use the comparison result as the index of pandas.DataFrame

Question

import pandas as pd
import numpy as np

df = pd.DataFrame([[1,2,3],[4,np.nan,6]])
whereNans = np.isnan(df)
print whereNans
print df[whereNans]

print "--"*30

print df>3
print df[df>3]

As above, whereNans is correct, but df[whereNans] doesn't get what I want. However, df[df>3] can get what I want.

Actually, the index stored in whereNans is same as the df>3 . What is the problem?

Answer 1

You seem to be confused by this, this is correct behaviour, where the mask is True it will display the result in that position, where False it will display NaN , so in effect you're going to display a df with all NaN s

Because you have a single NaN value it returns NaN for that position, where it's False you just get NaN

If you compare with df>3 version you observe the same behaviour:

In[49]:
df[df>3]

Out[49]: 
     0   1    2
0  NaN NaN  NaN
1  4.0 NaN  6.0

Also just to show this has nothing to do with numpy , using pandas isnull gives the same result:

In[50]:
df[df.isnull()]

Out[50]: 
    0   1   2
0 NaN NaN NaN
1 NaN NaN NaN