加速异常值检查熊猫系列

时间:2013-02-25 15:20:31

标签: python pandas outliers

我正在使用不同的标准偏差标准对两个传球进行异常值检查。但是,我使用两个循环,它运行速度非常慢。我想知道是否有任何大熊猫“伎俩”来加速这一步。

这是我正在使用的代码(警告非常难看的代码!):

def find_outlier(point, window, n):
    return np.abs(point - nanmean(window)) >= n * nanstd(window)

def despike(self, std1=2, std2=20, block=100, keep=0):
    res = self.values.copy()
    # First run with std1:
    for k, point in enumerate(res):
        if k <= block:
            window = res[k:k + block]
        elif k >= len(res) - block:
            window = res[k - block:k]
        else:
            window = res[k - block:k + block]
        window = window[~np.isnan(window)]
        if np.abs(point - window.mean()) >= std1 * window.std():
            res[k] = np.NaN
    # Second run with std2:
    for k, point in enumerate(res):
        if k <= block:
            window = res[k:k + block]
        elif k >= len(res) - block:
            window = res[k - block:k]
        else:
            window = res[k - block:k + block]
        window = window[~np.isnan(window)]
        if np.abs(point - window.mean()) >= std2 * window.std():
            res[k] = np.NaN
    return Series(res, index=self.index, name=self.name)

1 个答案:

答案 0 :(得分:12)

我不确定你对这个块块做了什么,但是在系列中找到异常值应该像以下一样简单:

In [1]: s > s.std() * 3

其中s是你的系列,3是超出异常状态的标准偏差。此表达式将返回一系列布尔值,然后您可以通过以下方式索引系列:

In [2]: s.head(10)
Out[2]:
0    1.181462
1   -0.112049
2    0.864603
3   -0.220569
4    1.985747
5    4.000000
6   -0.632631
7   -0.397940
8    0.881585
9    0.484691
Name: val

In [3]: s[s > s.std() * 3]
Out[3]:
5    4
Name: val

<强>更新

解决关于阻止的评论。我认为在这种情况下你可以使用pd.rolling_std()

In [53]: pd.rolling_std(s, window=5).head(10)
Out[53]:
0         NaN
1         NaN
2         NaN
3         NaN
4    0.871541
5    0.925348
6    0.920313
7    0.370928
8    0.467932
9    0.391485

In [55]: abs(s) > pd.rolling_std(s, window=5) * 3

Docstring:
Unbiased moving standard deviation

Parameters
----------
arg : Series, DataFrame
window : Number of observations used for calculating statistic
min_periods : int
    Minimum number of observations in window required to have a value
freq : None or string alias / date offset object, default=None
    Frequency to conform to before computing statistic
    time_rule is a legacy alias for freq

Returns
-------
y : type of input argument