Python pandas使用以矢量化方式滚动应用于groupby对象来计算机动车辆测试版
我有一个大数据框df，包含4列：
id period ret_1m mkt_ret_1m
131146 CAN00WG0 199609 -0.1538 0.047104
133530 CAN00WG0 199610 -0.0455 -0.014143
135913 CAN00WG0 199611 0.0000 0.040926
138334 CAN00WG0 199612 0.2952 0.008723
140794 CAN00WG0 199701 -0.0257 0.039916
143274 CAN00WG0 199702 -0.0038 -0.025442
145754 CAN00WG0 199703 -0.2992 -0.049279
148246 CAN00WG0 199704 -0.0919 -0.005948
150774 CAN00WG0 199705 0.0595 0.122322
153318 CAN00WG0 199706 -0.0337 0.045765
id period ret_1m mkt_ret_1m
160980 CAN00WH0 199709 0.0757 0.079293
163569 CAN00WH0 199710 -0.0741 -0.044000
166159 CAN00WH0 199711 0.1000 -0.014644
168782 CAN00WH0 199712 -0.0909 -0.007072
171399 CAN00WH0 199801 -0.0100 0.001381
174022 CAN00WH0 199802 0.1919 0.081924
176637 CAN00WH0 199803 0.0085 0.050415
179255 CAN00WH0 199804 -0.0168 0.018393
181880 CAN00WH0 199805 0.0427 -0.051279
184516 CAN00WH0 199806 -0.0656 -0.011516
id period ret_1m mkt_ret_1m
143275 CAN00WO0 199702 -0.1176 -0.025442
145755 CAN00WO0 199703 -0.0074 -0.049279
148247 CAN00WO0 199704 -0.0075 -0.005948
150775 CAN00WO0 199705 0.0451 0.122322
等
我正在尝试使用一个函数来计算一个称为beta的常见财务指标，该指标包含两个列，ret_1m，每月stock_return和ret_1m_mkt，同一时期的市场1个月回报（period_id）。我想应用一个函数（calc_beta）来计算12个月滚动基础上该函数的12个月结果。
为此，我创建了一个groupby对象：
grp = df.groupby('id')
我想做的是使用类似的东西：
period = 12
for stock, sub_df in grp:
arg = sub_df[['ret_1m', 'mkt_ret_1m']]
beta = pd.rolling_apply(arg, period, calc_beta, min_periods = period)
现在，这是第一个问题。根据文档，pd.rolling_apply arg可以是系列或数据框架。但是，似乎我提供的数据帧被转换为一个只能包含一列数据的numpy数组，而不是我试图提供的两个数据。所以我在下面的calc_beta代码不起作用，因为我需要传递股票和市场回报：
def calc_beta(np_array)
s = np_array[:,0] # stock returns are column zero from numpy array
m = np_array[:,1] # market returns are column one from numpy array
covariance = np.cov(s,m) # Calculate covariance between stock and market
beta = covariance[0,1]/covariance[1,1]
return beta
所以我的问题如下，我认为以这种方式列出它们是有意义的：
(i) How can I pass a data frame/multiple series/numpy array with more than one column to calc_beta using rolling_apply?
(ii) How can I return more than one value (e.g. the beta) from the calc_beta function?
(iii) Having calculated rolling quantities, how can I recombined with the original dataframe df so that I have the rolling quantities corresponding to the correct date in the period column?
(iv) Is there a better (vectorized) way of achieving this? I have seen some similar questions using e.g. df.apply(pd.rolling_apply,period,??) but I did not understand how these worked.
