集
df.rolling(3).corr(df['BM'])
…它获取了我的大集合中的每一列,并计算了与我的BM列的相关性
我试着为Beta版制作一个自定义函数,但因为它需要两个参数,所以我很挣扎。下面是我的自定义函数,以及我是如何通过向它提供两列返回来让它工作的
def beta(arr1,arr2):
#ddof = 0 gives population covar. the 0 and 1 coordinates take the arr1 vs arr2 covar from the matrix
return (np.cov(arr1,arr2,ddof=0)[0][1])/np.var(arr2)
beta_test = beta(df['PORT1'],df['BM'])
因此,这有助于我找到两列之间的测试版,我输入。。。问题是如何对我上面的数据和包含多个列/组合的数据执行此操作?那么如何在滚动的基础上进行呢?从上面我看到的相关性来看,下面应该是可能的,在每一列中运行每个滚动3个月的数据集,而不是一个指定的列
beta_data = df.rolling(3).agg(beta(df['BM']))
任何指向正确方向的指针都会受到欢迎IIUC,您可以
设置列的索引周期和BM,过滤包含端口的列(如果您有其他列,您不想应用beta
函数),然后使用滚动
def getbetas(df, market, window = 45):
""" given an unstacked pandas dataframe (columns instruments, rows
dates), compute the rolling betas vs the market.
"""
nmarket = market/market.rolling(window).var()
thebetas = df.rolling(window).cov(other=nmarket)
return thebetas
print (df.set_index(['PERIOD','BM']).filter(like='PORT')
.rolling(3).apply(lambda x: beta(x, x.index.get_level_values(1)))
.reset_index())
PERIOD BM PORT1 PORT2
0 201504 -0.013 NaN NaN
1 201505 0.022 NaN NaN
2 201506 -0.039 0.714514 0.898613
3 201507 0.017 0.814734 1.055798
4 201508 -0.081 0.736486 0.907336
5 201509 -0.032 0.724490 0.887755
6 201510 0.090 0.598332 0.736964
7 201511 0.038 0.715848 0.789221
8 201512 -0.044 0.787248 0.778703
9 201601 -0.057 0.658877 0.794949
10 201602 -0.011 0.412270 0.789567
11 201603 0.026 0.354829 0.690573
12 201604 0.010 0.562924 0.558083
13 201605 0.031 1.716066 1.530471
这很有帮助,谢谢。它解决了我的问题。将BM设置到索引中,然后将其放入索引中,这是我所缺少的。我花了几个小时才走到这一步!感谢您的帮助。谢谢您的回答-已经晚了,我明天会检查这个,如果它好的话,我会批准的。它看起来很优雅。再次感谢