我的算法在每天的时间范围内实施此功能时，运行时间从35秒提高到15分钟。该算法批量获取每日历史记录，并遍历数据帧的子集（从t0到tX，其中tX是迭代的当前行）。它这样做是为了模拟算法实时操作期间会发生的情况。我知道有几种方法可以通过在帧计算之间利用内存来改善它，但是我想知道是否还有更多的熊猫式实现会立即受益。
假设self.Step类似于0.00001，self.Precision为5；它们用于将Ohlc条信息分类为离散步骤，以便找到poc。 _frame是整个数据帧的行的子集，并且_low / _high分别对应于此。下面的代码块在整个_frame上执行，每当算法添加新行时（在按每日数据计算年度时间表时），该行可能会向上〜250行。我认为正是这种原因导致了主要的增长放缓。数据帧具有诸如high，low，open，close，volume的列。我正在计算时间价格机会和控制量点。
# Set the complete index of prices +/- 1 step due to weird floating point precision issues
volume_prices = pd.Series(0, index=np.around(np.arange(_low - self.Step, _high + self.Step, self.Step), decimals=self.Precision))
time_prices = volume_prices.copy()
for index, state in _frame.iterrows():
_prices = np.around(np.arange(state.low, state.high, self.Step), decimals=self.Precision)
# Evenly distribute the bar's volume over its range
volume_prices[_prices] += state.volume / _prices.size
# Increment time at price
time_prices[_prices] += 1
# Pandas only returns the 1st row of the max value,
# so we need to reverse the series to find the other side
# and then find the average price between those two extremes
volume_poc = (volume_prices.idxmax() + volume_prices.iloc[::-1].idxmax()) / 2)
time_poc = (time_prices.idxmax() + time_prices.iloc[::-1].idxmax()) / 2)
您可以将此功能用作基础并进行调整：
def f(x): #function to find the POC price and volume
a = x['tradePrice'].value_counts().index[0]
b = x.loc[x['tradePrice'] == a, 'tradeVolume'].sum()
return pd.Series([a,b],['POC_Price','POC_Volume'])
这是我的工作。我仍然不确定您的代码产生的答案是否正确，我认为您的行volume_prices[_prices] += state.Volume / _prices.size并未应用到volume_prices中的每条记录，但这里是基准测试。大约提高了9倍。
def vpOriginal():
Step = 0.00001
Precision = 5
_frame = getData()
_low = 85.0
_high = 116.4
# Set the complete index of prices +/- 1 step due to weird floating point precision issues
volume_prices = pd.Series(0, index=np.around(np.arange(_low - Step, _high + Step, Step), decimals=Precision))
time_prices = volume_prices.copy()
time_prices2 = volume_prices.copy()
for index, state in _frame.iterrows():
_prices = np.around(np.arange(state.Low, state.High, Step), decimals=Precision)
# Evenly distribute the bar's volume over its range
volume_prices[_prices] += state.Volume / _prices.size
# Increment time at price
time_prices[_prices] += 1
time_prices2 += 1
# Pandas only returns the 1st row of the max value,
# so we need to reverse the series to find the other side
# and then find the average price between those two extremes
# print(volume_prices.head(10))
volume_poc = (volume_prices.idxmax() + volume_prices.iloc[::-1].idxmax() / 2)
time_poc = (time_prices.idxmax() + time_prices.iloc[::-1].idxmax() / 2)
return volume_poc, time_poc
def vpNoDF():
Step = 0.00001
Precision = 5
_frame = getData()
_low = 85.0
_high = 116.4
# Set the complete index of prices +/- 1 step due to weird floating point precision issues
volume_prices = pd.Series(0, index=np.around(np.arange(_low - Step, _high + Step, Step), decimals=Precision))
time_prices = volume_prices.copy()
for index, state in _frame.iterrows():
_prices = np.around((state.High - state.Low) / Step , 0)
# Evenly distribute the bar's volume over its range
volume_prices.loc[state.Low:state.High] += state.Volume / _prices
# Increment time at price
time_prices.loc[state.Low:state.High] += 1
# Pandas only returns the 1st row of the max value,
# so we need to reverse the series to find the other side
# and then find the average price between those two extremes
volume_poc = (volume_prices.idxmax() + volume_prices.iloc[::-1].idxmax() / 2)
time_poc = (time_prices.idxmax() + time_prices.iloc[::-1].idxmax() / 2)
return volume_poc, time_poc
getData()
Out[8]:
Date Open High Low Close Volume Adj Close
0 2008-10-14 116.26 116.40 103.14 104.08 70749800 104.08
1 2008-10-13 104.55 110.53 101.02 110.26 54967000 110.26
2 2008-10-10 85.70 100.00 85.00 96.80 79260700 96.80
3 2008-10-09 93.35 95.80 86.60 88.74 57763700 88.74
4 2008-10-08 85.91 96.33 85.68 89.79 78847900 89.79
5 2008-10-07 100.48 101.50 88.95 89.16 67099000 89.16
6 2008-10-06 91.96 98.78 87.54 98.14 75264900 98.14
7 2008-10-03 104.00 106.50 94.65 97.07 81942800 97.07
8 2008-10-02 108.01 108.79 100.00 100.10 57477300 100.10
9 2008-10-01 111.92 112.36 107.39 109.12 46303000 109.12
vpOriginal()
Out[9]: (142.55000000000001, 142.55000000000001)
vpNoDF()
Out[10]: (142.55000000000001, 142.55000000000001)
%timeit vpOriginal()
2.79 s ± 24.3 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
%timeit vpNoDF()
300 ms ± 8.04 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
无论如何，我每天都将时间降为2分钟，而不是15分钟。在较短的时间范围内（2年内，每小时10分钟，对于股票而言，精度为2）仍然很慢。使用DataFrames而不是Series的速度要慢一些。我希望得到更多，但除了以下解决方案外，我不知道该怎么办：
# Upon class instantiation, I've created attributes for each timeframe
# related to `volume_at_price` and `time_at_price`. They serve as memory
# in between frame calculations
def _prices_at(self, frame, bars=0):
# Include 1 step above high as np.arange does not
# include the upper limit by default
state = frame.iloc[-min(bars + 1, frame.index.size)]
bins = np.around(np.arange(state.low, state.high + self.Step, self.Step), decimals=self.Precision)
return pd.Series(state.volume / bins.size, index=bins)
# SetFeature/Feature implement timeframed attributes (i.e., 'volume_at_price_D')
_v = 'volume_at_price'
_t = 'time_at_price'
# Add to x_at_price histogram
_p = self._prices_at(frame)
self.SetFeature(_v, self.Feature(_v).add(_p, fill_value=0))
self.SetFeature(_t, self.Feature(_t).add(_p * 0 + 1, fill_value=0))
# Remove old data from histogram
_p = self._prices_at(frame, self.Bars)
v = self.SetFeature(_v, self.Feature(_v).subtract(_p, fill_value=0))
t = self.SetFeature(_t, self.Feature(_t).subtract(_p * 0 + 1, fill_value=0))
self.SetFeature('volume_poc', (v.idxmax() + v.iloc[::-1].idxmax()) / 2)
self.SetFeature('time_poc', (t.idxmax() + t.iloc[::-1].idxmax()) / 2)