如何在Pandas的滚动窗口中计算波动率(标准差)

Question

我有一个时间序列"Ser",我想用滚动窗口计算波动率(标准偏差).我当前的代码正确地以这种形式执行:

w=10
for timestep in range(length):
    subSer=Ser[timestep:timestep+w]
    mean_i=np.mean(subSer)
    vol_i=(np.sum((subSer-mean_i)**2)/len(subSer))**0.5
    volList.append(w_i)

这在我看来非常低效.Pandas是否具有内置功能来执行此类操作？

Answer 1

看起来你正在寻找Series.rolling.您可以将std计算应用于结果对象:

roller = Ser.rolling(w)
volList = roller.std(ddof=0)

如果您不打算再次使用滚动窗口对象,可以编写一个单行程序:

volList = Ser.rolling(w).std(ddof=0)

请记住,ddof=0在这种情况下是必要的,因为标准差的标准化是由len(Ser)-ddof,并且ddof默认为1大熊猫.

Answer 2

即使在财务意义上，“波动性”也是模棱两可的。最常用的波动率类型是已实现波动率，它是已实现方差的平方根。与回报标准差的主要区别是：

• 使用日志返回（不是简单的返回）
• 该数字按年计算（通常假设每年有 252 至 260 个交易日）
• 在方差交换的情况下，对数回报不会贬低

有多种计算实际波动率的方法；但是，我已经实现了以下两个最常见的：

import numpy as np

window = 21  # trading days in rolling window
dpy = 252  # trading days per year
ann_factor = days_per_year / window

df['log_rtn'] = np.log(df['price']).diff()

# Var Swap (returns are not demeaned)
df['real_var'] = np.square(df['log_rtn']).rolling(window).sum() * ann_factor
df['real_vol'] = np.sqrt(df['real_var'])

# Classical (returns are demeaned, dof=1)
df['real_var'] = df['log_rtn'].rolling(window).var() * ann_factor
df['real_vol'] = np.sqrt(df['real_var'])

Answer 3

通常，[金融类型]人们以价格变动百分率的年度价格来引用波动率。

假设您在数据框中有每日价格，df并且一年中有252个交易日，则可能需要以下内容：

df.pct_change().rolling(window_size).std()*(252**0.5)

Answer 4

这是一种 NumPy 方法 -

# From http://stackoverflow.com/a/14314054/3293881 by @Jaime
def moving_average(a, n=3) :
    ret = np.cumsum(a, dtype=float)
    ret[n:] = ret[n:] - ret[:-n]
    return ret[n - 1:] / n

# From http://stackoverflow.com/a/40085052/3293881
def strided_app(a, L, S=1 ):  # Window len = L, Stride len/stepsize = S
    nrows = ((a.size-L)//S)+1
    n = a.strides[0]
    return np.lib.stride_tricks.as_strided(a, shape=(nrows,L), strides=(S*n,n))

def rolling_meansqdiff_numpy(a, w):
    A = strided_app(a, w)
    B = moving_average(a,w)
    subs = A-B[:,None]
    sums = np.einsum('ij,ij->i',subs,subs)
    return (sums/w)**0.5

样品运行 -

In [202]: Ser = pd.Series(np.random.randint(0,9,(20)))

In [203]: rolling_meansqdiff_loopy(Ser, w=10)
Out[203]: 
[2.6095976701399777,
 2.3000000000000003,
 2.118962010041709,
 2.022374841615669,
 1.746424919657298,
 1.7916472867168918,
 1.3000000000000003,
 1.7776388834631178,
 1.6852299546352716,
 1.6881943016134133,
 1.7578395831246945]

In [204]: rolling_meansqdiff_numpy(Ser.values, w=10)
Out[204]: 
array([ 2.60959767,  2.3       ,  2.11896201,  2.02237484,  1.74642492,
        1.79164729,  1.3       ,  1.77763888,  1.68522995,  1.6881943 ,
        1.75783958])

运行时测试

循环方法 -

def rolling_meansqdiff_loopy(Ser, w):
    length = Ser.shape[0]- w + 1
    volList= []
    for timestep in range(length):
        subSer=Ser[timestep:timestep+w]
        mean_i=np.mean(subSer)
        vol_i=(np.sum((subSer-mean_i)**2)/len(subSer))**0.5
        volList.append(vol_i)
    return volList

时间——

In [223]: Ser = pd.Series(np.random.randint(0,9,(10000)))

In [224]: %timeit rolling_meansqdiff_loopy(Ser, w=10)
1 loops, best of 3: 2.63 s per loop

# @Mad Physicist's vectorized soln
In [225]: %timeit Ser.rolling(10).std(ddof=0)
1000 loops, best of 3: 380 µs per loop

In [226]: %timeit rolling_meansqdiff_numpy(Ser.values, w=10)
1000 loops, best of 3: 393 µs per loop

7000x使用两种矢量化方法比循环方法更接近那里的加速！