BUG: None and np.nan behave differently for .replace()

[d-kz]

• I have checked that this issue has not already been reported.

• I have confirmed this bug exists on the latest version of pandas.
  pip freeze | grep pandas -> pandas==1.1.5

• (optional) I have confirmed this bug exists on the master branch of pandas.

---

Note: Please read this guide detailing how to provide the necessary information for us to reproduce your bug.

Code Sample, a copy-pastable example

In [9]: df = pd.DataFrame([1, None, np.nan])

In [10]: df
Out[10]:
     0
0  1.0
1  NaN
2  NaN

In [11]: df.replace(1, None)
Out[11]:
     0
0  1.0
1  NaN
2  NaN

In [12]: df.replace(1, np.nan)
Out[12]:
    0
0 NaN
1 NaN
2 NaN

Problem description

This is a problem because when creating a dataframe, both None and np.nan are converted into NaN. Therefore, one expect replacing will also treat them equivalently.

Expected Output

df = pd.DataFrame([1, None, np.nan])
df.replace(1, None)
     0
0  1.0
1  NaN
2  NaN

should return the same as

df = pd.DataFrame([1, None, np.nan])
df.replace(1, np.nan)
    0
0 NaN
1 NaN
2 NaN

Output of pd.show_versions()

[paste the output of pd.show_versions() here leaving a blank line after the details tag]

[simonjayhawkins]

This look like a duplicate of #36984 and can often cause confusion. see #36984 (comment)

>>> import numpy as np
>>> import pandas as pd
>>>
>>> pd.__version__
'1.3.0.dev0+83.g122d50246b'
>>>
>>> df = pd.DataFrame([1, None, np.nan, 1])
>>>
>>> df.replace(1, None)
     0
0  1.0
1  NaN
2  NaN
3  NaN
>>>
>>> df.replace({1:None})
      0
0  None
1   NaN
2   NaN
3  None
>>>
>>>
>>> df.replace(1, None, method="bfill")
     0
0  NaN
1  NaN
2  NaN
3  1.0
>>>

[mroeschke]

Agreed this looks related to #36984 and could be better documented