第 1 步:“当我应用测试时,P 值始终为“1.00”,统计信息的返回始终为“@987654322 @”
不,先生,不是。
print( 'Statistics\n(W)= %e,\n p = %e' % ( stat, p ) ) # will produce:
...
(W)= 9.438160e-01
p = 1.909053e-21
核心问题是,尊重事物的运作方式:
>>> df['PCT'] = df['Close'].pct_change() # this computes & stores .pct_change()
>>> df # read print( df['Close'].pct_change.__doc__ )
High ... Close Volume Adj Close PCT
Date
2010-01-04 113.389999 ... 113.330002 118944600.0 93.675278 NaN
2010-01-05 113.680000 ... 113.629997 111579900.0 93.923241 0.002647
2010-01-06 113.989998 ... 113.709999 116074400.0 93.989357 0.000704
2010-01-07 114.330002 ... 114.190002 131091100.0 94.386139 0.004221
2010-01-08 114.620003 ... 114.570000 126402800.0 94.700218 0.003328
...
显然,由于period == 1,单元格df['PCT'][0] 是并且必须是NaN
所以,宁可调用 W_stat, p_value = shapiro( df['PCT'][1:] ),不要包含一个对 w.r.t 没有意义的值。 shapiro()
print( shapiro.__doc__ ) # for more details
将值与参考样本进行比较 - 正态分布检验,其中 没有 NaN-s 导致必须直接拒绝零假设 p == 1 是绝对肯定的拒绝(从比较的两个“incomparable-due-to-NaN(s)”集合的角度来看,这显然是正确的)。
类似{ SPY | QQQ | AAPL | AMZN | ... }:
>>> shapiro( dr.data.get_data_yahoo( 'SPY',
start = '2010-01-01',
end = '2015-01-01'
)['Close'].pct_change()[1:]
)
(0.943816065788269, 1.9090532861060437e-21)
>>> shapiro( dr.data.get_data_yahoo( 'QQQ',
start = '2010-01-01',
end = '2015-01-01'
)['Close'].pct_change()[1:]
)
(0.9631340503692627, 2.548133516564297e-17)
>>> shapiro( dr.data.get_data_yahoo( 'AAPL',
start = '2010-01-01',
end = '2015-01-01'
)['Close'].pct_change()[1:]
)
(0.9560988545417786, 5.674560331738808e-19)
>>> shapiro( dr.data.get_data_yahoo( 'AMZN',
start = '2010-01-01',
end = '2015-01-01'
)['Close'].pct_change()[1:]
)
(0.9394155740737915, 3.106424182886848e-22)