python pandas数据帧趋势指示器在for循环中运行如何避免循环并优化这个cod

Question

这是一个趋势指标称为超级趋势，显然给出了非常准确的结果，结果是匹配的主要网站，其中支持。在

这种方法实际上使用计算出的ATR值作为输入，因为整个计算都在for循环中，因此需要花费大量的时间。在

stockdata是一个数据帧，包含Date、Open、High、Low和Close值，并将其作为输入传递。在

输入结构（从atr计算出的输出—supert、supert1、supert3）是通过调用ST三次而得到的输出，如果这是一个只返回可以很容易地附加到现有数据帧中的ST值的函数。在

            open    high     low    close   atr lband2  finaluband  finaluband1 finaluband2 finallband  finallband1 finallband2 supert  supert1 supert2
date                                                            
01-07-02    5.2397  5.5409  5.2397  5.4127  NaN 0   0   0   0   0   0   0   0   0   0
02-07-02    5.5234  5.537   5.4214  5.4438  NaN 0   0   0   0   0   0   0   0   0   0
03-07-02    5.506   5.5458  5.3281  5.4661  NaN 0   0   0   0   0   0   0   0   0   0
04-07-02    5.5011  5.572   5.4175  5.5283  NaN 0   0   0   0   0   0   0   0   0   0
05-07-02    5.5633  5.6566  5.4749  5.5905  NaN 0   0   0   0   0   0   0   0   0   0
08-07-02    5.5011  5.7187  5.5011  5.6255  NaN 0   0   0   0   0   0   0   0   0   0
09-07-02    5.5905  5.7586  5.5681  5.6167  NaN 0   0   0   0   0   0   0   0   0   0
10-07-02    5.4885  5.4885  5.1465  5.2222  NaN 0   0   0   0   0   0   0   0   0   0
11-07-02    4.9784  5.2135  4.9784  5.1863  NaN 0   0   0   0   0   0   0   0   0   0
12-07-02    5.5011  5.5011  5.2446  5.3194  NaN 0   0   0   0   0   0   0   0   0   0
15-07-02    5.3243  5.4797  5.1912  5.333   0.24036 4.61437 5.57581 5.81617 6.05653 5.09509 4.85473 4.61437 5.57581 5.81617 6.05653
16-07-02    5.1999  5.4389  5.1999  5.3155  0.240224    4.598728    5.559624    5.799848    6.040072    5.09509 4.85473 4.61437 5.559624    5.799848    6.040072
17-07-02    4.7024  5.1377  4.6189  5.0445  0.285862    4.020715    5.164162    5.450023    5.735885    5.09509 4.85473 4.61437 5.164162    5.450023    5.735885
18-07-02    4.8803  5.1465  4.8356  5.0804  0.288365    4.125954    5.164162    5.450023    5.735885    4.702685    4.85473 4.61437 5.164162    5.450023    5.735885
19-07-02    5.027   5.2038  5.0221  5.1513  0.277699    4.279853    5.164162    5.450023    5.735885    4.835251    4.85473 4.61437 5.164162    5.450023    5.735885
22-07-02    5.0804  5.1465  4.9687  4.9735  0.268189    4.253033    5.164162    5.450023    5.735885    4.835251    4.85473 4.61437 5.164162    5.450023    5.735885
23-07-02    4.8181  5.0843  4.8181  5.0619  0.26799 4.14723 5.164162    5.450023    5.735885    4.835251    4.85473 4.61437 5.164162    5.450023    5.735885
24-07-02    5.058   5.129   4.9376  5.0619  0.260331    4.252307    5.164162    5.450023    5.735885    4.835251    4.85473 4.61437 5.164162    5.450023    5.735885
25-07-02    5.058   5.058   4.7918  4.8492  0.261308    4.140976    5.164162    5.447516    5.708824    4.835251    4.85473 4.61437 5.164162    5.447516    5.708824
26-07-02    5.1465  5.1465  4.7297  4.8715  0.276857    4.107528    5.164162    5.447516    5.708824    4.835251    4.384386    4.61437 5.164162    5.447516    5.708824
29-07-02    4.2235  4.2507  3.9038  4.2507  0.345941    3.039426    4.423191    4.769133    5.115074    4.835251    4.384386    4.61437 4.423191    4.769133    5.115074
30-07-02    4.17    4.3255  4.0369  4.2769  0.340207    3.160578    4.423191    4.769133    5.115074    3.840993    3.500785    3.160578    4.423191    4.769133    5.115074
31-07-02    4.1797  4.3031  4.1486  4.2108  0.321637    3.26094 4.423191    4.769133    5.115074    3.904213    3.582577    3.26094 4.423191    4.769133    5.115074
01-08-02    4.2507  4.5169  4.2507  4.4586  0.320083    3.423551    4.423191    4.769133    5.115074    4.063717    3.743634    3.423551    4.063717    4.769133    5.115074
02-08-02    4.068   4.6101  4.068   4.4984  0.342285    3.312196    4.681335    4.769133    5.115074    4.063717    3.743634    3.423551    4.063717    4.769133    5.115074
05-08-02    4.5965  4.7918  4.5256  4.7918  0.337396    3.646511    4.681335    4.769133    5.115074    4.321304    3.983908    3.646511    4.321304    3.983908    5.115074
06-08-02    4.6276  4.6811  4.6276  4.65    0.320077    3.69412 4.974427    5.294503    5.115074    4.334273    4.014197    3.69412 4.334273    4.014197    5.115074
07-08-02    4.8006  4.854   4.6587  4.752   0.308469    3.830943    4.974427    5.294503    5.115074    4.447881    4.139412    3.830943    4.447881    4.139412    5.115074
08-08-02    4.7918  4.7918  4.614   4.65    0.295402    3.816694    4.974427    5.293704    5.115074    4.447881    4.139412    3.830943    4.447881    4.139412    5.115074
09-08-02    4.6587  4.7024  4.5742  4.7024  0.278682    3.802255    4.916982    5.195664    5.115074    4.447881    4.139412    3.830943    4.447881    4.139412    5.115074

