Question

我想过滤掉NaN值，并将其余行保留在Label列中。

df：

        Timestamp               Label
157505  2010-09-21 23:13:21.090 1
321498  2010-09-22 00:44:14.890 1
332687  2010-09-22 00:44:15.890 1
330028  2010-09-22 00:44:17.890 NaN
293410  2010-09-22 00:44:18.440 2
23093   2010-09-22 00:44:19.890 2
282054  2010-09-22 00:44:23.440 2
158381  2010-09-22 01:04:33.440 NaN
317397  2010-09-22 01:27:01.790 NaN
170770  2010-09-22 02:18:52.850 NaN

可复制的示例：

from pandas import *
import numpy as np 
import pandas as pd 

df = pd.DataFrame({'Timestamp': {157505: Timestamp('2010-09-21 23:13:21.090000'),
  321498: Timestamp('2010-09-22 00:44:14.890000'),
  332687: Timestamp('2010-09-22 00:44:15.890000'),
  330028: Timestamp('2010-09-22 00:44:17.890000'),
  293410: Timestamp('2010-09-22 00:44:18.440000'),
  23093: Timestamp('2010-09-22 00:44:19.890000'),
  282054: Timestamp('2010-09-22 00:44:23.440000'),
  158381: Timestamp('2010-09-22 01:04:33.440000'),
  317397: Timestamp('2010-09-22 01:27:01.790000'),
  170770: Timestamp('2010-09-22 02:18:52.850000')},
 'Label': {157505: 1,
  321498: 1,
  332687: 1,
  330028: 'NaN',
  293410: 2,
  23093: 2,
  282054: 2,
  158381: 'NaN',
  317397: 'NaN',
  170770: 'NaN'}})
df

我尝试过：

df[df.Label.notnull()]

并获得完全相同的表：


        Timestamp               Label
157505  2010-09-21 23:13:21.090 1
321498  2010-09-22 00:44:14.890 1
332687  2010-09-22 00:44:15.890 1
330028  2010-09-22 00:44:17.890 NaN
293410  2010-09-22 00:44:18.440 2
23093   2010-09-22 00:44:19.890 2
282054  2010-09-22 00:44:23.440 2
158381  2010-09-22 01:04:33.440 NaN
317397  2010-09-22 01:27:01.790 NaN
170770  2010-09-22 02:18:52.850 NaN

怎么了？什么是最好的方法？

Answer 1

请从dtype float将Label转换为object并使用notna()或isna()

df=df[df.Label.astype(float).notna()]
print(df)




                   Timestamp  Label
157505 2010-09-21 23:13:21.090    1.0
321498 2010-09-22 00:44:14.890    1.0
332687 2010-09-22 00:44:15.890    1.0
293410 2010-09-22 00:44:18.440    2.0
23093  2010-09-22 00:44:19.890    2.0
282054 2010-09-22 00:44:23.440    2.0

Answer 2

您可以这样做：

df['Label'] = df['Label'].replace('NaN', np.nan)
df.dropna(inplace=True)
print(df)

或

df = df[df['Label'].notna()]
print(df)

                     Timestamp  Label
157505 2010-09-21 23:13:21.090    1.0
321498 2010-09-22 00:44:14.890    1.0
332687 2010-09-22 00:44:15.890    1.0
293410 2010-09-22 00:44:18.440    2.0
23093  2010-09-22 00:44:19.890    2.0
282054 2010-09-22 00:44:23.440    2.0

Answer 3

我了解您正在尝试过滤Nan值。但是notnull（）过滤器不会过滤字符串'NaN'。用np.nan替换它会得到您期望的结果。另外，您可以选择删除它。

from pandas import *
import numpy as np 
import pandas as pd 

df = pd.DataFrame({'Timestamp': {157505: Timestamp('2010-09-21 23:13:21.090000'),
  321498: Timestamp('2010-09-22 00:44:14.890000'),
  332687: Timestamp('2010-09-22 00:44:15.890000'),
  330028: Timestamp('2010-09-22 00:44:17.890000'),
  293410: Timestamp('2010-09-22 00:44:18.440000'),
  23093: Timestamp('2010-09-22 00:44:19.890000'),
  282054: Timestamp('2010-09-22 00:44:23.440000'),
  158381: Timestamp('2010-09-22 01:04:33.440000'),
  317397: Timestamp('2010-09-22 01:27:01.790000'),
  170770: Timestamp('2010-09-22 02:18:52.850000')},
 'Label': {157505: 1,
  321498: 1,
  332687: 1,
  330028: np.nan,
  293410: 2,
  23093: 2,
  282054: 2,
  158381: np.nan,
  317397: np.nan,
  170770: np.nan}})

df[df.Label.notnull()]

将得到：


Timestamp   Label
157505  2010-09-21 23:13:21.090 1.0
321498  2010-09-22 00:44:14.890 1.0
332687  2010-09-22 00:44:15.890 1.0
293410  2010-09-22 00:44:18.440 2.0
23093   2010-09-22 00:44:19.890 2.0
282054  2010-09-22 00:44:23.440 2.0

或

df.dropna()

它将给出相同的结果：

    Timestamp   Label
157505  2010-09-21 23:13:21.090 1.0
321498  2010-09-22 00:44:14.890 1.0
332687  2010-09-22 00:44:15.890 1.0
293410  2010-09-22 00:44:18.440 2.0
23093   2010-09-22 00:44:19.890 2.0
282054  2010-09-22 00:44:23.440 2.0

按熊猫中的列过滤非NaN值

3 个答案: