数据:
{'Open': {0: 159.18000000000001, 1: 157.99000000000001, 2: 157.66, 3: 157.53999999999999, 4: 155.03999999999999, 5: 155.47999999999999, 6: 155.44999999999999, 7: 155.93000000000001, 8: 155.0, 9: 157.72999999999999},
'Close': {0: 157.97999999999999, 1: 157.66, 2: 157.53999999999999, 3: 155.03999999999999, 4: 155.47999999999999, 5: 155.44999999999999, 6: 155.87, 7: 155.0, 8: 157.72999999999999, 9: 157.31}}
码:
import pandas as pd
d = #... data above.
df = pd.DataFrame.from_dict(d)
df['Close_Stdev'] = pd.rolling_std(df[['Close']],window=5)
print df
# Close Open Close_Stdev
# 0 157.98 159.18 NaN
# 1 157.66 157.99 NaN
# 2 157.54 157.66 NaN
# 3 155.04 157.54 NaN
# 4 155.48 155.04 1.369452
# 5 155.45 155.48 1.259754
# 6 155.87 155.45 0.975464
# 7 155.00 155.93 0.358567
# 8 157.73 155.00 1.065190
# 9 157.31 157.73 1.189378
问题:
上面的代码没有问题。但是，rolling_std是否有可能在其观察窗中考虑Close中的前四个值和Open中的第五个值？基本上，我希望rolling_std为第一个Stdev计算以下内容:
157.98 # From Close
157.66 # From Close
157.54 # From Close
155.04 # From Close
155.04 # Bzzt, from Open.
从技术上讲，这意味着观察列表的最后一个值始终是最后一个Close值。
逻辑/原因:
显然，这是库存数据。我正在尝试检查在计算标准差时是否最好将当前交易日股票的Open价格考虑在内，而不是仅仅检查先前的Close。
预期结果:
# Close Open Close_Stdev Desired_Stdev
# 0 157.98 159.18 NaN NaN
# 1 157.66 157.99 NaN NaN
# 2 157.54 157.66 NaN NaN
# 3 155.04 157.54 NaN NaN
# 4 155.48 155.04 1.369452 1.480311
# 5 155.45 155.48 1.259754 1.255149
# 6 155.87 155.45 0.975464 0.994017
# 7 155.00 155.93 0.358567 0.361151
# 8 157.73 155.00 1.065190 0.368035
# 9 157.31 157.73 1.189378 1.291464
额外详情:
通过使用公式STDEV.S并选择数字，可以在Excel中轻松完成此操作，如下面的屏幕快照所示。但是，出于个人原因，我希望在Python和pandas中完成此操作(我强调了F6，由于Snagit的作用，它不仅仅可见)。
python大神给出的解决方案
您可以使用Welford's method计算标准偏差。
这样做的好处是，它只需5次迭代就可以表示为整个列上的矢量化算术。
这应该比逐行计算并且必须为每一行组成窗口要快。
首先，这是一个健全性检查，表明Welford的方法可以重现与
df['Close_Stdev'] = pd.rolling_std(df[['Close']],window=5)
import numpy as np
import pandas as pd
class OnlineVariance(object):
"""
Welford's algorithm computes the sample variance incrementally.
"""
def __init__(self, iterable=None, ddof=1):
self.ddof, self.n, self.mean, self.M2 = ddof, 0, 0.0, 0.0
if iterable is not None:
for datum in iterable:
self.include(datum)
def include(self, datum):
self.n += 1
self.delta = datum - self.mean
self.mean += self.delta / self.n
self.M2 += self.delta * (datum - self.mean)
self.variance = self.M2 / (self.n-self.ddof)
@property
def std(self):
return np.sqrt(self.variance)
d = {'Open': {0: 159.18000000000001, 1: 157.99000000000001, 2: 157.66, 3:
157.53999999999999, 4: 155.03999999999999, 5: 155.47999999999999, 6:
155.44999999999999, 7: 155.93000000000001, 8: 155.0, 9: 157.72999999999999},
'Close': {0: 157.97999999999999, 1: 157.66, 2: 157.53999999999999, 3:
155.03999999999999, 4: 155.47999999999999, 5: 155.44999999999999, 6: 155.87, 7:
155.0, 8: 157.72999999999999, 9: 157.31}}
df = pd.DataFrame.from_dict(d)
df['Close_Stdev'] = pd.rolling_std(df[['Close']],window=5)
ov = OnlineVariance()
for n in range(5):
ov.include(df['Close'].shift(n))
df['std'] = ov.std
print(df)
assert np.isclose(df['Close_Stdev'], df['std'], equal_nan=True).all()
产量
Close Open Close_Stdev std
0 157.98 159.18 NaN NaN
1 157.66 157.99 NaN NaN
2 157.54 157.66 NaN NaN
3 155.04 157.54 NaN NaN
4 155.48 155.04 1.369452 1.369452
5 155.45 155.48 1.259754 1.259754
6 155.87 155.45 0.975464 0.975464
7 155.00 155.93 0.358567 0.358567
8 157.73 155.00 1.065190 1.065190
9 157.31 157.73 1.189378 1.189378
因此，要在计算中加入开盘价，
ov = OnlineVariance()
ov.include(df['Open'])
for n in range(1, 5):
ov.include(df['Close'].shift(n))
df['std'] = ov.std
print(df)
产量
Close Open std
0 157.98 159.18 NaN
1 157.66 157.99 NaN
2 157.54 157.66 NaN
3 155.04 157.54 NaN
4 155.48 155.04 1.480311
5 155.45 155.48 1.255149
6 155.87 155.45 0.994017
7 155.00 155.93 0.361151
8 157.73 155.00 0.368035
9 157.31 157.73 1.291464