我认为rolling_apply以前无法处理数据框，但文档表明它现在能够这样做。我的熊猫。版本是0.16.1。
感谢您的帮助！我已经失去了1.5天试图解决这个问题并且完全被难倒了。
最终，我想要的是这样的：
id period ret_1m mkt_ret_1m beta other_quantities
131146 CAN00WG0 199609 -0.1538 0.047104 0.521 xxx
133530 CAN00WG0 199610 -0.0455 -0.014143 0.627 xxxx
135913 CAN00WG0 199611 0.0000 0.040926 0.341 xxx
138334 CAN00WG0 199612 0.2952 0.008723 0.567 xx
140794 CAN00WG0 199701 -0.0257 0.039916 0.4612 xxx
143274 CAN00WG0 199702 -0.0038 -0.025442 0.215 xxx
145754 CAN00WG0 199703 -0.2992 -0.049279 0.4678 xxx
148246 CAN00WG0 199704 -0.0919 -0.005948 -0.4225 xxx
150774 CAN00WG0 199705 0.0595 0.122322 0.780 xxx
153318 CAN00WG0 199706 -0.0337 0.045765 0.623 xxx
id period ret_1m mkt_ret_1m beta other_quantities
160980 CAN00WH0 199709 0.0757 0.079293 -0.913 xx
163569 CAN00WH0 199710 -0.0741 -0.044000 0.894 xxx
166159 CAN00WH0 199711 0.1000 -0.014644 0.563 xxx
168782 CAN00WH0 199712 -0.0909 -0.007072 0.734 xxx
171399 CAN00WH0 199801 -0.0100 0.001381 0.894 xxxx
174022 CAN00WH0 199802 0.1919 0.081924 0.789 xx
176637 CAN00WH0 199803 0.0085 0.050415 0.1563 xxxx
179255 CAN00WH0 199804 -0.0168 0.018393 -0.64 xxxx
181880 CAN00WH0 199805 0.0427 -0.051279 -0.742 xxx
184516 CAN00WH0 199806 -0.0656 -0.011516 0.925 xxx
id period ret_1m mkt_ret_1m beta
143275 CAN00WO0 199702 -0.1176 -0.025442 -1.52 xx
145755 CAN00WO0 199703 -0.0074 -0.049279 -0.632 xxx
148247 CAN00WO0 199704 -0.0075 -0.005948 1.521 xx
150775 CAN00WO0 199705 0.0451 0.122322 0.0321 xxx
等
5 个答案:
答案 0 :(得分：3)
我猜pd.rolling_apply在这种情况下没有帮助，因为在我看来它基本上只需要Series（即使数据帧通过，它也会处理一个列一次）。但是你总是可以自己编写一个带有数据帧的rolling_apply。
import pandas as pd
import numpy as np
from StringIO import StringIO
df = pd.read_csv(StringIO(''' id period ret_1m mkt_ret_1m
131146 CAN00WG0 199609 -0.1538 0.047104
133530 CAN00WG0 199610 -0.0455 -0.014143
135913 CAN00WG0 199611 0.0000 0.040926
138334 CAN00WG0 199612 0.2952 0.008723
140794 CAN00WG0 199701 -0.0257 0.039916
143274 CAN00WG0 199702 -0.0038 -0.025442
145754 CAN00WG0 199703 -0.2992 -0.049279
148246 CAN00WG0 199704 -0.0919 -0.005948
150774 CAN00WG0 199705 0.0595 0.122322
153318 CAN00WG0 199706 -0.0337 0.045765
160980 CAN00WH0 199709 0.0757 0.079293
163569 CAN00WH0 199710 -0.0741 -0.044000
166159 CAN00WH0 199711 0.1000 -0.014644
168782 CAN00WH0 199712 -0.0909 -0.007072
171399 CAN00WH0 199801 -0.0100 0.001381
174022 CAN00WH0 199802 0.1919 0.081924
176637 CAN00WH0 199803 0.0085 0.050415
179255 CAN00WH0 199804 -0.0168 0.018393
181880 CAN00WH0 199805 0.0427 -0.051279
184516 CAN00WH0 199806 -0.0656 -0.011516
143275 CAN00WO0 199702 -0.1176 -0.025442
145755 CAN00WO0 199703 -0.0074 -0.049279
148247 CAN00WO0 199704 -0.0075 -0.005948
150775 CAN00WO0 199705 0.0451 0.122322'''), sep='\s+')
def calc_beta(df):
np_array = df.values
s = np_array[:,0] # stock returns are column zero from numpy array
