如何在 Pandas 的应用函数中测试 nan？
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How to test for nan's in an apply function in pandas?
提问by Hunle
I have a simple applyfunction that I execute on some of the columns. But, it keeps getting tripped up by NaNvalues in pandas.
我有一个apply在某些列上执行的简单函数。但是，它不断得到由绊倒NaN价值观pandas。
input_data = np.array(
[
[random.randint(0,9) for x in range(2)]+['']+['g'],
[random.randint(0,9) for x in range(3)]+['g'],
[random.randint(0,9) for x in range(3)]+['a'],
[random.randint(0,9) for x in range(3)]+['b'],
[random.randint(0,9) for x in range(3)]+['b']
]
)
input_df = pd.DataFrame(data=input_data, columns=['B', 'C', 'D', 'label'])
I have a simple lambda like this:
我有一个像这样的简单 lambda：
input_df['D'].apply(lambda aCode: re.sub('\.', '', aCode) if not np.isnan(aCode) else aCode)
And it gets tripped up by the NaN values:
它被 NaN 值绊倒了：
File "<pyshell#460>", line 1, in <lambda>
input_df['D'].apply(lambda aCode: re.sub('\.', '', aCode) if not np.isnan(aCode) else aCode)
TypeError: Not implemented for this type
So, I tried just testing for nan values that Pandas adds:
因此，我尝试只测试 Pandas 添加的 nan 值：
np.isnan(input_df['D'].values[0])
np.isnan(input_df['D'].iloc[0])
Both get the same error.
两者都得到相同的错误。
I do not know how to test for nan values other than np.isnan. Is there an easier way to do this? Thanks.
我不知道如何测试 nan 值以外的np.isnan. 有没有更简单的方法来做到这一点？谢谢。
回答by EdChum
your code fails because your first entry is an empty string and np.isnandoesn't understand empty strings:
您的代码失败，因为您的第一个条目是一个空字符串并且np.isnan不理解空字符串：
In [55]:
input_df['D'].iloc[0]
Out[55]:
''
In [56]:
np.isnan('')
---------------------------------------------------------------------------
TypeError Traceback (most recent call last)
<ipython-input-56-a9f139a0c5b8> in <module>()
----> 1 np.isnan('')
TypeError: Not implemented for this type
ps.notnulldoes work:
ps.notnull确实有效：
In [57]:
import re
input_df['D'].apply(lambda aCode: re.sub('\.', '', aCode) if pd.notnull(aCode) else aCode)
Out[57]:
0
1 3
2 3
3 0
4 3
Name: D, dtype: object
However, if you just want to replace something then just use .str.replace:
但是，如果您只想替换某些内容，则只需使用.str.replace：
In [58]:
input_df['D'].str.replace('\.','')
Out[58]:
0
1 3
2 3
3 0
4 3
Name: D, dtype: object
