pandas 高效VWAP计算 [英] Pandas Efficient VWAP Calculation

问题描述

我有下面的代码，通过它我可以通过三行熊猫代码来计算交易量加权平均价格.

I have the below code, using which I can calculate the volume-weighted average price by three lines of Pandas code.

import numpy as np
import pandas as pd
from pandas.io.data import DataReader
import datetime as dt

df = DataReader(['AAPL'], 'yahoo', dt.datetime(2013, 12, 30), dt.datetime(2014, 12, 30))
df['Cum_Vol'] = df['Volume'].cumsum()
df['Cum_Vol_Price'] = (df['Volume'] * (df['High'] + df['Low'] + df['Close'] ) /3).cumsum()
df['VWAP'] = df['Cum_Vol_Price'] / df['Cum_Vol']

我正在尝试找到一种无需使用 cumsum()作为练习的方式对此进行编码的方法.我正在尝试找到一种解决方案，它可以一次性给出 VWAP 列.我已经尝试过使用 .apply()在下面的行.逻辑在那里，但是问题是我无法在第n行中存储值以便在第(n + 1)行中使用.您如何在 pandas 中解决此问题-仅使用外部连音符或字典临时存储累积值?

I am trying to find a way to code this without using cumsum() as an exercise. I am trying to find a solution which gives the VWAP column in one pass. I have tried the below line, using .apply(). The logic is there, but the issue is I am not able to store values in row n in order to use in row (n+1). How do you approach this in pandas - just use an external tuplet or dictionary for temporary storage of cumulative values?

df['Cum_Vol']= np.nan
df['Cum_Vol_Price'] = np.nan
# calculate running cumulatives by apply - assume df row index is 0 to N
df['Cum_Vol'] = df.apply(lambda x: df.iloc[x.name-1]['Cum_Vol'] + x['Volume'] if int(x.name)>0 else x['Volume'], axis=1)

是否存在针对上述问题的一站式解决方案?

Is there a one-pass solution to the above problem?

我的主要动机是了解幕后发生的事情.因此，这主要是出于锻炼而非任何正当理由.我相信大小为N的序列上的每个累加都具有时间复杂度N(?).因此，我想知道，我们是否可以一次计算两次而不是运行两个单独的累加数-沿此的行进行计算.很高兴接受这个答案-而不是有效的代码.

My main motivation is to understand what is happening under the hood. So, it is mainly for exercise than any valid reason. I believe each cumsum on a Series of size N has time complexity N (?). So I was wondering, instead of running two separate cumsum's, can we calculate both in one pass - along the lines of this. Very happy to accept an answer to this - rather than working code.

推荐答案

单行通行与单行通行在语义上有所不同.这样做有何不同:您可以用1行熊猫，1行numpy或多行numba来做到这一点.

Getting into one pass vs one line starts to get a little semantical. How about this for a distinction: you can do it with 1 line of pandas, 1 line of numpy, or several lines of numba.

from numba import jit

df=pd.DataFrame( np.random.randn(10000,3), columns=['v','h','l'] )

df['vwap_pandas'] = (df.v*(df.h+df.l)/2).cumsum() / df.v.cumsum()

@jit
def vwap():
    tmp1 = np.zeros_like(v)
    tmp2 = np.zeros_like(v)
    for i in range(0,len(v)):
        tmp1[i] = tmp1[i-1] + v[i] * ( h[i] + l[i] ) / 2.
        tmp2[i] = tmp2[i-1] + v[i]
    return tmp1 / tmp2

v = df.v.values
h = df.h.values
l = df.l.values

df['vwap_numpy'] = np.cumsum(v*(h+l)/2) / np.cumsum(v)

df['vwap_numba'] = vwap()

时间:

%timeit (df.v*(df.h+df.l)/2).cumsum() / df.v.cumsum()  # pandas
1000 loops, best of 3: 829 µs per loop

%timeit np.cumsum(v*(h+l)/2) / np.cumsum(v)            # numpy
10000 loops, best of 3: 165 µs per loop

%timeit vwap()                                         # numba
10000 loops, best of 3: 87.4 µs per loop

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