m = np_array[:,1] # market returns are column one from numpy array
covariance = np.cov(s,m) # Calculate covariance between stock and market
beta = covariance[0,1]/covariance[1,1]
return beta
def rolling_apply(df, period, func, min_periods=None):
if min_periods is None:
min_periods = period
result = pd.Series(np.nan, index=df.index)
for i in range(1, len(df)+1):
sub_df = df.iloc[max(i-period, 0):i,:] #I edited here
if len(sub_df) >= min_periods:
idx = sub_df.index[-1]
result[idx] = func(sub_df)
return result
df['beta'] = np.nan
grp = df.groupby('id')
period = 6 #I'm using 6 to see some not NaN values, since sample data don't have longer than 12 groups
for stock, sub_df in grp:
beta = rolling_apply(sub_df[['ret_1m','mkt_ret_1m']], period, calc_beta, min_periods = period)
beta.name = 'beta'
df.update(beta)
print df
输出
id period ret_1m mkt_ret_1m beta
131146 CAN00WG0 199609 -0.1538 0.047104 NaN
133530 CAN00WG0 199610 -0.0455 -0.014143 NaN
135913 CAN00WG0 199611 0.0000 0.040926 NaN
138334 CAN00WG0 199612 0.2952 0.008723 NaN
140794 CAN00WG0 199701 -0.0257 0.039916 NaN
143274 CAN00WG0 199702 -0.0038 -0.025442 -1.245908
145754 CAN00WG0 199703 -0.2992 -0.049279 2.574464
148246 CAN00WG0 199704 -0.0919 -0.005948 2.657887
150774 CAN00WG0 199705 0.0595 0.122322 1.371090
153318 CAN00WG0 199706 -0.0337 0.045765 1.494095
... ... ... ... ... ...
171399 CAN00WH0 199801 -0.0100 0.001381 NaN
174022 CAN00WH0 199802 0.1919 0.081924 1.542782
176637 CAN00WH0 199803 0.0085 0.050415 1.605407
179255 CAN00WH0 199804 -0.0168 0.018393 1.571015
181880 CAN00WH0 199805 0.0427 -0.051279 1.139972
184516 CAN00WH0 199806 -0.0656 -0.011516 1.101890
143275 CAN00WO0 199702 -0.1176 -0.025442 NaN
145755 CAN00WO0 199703 -0.0074 -0.049279 NaN
148247 CAN00WO0 199704 -0.0075 -0.005948 NaN
150775 CAN00WO0 199705 0.0451 0.122322 NaN
答案 1 :(得分：3)
尝试pd.rolling_cov（）和pd.rolling.var（），如下所示：
import pandas as pd
import numpy as np
from StringIO import StringIO
df = pd.read_csv(StringIO(''' id period ret_1m mkt_ret_1m
131146 CAN00WG0 199609 -0.1538 0.047104
133530 CAN00WG0 199610 -0.0455 -0.014143
135913 CAN00WG0 199611 0.0000 0.040926
138334 CAN00WG0 199612 0.2952 0.008723
140794 CAN00WG0 199701 -0.0257 0.039916
143274 CAN00WG0 199702 -0.0038 -0.025442
145754 CAN00WG0 199703 -0.2992 -0.049279
148246 CAN00WG0 199704 -0.0919 -0.005948
150774 CAN00WG0 199705 0.0595 0.122322
153318 CAN00WG0 199706 -0.0337 0.045765
160980 CAN00WH0 199709 0.0757 0.079293
163569 CAN00WH0 199710 -0.0741 -0.044000
166159 CAN00WH0 199711 0.1000 -0.014644
168782 CAN00WH0 199712 -0.0909 -0.007072
171399 CAN00WH0 199801 -0.0100 0.001381
174022 CAN00WH0 199802 0.1919 0.081924
176637 CAN00WH0 199803 0.0085 0.050415
179255 CAN00WH0 199804 -0.0168 0.018393
181880 CAN00WH0 199805 0.0427 -0.051279
184516 CAN00WH0 199806 -0.0656 -0.011516
