Tradingview pinescript在python中的RMA（RSI中使用的移动平均值。它是指数加权移动平均值，alpha=1/长度），pandas

Question

我一直试图从tradingviews RMA方法中获得相同的结果，但我不知道如何实现它

在其页面中，RMA计算如下：

plot(rma(close, 15))

//the same on pine
pine_rma(src, length) =>
    alpha = 1/length
    sum = 0.0
    sum := na(sum[1]) ? sma(src, length) : alpha * src + (1 - alpha) * nz(sum[1])
plot(pine_rma(close, 15))

为了测试我使用的输入及其结果，这是输入列，也是应用tradingview的rma(input,14)后的相同输入：

data = [[588.0,519.9035093599585],
[361.98999999999984,508.62397297710436],
[412.52000000000055,501.7594034787397],
[197.60000000000042,480.0337318016869],
[208.71999999999932,460.6541795301378],
[380.1100000000006,454.90102384941366],
[537.6599999999999,460.8123792887413],
[323.5600000000013,451.0086379109742],
[431.78000000000077,449.6351637744761],
[299.6299999999992,438.9205092191563],
[225.1900000000005,423.65404427493087],
[292.42000000000013,414.28018396957873],
[357.64999999999964,410.23517082889435],
[692.5100000000003,430.3976586268306],
[219.70999999999916,415.34854015348543],
[400.32999999999987,414.2757872853794],
[604.3099999999995,427.849659622138],
[204.29000000000087,411.8811125062711],
[176.26000000000022,395.0510330415374],
[204.1800000000003,381.41738782428473],
[324.0,377.3161458368358],
[231.67000000000007,366.91284970563316],
[184.21000000000092,353.8626461552309],
[483.0,363.08674285842864],
[290.6399999999994,357.911975511398],
[107.10000000000036,339.996834403441],
[179.0,328.49706051748086],
[182.36000000000058,318.05869905194663],
[275.0,314.98307769109323],
[135.70000000000073,302.17714357030087],
[419.59000000000015,310.56377617242225],
[275.6399999999994,308.06922073153487],
[440.48999999999984,317.5278478221396],
[224.0,310.8472872634153],
[548.0100000000001,327.78748103031415],
[257.0,322.73123238529183],
[267.97999999999956,318.82043007205664],
[366.51000000000016,322.2268279240526],
[341.14999999999964,323.57848307233456],
[147.4200000000001,310.9957342814536],
[158.78000000000063,300.12318183277836],
[416.03000000000077,308.4022402732943],
[360.78999999999917,312.14422311091613],
[1330.7299999999996,384.90035003156487],
[506.92000000000013,393.61603931502464],
[307.6100000000006,387.4727507925229],
[296.7299999999996,380.991125735914],
[462.0,386.7774738976345],
[473.8099999999995,392.9940829049463],
[769.4200000000002,419.88164841173585],
[971.4799999999997,459.2815306680404],
[722.1399999999994,478.0571356203232],
[554.9799999999996,483.5516259331572],
[688.5,498.19079550936027],
[292.0,483.462881544406],
[634.9500000000007,494.2833900055199]]

# Create the pandas DataFrame
dfRMA = pd.DataFrame(data, columns = ['input', 'wantedrma'])
dfRMA['try1'] = dfRMA['input'].ewm( alpha=1/14, adjust=False).mean()
dfRMA['try2'] = numpy_ewma_vectorized(dfRMA['input'],14)
dfRMA

ewm并没有给我相同的结果，所以我搜索并找到了这个，但我只是复制了ewma

def numpy_ewma_vectorized(data, window):

    alpha = 1/window
    alpha_rev = 1-alpha

    scale = 1/alpha_rev
    n = data.shape[0]

    r = np.arange(n)
    scale_arr = scale**r
    offset = data[0]*alpha_rev**(r+1)
    pw0 = alpha*alpha_rev**(n-1)

    mult = data*pw0*scale_arr
    cumsums = mult.cumsum()
    out = offset + cumsums*scale_arr[::-1]
    return out

我得到了这些结果

你知道如何翻译熊猫的rma方法吗

我意识到使用pandas ewm似乎会收敛，最后一行越来越接近该值，对吗

。。。