我将如何计算指数移动平均线？

Question

我从thread 发现了一个有用的sn-p。这个帖子已经有十多年的历史了，没有太多的讨论。然而，它继续获得大量观点 - 毫无疑问，它对未来的读者很有用。

def ema(s, n):
    """
    returns an n period exponential moving average for
    the time series s

    s is a list ordered from oldest (index 0) to most
    recent (index -1)
    n is an integer

    returns a numeric array of the exponential
    moving average
    """
    ema = []
    j = 1

    #get n sma first and calculate the next n period ema
    sma = sum(s[:n]) / n
    multiplier = 2 / float(1 + n)
    ema.append(sma)

    #EMA(current) = ( (Price(current) - EMA(prev) ) x Multiplier) + EMA(prev)
    ema.append(( (s[n] - sma) * multiplier) + sma)

    #now calculate the rest of the values
    for i in s[n+1:]:
        tmp = ( (i - ema[j]) * multiplier) + ema[j]
        j = j + 1
        ema.append(tmp)

    return ema

问题是 EMA 值实际上是附加的 SMA 数字。我们应该如何着手修复这个功能？

Answer 1

使用一个临时数组进行ema 计算，并使用不同的数组返回，

def ema(s, n):
    """
    returns an n period exponential moving average for
    the time series s

    s is a list ordered from oldest (index 0) to most
    recent (index -1)
    n is an integer

    returns a numeric array of the exponential
    moving average
    """
    ema1 = []
    ema2 = []
    j = 1

    #get n sma first and calculate the next n period ema
    sma = sum(s[:n]) / n
    multiplier = 2 / float(1 + n)
    ema1.append(sma)

    #EMA(current) = ( (Price(current) - EMA(prev) ) x Multiplier) + EMA(prev)
    ema1.append(( (s[n] - sma) * multiplier) + sma)
    ema2.append(( (s[n] - sma) * multiplier) + sma)
    #now calculate the rest of the values
    for i in s[n+1:]:
        tmp = ( (i - ema1[j]) * multiplier) + ema1[j]
        j = j + 1
        ema1.append(tmp)
        ema2.append(tmp)

    return ema2