python - 使用前 10 个值和下一个值之间的平均值替换 pandas 数据框中的特定值

Question

假设我有以下数据框

df.Consumption

0        16.208
1        11.193
2         9.845
3         9.348
4         9.091
          ...  
19611     0.000
19612     0.000
19613     0.000
19614     0.000
19615     0.000
Name: Consumption, Length: 19616, dtype: float64

我想用前 10 个和下一个不为 0.00 的值的平均值替换 0 值

有什么好的方法吗？我正在考虑使用替换和插值方法，但我不知道如何有效地编写它

Answer 1

您可以使用Series.rolling()与 center=True 以及 Rolling.mean() 一起使用获取前一个值和下一个值的平均值。

如果您想从平均值计算中排除 0，请将 0 替换为 NaN。

设置center=True，以便滚动窗口查找上一个和下一个条目。

最后，使用 .loc 将值为 0 的条目设置为平均值，如下所示:

n = 10     # check previous and next 10 entries

# rolling window size is (2n + 1)
Consumption_mean = (df['Consumption'].replace(0, np.nan)
                                     .rolling(n * 2 + 1, min_periods=1, center=True)
                                     .mean())

df.loc[df['Consumption'] == 0, 'Consumption'] = Consumption_mean

演示

使用较小的窗口大小n = 3来演示:

df


    Consumption
0        16.208
1        11.193
2         9.845
3         9.348
4         9.091
5         8.010
6         0.000              <====   target entry
7         7.100
8         0.000              <====   target entry
9         6.800
10        6.500
11        6.300
12        5.900
13        5.800
14        5.600

#n = 10     # check previous and next 10 entries
n = 3     # smaller window size for demo

# rolling window size is (2n + 1)
Consumption_mean = (df['Consumption'].replace(0, np.nan)
                                     .rolling(n * 2 + 1, min_periods=1, center=True)
                                     .mean())

# Update into a new column `Consumption_New` for demo purpose
df['Consumption_New'] = df['Consumption']    
df.loc[df['Consumption'] == 0, 'Consumption_New'] = Consumption_mean

演示结果:

print(df)

    Consumption  Consumption_New
0        16.208          16.2080
1        11.193          11.1930
2         9.845           9.8450
3         9.348           9.3480
4         9.091           9.0910
5         8.010           8.0100
6         0.000           8.0698   # 8.0698 = (9.348 + 9.091 + 8.01 + 7.1 + 6.8) / 5 with skipping 0.000 between 7.100 and 6.800
7         7.100           7.1000
8         0.000           6.9420   # 6.942 = (8.01 + 7.1 + 6.8 + 6.5 + 6.3) / 5 with skipping 0.000 between 8.010 and 7.100
9         6.800           6.8000
10        6.500           6.5000
11        6.300           6.3000
12        5.900           5.9000
13        5.800           5.8000
14        5.600           5.6000