Check for None in pandas dataframe

Question

I would like to find where None is found in the dataframe.

pd.DataFrame([None,np.nan]).isnull()
OUT: 
      0
0  True
1  True

isnull() finds bo

Answer 1

If you want to get True/False for each line, you can use the following code. Here is an example as a result for the following DataFrame:

df = pd.DataFrame([[None, 3], ["", np.nan]])

df
#      0      1
#0  None    3.0
#1          NaN

How to check None

Available: .isnull()

>>> df[0].isnull()
0     True
1    False
Name: 0, dtype: bool

Available: .apply == or is None

>>> df[0].apply(lambda x: x == None)
0     True
1    False
Name: 0, dtype: bool

>>> df[0].apply(lambda x: x is None)
0     True
1    False
Name: 0, dtype: bool

Available: .values == None

>>> df[0].values == None
array([ True, False])

Unavailable: is or ==

>>> df[0] is None
False

>>> df[0] == None
0    False
1    False
Name: 0, dtype: bool

Unavailable: .values is None

>>> df[0].values is None
False

How to check np.nan

Available: .isnull()

>>> df[1].isnull()
0    False
1     True
Name: 1, dtype: bool

Available: np.isnan

>>> np.isnan(df[1])
0    False
1     True
Name: 1, dtype: bool

>>> np.isnan(df[1].values)
array([False,  True])

>>> df[1].apply(lambda x: np.isnan(x))
0    False
1     True
Name: 1, dtype: bool

Unavailable: is or == np.nan

>>> df[1] is np.nan
False

>>> df[1] == np.nan
0    False
1    False
Name: 1, dtype: bool

>>> df[1].values is np.nan
False

>>> df[1].values == np.nan
array([False, False])

>>> df[1].apply(lambda x: x is np.nan)
0    False
1    False
Name: 1, dtype: bool

>>> df[1].apply(lambda x: x == np.nan)
0    False
1    False
Name: 1, dtype: bool

Answer 2

You could use applymap with a lambda to check if an element is None as follows, (constructed a different example, as in your original one, None is coerced to np.nan because the data type is float, you will need an object type column to hold None as is, or as commented by @Evert, None and NaN are indistinguishable in numeric type columns):

df = pd.DataFrame([[None, 3], ["", np.nan]])

df
#      0      1
#0  None    3.0
#1          NaN

df.applymap(lambda x: x is None)

#       0       1
#0   True   False
#1  False   False