Python的高效滚动修剪意味着

Question

用Python计算滚动(又称移动窗口)修剪平均值的最有效方法是什么？

例如,对于50K行和窗口大小为50的数据集,对于每行我需要取最后50行,删除顶部和底部3个值(窗口大小的5%,向上舍入),并获取其余44个值的平均值.

目前我正在为每一行切片以获取窗口,对窗口进行排序然后切片以修剪它.它运作缓慢,但必须有一个更有效的方式.

例

[10,12,8,13,7,18,19,9,15,14] # data used for example, in real its a 50k lines df

窗口大小为5.对于每一行,我们查看最后5行,对它们进行排序并丢弃1个顶部和1个底部行(5%的5 = 0.25,向上舍入为1).然后我们平均剩下的中间行.

生成此示例的代码设置为DataFrame

pd.DataFrame({
    'value': [10, 12, 8, 13, 7, 18, 19, 9, 15, 14],
    'window_of_last_5_values': [
        np.NaN, np.NaN, np.NaN, np.NaN, '10,12,8,13,7', '12,8,13,7,18',
        '8,13,7,18,19', '13,7,18,19,9', '7,18,19,9,15', '18,19,9,15,14'
    ],
    'values that are counting for average': [
        np.NaN, np.NaN, np.NaN, np.NaN, '10,12,8', '12,8,13', '8,13,18',
        '13,18,9', '18,9,15', '18,15,14'
    ],
    'result': [
        np.NaN, np.NaN, np.NaN, np.NaN, 10.0, 11.0, 13.0, 13.333333333333334,
        14.0, 15.666666666666666
    ]
})

天真实现的示例代码

window_size = 5
outliers_to_remove = 1

for index in range(window_size - 1, len(df)):
    current_window = df.iloc[index - window_size + 1:index + 1]
    trimmed_mean = current_window.sort_values('value')[
        outliers_to_remove:window_size - outliers_to_remove]['value'].mean()
    # save the result and the window content somewhere

关于DataFrame vs list vs NumPy数组的注释

只需将数据从DataFrame移动到列表,我就可以使用相同的算法获得3.5倍的速度提升.有趣的是,使用NumPy阵列也可以提供几乎相同的速度提升.但是,必须有更好的方法来实现这一目标并实现数量级的提升.

Answer 1

一个可以派上用场的观察是你不需要在每一步中对所有值进行排序.相反,如果您确保窗口始终排序,您需要做的就是在相关位置插入新值,并从原来的位置删除旧值,这两个操作都可以在O中完成(log_2 (window_size))使用bisect.在实践中,这看起来像

def rolling_mean(data):
    x = sorted(data[:49])
    res = np.repeat(np.nan, len(data))
    for i in range(49, len(data)):
        if i != 49:
            del x[bisect.bisect_left(x, data[i - 50])]
        bisect.insort_right(x, data[i])
        res[i] = np.mean(x[3:47])
    return res

现在,在这种情况下,额外的好处是比scipy.stats.trim_mean依赖的矢量化所获得的更少,因此特别是,它仍然比@ ChrisA的解决方案慢,但它是进一步性能优化的有用起点.

> data = pd.Series(np.random.randint(0, 1000, 50000))
> %timeit data.rolling(50).apply(lambda w: trim_mean(w, 0.06))
727 ms ± 34.7 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
> %timeit rolling_mean(data.values)
812 ms ± 42.1 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

值得注意的是,Numba的抖动在这种情况下通常很有用,也没有任何好处:

> from numba import jit
> rolling_mean_jit = jit(rolling_mean)
> %timeit rolling_mean_jit(data.values)
1.05 s ± 183 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

以下似乎远非最优的方法优于上述两种方法:

def rolling_mean_np(data):
    res = np.repeat(np.nan, len(data))
    for i in range(len(data)-49):
        x = np.sort(data[i:i+50])
        res[i+49] = x[3:47].mean()
    return res

定时:

> %timeit rolling_mean_np(data.values)
564 ms ± 4.44 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

更重要的是,这一次,JIT编译确实有帮助:

> rolling_mean_np_jit = jit(rolling_mean_np)
> %timeit rolling_mean_np_jit(data.values)
94.9 ms ± 605 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

虽然我们正在努力,但我们只是快速验证这实际上是否符合我们的预期:

