在一维观测数据中检测异常值的Pythonic方法

Question

对于给定的数据，我想将异常值（由95％confidense级别或95％分位数函数或任何需要的东西定义）设置为nan值。以下是我现在使用的数据和代码。如果有人能够进一步解释我，我会很高兴。

import numpy as np, matplotlib.pyplot as plt

data = np.random.rand(1000)+5.0

plt.plot(data)
plt.xlabel('observation number')
plt.ylabel('recorded value')
plt.show()

Answer 1

使用percentile的问题在于，标识为异常值的点是样本量的函数。

有很多方法可以测试异常值，你应该考虑一下如何对它们进行分类。理想情况下，您应该使用先验信息（例如＆＃34;高于/低于此值的任何内容都是不切实际的，因为......＆＃34;）

然而，一个常见的，不太不合理的异常值测试是根据他们的＆＃34;中位数绝对偏差＆＃34;去除积分。

以下是N维案例的实现（来自此处论文的某些代码：https://github.com/joferkington/oost_paper_code/blob/master/utilities.py）：

def is_outlier(points, thresh=3.5):
    """
    Returns a boolean array with True if points are outliers and False 
    otherwise.

    Parameters:
    -----------
        points : An numobservations by numdimensions array of observations
        thresh : The modified z-score to use as a threshold. Observations with
            a modified z-score (based on the median absolute deviation) greater
            than this value will be classified as outliers.

    Returns:
    --------
        mask : A numobservations-length boolean array.

    References:
    ----------
        Boris Iglewicz and David Hoaglin (1993), "Volume 16: How to Detect and
        Handle Outliers", The ASQC Basic References in Quality Control:
        Statistical Techniques, Edward F. Mykytka, Ph.D., Editor. 
    """
    if len(points.shape) == 1:
        points = points[:,None]
    median = np.median(points, axis=0)
    diff = np.sum((points - median)**2, axis=-1)
    diff = np.sqrt(diff)
    med_abs_deviation = np.median(diff)

    modified_z_score = 0.6745 * diff / med_abs_deviation

    return modified_z_score > thresh

这与one of my previous answers非常相似，但我想详细说明样本量效应。

让我们将基于百分位数的离群值测试（类似于@ CTZhu的答案）与各种不同样本量的中位数绝对偏差（MAD）测试进行比较：

import numpy as np
import matplotlib.pyplot as plt
import seaborn as sns

def main():
    for num in [10, 50, 100, 1000]:
        # Generate some data
        x = np.random.normal(0, 0.5, num-3)

        # Add three outliers...
        x = np.r_[x, -3, -10, 12]
        plot(x)

    plt.show()

def mad_based_outlier(points, thresh=3.5):
    if len(points.shape) == 1:
        points = points[:,None]
    median = np.median(points, axis=0)
    diff = np.sum((points - median)**2, axis=-1)
    diff = np.sqrt(diff)
    med_abs_deviation = np.median(diff)

    modified_z_score = 0.6745 * diff / med_abs_deviation

    return modified_z_score > thresh

def percentile_based_outlier(data, threshold=95):
    diff = (100 - threshold) / 2.0
    minval, maxval = np.percentile(data, [diff, 100 - diff])
    return (data < minval) | (data > maxval)

def plot(x):
    fig, axes = plt.subplots(nrows=2)
    for ax, func in zip(axes, [percentile_based_outlier, mad_based_outlier]):
        sns.distplot(x, ax=ax, rug=True, hist=False)
        outliers = x[func(x)]
        ax.plot(outliers, np.zeros_like(outliers), 'ro', clip_on=False)

    kwargs = dict(y=0.95, x=0.05, ha='left', va='top')
    axes[0].set_title('Percentile-based Outliers', **kwargs)
    axes[1].set_title('MAD-based Outliers', **kwargs)
    fig.suptitle('Comparing Outlier Tests with n={}'.format(len(x)), size=14)

main()

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请注意，无论样本大小如何，基于MAD的分类器都能正常工作，而基于百分位数的分类器将样本大小越大，分类越多，无论它们是否实际为异常值。

Answer 2

检测一维数据中的异常值取决于其分布

1 - 正常分布：

1. 数据值在预期范围内几乎平均分布： 在这种情况下，您可以轻松地使用包含均值的所有方法，例如3或2个标准差的置信区间（95％或99.7％）相应的正态分布数据（中心极限定理和样本均值的采样分布）.I是一种非常有效的方法。 在可汗学院统计和概率 - 采样分布库中进行了解释。

另一种方法是预测间隔，如果你想要数据点的置信区间而不是平均值。

1. 数据值随机分布在范围内： mean可能不是数据的公平表示，因为平均值很容易受到异常值的影响（数据集中的非常小或大的值不是典型的） 中位数是衡量数值数据集中心的另一种方法。

