python - Pandas Rolling vs Scipy kurtosis - 严重的数值不准确

Question

首先，对于我在下面列出的明显不简单的示例，我深表歉意。我完全知道这不符合 SO 的最小可重现性约束，但是，现在已经进行了数小时的实验，试图重现该问题，在我看来，它确实只在对至少数百个值执行计算时才会出现。

我有一个包含数百万个值的数据框，我想在其中滚动计算每一列的峰度。最初我使用 pd.rolling.kurt:

df.rolling(20, min_periods=3).kurt(bias=False)

但注意到该方法存在两个严重问题:

1. 准确性不令人满意；尽管 pandas 的方法给出了大致还可以的结果，但对于我的用例来说，1e-4 数量级的偏差很难接受；
2. 更令人担忧的是经常“爆炸”的峰度值:峰度值突然开始偏离 +/-10,000，完全扭曲了预期的输出，没有明显的原因。

我创建了三个系列，s1、s2 和 s3，分别具有 300、600 和 900 个值。 (在这篇文章的末尾添加了具有确切值的赋值，以免在我的文章之后造成太多麻烦。)这三个系列是数据框一列的切片。切片以最后位置固定的方式创建，即 s1 的值从 N-299 到 N， s2 从 N-599 到 N 和 s3 从 N-899 到 N 。在这三个系列上运行 pd.rolling.kurt 并打印数据帧的尾部(出现我要讨论的问题的地方)给出以下结果:

>>> s1.rolling(20,min_periods=3).kurt().tail(10)
290     9.591067
291     9.591067
292     9.591067
293     9.591067
294    19.663666
295    14.872262
296    14.147157
297    16.716964
298     7.032522
299    19.983796
>>> s2.rolling(20,min_periods=3).kurt().tail(10)
590     9.591067
591     9.591067
592     9.591067
593     9.591067
594    19.663666
595    14.872262
596    14.147157
597    16.716964
598     7.032522
599    19.983796
>>> s3.rolling(20,min_periods=3).kurt().tail(10)
890         9.591071
891         9.591071
892         9.591071
893         9.591071
894        19.663685
895        15.248361
896        40.444894
897      1368.233241
898    251407.375343
899    902540.031652

我在 Excel 中执行了相同的计算，对于最后十个指数，峰度值应该如下(我使用符号 290/590/890 来节省一些空间:三个输出系列索引值 290-299、590-599 和 890-899 具有相同的值):

290 / 590 / 890      9.591067361
291 / 591 / 891      9.591067361
292 / 592 / 892      9.591067361
293 / 593 / 893      9.591067361
294 / 594 / 894      19.66366573
295 / 595 / 895      14.87226197
296 / 596 / 896      14.14715754
297 / 597 / 897      16.7169886
298 / 598 / 898      7.037037037
299 / 599 / 899      20

观察 pd.rolling.kurt 提供的输出，我们看到前两个输出是相同的，尽管它们与我使用 Excel 计算的实际输出不匹配。然而，更大的问题发生在第三个输出中，其中值爆炸，就好像系列中值的总数会以某种方式影响峰度值，即使对于所有三种情况我都使用了 20 的滚动窗口和最小所需数量3. 从理论上讲，如果我的理解是正确的，这意味着除了当前行和最后 19 行之外，没有其他东西会干扰峰度输出。我很困惑这些“爆炸性”值是如何出现的。

然后我使用 scipy.stats.kurtosis 重新计算了同一系列的峰度值。这给了我以下输出:

>>> s1.rolling(20,min_periods=3).apply(lambda x: kurtosis(x, bias=False)).tail(10)
290     9.591067
291     9.591067
292     9.591067
293     9.591067
294    19.663666
295    14.872262
296    14.147158
297    16.716989
298     7.037037
299    20.000000
>>> s2.rolling(20,min_periods=3).apply(lambda x: kurtosis(x, bias=False)).tail(10)
590     9.591067
591     9.591067
592     9.591067
593     9.591067
594    19.663666
595    14.872262
596    14.147158
597    16.716989
598     7.037037
599    20.000000
>>> s3.rolling(20,min_periods=3).apply(lambda x: kurtosis(x, bias=False)).tail(10)
890     9.591067
891     9.591067
892     9.591067
893     9.591067
894    19.663666
895    14.872262
896    14.147158
897    16.716989
898     7.037037
899    20.000000

