DataFrame.interpolate()推断尾随丢失的数据

Question

考虑以下示例,其中我们设置示例数据集,创建MultiIndex,取消堆叠数据帧,然后执行线性插值,我们逐行填充:

import pandas as pd  # version 0.14.1
import numpy as np  # version 1.8.1

df = pd.DataFrame({'location': ['a', 'b'] * 5,
                   'trees': ['oaks', 'maples'] * 5,
                   'year': range(2000, 2005) * 2,
                   'value': [np.NaN, 1, np.NaN, 3, 2, np.NaN, 5, np.NaN, np.NaN, np.NaN]})
df.set_index(['trees', 'location', 'year'], inplace=True)
df = df.unstack()
df = df.interpolate(method='linear', axis=1)

未堆叠数据集的位置如下所示:

                 value                        
year              2000  2001  2002  2003  2004
trees  location                               
maples b           NaN     1   NaN     3   NaN
oaks   a           NaN     5   NaN   NaN     2

作为插值方法,我期望输出:

                 value                        
year              2000  2001  2002  2003  2004
trees  location                               
maples b           NaN     1     2     3   NaN
oaks   a           NaN     5     4     3     2

但相反,该方法产生(注意外推值):

                 value                        
year              2000  2001  2002  2003  2004
trees  location                               
maples b           NaN     1     2     3     3
oaks   a           NaN     5     4     3     2

有没有办法指示大熊猫不要推断出系列中最后一个非缺失值？

编辑:

我仍然喜欢在熊猫中看到这个功能,但是现在我已经将它实现为numpy中的一个函数,然后我df.apply()用来修改它df.这是的功能left和right参数np.interp(),我是错过了在大熊猫.

def interpolate(a, dec=None):
    """
    :param a: a 1d array to be interpolated
    :param dec: the number of decimal places with which each
                value should be returned
    :return: returns an array of integers or floats
    """

    # default value is the largest number of decimal places in the input array
    if dec is None:
        dec = max_decimal(a)

    # detect array format convert to numpy as necessary
    if type(a) == list:
        t = 'list'
        b = np.asarray(a, dtype='float')
    if type(a) in [pd.Series, np.ndarray]:
        b = a

    # return the row if it's all nan's
    if np.all(np.isnan(b)):
        return a

    # interpolate
    x = np.arange(b.size)
    xp = np.where(~np.isnan(b))[0]
    fp = b[xp]
    interp = np.around(np.interp(x, xp, fp, np.nan, np.nan), decimals=dec)

    # return with proper numerical type formatting
    # check to make sure there aren't nan's before converting to int
    if dec == 0 and np.isnan(np.sum(interp)) == False:
        interp = interp.astype(int)
    if t == 'list':
        return interp.tolist()
    else:
        return interp


# two little helper functions
def count_decimal(i):
    try:
        return int(decimal.Decimal(str(i)).as_tuple().exponent) * -1
    except ValueError:
        return 0


def max_decimal(a):
    m = 0
    for i in a:
        n = count_decimal(i)
        if n > m:
            m = n
    return m

像示例数据集上的魅力一样工作:

In[1]: df.apply(interpolate, axis=1)
Out[1]:
                 value                        
year              2000  2001  2002  2003  2004
trees  location                               
maples b           NaN     1     2     3   NaN
oaks   a           NaN     5     4     3     2

Answer 1

替换以下行:

df = df.interpolate(method='linear', axis=1)

有了这个:

df = df.interpolate(axis=1).where(df.bfill(axis=1).notnull())

它通过使用回填找到尾随NaN的掩码.它不是非常有效,因为它执行两次NaN填充操作,但这些问题通常可能不是问题.

Answer 2

这确实是一个令人费解的功能。这是一个更紧凑的解决方案，可以在初始插值后应用。

def de_extrapolate(row):  
    extrap = row[row==row[-1]]    
    if extrap.size > 1:
        first_index = extrap.index[1]
        row[first_index:] = np.nan
    return row

和以前一样，我们有：

In [1]: df.interpolate(axis=1).apply(de_extrapolate, axis=1)
Out[1]: 
                value                    
year             2000 2001 2002 2003 2004
trees  location                          
maples b          NaN    1    2    3  NaN
oaks   a          NaN    5    4    3    2