   中位数绝对偏差 - 根据中位距离测量所有点距中位数的距离的方法 http://www.itl.nist.gov/div898/handbook/eda/section3/eda35h.htm - 有一个很好的解释，正如Joe Kington在上面的回答中所解释的

2 - 对称分布：如果z分数计算和阈值相应更改，中位数绝对偏差是一个很好的方法

说明： http://eurekastatistics.com/using-the-median-absolute-deviation-to-find-outliers/

3 - 不对称分布：双重MAD - 双重中位数绝对偏差 上述链接中的说明

附加我的python代码以供参考：

 def is_outlier_doubleMAD(self,points):
    """
    FOR ASSYMMETRIC DISTRIBUTION
    Returns : filtered array excluding the outliers

    Parameters : the actual data Points array

    Calculates median to divide data into 2 halves.(skew conditions handled)
    Then those two halves are treated as separate data with calculation same as for symmetric distribution.(first answer) 
    Only difference being , the thresholds are now the median distance of the right and left median with the actual data median
    """

    if len(points.shape) == 1:
        points = points[:,None]
    median = np.median(points, axis=0)
    medianIndex = (points.size/2)

    leftData = np.copy(points[0:medianIndex])
    rightData = np.copy(points[medianIndex:points.size])

    median1 = np.median(leftData, axis=0)
    diff1 = np.sum((leftData - median1)**2, axis=-1)
    diff1 = np.sqrt(diff1)

    median2 = np.median(rightData, axis=0)
    diff2 = np.sum((rightData - median2)**2, axis=-1)
    diff2 = np.sqrt(diff2)

    med_abs_deviation1 = max(np.median(diff1),0.000001)
    med_abs_deviation2 = max(np.median(diff2),0.000001)

    threshold1 = ((median-median1)/med_abs_deviation1)*3
    threshold2 = ((median2-median)/med_abs_deviation2)*3

    #if any threshold is 0 -> no outliers
    if threshold1==0:
        threshold1 = sys.maxint
    if threshold2==0:
        threshold2 = sys.maxint
    #multiplied by a factor so that only the outermost points are removed
    modified_z_score1 = 0.6745 * diff1 / med_abs_deviation1
    modified_z_score2 = 0.6745 * diff2 / med_abs_deviation2

    filtered1 = []
    i = 0
    for data in modified_z_score1:
        if data < threshold1:
            filtered1.append(leftData[i])
        i += 1
    i = 0
    filtered2 = []
    for data in modified_z_score2:
        if data < threshold2:
            filtered2.append(rightData[i])
        i += 1

    filtered = filtered1 + filtered2
    return filtered

Answer 3

我已经改编了http://eurekastatistics.com/using-the-median-absolute-deviation-to-find-outliers的代码，它提供了与Joe Kington相同的结果，但是使用了L1距离而不是L2距离，并且支持非对称分布。原始的R代码没有Joe的0.6745乘数，所以我还在此线程中添加了一致性。不是100％确定是否有必要，而是进行比较。

def doubleMADsfromMedian(y,thresh=3.5):
    # warning: this function does not check for NAs
    # nor does it address issues when 
    # more than 50% of your data have identical values
    m = np.median(y)
    abs_dev = np.abs(y - m)
    left_mad = np.median(abs_dev[y <= m])
    right_mad = np.median(abs_dev[y >= m])
    y_mad = left_mad * np.ones(len(y))
    y_mad[y > m] = right_mad
    modified_z_score = 0.6745 * abs_dev / y_mad
    modified_z_score[y == m] = 0
    return modified_z_score > thresh

Answer 4

使用np.percentile作为@Martin建议：

percentiles = np.percentile(data, [2.5, 97.5])

# or =>, <= for within 95%
data[(percentiles[0]<data) & (percentiles[1]>data)]

# set the outliners to np.nan
data[(percentiles[0]>data) | (percentiles[1]<data)] = np.nan

Answer 5

一个简单的解决方案也可以是，删除超出2个标准偏差（或1.96）的东西：

import random
def outliers(tmp):
    """tmp is a list of numbers"""
    outs = []
    mean = sum(tmp)/(1.0*len(tmp))
    var = sum((tmp[i] - mean)**2 for i in range(0, len(tmp)))/(1.0*len(tmp))
    std = var**0.5
    outs = [tmp[i] for i in range(0, len(tmp)) if abs(tmp[i]-mean) > 1.96*std]
    return outs


lst = [random.randrange(-10, 55) for _ in range(40)]
print lst
print outliers(lst)