BUG: Inconsistent behavior for df.replace over pd.Period columns

[justinessert]

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

• I have confirmed this bug exists on the latest version of pandas.

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

---

Code Sample, a copy-pastable example

import pandas as pd

df = pd.DataFrame({
    'Int': [1, 2, 3],
    'Flo': [1.0, 6.0, 1.0],
    'Per': [pd.Period('2020-01')]*3,
    'Tim': [pd.Timestamp('2020-01')]*3
})

# TypeError: 'value' should be a 'Period', 'NaT', or array of those. Got 'float' instead.
df.replace(1.0, 0.0)

# Works
df[['Int', 'Flo', 'Tim']].replace(1.0, 0.0)

# Works
df['Per'].replace(1.0, pd.Period('2020-02'))

Problem description

df.replace(1.0, 0.0) fails if any columns of df are Period types. The error is saying that 0.0 is an invalid value to put in a Pandas Period column. This should never be an issue though, 1.0 will never exist in a Pandas Period column, so nothing will ever be replaced for 0.0.

Expected Output

The expected behavior would be consistent with the df[['Int', 'Flo', 'Tim']].replace(1.0, 0.0) in the example above. This does work, the 'Flo' column has all of it's 1.0s replaced with 0.0s and the 'Int' and 'Tim' columns (which are not Float types) remain unaffected. I would expect that a Pandas Period column would also be skipped in the same way.

Output of pd.show_versions()

INSTALLED VERSIONS

commit : None
python : 3.7.7.final.0
python-bits : 64
OS : Darwin
OS-release : 19.5.0
machine : x86_64
processor : i386
byteorder : little
LC_ALL : None
LANG : en_US.UTF-8
LOCALE : en_US.UTF-8

pandas : 1.0.4
numpy : 1.18.5
pytz : 2020.1
dateutil : 2.8.0
pip : 20.1.1
setuptools : 47.3.0.post20200616
Cython : None
pytest : 4.4.1
hypothesis : None
sphinx : 3.1.1
blosc : None
feather : None
xlsxwriter : 1.2.9
lxml.etree : 4.5.1
html5lib : None
pymysql : None
psycopg2 : None
jinja2 : 2.11.2
IPython : 7.15.0
pandas_datareader: None
bs4 : None
bottleneck : None
fastparquet : None
gcsfs : None
lxml.etree : 4.5.1
matplotlib : 3.2.1
numexpr : None
odfpy : None
openpyxl : None
pandas_gbq : None
pyarrow : None
pytables : None
pytest : 4.4.1
pyxlsb : None
s3fs : None
scipy : 1.4.1
sqlalchemy : None
tables : None
tabulate : None
xarray : None
xlrd : 1.2.0
xlwt : None
xlsxwriter : 1.2.9
numba : None

[dsaxton]

This looks to be fixed on master (but could warrant a test if there isn't one already). Interesting though that the 1 in the first column appears not to be replaced.

>>> import pandas as pd
>>> 
>>> df = pd.DataFrame({
...     'Int': [1, 2, 3],
...     'Flo': [1.0, 6.0, 1.0],
...     'Per': [pd.Period('2020-01')]*3,
...     'Tim': [pd.Timestamp('2020-01')]*3
... })
>>> print(df.replace(1.0, 0.0))
   Int  Flo      Per        Tim
0    1  0.0  2020-01 2020-01-01
1    2  6.0  2020-01 2020-01-01
2    3  0.0  2020-01 2020-01-01
>>> print(pd.__version__)
1.1.0.dev0+1900.g80ba4c429

[justinessert]

I just realized that Pandas 1.0.5 was released, I was running this on version 1.0.4. I believe everything is fixed now!

[nwweber]

take

[nwweber]

Hi all,

I can confirm that this failed in 1.0.4 and indeed does not raise an exception anymore in 1.10.

However, the new behavior in 1.10 is still not what I would expect it to be. Concretely, calling df.replace(1.0, 0.0) on a dataframe with a Period column now has the following effects:

• the values in a Period column are now left unchanged
• but the dtype of that same column is changed from Period to Object

I think a user would be surprised by this change in dtype (I certainly am). This also makes tm.assert_frame_equal() fail when trying to write a test for this.

For now I have created a tiny regression test that makes sure that calling df.replace() does not raise an exception anymore, see this pull request: #34904

I think this issue should stay open and df.replace() should get changed to return a Period column in the scenario described. That's a bit beyond my current insights into the Pandas code, it'd be great if someone else could have a look at that.

Minimal code to reproduce (adapted from original submitter):

df = pd.DataFrame({"Per": [pd.Period("2020-01")] * 3})
new_df = df.replace(1.0, 0.0)

# this is 'Period'
df.Per.dtype
# this is 'Object' but should also be 'Period'
new_df.Per.dtype

[nwweber]

Unit test for desired behavior implemented here: #34904

Will have a look at underlying issue causing modification of dtype from period to object

[tsibley]

Will have a look at underlying issue causing modification of dtype from period to object

I believe this is the same problem described in #35268.