Check if single cell value is NaN in Pandas

Just encountered this problem myself and found a solution, imperfect, but works. As noted above, none of these 3 answers are addressing OP's question. Here's an example of my problem which I feel is the same.

# fill null values of one column with that of another
f = lambda row: row['A'] if (row['B'].isnull()) else row['B']
df['B'] = df.apply(f, axis=1)

>>> AttributeError: 'str' object has no attribute 'isnull'

Because the value within a cell of a dataframe is just a primative datatype, you can't use any of pandas built-in methods. So this is what I did.

f = lambda row: row['A'] if (str(row['B'])=='nan') else row['B']

This actually the only thing I could get to work!

Try this:

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

>>> L = [4, nan ,6]
>>> df = Series(L)

>>> df
0     4
1   NaN
2     6

>>> if(pd.isnull(df[1])):
        print "Found"

Found

>>> if(np.isnan(df[1])):
        print "Found"

Found

You can use "isnull" with "at" to check a specific value in a dataframe.

For example:

import pandas as pd
import numpy as np

df = pd.DataFrame([[np.nan, 2], [1, 3], [4, 6]], columns=['A', 'B'])

Yeilds:

    A   B
0   NaN 2
1   1.0 3
2   4.0 6

To check the values:

pd.isnull(df.at[0,'A'])

-> True

pd.isnull(df.at[0,'B'])

-> False

STEP 1.)

df[df.isnull().any(1)]

----> Will give you dataframe with rows and column, if any value there is nan.

STEP 2.)

this will give you location in dataframe where exactly value is nan. then you could do

if(**df.iloc[loc_row,loc_colum]==np.nan**):
    print"your code here"