这完美地计算了峰度。然而，.apply(lambda x: kurtosis(x,...) 构造与矢量化 pandas 方法相比效率低得惊人，将整个数据帧的总处理时间从几分钟推到所有一个多小时!我完全意识到，在许多情况下，内置矢量化解决方案往往更喜欢速度而不是数值精度，这可以解释我上面列出的第一个问题；但是，至于第二个问题(即“爆炸”值) 我根本看不出有什么理由。

有没有什么方法可以有效地计算峰度，而不会出现值发散和使我的整个输出无效的情况？

---

系列定义

这是我用来计算上述输出的确切值:

s1 = pd.Series([0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0001499887511247459,-7.499156348433101e-05,-3.699790962233055e-05,-1.899945851585629e-05,-8.999869502079515e-06,-4.999962500264377e-06,-1.999992000039351e-06,-9.999974999814318e-07,-9.999984999603102e-07,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,1.699983850190338e-05,-8.999878501628346e-06,-3.999972000122605e-06,-1.999992000039351e-06,-9.999974999814318e-07,0.0003669319382432873,-0.0001849488621671012,-9.198730581664589e-05,-4.499687272496313e-05,0.0009075453820856781,0.0004854184782060238,-0.000720221831477389,-0.000359805708801156,-0.0001799514136040646,-8.998785170075082e-05,-5.999640023402946e-05,-1.9999600008734e-05,-6.999954500263924e-06,-1.999995999958864e-06,-9.999994999391884e-07,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.001201278176363365,-0.0008013581550363867,-0.0002669288650428971,-8.89921242557729e-05,-2.899914452727788e-05,-9.99990000099588e-06,-2.999989500049026e-06,-9.999984999603102e-07,-9.999994999391884e-07,0.0,0.0,0.0,0.0,0.0,0.0,0.0005218638053935734,-0.0004638654873286288,-3.799851806232993e-05,-1.299982450270071e-05,-4.999977500118572e-06,-9.999984999603102e-07,-9.999994999391884e-07,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0])

s2 = pd.Series([0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0001499887511247459,-7.499156348433101e-05,-3.699790962233055e-05,-1.899945851585629e-05,-8.999869502079515e-06,-4.999962500264377e-06,-1.999992000039351e-06,-9.999974999814318e-07,-9.999984999603102e-07,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,1.699983850190338e-05,-8.999878501628346e-06,-3.999972000122605e-06,-1.999992000039351e-06,-9.999974999814318e-07,0.0003669319382432873,-0.0001849488621671012,-9.198730581664589e-05,-4.499687272496313e-05,0.0009075453820856781,0.0004854184782060238,-0.000720221831477389,-0.000359805708801156,-0.0001799514136040646,-8.998785170075082e-05,-5.999640023402946e-05,-1.9999600008734e-05,-6.999954500263924e-06,-1.999995999958864e-06,-9.999994999391884e-07,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.001201278176363365,-0.0008013581550363867,-0.0002669288650428971,-8.89921242557729e-05,-2.899914452727788e-05,-9.99990000099588e-06,-2.999989500049026e-06,-9.999984999603102e-07,-9.999994999391884e-07,0.0,0.0,0.0,0.0,0.0,0.0,0.0005218638053935734,-0.0004638654873286288,-3.799851806232993e-05,-1.299982450270071e-05,-4.999977500118572e-06,-9.999984999603102e-07,-9.999994999391884e-07,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0])

s3 = 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Answer 1

它看起来像是旧 Pandas 版本中的错误。我可以在 win32、Pandas 1.0.3、numpy 1.15.4 上的旧安装 Python 3.6.2 64 位上重现:

>>> s3.rolling(20,min_periods=3).kurt().tail(10)
890         9.591071
891         9.591071
892         9.591071
893         9.591071
894        19.663685
895        15.248361
896        40.444894
897      1368.233241
898    251407.375343
899    902540.031652
dtype: float64

它似乎在我的新版本 Python 3.8.4 64 位、Pandas 1.2.2、numpy 1.20.1 上得到修复:

>>> s3.rolling(20,min_periods=3).kurt().tail(10)
890     9.591067
891     9.591067
892     9.591067
893     9.591067
894    19.663666
895    14.872262
896    14.147158
897    16.716989
898     7.037037
899    20.000000
dtype: float64

两个安装都在同一台 Windows 10 机器上。

我不能说是哪个组件(Pandas 或 numpy)引起的。由于您使用 numpy.stats.kurtosis 的测试给出了正确的结果，我会怀疑 Pandas，但如果没有 Pandas 专家(我不是专家)的进一步分析，我不能肯定。

恕我直言，最合理的解决方案是升级您的系统，或者使用最新的 Pandas 版本添加全新的独立 Python 安装。