Question

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}}

代码：

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是否可以将Close中的前四个值和Open中的第五个值计入其观察窗口？基本上，我希望rolling_std为其第一个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Close值。

逻辑/原因：

显然，这是库存数据。我试图检查在计算标准差时是否更好地考虑当前交易日股票的Open价格，而不是仅仅检查前一个{{1 }}第

期望的结果：

Close

额外详情：

这可以通过使用公式# 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并选择下面的屏幕截图中显示的数字在Excel中轻松完成。但是，由于个人原因，我希望在Python和STDEV.S中完成此操作（我突出显示pandas，但由于Snagit的影响，它不仅仅是可见的）。

enter image description here

Answer 1

您可以使用Welford's method来计算标准偏差。这样做的好处是它可以在整个列上表示为矢量化算术，只需5次迭代。这应该比逐行计算更快，并且必须为每一行组成窗口。

首先，这是一个健全性检查，显示Welford的方法可以重现与

相同的结果

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，要在计算中加入开头价值，

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

Answer 2

我和numpy一起玩，直到我得到了我想要的东西。它速度非常快，但它不是 pandaic ，并且在很多层面上都可能不安全。我对这个更漂亮的答案持开放态度。与此同时，这对我的事业来说足够好。

import numpy
...

new_std = []
for i in range(df2.shape[0]+1):
    print df2['Close'].iloc[i-5:i]
    try:
        close_ = np.array(df2['Close'].iloc[i-5:i])
        open_ = np.array(df2['Open'].iloc[i-5:i])
        # Change the close from last date in list to the open
        # of that same date to simulate before-end-of-day trading.
        close_[-1] = open_[-1]
        new_std.append(np.std(close_, ddof=1))
    except:
        new_std.append(np.NAN)

df2['Desired_Stdev'] = new_std[1:] # Truncate to fit index.
print df2

#     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

如何使用pandas的rolling_std来考虑观察中的两列？

2 个答案: