Python Pandas DataFrame的指数衰减 [英] Exponential Decay on Python Pandas DataFrame

问题描述

我正在尝试有效地计算Pandas DataFrame每一列的运行总和，并且呈指数衰减. DataFrame包含世界上每个国家/地区的每日得分. DataFrame看起来像这样:

I'm trying to efficiently compute a running sum, with exponential decay, of each column of a Pandas DataFrame. The DataFrame contains a daily score for each country in the world. The DataFrame looks like this:

                AF        UK        US
2014-07-01  0.998042  0.595720  0.524698
2014-07-02  0.380649  0.838436  0.355149
2014-07-03  0.306240  0.274755  0.964524
2014-07-04  0.396721  0.836027  0.225848
2014-07-05  0.151291  0.677794  0.603548
2014-07-06  0.558846  0.050535  0.551785
2014-07-07  0.463514  0.552748  0.265537
2014-07-08  0.240282  0.278825  0.116432
2014-07-09  0.309446  0.096573  0.246021
2014-07-10  0.800977  0.583496  0.713893

我不确定如何在不迭代数据帧的情况下计算滚动总和(带有衰减)，因为我需要知道昨天的分数才能计算出今天的分数.但是要计算昨天的分数，我需要知道昨天的分数的前一天，等等.这是我一直在使用的代码，但是我想找到一种更有效的方法.

I'm not sure how to calculate the rolling sum (with decay) without iterating through the dataframe, since I need to know yesterday's score to calculate today's score. But to calculate yesterday's score, I need to know the day before yesterday's score, etc. This is the code that I've been using, but I'd like a more efficient way to go about it.

for j, val in df.iteritems():
    for i, row in enumerate(val):
        df[j].iloc[i] = row + val[i-1]*np.exp(-0.05)

推荐答案

您可以使用以下事实:当指数乘以它们的指数时，会添加:

You can use the fact that when exponentials multiply their exponents add:

例如:

N(2) = N(2) + N(1) * exp(-0.05)
N(3) = N(3) + (N(2) + N(1) * exp(-0.05))*exp(-0.05)
N(3) = N(3) + N(2)*exp(-0.05) + N(1)*exp(-0.1)
N(4) = ...and so on

然后可以使用numpy将其矢量化:

This can then be vectorized using numpy:

dataset = pd.DataFrame(np.random.rand(1000,3), columns=["A", "B","C"])

weightspace = np.exp(np.linspace(len(dataset), 0, num=len(dataset))*-0.05)
def rollingsum(array):
    weights = weightspace[0-len(array):]
    # Convolve the array and the weights to obtain the result
    a = np.dot(array, weights).sum()
    return a


a = pd.expanding_apply(dataset, rollingsum)

pd.expanding_apply将rollingsum函数向后应用到每一行，并调用len(dataset)次. np.linspace生成大小为len(dataset)的数据集，并计算当前行每行乘以exp(-0.05)的次数.

pd.expanding_apply applies the rollingsum function backwards to each row, calling it len(dataset) times. np.linspace generates a dataset of size len(dataset) and calculates how many times each row is multiplied by exp(-0.05) for the current row.

因为它是矢量化的，所以应该很快:

Because it is vectorized, it should be fast:

%timeit a = pd.expanding_apply(dataset, rollingsum)
10 loops, best of 3: 25.5 ms per loop

与之相比(请注意，我使用的是python 3，必须对第一行的行为进行更改...):

This compares with (note I'm using python 3 and had to make a change to the behaviour on the first row...):

def multipleApply(df):
    for j, val in df.iteritems():
        for i, row in enumerate(val):
            if i == 0:
                continue
            df[j].iloc[i] = row + val[i-1]*np.exp(-0.05)

结果显示为:

In[68]: %timeit multipleApply(dataset)
1 loops, best of 3: 414 ms per loop

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