143275 CAN00WO0 199702 -0.1176 -0.025442
145755 CAN00WO0 199703 -0.0074 -0.049279
148247 CAN00WO0 199704 -0.0075 -0.005948
150775 CAN00WO0 199705 0.0451 0.122322'''), sep='\s+')
df['beta'] = pd.rolling_cov(df['ret_1m'], df['mkt_ret_1m'], window=6) / pd.rolling_var(df['mkt_ret_1m'], window=6)
print df
输出：
id period ret_1m mkt_ret_1m beta
131146 CAN00WG0 199609 -0.1538 0.047104 NaN
133530 CAN00WG0 199610 -0.0455 -0.014143 NaN
135913 CAN00WG0 199611 0.0000 0.040926 NaN
138334 CAN00WG0 199612 0.2952 0.008723 NaN
140794 CAN00WG0 199701 -0.0257 0.039916 NaN
143274 CAN00WG0 199702 -0.0038 -0.025442 -1.245908
145754 CAN00WG0 199703 -0.2992 -0.049279 2.574464
148246 CAN00WG0 199704 -0.0919 -0.005948 2.657887
150774 CAN00WG0 199705 0.0595 0.122322 1.371090
153318 CAN00WG0 199706 -0.0337 0.045765 1.494095
160980 CAN00WH0 199709 0.0757 0.079293 1.616520
163569 CAN00WH0 199710 -0.0741 -0.044000 1.630411
166159 CAN00WH0 199711 0.1000 -0.014644 0.651220
168782 CAN00WH0 199712 -0.0909 -0.007072 0.652148
171399 CAN00WH0 199801 -0.0100 0.001381 0.724120
174022 CAN00WH0 199802 0.1919 0.081924 1.542782
176637 CAN00WH0 199803 0.0085 0.050415 1.605407
179255 CAN00WH0 199804 -0.0168 0.018393 1.571015
181880 CAN00WH0 199805 0.0427 -0.051279 1.139972
184516 CAN00WH0 199806 -0.0656 -0.011516 1.101890
143275 CAN00WO0 199702 -0.1176 -0.025442 1.372437
145755 CAN00WO0 199703 -0.0074 -0.049279 0.031939
148247 CAN00WO0 199704 -0.0075 -0.005948 -0.535855
150775 CAN00WO0 199705 0.0451 0.122322 0.341747
答案 2 :(得分：1)
def rolling_apply(df, period, func, min_periods=None):
if min_periods is None:
min_periods = period
result = pd.Series(np.nan, index=df.index)
for i in range(1, len(df)):
sub_df = df.iloc[max(i-period, 0):i,:] #get a subsample to run
if len(sub_df) >= min_periods:
idx = sub_df.index[-1]+1 # mind the forward looking bias,your return in time t should not be inclued in the beta calculating in time t
result[idx] = func(sub_df)
return result
我修复了Happy001's code的前瞻性偏见。这是一个财务问题，所以应该谨慎。
我发现vlmercado的答案很错误。如果仅使用pd.rolling_cov和pd.rolling_var，那么您在财务上会犯错误。首先，很明显第二个股票CAN00WH0没有任何NaN beta，因为它使用了CAN00WG0的返回值，这是完全错误的。其次，考虑这样一种情况：股票停产十年，您也可以将该样本纳入beta计算。
我发现pandas.rolling也适用于Timestamp，但是groupby似乎不行。因此，我更改了Happy001's code的代码。这不是最快的方法，但至少比原始代码快20倍。
crsp_daily['date']=pd.to_datetime(crsp_daily['date'])
crsp_daily=crsp_daily.set_index('date') # rolling needs a time serie index
crsp_daily.index=pd.DatetimeIndex(crsp_daily.index)
calc=crsp_daily[['permno','ret','mkt_ret']]
grp = calc.groupby('permno') #rolling beta for each stock
beta=pd.DataFrame()
for stock, sub_df in grp:
sub2_df=sub_df[['ret','mkt_ret']].sort_index()
beta_m = sub2_df.rolling('1825d',min_periods=150).cov() # 5yr rolling beta , note that d for day, and you cannot use w/m/y, s/d are availiable.
