为 pandas 创建自定义插值函数 [英] Creating a custom interpolation function for pandas

问题描述

我目前正在尝试使用熊猫清理并填写一些缺少的时间序列数据.插值函数工作得很好，但是它没有我的数据集所需的一些(使用较少的插值函数).有几个例子是一个简单的最后一个"有效数据点，它将创建类似于阶跃函数的东西，或者诸如对数或几何插值之类的东西.

I am currently trying to clean up and fill in some missing time-series data using pandas. The interpolate function works quite well, however it doesn't have a few (less widely used) interpolation functions that I require for my data set. A couple examples would be a simple "last" valid data point which would create something akin to a step function, or something like a logarithmic or geometric interpolation.

浏览文档时，似乎没有一种方法可以传递自定义插值函数.这样的功能是否直接存在于熊猫中?如果不是，那么有没有人做过pandas-fu来通过其他方式有效应用自定义插值?

Browsing through the docs, it didn't appear there is a way to pass a custom interpolation function. Does such functionality exist directly within pandas? And if not, has anyone done any pandas-fu to efficiently apply custom interpolations through other means?

推荐答案

熊猫提供的插值方法是 numpy-quaternion )，有效地做到这一点.我将在此处复制代码，以希望您可以对其进行调整:

The interpolation methods offered by Pandas are those offered by scipy.interpolate.interp1d - which, unfortunately, do not seem to be extendable in any way. I had to do something like that to apply SLERP quaternion interpolation (using numpy-quaternion), and I managed to do it quite efficiently. I'll copy the code here in the hope that you can adapt it for your purposes:

def interpolate_slerp(data):
    if data.shape[1] != 4:
        raise ValueError('Need exactly 4 values for SLERP')
    vals = data.values.copy()
    # quaternions has size Nx1 (each quaternion is a scalar value)
    quaternions = quaternion.as_quat_array(vals)
    # This is a mask of the elements that are NaN
    empty = np.any(np.isnan(vals), axis=1)
    # These are the positions of the valid values
    valid_loc = np.argwhere(~empty).squeeze(axis=-1)
    # These are the indices (e.g. time) of the valid values
    valid_index = data.index[valid_loc].values
    # These are the valid values
    valid_quaternions = quaternions[valid_loc]
    # Positions of the missing values
    empty_loc = np.argwhere(empty).squeeze(axis=-1)
    # Missing values before first or after last valid are discarded
    empty_loc = empty_loc[(empty_loc > valid_loc.min()) & (empty_loc < valid_loc.max())]
    # Index value for missing values
    empty_index = data.index[empty_loc].values
    # Important bit! This tells you the which valid values must be used as interpolation ends for each missing value
    interp_loc_end = np.searchsorted(valid_loc, empty_loc)
    interp_loc_start = interp_loc_end - 1
    # These are the actual values of the interpolation ends
    interp_q_start = valid_quaternions[interp_loc_start]
    interp_q_end = valid_quaternions[interp_loc_end]
    # And these are the indices (e.g. time) of the interpolation ends
    interp_t_start = valid_index[interp_loc_start]
    interp_t_end = valid_index[interp_loc_end]
    # This performs the actual interpolation
    # For each missing value, you have:
    #   * Initial interpolation value
    #   * Final interpolation value
    #   * Initial interpolation index
    #   * Final interpolation index
    #   * Missing value index
    interpolated = quaternion.slerp(interp_q_start, interp_q_end, interp_t_start, interp_t_end, empty_index)
    # This puts the interpolated values into place
    data = data.copy()
    data.iloc[empty_loc] = quaternion.as_float_array(interpolated)
    return data

诀窍在np.searchsorted中，它可以很快找到每个值的正确内插结束点.这种方法的局限性在于:

The trick is in np.searchsorted, which very quickly finds the right interpolation ends for each value. The limitation of this method is that:

• 您的插值功能必须像quaternion.slerp一样工作 (这应该很奇怪，因为它具有常规的ufunc广播行为).
• 它仅适用于两端仅需要一个值的插值方法，因此，例如类似于三次插值(您不会因为已经提供了它)之类的东西就行不通了.

• Your interpolation function must work somewhat like quaternion.slerp (which should not be strange since it has regular ufunc broadcasting behaviour).
• It only works for interpolation methods that require only one value on each end, so if you want e.g. something like a cubic interpolation (which you don't because that one is already provided) this wouldn't work.

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