如何用Python计算赫尔移动平均线？

Question

我很高兴能与大家分享我的问题，并期待向大家学习。我当前的问题是def calculating_hma无法获得正确的结果：#python27#inputs period = 9Coin_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_pricesdef 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 hmaema = 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)

Answer 1

使用 Pandas 系列可以轻松解决这个问题。整个公式：

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

给定一个输入序列 s 和一个句点可以打包成一行：

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)

但为了清晰和方便起见，最好定义两个函数：

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)))

Answer 2

移动平均线通常用 的签名来定义ma(series) -> series。我认为您的困惑很大一部分根源在于 WMA 被定义为返回一个系列，而不是您所期望的单个值。

这是单点 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)

Answer 3

解决了

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