如何使用熊猫从OHLC数据计算三折线图

Question

我在Pandas数据框中有OHLC（打开，高，低，关闭）数据。我想从中获取三行断行数据。我已经为此编写了一个函数。但是，从具有近1400行的OHLC数据帧中创建Line Break数据帧大约需要2.5秒。我可以理解，由于我要对每一行进行迭代以进行计算，因此花费了时间。我需要帮助来转换函数，而无需迭代每一行或通过Pandas实现。

我写了下面提到的函数。

def get_lbdata(data, lines = 3):
    data = data.reset_index().drop(['index'], axis=1)
    lbdata = pd.DataFrame(columns = ['open', 'high', 'low', 'close', 'timestamp', 'date', 'month', 'year', 'weekday', 'hour', 'minute', 'second'])
    lbdata.at[0, "open"] = data.open.iloc[0]
    lbdata.at[0, "close"] = data.close.iloc[0]
    lbdata.at[0, "timestamp"] = data.timestamp.iloc[0]
    lbdata.at[0, "date"] = data.date.iloc[0]
    lbdata.at[0, "month"] = data.month.iloc[0]
    lbdata.at[0, "year"] = data.year.iloc[0]
    lbdata.at[0, "weekday"] = data.weekday.iloc[0]
    lbdata.at[0, "hour"] = data.hour.iloc[0]
    lbdata.at[0, "minute"] = data.minute.iloc[0]
    lbdata.at[0, "second"] = data.second.iloc[0]
    if lbdata.close.iloc[-1] > lbdata.open.iloc[-1]:
        lbdata.at[0, "high"] = data.close.iloc[0]
        lbdata.at[0, "low"] = data.open.iloc[0]
        uptrend = 1
    else:
        lbdata.at[0, "high"] = data.open.iloc[0]
        lbdata.at[0, "low"] = data.close.iloc[0]
        uptrend = 0
    for ri,row in data.iterrows():
        if ri == 0:
            continue
        i = len(lbdata)
        vals = []
        for x in range(-(min(lines, i)), 0):
            vals.insert(len(vals), lbdata.open.iloc[x])
            vals.insert(len(vals), lbdata.close.iloc[x])
        minbreak = min(vals)
        maxbreak = max(vals)
        if row.close > maxbreak:
            if uptrend:
                lbdata.at[i, "open"] = lbdata.close.iloc[-1]
            else:
                lbdata.at[i, "open"] = lbdata.open.iloc[-1]
                uptrend = 1
            lbdata.at[i, "high"] = row.close
            lbdata.at[i, "low"] = lbdata.open.iloc[i]
            lbdata.at[i, "close"] = row.close
            lbdata.at[i, "timestamp"] = row.timestamp
            lbdata.at[i, "date"] = row.date
            lbdata.at[i, "month"] = row.month
            lbdata.at[i, "year"] = row.year
            lbdata.at[i, "weekday"] = row.weekday
            lbdata.at[i, "hour"] = row.hour
            lbdata.at[i, "minute"] = row.minute
            lbdata.at[i, "second"] = row.second
        elif row.close < minbreak:
            if not uptrend:
                lbdata.at[i, "open"] = lbdata.close.iloc[-1]
            else:
                lbdata.at[i, "open"] = lbdata.open.iloc[-1]
                uptrend = 0
            lbdata.at[i, "high"] = lbdata.open.iloc[i]
            lbdata.at[i, "low"] = row.close
            lbdata.at[i, "close"] = row.close
            lbdata.at[i, "timestamp"] = row.timestamp
            lbdata.at[i, "date"] = row.date
            lbdata.at[i, "month"] = row.month
            lbdata.at[i, "year"] = row.year
            lbdata.at[i, "weekday"] = row.weekday
            lbdata.at[i, "hour"] = row.hour
            lbdata.at[i, "minute"] = row.minute
            lbdata.at[i, "second"] = row.second
    return lbdata

我需要熊猫对应的代码来大大减少计算时间。