NumPy中的Arnaud Legoux移动平均(ALMA)

Question

NumPy中的Arnaud Legoux移动平均(ALMA)

5

我想用NumPy（或Pandas）编写计算Arnaud Legoux移动平均（ALMA）的向量化代码。你能帮我完成这个吗？谢谢。

非向量化版本如下所示（见下文）。

def NPALMA(pnp_array, **kwargs) :
    '''
    ALMA - Arnaud Legoux Moving Average,
    http://www.financial-hacker.com/trend-delusion-or-reality/
    https://github.com/darwinsys/Trading_Strategies/blob/master/ML/Features.py
    '''
    length = kwargs['length']
    # just some number (6.0 is useful)
    sigma = kwargs['sigma']
    # sensisitivity (close to 1) or smoothness (close to 0)
    offset = kwargs['offset']

    asize = length - 1
    m = offset * asize
    s = length  / sigma
    dss = 2 * s * s
    
    alma = np.zeros(pnp_array.shape)
    wtd_sum = np.zeros(pnp_array.shape)

    for l in range(len(pnp_array)):
        if l >= asize:
            for i in range(length):
                im = i - m
                wtd = np.exp( -(im * im) / dss)
                alma[l] += pnp_array[l - length + i] * wtd
                wtd_sum[l] += wtd
            alma[l] = alma[l] / wtd_sum[l]

    return alma

- Prokhozhii

2个回答

1

在Divakar之前的回答中，没有一个与TradingView的ALMA v26.0产生的数字完全匹配。重要的是数字能够与参考值相匹配。考虑到这个目标，Pine Script v5实现的ta.alma可以作为一个参考，尽管它不是向量化的。

下面是一个使用NumPy的版本，它产生与TradingView相同的输出：

import numpy as np

def alma_np(prices: np.ndarray, window: int = 9, sigma: float = 6, offset: float = 0.85) -> np.ndarray:
    # Ref: https://dev59.com/cKbja4cB1Zd3GeqPeVLK#76031340/
    m = offset * (window - 1)
    s = window / sigma
    i = np.arange(window)
    weights = np.exp(-1 * np.square(i - m) / (2 * np.square(s)))
    norm_weights = weights / np.sum(weights)
    padded_prices = np.pad(prices, (window - 1, 0), mode='edge')
    alma_values = np.convolve(padded_prices, norm_weights[::-1], mode='valid')
    return alma_values

这里是一个使用上述NumPy版本的Pandas版本。

def alma_pd(prices: pd.Series, window: int = 9, *, sigma: float = 6, offset: float = 0.85) -> pd.Series:
    # Ref: https://dev59.com/cKbja4cB1Zd3GeqPeVLK#76031340/
    prices_series = prices
    prices = prices.to_numpy()
    alma_values = alma_np(prices, window, sigma=sigma, offset=offset)
    alma_series = pd.Series(alma_values, index=prices_series.index)
    return alma_series

通过2023-04-14的$SPY的5分钟数据进行交叉核对，匹配到小数点后第五位。

- Asclepius

网页内容由stack overflow 提供, 点击上面的

可以查看英文原文，
原文链接

- Divakar · Accepted Answer

开始方法

我们可以沿着第一个轴创建滑动窗口，然后使用wtd值的范围进行张量乘法来进行求和约简。

实现大致如下 -

# Get all wtd values in an array
wtds = np.exp(-(np.arange(length) - m)**2/dss)

# Get the sliding windows for input array along first axis
pnp_array3D = strided_axis0(pnp_array,len(wtds))

# Initialize o/p array
out = np.zeros(pnp_array.shape)

# Get sum-reductions for the windows which don't need wrapping over
out[length:] = np.tensordot(pnp_array3D,wtds,axes=((1),(0)))[:-1]

# Last element of the output needed wrapping. So, do it separately.
out[length-1] = wtds.dot(pnp_array[np.r_[-1,range(length-1)]])

# Finally perform the divisions
out /= wtds.sum()

获取滑动窗口的函数：strided_axis0是来自这里。

使用1D卷积提高性能

那些与wtds值相乘，然后进行求和约简的操作基本上是沿着第一个轴进行的卷积。因此，我们可以在axis=0上使用scipy.ndimage.convolve1d。由于内存效率很高，这样会快得多，因为我们不会创建巨大的滑动窗口。

实现方法如下：

from scipy.ndimage import convolve1d as conv

avgs = conv(pnp_array, weights=wtds/wtds.sum(),axis=0, mode='wrap')

因此，out[length-1:]，即非零行与 avgs[:-length+1] 相同。

如果我们使用非常小的核数从 wtds 进行操作，则可能存在一些精度差异。因此，在使用此 convolution 方法时，请注意这一点。

运行时间测试

方法 -

def original_app(pnp_array, length, m, dss):
    alma = np.zeros(pnp_array.shape)
    wtd_sum = np.zeros(pnp_array.shape)

    for l in range(len(pnp_array)):
        if l >= asize:
            for i in range(length):
                im = i - m
                wtd = np.exp( -(im * im) / dss)
                alma[l] += pnp_array[l - length + i] * wtd
                wtd_sum[l] += wtd
            alma[l] = alma[l] / wtd_sum[l]
    return alma

def vectorized_app1(pnp_array, length, m, dss):
    wtds = np.exp(-(np.arange(length) - m)**2/dss)
    pnp_array3D = strided_axis0(pnp_array,len(wtds))
    out = np.zeros(pnp_array.shape)
    out[length:] = np.tensordot(pnp_array3D,wtds,axes=((1),(0)))[:-1]
    out[length-1] = wtds.dot(pnp_array[np.r_[-1,range(length-1)]])
    out /= wtds.sum()
    return out

def vectorized_app2(pnp_array, length, m, dss):
    wtds = np.exp(-(np.arange(length) - m)**2/dss)
    return conv(pnp_array, weights=wtds/wtds.sum(),axis=0, mode='wrap')

计时 -

In [470]: np.random.seed(0)
     ...: m,n = 1000,100
     ...: pnp_array = np.random.rand(m,n)
     ...: 
     ...: length = 6
     ...: sigma = 0.3
     ...: offset = 0.5
     ...: 
     ...: asize = length - 1
     ...: m = np.floor(offset * asize)
     ...: s = length  / sigma
     ...: dss = 2 * s * s
     ...: 

In [471]: %timeit original_app(pnp_array, length, m, dss)
     ...: %timeit vectorized_app1(pnp_array, length, m, dss)
     ...: %timeit vectorized_app2(pnp_array, length, m, dss)
     ...: 
10 loops, best of 3: 36.1 ms per loop
1000 loops, best of 3: 1.84 ms per loop
1000 loops, best of 3: 684 µs per loop

In [472]: np.random.seed(0)
     ...: m,n = 10000,1000 # rest same as previous one

In [473]: %timeit original_app(pnp_array, length, m, dss)
     ...: %timeit vectorized_app1(pnp_array, length, m, dss)
     ...: %timeit vectorized_app2(pnp_array, length, m, dss)
     ...: 
1 loop, best of 3: 503 ms per loop
1 loop, best of 3: 222 ms per loop
10 loops, best of 3: 106 ms per loop