如何重写这段代码，以便在不使用for循环的情况下实现相同的结果，而无需计算精度。我希望通过一个函数调用将pandas dataframe格式的“supert”值添加到stockdata中。在

^{pr2}$

我正在尝试这个优化，以便它运行得更快。我可以将同样多的参数传递给函数。如果你有一个解决方案，请张贴。在

Answer 1

更新3

我无法用给定的ST实现在给定的示例中重现结果。对不起的。在

但我更新了代码以返回和示例中相同的形状，并跳过了NaN。在

---

矢量化将给你巨大的改进。比如31.4 ms矢量化版本vs14.9秒原始的10**5值样本。在

#generate sample
n = 10**5
stockdata = pd.DataFrame(np.random.randint(0,100,size=(n,7)),

            columns = ['atr','high', 'low','close','finaluband','finallband','supert'])
print (stockdata.shape)
stockdata.head()

SuperST将返回stockdata上的所有DataFramefor i in range(3)这正是不同乘法的循环，您可以将其重写为for i in [7,8,9]: SuperST(stockdata,estim,i)

你的ST实现取决于之前的通话结果。它使用'finaluband','finallband','supert'并自行返回它们。最初它们是零，但在第一次调用之后它们将有一些值。所以estim让你控制这种情况，你可以用上一次调用的结果或初始零来调用SuperST。只需用零重命名初始数据帧并将其传递给SuperST。 在

^{pr2}$

超级明星是同一个ST 在

def SuperST(stockdata,estim,multiplier = 7):
    #first we need next row value
    df_prev = estim[['finaluband','finallband','supert']].join(stockdata[['close']]).shift(1)
    # name colums as it was in paramas
    df_prev.columns = ['prevfinaluband', 'prevfinallband', 'prevsupertrend', 'prevclose']
    # working matrix parmas with shifed values w/o NaN in ATR
    df = stockdata.dropna().join(df_prev).fillna(0)

    #calulate vectors 
    upperbasicband = ((df.high+df.low)/2 + (multiplier * df.atr))
    lowerbasicband = ((df.high+df.low)/2 - (multiplier * df.atr))
    # np.where is a vector version of if
    upperband = np.where((upperbasicband < df.prevfinaluband) | (df.prevclose > df.prevfinaluband),
                        upperbasicband,df.prevfinaluband)
    lowerband = np.where ((lowerbasicband > df.prevfinallband) | (df.prevclose < df.prevfinallband),
                         lowerbasicband,df.prevfinallband)

    supertrend = np.where (df.prevsupertrend == df.prevfinaluband,
                             np.where(df.close <= upperband,
                                      upperband,lowerband),
                          np.where(df.prevsupertrend == df.prevfinallband,
                                      np.where(df.close >= lowerband,
                                              lowerband,upperband),
                                    np.zeros_like(lowerband)))


    return upperbasicband.to_frame('uband').assign( 
                    lband = lowerbasicband).assign(
                    finaluband = upperband).assign(
                    finallband = lowerband).assign(
                    supert = supertrend)