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 = 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)

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 pdimport numpy as npHMA = 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):&nbsp; &nbsp; HMA = ((calculate_wma(coin_pair, int(period / 2), unit) * 2 - calculate_wma(coin_pair, period, unit)) + (&nbsp; &nbsp; &nbsp; &nbsp; calculate_wma(coin_pair, int(math.sqrt(period)), unit))) / 2