我很高興能與大家分享我的問題，并期待向大家學習。我當前的問題是def calculating_hma無法獲得正確的結果：#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)
3 回答
躍然一笑
TA貢獻1826條經驗 獲得超6個贊
使用 Pandas 系列可以輕松解決這個問題。整個公式：
HMA = WMA(2*WMA(period/2) - WMA(period)), sqrt(period))
給定一個輸入序列 s 和一個句點可以打包成一行：
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)
但為了清晰和方便起見，最好定義兩個函數：
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)))
波斯汪
TA貢獻1811條經驗 獲得超4個贊
移動平均線通常用 的簽名來定義ma(series) -> series。我認為您的困惑很大一部分根源在于 WMA 被定義為返回一個系列，而不是您所期望的單個值。
這是單點 HMA 的 python 實現：
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)
慕虎7371278
TA貢獻1802條經驗 獲得超4個贊
解決了
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
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