如何访问滚动操作员中的多列

Question

我想在熊猫中进行一些滚动窗口计算，需要同时处理两列。我将采用一个简单的实例来清楚地表达问题：

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)

有没有办法在熊猫中没有循环解决问题？任何帮助都将受到赞赏

Answer 1

您可以将滚动窗口技巧用于numpy数组，然后将其应用于数据框架的阵列。

import pandas as pd
import numpy as np
def rolling_window(a, window):
    shape = a.shape[:-1] + (a.shape[-1] - window + 1, window)
    strides = a.strides + (a.strides[-1],)
    return np.lib.stride_tricks.as_strided(a, shape=shape, strides=strides)
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    
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

结果缺乏领先的None值，但是它们应该易于附加(以np.nan的形式或将结果转换为列表(。

这可能不是您正在寻找的，与Pandas一起工作，但是它将完成工作而无需循环。

Answer 2

这是使用NumPy工具的一种矢量化方法 -

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

strided_app取自here。

运行时测试 -

接近 -

# @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

时间 -

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

要挤压更多性能，我们可以计算Y.mean(1)部分，这基本上是用Scipy's 1D uniform filter的窗口求和。因此，可以将M用于windowSize=4作为 -

from scipy.ndimage.filters import uniform_filter1d as unif1d
M = unif1d(a[:,1].astype(float),windowSize)[2:-1]

性能增长很大 -

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