> np.all(rolling_mean_np_jit(data.values)[49:] == data.rolling(50).apply(lambda w: trim_mean(w, 0.06)).values[49:])
True

事实上,通过帮助分拣机一点点,我们可以挤出另一个因子2,将总时间缩短到57毫秒:

def rolling_mean_np_manual(data):
    x = np.sort(data[:50])
    res = np.repeat(np.nan, len(data))
    for i in range(50, len(data)+1):
        res[i-1] = x[3:47].mean()
        if i != len(data):
            idx_old = np.searchsorted(x, data[i-50])
            x[idx_old] = data[i]
            x.sort()
    return res

> %timeit rolling_mean_np_manual(data.values)
580 ms ± 23 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
> rolling_mean_np_manual_jit = jit(rolling_mean_np_manual)
> %timeit rolling_mean_np_manual_jit(data.values)
57 ms ± 5.89 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
> np.all(rolling_mean_np_manual_jit(data.values)[49:] == data.rolling(50).apply(lambda w: trim_mean(w, 0.06)).values[49:])
True

现在,在这个例子中正在进行的"排序"当然只是归结为将新元素放在正确的位置,同时将所有内容移到一个之间.手动执行此操作将使纯Python代码变慢,但jitted版本获得另一个因子2,使我们低于30毫秒:

def rolling_mean_np_shift(data):
    x = np.sort(data[:50])
    res = np.repeat(np.nan, len(data))
    for i in range(50, len(data)+1):
        res[i-1] = x[3:47].mean()
        if i != len(data):
            idx_old, idx_new = np.searchsorted(x, [data[i-50], data[i]])
            if idx_old < idx_new:
                x[idx_old:idx_new-1] = x[idx_old+1:idx_new]
                x[idx_new-1] = data[i]
            elif idx_new < idx_old:
                x[idx_new+1:idx_old+1] = x[idx_new:idx_old]
                x[idx_new] = data[i]
            else:
                x[idx_new] = data[i]
    return res

> %timeit rolling_mean_np_shift(data.values)
937 ms ± 97.8 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
> rolling_mean_np_shift_jit = jit(rolling_mean_np_shift)
> %timeit rolling_mean_np_shift_jit(data.values)
26.4 ms ± 693 µs per loop (mean ± std. dev. of 7 runs, 1 loop each)
> np.all(rolling_mean_np_shift_jit(data.values)[49:] == data.rolling(50).apply(lambda w: trim_mean(w, 0.06)).values[49:])
True

在这一点上,大部分时间都花在了np.searchsorted,所以让我们让搜索本身对JIT友好.采用源代码bisect,我们让

@jit
def binary_search(a, x):
    lo = 0
    hi = 50
    while lo < hi:
        mid = (lo+hi)//2
        if a[mid] < x: lo = mid+1
        else: hi = mid
    return lo

@jit
def rolling_mean_np_jitted_search(data):
    x = np.sort(data[:50])
    res = np.repeat(np.nan, len(data))
    for i in range(50, len(data)+1):
        res[i-1] = x[3:47].mean()
        if i != len(data):
            idx_old = binary_search(x, data[i-50])
            idx_new = binary_search(x, data[i])
            if idx_old < idx_new:
                x[idx_old:idx_new-1] = x[idx_old+1:idx_new]
                x[idx_new-1] = data[i]
            elif idx_new < idx_old:
                x[idx_new+1:idx_old+1] = x[idx_new:idx_old]
                x[idx_new] = data[i]
            else:
                x[idx_new] = data[i]
    return res

这将我们降低到12毫秒,比原始熊猫+ SciPy方法提高了x60:

> %timeit rolling_mean_np_jitted_search(data.values)
12 ms ± 210 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

Answer 2

您可以尝试使用scipy.stats.trim_mean:

from scipy.stats import trim_mean

df['value'].rolling(5).apply(lambda x: trim_mean(x, 0.2))

[输出]

0          NaN
1          NaN
2          NaN
3          NaN
4    10.000000
5    11.000000
6    13.000000
7    13.333333
8    14.000000
9    15.666667

请注意,我必须使用rolling(5)和proportiontocut=0.2您的玩具数据集.

对于您的真实数据,您应该使用rolling(50)和trim_mean(x, 0.06)从滚动窗口中删除顶部和底部3值.