pandas 对某些方法和某些大小的数据帧进行插值返回ValueErrors [英] Pandas Interpolate returning ValueErrors for some methods and some sizes of dataframes

问题描述

我在插入熊猫数据框时遇到一些问题.

I am having some issues with interpolation of a Pandas dataframe.

基本上，我有一个295339行的数据框，并人工生成了nan，以研究不同的采样率和完成方法.

Basically, I have a dataframe of 295339 rows and have artificially generated nan's to study different sampling rates and completion methods.

问题是，当我对采样率和完成方法进行某种组合时，所有方法都可以解决，而对于其他人，我会收到以下错误消息，

The issue is that when I do some combinations of my sampling rates and completion methods it all works out while for others I get the following error message,

ValueError: The number of derivatives at boundaries does not match: expected. 1, got 0+0.

ValueError的类型取决于我使用的采样率和完成方法的组合.

The type of ValueError depends on the combination of sampling rate and completion method I'm using.

因此，例如，如果我每位客户每小时赚一纳，然后使用线性或三次方法进行插值，那么它将起作用.但是，如果我每位客户每四个小时采样一次，则它适用于线性方法，而不适用于三次方法(插值波纹管的代码):

So for example, if I make one nan per hour per customer and then interpolate using either the linear or the cubic method it works. But if I sample once every four hours per customer it works for the linear method but not for the cubic method (code for the interpolation bellow):

latitude = my_frame.filter(['Customer_id', 'Lat'], axis=1)
latitude = latitude.groupby('Customer_id').apply(lambda group: group.interpolate(method= 'cubic')

奇怪的是，在测试过程中，出于速度目的，我仅将方法限制在3个客户(代表8500行)上，没有出现任何问题.

The weird thing is that during my tests I limited my approach to 3 customers (representing 8500 rows) for speed purposes and no issues were raised.

所以，我的问题是为什么会发生这种情况，并且有任何解决方法.

So, my question is why does this happen and is there any workaround.

推荐答案

我发现问题是，对于记录较少的客户，我无法使用三次方法进行插值，因为他们没有至少4个已知点

I found that the issue was that for customers with fewer records I wasn't capable to interpolate using the cubic method because they did not have at least 4 known points.

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