Question

我需要对我的数据使用Hampel过滤器，剥离异常值。

我还没能在Python中找到现有的;仅在Matlab和R中。

[matlab功能描述] [1]

[Mats Hampel函数的统计交换讨论] [2]

[R pracma package vignette;包含hampel函数] [3]

我已经编写了以下函数，将其建模在R pracma包中的函数中;然而，它远远慢于Matlab版本。这不理想;我会很感激如何加快速度。

该功能如下所示 -

def hampel(x,k, t0=3):
    '''adapted from hampel function in R package pracma
    x= 1-d numpy array of numbers to be filtered
    k= number of items in window/2 (# forward and backward wanted to capture in median filter)
    t0= number of standard deviations to use; 3 is default
    '''
    n = len(x)
    y = x #y is the corrected series
    L = 1.4826
    for i in range((k + 1),(n - k)):
        if np.isnan(x[(i - k):(i + k+1)]).all():
            continue
        x0 = np.nanmedian(x[(i - k):(i + k+1)])
        S0 = L * np.nanmedian(np.abs(x[(i - k):(i + k+1)] - x0))
        if (np.abs(x[i] - x0) > t0 * S0):
            y[i] = x0
    return(y)

＆＃34;实践中的R实施＆＃34;包，我用作模型：

function (x, k, t0 = 3) 
{
    n <- length(x)
    y <- x
    ind <- c()
    L <- 1.4826
    for (i in (k + 1):(n - k)) {
        x0 <- median(x[(i - k):(i + k)])
        S0 <- L * median(abs(x[(i - k):(i + k)] - x0))
        if (abs(x[i] - x0) > t0 * S0) {
            y[i] <- x0
            ind <- c(ind, i)
        }
    }
    list(y = y, ind = ind)
}

任何帮助提高函数效率的帮助，或者指向现有Python模块中现有实现的指针都将非常受欢迎。以下示例数据; Jupyter中的%% timeit cell magic表示它目前需要15秒才能运行：

vals=np.random.randn(250000)
vals[3000]=100
vals[200]=-9000
vals[-300]=8922273
%%timeit
hampel(vals, k=6)

[1]：https://www.mathworks.com/help/signal/ref/hampel.html [2]：https://dsp.stackexchange.com/questions/26552/what-is-a-hampel-filter-and-how-does-it-work [3]：https://cran.r-project.org/web/packages/pracma/pracma.pdf

Answer 1

熊猫解决方案的速度提高了几个数量级：

def hampel(vals_orig, k=7, t0=3):
    '''
    vals: pandas series of values from which to remove outliers
    k: size of window (including the sample; 7 is equal to 3 on either side of value)
    '''
    #Make copy so original not edited
    vals=vals_orig.copy()    
    #Hampel Filter
    L= 1.4826
    rolling_median=vals.rolling(k).median()
    difference=np.abs(rolling_median-vals)
    median_abs_deviation=difference.rolling(k).median()
    threshold= t0 *L * median_abs_deviation
    outlier_idx=difference>threshold
    vals[outlier_idx]=np.nan
    return(vals)

此时间为11 ms vs 15秒;巨大的进步。

我在this post.

中找到了类似过滤器的解决方案

Answer 2

上面的@EHB解决方案很有帮助，但这是不正确的。具体来说，在 median_abs_deviation 中计算的滚动平均值是 difference ，这本身就是每个数据点与在 rolling_median 中计算的滚动平均值之间的差，但是它应该是滚动窗口中的数据与窗口上的数据之间的差的中值。我将上面的代码修改为：

def hampel(vals_orig, k=7, t0=3):
    '''
    vals: pandas series of values from which to remove outliers
    k: size of window (including the sample; 7 is equal to 3 on either side of value)
    '''

    #Make copy so original not edited
    vals = vals_orig.copy()

    #Hampel Filter
    L = 1.4826
    rolling_median = vals.rolling(window=k, center=True).median()
    MAD = lambda x: np.median(np.abs(x - np.median(x)))
    rolling_MAD = vals.rolling(window=k, center=True).apply(MAD)
    threshold = t0 * L * rolling_MAD
    difference = np.abs(vals - rolling_median)

    '''
    Perhaps a condition should be added here in the case that the threshold value
    is 0.0; maybe do not mark as outlier. MAD may be 0.0 without the original values
    being equal. See differences between MAD vs SDV.
    '''

    outlier_idx = difference > threshold
    vals[outlier_idx] = np.nan
    return(vals)

过滤异常值 - 如何更快地使基于中值的Hampel函数？

2 个答案: