如何访问滚动运算符中的多列?

Eas*_*sun 5 python numpy vectorization pandas

我想在pandas中做一些滚动窗口计算,需要同时处理两列.我将用一个简单的例子来清楚地表达问题:

import pandas as pd

df = pd.DataFrame({
    'x': [1, 2, 3, 2, 1, 5, 4, 6, 7, 9],
    'y': [4, 3, 4, 6, 5, 9, 1, 3, 1, 2]
})

windowSize = 4
result = []

for i in range(1, len(df)+1):
    if i < windowSize:
        result.append(None)
    else:
        x = df.x.iloc[i-windowSize:i]
        y = df.y.iloc[i-windowSize:i]
        m = y.mean()
        r = sum(x[y > m]) / sum(x[y <= m])
        result.append(r)

print(result)
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有没有办法在没有for pringas循环来解决问题?任何帮助表示赞赏

Div*_*kar 1

这是一种使用NumPy工具的矢量化方法 -

windowSize = 4
a = df.values
X = strided_app(a[:,0],windowSize,1)
Y = strided_app(a[:,1],windowSize,1)
M = Y.mean(1)
mask = Y>M[:,None]
sums = np.einsum('ij,ij->i',X,mask)
rest_sums = X.sum(1) - sums
out = sums/rest_sums
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strided_app取自here.

运行时测试 -

方法 -

# @kazemakase's solution
def rolling_window_sum(df, windowSize=4):
    rw = rolling_window(df.values.T, windowSize)
    m = np.mean(rw[1], axis=-1, keepdims=True)
    a = np.sum(rw[0] * (rw[1] > m), axis=-1)
    b = np.sum(rw[0] * (rw[1] <= m), axis=-1)
    result = a / b
    return result    

# Proposed in this post    
def strided_einsum(df, windowSize=4):
    a = df.values
    X = strided_app(a[:,0],windowSize,1)
    Y = strided_app(a[:,1],windowSize,1)
    M = Y.mean(1)
    mask = Y>M[:,None]
    sums = np.einsum('ij,ij->i',X,mask)
    rest_sums = X.sum(1) - sums
    out = sums/rest_sums
    return out
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时间安排 -

In [46]: df = pd.DataFrame(np.random.randint(0,9,(1000000,2)))

In [47]: %timeit rolling_window_sum(df)
10 loops, best of 3: 90.4 ms per loop

In [48]: %timeit strided_einsum(df)
10 loops, best of 3: 62.2 ms per loop
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为了获得更多性能,我们可以计算该Y.mean(1)部分,这基本上是一个带有 的窗口求和Scipy's 1D uniform filter。因此,M也可以计算为windowSize=4-

from scipy.ndimage.filters import uniform_filter1d as unif1d

M = unif1d(a[:,1].astype(float),windowSize)[2:-1]
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性能提升非常显着 -

In [65]: %timeit strided_einsum(df)
10 loops, best of 3: 61.5 ms per loop

In [66]: %timeit strided_einsum_unif_filter(df)
10 loops, best of 3: 49.4 ms per loop
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