我有以下数据框，所有空白区域均为np.nan。
coupler_id 25 26 28 29
timestamp
2015-12-05 03:02:29 12017.0 12008.0
2015-12-05 03:04:47 12017.0 12008.0
2015-12-05 03:09:14 12017.0 12008.0
2015-12-05 03:12:12 12017.0 12008.0
2015-12-05 03:23:06 12008.0
2015-12-05 03:24:45 12017.0
2015-12-05 06:31:20 12017.0
2015-12-05 09:36:29 12011.0
2015-12-05 23:59:35 12017.0
2015-12-06 23:59:38 12017.0
我想转发填充的缺失值（限制1）在行中不重复值。因此，以上数据框应如下所示：
coupler_id 25 26 28 29
timestamp
2015-12-05 03:02:29 12017.0 12008.0
2015-12-05 03:04:47 12017.0 12008.0
2015-12-05 03:09:14 12017.0 12008.0
2015-12-05 03:12:12 12017.0 12008.0
2015-12-05 03:23:06 12017.0 12008.0
2015-12-05 03:24:45 12017.0
2015-12-05 06:31:20 12017.0
2015-12-05 09:36:29 12011.0
2015-12-05 23:59:35 12011.0 12017.0
2015-12-06 23:59:38 12017.0
编辑：
如果第25列和第26列中有数据，而第28列索引中没有前面的nan则为2015-12-05 03:24:45。
coupler_id 25 26 28 29
timestamp
2015-12-05 03:02:29 12017.0 12008.0
2015-12-05 03:04:47 12017.0 12008.0
2015-12-05 03:09:14 12017.0 12008.0
2015-12-05 03:12:12 12017.0 12008.0
2015-12-05 03:23:06 12007.0 12018.0 12008.0
2015-12-05 03:24:45 12033.0 12050.0 12025.0 12017.0
2015-12-05 06:31:20 12033.0 12017.0
2015-12-05 09:36:29 12008.0 12011.0
2015-12-05 23:59:35 12017.0
2015-12-06 23:59:38 12017.0
答案 0 :(得分：1)
更新后的答案
这是一种比较通用的情况，可以检查所有列：
def remove_duplicates(data, ix, names):
# if only 1 entry, no comparison needed
if data.notnull().sum() == 1:
return data
# mark all duplicates
dupes = data.dropna().duplicated(keep=False)
if dupes.any():
for name in names:
# if previous value was NaN AND current is duplicate, replace with NaN
if np.isnan(df.loc[ix, name]) & dupes[name]:
data[name] = np.nan
return data
filled = df.ffill(limit=1)
filled.apply(lambda row: remove_duplicates(row, row.name, row.index), axis=1)
25 26 28 29
2015-12-05 03:02:29 NaN NaN 12017.0 12008.0
2015-12-05 03:04:47 NaN NaN 12017.0 12008.0
2015-12-05 03:09:14 NaN NaN 12017.0 12008.0
2015-12-05 03:12:12 NaN NaN 12017.0 12008.0
2015-12-05 03:23:06 12007.0 12018.0 12017.0 12008.0
2015-12-05 03:24:45 12033.0 12050.0 12025.0 12017.0
2015-12-05 06:31:20 NaN 12033.0 12017.0 NaN
2015-12-05 09:36:29 12008.0 12033.0 12011.0 NaN
2015-12-05 23:59:35 12008.0 NaN 12011.0 12017.0
2015-12-06 23:59:38 NaN NaN NaN 12017.0
上一个答案
您可以使用ffill(limit=1)，然后检查是否存在重复的和，如果前面的列之一是NaN。
import numpy as np
def remove_duplicates(data, ix, names):
if data[0] - data[1] != 0:
return data
if np.isnan(filled.loc[ix-1, names[0]]):
return [data[0], np.nan]
elif np.isnan(filled.loc[ix-1, names[1]]):
return [np.nan, data[1]]
return data
filled = df[["28","29"]].ffill(limit=1)
df[["28","29"]] = filled.apply(
lambda row: remove_duplicates(row, row.name, row.index), axis=1
)
df
coupler_id 25 26 28 29
0 2015-12-05 03:02:29 NaN NaN 12017.0 12008.0
1 2015-12-05 03:04:47 NaN NaN 12017.0 12008.0
2 2015-12-05 03:09:14 NaN NaN 12017.0 12008.0
3 2015-12-05 03:12:12 NaN NaN 12017.0 12008.0
4 2015-12-05 03:23:06 NaN NaN 12017.0 12008.0
5 2015-12-05 03:24:45 NaN NaN NaN 12017.0
6 2015-12-05 06:31:20 NaN NaN 12017.0 NaN
7 2015-12-05 09:36:29 NaN NaN 12011.0 NaN
8 2015-12-05 23:59:35 NaN NaN 12011.0 12017.0
9 2015-12-06 23:59:38 NaN NaN NaN 12017.0
答案 1 :(得分：0)
根据docs填充是DataFrame.fillna（method ='ffill'）的同义词，因此在填充上使用限制arg将限制填充的数量。
df = df.ffill(limit=1)
示例：
temp
Out[224]:
X Y Z
0 0.0 0.0 0.0
1 1.0 2.0 2.0
2 NaN NaN NaN
3 NaN 3.0 3.0
4 1.0 NaN NaN
5 NaN NaN NaN
temp.ffill(limit=1)
Out[225]:
X Y Z
0 0.0 0.0 0.0
1 1.0 2.0 2.0
2 1.0 2.0 2.0
3 NaN 3.0 3.0
4 1.0 3.0 3.0
5 1.0 NaN NaN