根据另一个数据帧中的最近位置填充数据帧中的缺失值

Question

我有一个类似于以下内容的数据帧：

import pandas as pd
import numpy as np
date = pd.date_range(start='2020-01-01', freq='H', periods=4) 
locations = ["AA3", "AB1", "AD1", "AC0"] 
x = [5.5, 10.2, np.nan, 2.3, 11.2, np.nan, 2.1, 4.0, 6.1, np.nan, 20.3, 11.3, 4.9, 15.2, 21.3, np.nan] 

df = pd.DataFrame({'x': x}) 
df.index = pd.MultiIndex.from_product([locations, date], names=['location', 'date']) 
df = df.sort_index() 
df

                                 x
location date                     
AA3      2020-01-01 00:00:00   5.5
         2020-01-01 01:00:00  10.2
         2020-01-01 02:00:00   NaN
         2020-01-01 03:00:00   2.3
AB1      2020-01-01 00:00:00  11.2
         2020-01-01 01:00:00   NaN
         2020-01-01 02:00:00   2.1
         2020-01-01 03:00:00   4.0
AC0      2020-01-01 00:00:00   4.9
         2020-01-01 01:00:00  15.2
         2020-01-01 02:00:00  21.3
         2020-01-01 03:00:00   NaN
AD1      2020-01-01 00:00:00   6.1
         2020-01-01 01:00:00   NaN
         2020-01-01 02:00:00  20.3
         2020-01-01 03:00:00  11.3

索引值是位置代码和一天中的小时数。我想用同一天和同一小时内最近位置的同一列的有效值来填充x列缺少的值，其中每个位置到其他位置的距离定义为

nearest = pd.DataFrame({"AA3": ["AA3", "AB1", "AD1", "AC0"],
                        "AB1": ["AB1", "AA3", "AC0", "AD1"],
                        "AD1": ["AD1", "AC0", "AB1", "AA3"],
                        "AC0": ["AC0", "AD1", "AA3", "AB1"]})
nearest

   AA3  AB1  AD1  AC0
0  AA3  AB1  AD1  AC0
1  AB1  AA3  AC0  AD1
2  AD1  AC0  AB1  AA3
3  AC0  AD1  AA1  AB1

在此数据集中，列名是位置代码，每列下的行值按其与名称为列名的位置的接近程度指示其他位置

如果最近的位置在同一天和同一小时也缺少值，那么我将取第二个最近的位置在同一天和同一小时的值。如果第二个最近的位置丢失，则第三个最近的位置在同一天和同一小时，依此类推

期望输出：

                                 x
location date                     
AA3      2020-01-01 00:00:00   5.5
         2020-01-01 01:00:00  10.2
         2020-01-01 02:00:00   2.1
         2020-01-01 03:00:00   2.3
AB1      2020-01-01 00:00:00  11.2
         2020-01-01 01:00:00  10.2
         2020-01-01 02:00:00   2.1
         2020-01-01 03:00:00   4.0
AC0      2020-01-01 00:00:00   4.9
         2020-01-01 01:00:00  15.2
         2020-01-01 02:00:00  21.3
         2020-01-01 03:00:00  11.3
AD1      2020-01-01 00:00:00   6.1
         2020-01-01 01:00:00  15.2
         2020-01-01 02:00:00  20.3
         2020-01-01 03:00:00  11.3

以下基于@kiona1018的建议按预期工作，但速度较慢

def fillna_by_nearest(x: pd.Series, nn_data: pd.DataFrame):
    out = x.copy()
    for index, value in x.iteritems():
        if np.isnan(value) and (index[0] in nn_data.columns):
            location, date = index
            for near_location in nn_data[location]:
                if ((near_location, date) in x.index) and pd.notna(x.loc[near_location, date]):
                    out.loc[index] = x.loc[near_location, date]
                    break
    return out

fillna_by_nearest(df['x'], nearest)

Answer 1

我同意Serial Lazer的说法，没有对熊猫/裸体进行更整洁的处决。要求有点复杂。在这种情况下，您应该创建自己的函数。下面是一个例子

nearest = pd.DataFrame({"AA3": ["AA3", "AB1", "AD1", "AC0"],
                        "AB1": ["AB1", "AA3", "AC0", "AD1"],
                        "AD1": ["AD1", "AC0", "AB1", "AA3"],
                        "AC0": ["AC0", "AD1", "AA3", "AB1"]})


def fill_by_nearest(sr: pd.Series):
    if not np.isnan(sr['x']):
        return sr

    location = sr.name[0]
    date = sr.name[1]
    for near_location in nearest[location]:
        if not np.isnan(df.loc[near_location, date]['x']):
            sr['x'] = df.loc[near_location, date]['x']
            return sr
    return sr

df = df.apply(fill_by_nearest, axis=1)