如何在Python中计算Hull移动平均？

Question

我很高兴能和大家分享我的问题，并期待从你们中学到更多。我的问题在于def calculating_hma，我无法获得正确的结果：

#python27

#inputs 
period = 9
Coin_pair = "USD-BTC"
Unit = thirtyMin''

def getClosingPrices(coin_pair, period, unit):
    historical_data = api.getHistoricalData(coin_pair, period, unit)
    closing_prices = []
    for i in historical_data:
        closing_prices.append(i['C'])
    return closing_prices


def calculate_sma(coin_pair, period, unit):

    total_closing = sum(getClosingPrices(coin_pair, period, unit))
    return (total_closing / period)


def calculate_ema(coin_pair, period, unit):

    closing_prices = getClosingPrices(coin_pair, period, unit)
    previous_EMA = calculate_sma(coin_pair, period, unit)
    constant = (2 / (period + 1))
    current_EMA = (closing_prices[-1] * (2 / (1 + period))) + (previous_EMA * (1 - (2 / (1 + period))))

def calculate_hma(coin_pair, period, unit):
    """
    Hull Moving Average.
    
    Formula:
    HMA = WMA(2*WMA(n/2) - WMA(n)), sqrt(n)
    """
    
    # MY Try of calculation ?
    ma = calculate_sma(coin_pair, period, unit)
    HMA = ma(2*ma(period/2) - ma(period)), sqrt(period)
    
    # my question  ?
    # where to use the unit and pierod and coin_pair in the calculation ?  

    # check inputs above
    return hma

ema = calculate_ema(market, period=9, unit=timeframe)
sma = calculate_sma(market, period=9, unit=timeframe)
hma = calculate_sma(market, period=9, unit=timeframe) ? 

print (ema)
print (sma)
print (hma)

- MOIQ

Answer 1

这可以很容易地使用Pandas序列解决。整个公式：

HMA = WMA(2*WMA(period/2) - WMA(period)), sqrt(period))

给定输入序列 s 和一个周期，可以将其压缩成一行：

import pandas as pd
import numpy as np

HMA = s.rolling(period//2).apply(lambda x: ((np.arange(period//2) + 1)*x).sum()/(np.arange(period//2) + 1).sum(), raw=True).multiply(2).sub(
                        s.rolling(period).apply(lambda x: ((np.arange(period) + 1)*x).sum()/(np.arange(period) + 1).sum(), raw=True)
                ).rolling(int(np.sqrt(period))).apply(lambda x: ((np.arange(int(np.sqrt(period))) + 1)*x).sum()/(np.arange(int(np.sqrt(period))) + 1).sum(), raw=True)

为了清晰和方便起见，最好定义2个函数：

def WMA(s, period):
       return s.rolling(period).apply(lambda x: ((np.arange(period)+1)*x).sum()/(np.arange(period)+1).sum(), raw=True)

def HMA(s, period):
       return WMA(WMA(s, period//2).multiply(2).sub(WMA(s, period)), int(np.sqrt(period)))

- mac13k

---

这看起来是正确的，但如果可以通过跳过lambda并使用所有向量化操作来加速（同时保持正确性），那就更好了。 - Asclepius

在SO上有一些关于使用Numpy更高效地计算WMA而不使用“lambda”的答案。也许它们可以被利用。 - Asclepius

@Asclepius，你可以使用技术分析库来加快进程。Tulip或"Tulipy"是最快的，同时HMA和WMA都内置在其中：https://tulipindicators.org/benchmark。 - Francesco Battisti

Answer 2

移动平均通常用 ma(series) -> series 的方式定义。我认为您困惑的一大部分在于WMA被定义为返回一个系列，而不是您预期的单个值。

这里可以找到有关问题的良好正式定义： https://oxfordstrat.com/trading-strategies/hull-moving-average/

以下是单点HMA的Python实现：

def weighted_moving_average(series: List[float], lookback: Optional[int] = None) -> float:
    if not lookback:
        lookback = len(series)
    if len(series) == 0:
        return 0
    assert 0 < lookback <= len(series)

    wma = 0
    lookback_offset = len(series) - lookback
    for index in range(lookback + lookback_offset - 1, lookback_offset - 1, -1):
        weight = index - lookback_offset + 1
        wma += series[index] * weight
    return wma / ((lookback ** 2 + lookback) / 2)


def hull_moving_average(series: List[float], lookback: int) -> float:
    assert lookback > 0
    hma_series = []
    for k in range(int(lookback ** 0.5), -1, -1):
        s = series[:-k or None]
        wma_half = weighted_moving_average(s, min(lookback // 2, len(s)))
        wma_full = weighted_moving_average(s, min(lookback, len(s)))
        hma_series.append(wma_half * 2 - wma_full)
    return weighted_moving_average(hma_series)

- freaker

Answer 3

已解决

def calculate_hma(coin_pair, period, unit):

    HMA = ((calculate_wma(coin_pair, int(period / 2), unit) * 2 - calculate_wma(coin_pair, period, unit)) + (
        calculate_wma(coin_pair, int(math.sqrt(period)), unit))) / 2

- MOIQ

---

此函数不符合壳移动平均线的定义。 - freaker