长度较大的RSI

Question

我想将1D RSI合并到4H时间框架中。我尝试过使用分辨率或安全函数，但这两个函数都在每日时间框架接近尾声时更新，并一直保持到第二天。为了从1D图中得到大约20SMA，我会在4H图中绘制120SMA。它并不完全相同，因为输入更多，但它更精确，更具反应性。但是，如果我在4H图中绘制一个1D 14RSI作为84RSI，我只会得到一个无用的指标，它总是在50区域上下浮动。

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是否有方法可以获得与SMA相似的结果？

//@version=4
study(title="Relative Strength Index", shorttitle="RSI", format=format.price, precision=2, resolution="")
len = input(14, minval=1, title="Length")
src = input(close, "Source", type = input.source)

rma_(src, length) =>
    alpha = 1/length
    sum = 0.0
    sum := na(sum[1]) ? sma(src, length) : alpha * src + (1 - alpha) * nz(sum[1])

rsi_(x, y) => 
    u = max(x - x[1], 0)
    d = max(x[1] - x, 0)
    rs = rma_(u, y) / rma_(d, y)
    res = 100 - 100 / (1 + rs)
    res

rsi = rsi_(src, len)
plot(rsi, "RSI", color=color.blue, linewidth = 1)
band1 = hline(70, "Upper Band", color=#C0C0C0)
band0 = hline(30, "Lower Band", color=#C0C0C0)
fill(band1, band0, color=#9915FF, transp=90, title="Background")

提亚

Answer 1

//@version=4
study(title="Help (Relative Strength Index)", shorttitle="RSI", format=format.price, precision=2, resolution="")
len = input(14, minval=1, title="Length")
src = input(close, "Source", type = input.source)

rma_(src, length) =>
    alpha = 1/length
    sum = 0.0
    sum := na(sum[1]) ? sma(src, length) : alpha * src + (1 - alpha) * nz(sum[1])

rsi_(x, y) => 
    u = max(x - x[1], 0)
    d = max(x[1] - x, 0)
    rs = rma_(u, y) / rma_(d, y)
    res = 100 - 100 / (1 + rs)
    res

rsi = rsi_(src, len) //currend TF
rsiD = security(syminfo.tickerid, "D", rsi_(src, len))  // TF='D'
plot(rsi, "RSI", color=color.blue, linewidth = 1)
plot(rsiD, "RSI", color=color.green, linewidth = 1)
band1 = hline(70, "Upper Band", color=#C0C0C0)
band0 = hline(30, "Lower Band", color=#C0C0C0)
fill(band1, band0, color=#9915FF, transp=90, title="Background")

上面的窗口是时间范围D，下面的窗口是4H。RSI的值是相同的，并且实时同步变化。这有什么问题吗？

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