查询panda-df筛选列不是Nan的行

Question

我不熟悉Python和熊猫。在

我想查询一个数据帧并过滤其中一列不是NaN的行。在

我试过：

a=dictionarydf.label.isnull()

但a填充了true或false。 试过这个

^{pr2}$

但却如我所料给出了一个错误

样本数据：

      reference_word         all_matching_words  label review
0           account             fees - account    NaN      N
1           account           mobile - account    NaN      N
2           account          monthly - account    NaN      N
3    administration  delivery - administration    NaN      N
4    administration      fund - administration    NaN      N
5           advisor             fees - advisor    NaN      N
6           advisor          optimum - advisor    NaN      N
7           advisor              sub - advisor    NaN      N
8             aichi           delivery - aichi    NaN      N
9             aichi               pref - aichi    NaN      N
10          airport              biz - airport    travel      N
11          airport              cfo - airport    travel      N
12          airport           cfomtg - airport    travel      N
13          airport          meeting - airport    travel      N
14          airport           summit - airport    travel      N
15          airport             taxi - airport    travel      N
16          airport            train - airport    travel      N
17          airport         transfer - airport    travel      N
18          airport             trip - airport    travel      N
19              ais                admin - ais    NaN      N
20              ais               alpine - ais    NaN      N
21              ais                 fund - ais    NaN      N
22       allegiance       custody - allegiance    NaN      N
23       allegiance          fees - allegiance    NaN      N
24            alpha               late - alpha    NaN      N
25            alpha               meal - alpha    NaN      N
26            alpha               taxi - alpha    NaN      N
27           alpine             admin - alpine    NaN      N
28           alpine               ais - alpine    NaN      N
29           alpine              fund - alpine    NaN      N

我想过滤label不是NaN的数据

预期产量：

     reference_word         all_matching_words   label    review
0          airport              biz - airport    travel      N
1          airport              cfo - airport    travel      N
2          airport           cfomtg - airport    travel      N
3          airport          meeting - airport    travel      N
4          airport           summit - airport    travel      N
5          airport             taxi - airport    travel      N
6          airport            train - airport    travel      N
7          airport         transfer - airport    travel      N
8          airport             trip - airport    travel      N

Answer 1

您可以使用^{}：

df = df.dropna(subset=['label'])

print (df)
   reference_word  all_matching_words   label review
10        airport       biz - airport  travel      N
11        airport       cfo - airport  travel      N
12        airport    cfomtg - airport  travel      N
13        airport   meeting - airport  travel      N
14        airport    summit - airport  travel      N
15        airport      taxi - airport  travel      N
16        airport     train - airport  travel      N
17        airport  transfer - airport  travel      N
18        airport      trip - airport  travel      N

另一个解决方案-^{}和^{}：

^{pr2}$