python - Pandas numpy.where() 使用 - 没有得到想要的结果

Question

我正在尝试根据 NaN 值将两列合并为第三列

df['code2'] = np.where(df['code']==np.nan, df['code'], df['code1'])

我只得到 code2 中 code1 列的值。结果如图所示 输出图像

请告诉我我写的代码有什么问题。谢谢

Answer 1

我想你需要isnull用于比较 NaN:

df['code2'] = np.where(df['code'].isnull(), df['code'], df['code1'])

Docs :

Warning

One has to be mindful that in python (and numpy), the nan's don’t compare equal, but None's do. Note that Pandas/numpy uses the fact that np.nan != np.nan, and treats None like np.nan.

In [11]: None == None
Out[11]: True

In [12]: np.nan == np.nan
Out[12]: False

So as compared to above, a scalar equality comparison versus a None/np.nan doesn’t provide useful information.

In [13]: df2['one'] == np.nan
Out[13]: 
a    False
b    False
c    False
d    False
e    False
f    False
g    False
h    False
Name: one, dtype: bool