beta_m['beta']=beta_m['ret']/beta_m['mkt_ret']
beta_m=beta_m.xs('mkt_ret',level=1,axis=0)
beta=beta.append(pd.merge(sub_df,pd.DataFrame(beta_m['beta'])))
beta=beta.reset_index()
beta=beta[['date','permno','beta']]
答案 3 :(得分：1)
顺便说一句，由于没有人问python中的多变量滚动回归，我也找到了解决这个问题的方法。关键是先把它们堆成一列，然后在函数中重塑dataframe。
这是代码
import pandas as pd
import numpy as np
import timeit
from numba import jit
@jit(nopython=True, cache=True,fastmath=True) # numba only support numpy but pandas, so pd.DataFrame is forbiddened.
def coefcalc(df,coefpos,varnum):
# coefpos: which coef you need, for example I want alpha, so I would set the coefpos as 5 to obtain the alpha since the constant is in the last column in X. varnum: how many variables you put in df except "stkcd and date", in this sample, it's 7 (return,five fama factors, and a constant)
if np.mod(df.shape[0],varnum)==0:
df=df.reshape(df.shape[0]//varnum,varnum) # reshape the one column to n column for reg.
Y=np.ascontiguousarray(df[:,0]) # rebuild a contigous numpy array for faster reg
X=np.ascontiguousarray(df[:,1:])
try:
b=(np.linalg.inv(X.T@X))@X.T@Y
result=b[coefpos]
except:
result=np.nan
return result
else:
return np.nan
calc2=pd.read_csv(r'sample.csv')
# A way for rolling beta/alpha
calc2=calc2.set_index(['date','F_INFO_WINDCODE'])
calc2=calc2.dropna() # regression will drop Nan automatically
calc2=calc2.stack().reset_index().set_index('date') # put n columns into one columns, and let datetime64 variable (date) to be the index.
localtime = time.asctime( time.localtime(time.time()) )
print(localtime)
order_of_variable=5 # expect for y (return), start from zero.
total_number_variable=7 # stkcd/date/return/fama5/constant
required_sample=30*total_number_variable # Monthly data
# the parallel kwarg may require loops in def so I turn it off.
alphaest=calc2.groupby('F_INFO_WINDCODE').rolling('1095d',min_periods=required_sample)[0].apply(lambda x:coefcalc(x,5,7),engine='numba', raw=True,engine_kwargs={'nopython': True, 'nogil': True, 'parallel': False})
# as the pandas's document numba engine is faster than cpython when obs is over 1 million.
# you can check in https://pandas.pydata.org/pandas-docs/stable/user_guide/window.html , the numba engine part.
localtime = time.asctime( time.localtime(time.time()) )
print(localtime)
答案 4 :(得分：1)
好消息！在pandas 1.3.0中，rolling.apply增加了一个新的方法“Table”，一切都解决了！
这是一个示例代码。
def coefcalc(df):
Y=np.ascontiguousarray(df[:,0]) # rebuild a contigous numpy array for faster reg
X=np.ascontiguousarray(df[:,1:])
try:
b=(np.linalg.inv(X.T@X))@X.T@Y
return np.nan,b[0],b[1],b[2],b[3],b[4]
except:
return np.nan,np.nan,np.nan,np.nan,np.nan,np.nan
fama5=pd.read_csv('F-F_Research_Data_5_Factors_2x3.csv')
fama5=fama5.iloc[0:695,:].rename(columns={'Mkt-RF':'Mkt_Rf'})
fama5['date']=pd.to_datetime(fama5.date,format='%Y%m',errors='ignore')
for var in ['Mkt_Rf','SMB','HML','RMW','CMA','RF']:
fama5[var]=pd.to_numeric(fama5[var])/100
fama5[var]=fama5[var].astype('float64')
fama5=fama5.set_index('date')
fama5=fama5.drop('RF',axis=1)
beta=fama5.rolling('100d', method="table", min_periods=0).apply(coefcalc, raw=True, engine="numba")
您可以从 Ken French 的主页下载 F-F_Research_Data_5_Factors_2x3.csv https://mba.tuck.dartmouth.edu/pages/faculty/ken.french/ftp/F-F_Research_Data_5_Factors_2x3_CSV.zip
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