Series.replace not working on slices of heterogenoues types

[bluenote10]

Code Sample

import pandas as pd
import numpy as np 

c = pd.Series([
    np.nan,
    1,
    "hello",
])
c_replaced_1 = c[0:3].replace({np.nan: None})
c_replaced_2 = c[0:2].replace({np.nan: None})
print(c_replaced_1)
print(c_replaced_2)

Problem description

Series.replace does not work in the second case. In both cases one would expect that the np.nan gets replaced by None. However, the replace fails in the second case.

The actual output is:

0     None
1        1
2    hello
dtype: object
0    NaN
1    1.0
dtype: float64

Expected Output

0     None
1        1
2    hello
dtype: object
0    None
1    1.0
dtype: object

Output of pd.show_versions()

INSTALLED VERSIONS
------------------
commit: None
python: 2.7.15.candidate.1
python-bits: 64
OS: Linux
OS-release: 4.15.0-70-generic
machine: x86_64
processor: x86_64
byteorder: little
LC_ALL: None
LANG: en_US.UTF-8
LOCALE: en_US.UTF-8

pandas: 0.24.2
pytest: None
pip: 19.2.3
setuptools: 41.2.0
Cython: None
numpy: 1.15.2
scipy: 1.2.2
pyarrow: None
xarray: None
IPython: 5.8.0
sphinx: None
patsy: 0.5.1
dateutil: 2.7.3
pytz: 2018.5
blosc: None
bottleneck: None
tables: None
numexpr: None
feather: None
matplotlib: 2.2.3
openpyxl: None
xlrd: 1.2.0
xlwt: None
xlsxwriter: 1.1.2
lxml.etree: 4.2.1
bs4: 4.6.0
html5lib: 0.999999999
sqlalchemy: None
pymysql: None
psycopg2: None
jinja2: 2.10.3
s3fs: None
fastparquet: None
pandas_gbq: None
pandas_datareader: None
gcsfs: None

[kurzacz]

I have the same problem. I'm going to clean Kaggle dataset from missing values represented by question marks ?. Issue happens when I try to do this across all dataset.
Steps to reproduce:

1. Find and download Kaggle dataset geoplaces2.csv:
   https://www.kaggle.com/uciml/restaurant-data-with-consumer-ratings#geoplaces2.csv
2. Code step by step:

import pandas as pd
df = pd.read_csv(
        'restaurant-data-with-consumer-ratings/geoplaces2.csv')
df.replace('?', None, inplace=True)
print(df.fax[0])

Output: ?
Expected output: None

Output of pd.show_versions()

>>> pd.show_versions()
INSTALLED VERSIONS
------------------
commit           : None
python           : 3.7.4.final.0
python-bits      : 64
OS               : Linux
OS-release       : 4.15.0-70-generic
machine          : x86_64
processor        : x86_64
byteorder        : little
LC_ALL           : None
LANG             : en_US.UTF-8
LOCALE           : en_US.UTF-8
pandas           : 0.25.3
numpy            : 1.17.4
pytz             : 2019.3
dateutil         : 2.8.1
pip              : 19.0.3
setuptools       : 40.8.0
Cython           : None
pytest           : None
hypothesis       : None
sphinx           : None
blosc            : None
feather          : None
xlsxwriter       : None
lxml.etree       : None
html5lib         : None
pymysql          : None
psycopg2         : 2.8.4 (dt dec pq3 ext lo64)
jinja2           : None
IPython          : None
pandas_datareader: None
bs4              : None
bottleneck       : None
fastparquet      : None
gcsfs            : None
lxml.etree       : None
matplotlib       : None
numexpr          : None
odfpy            : None
openpyxl         : None
pandas_gbq       : None
pyarrow          : None
pytables         : None
s3fs             : None
scipy            : None
sqlalchemy       : 1.3.11
tables           : None
xarray           : None
xlrd             : None
xlwt             : None
xlsxwriter       : None

[Skitionek]

Facing same issue for a while now, can you post a workaround you are currently using?

I used to call it twice so it appears to work on this example:

import pandas as pd
import numpy as np 

c = pd.Series([
    np.nan,
    1,
    "hello",
])
c_replaced_2 = c[0:2].replace({np.nan: None})
c_replaced_2_2 = c_replaced_2.replace({np.nan: None})
print(c_replaced_2)
print(c_replaced_2_2)

Outputs

0    NaN
1    1.0
dtype: float64
0    None
1       1
dtype: object

However, I have just run into problem that this also not always works.

[arnocandel]

Seems like can't replace with None

Python 3.6.9 (default, Nov  7 2019, 10:44:02) 
[GCC 8.3.0] on linux
Type "help", "copyright", "credits" or "license" for more information.
>>> import pandas as pd
>>> pd.Series(["a"]).replace(["a"], None)
0    a
dtype: object
>>> pd.Series([1]).replace([1], None)
0    1
dtype: int64
>>> pd.Series([1.]).replace([1.], None)
0    1.0
dtype: float64

>>> pd.show_versions()

INSTALLED VERSIONS
------------------
commit           : None
python           : 3.6.9.final.0
python-bits      : 64
OS               : Linux
OS-release       : 5.0.0-37-generic
machine          : x86_64
processor        : x86_64
byteorder        : little
LC_ALL           : None
LANG             : en_US.UTF-8
LOCALE           : en_US.UTF-8

pandas           : 1.0.1
numpy            : 1.18.1
pytz             : 2019.3
dateutil         : 2.8.1
pip              : 20.0.2
setuptools       : 45.2.0
Cython           : None
pytest           : None
hypothesis       : None
sphinx           : None
blosc            : None
feather          : None
xlsxwriter       : None
lxml.etree       : None
html5lib         : None
pymysql          : None
psycopg2         : None
jinja2           : None
IPython          : None
pandas_datareader: None
bs4              : None
bottleneck       : None
fastparquet      : None
gcsfs            : None
lxml.etree       : None
matplotlib       : None
numexpr          : None
odfpy            : None
openpyxl         : None
pandas_gbq       : None
pyarrow          : None
pytables         : None
pytest           : None
pyxlsb           : None
s3fs             : None
scipy            : None
sqlalchemy       : None
tables           : None
tabulate         : None
xarray           : None
xlrd             : None
xlwt             : None
xlsxwriter       : None
numba            : None

[giacomomarchioro]

Hello, I am also experiencing this critical bug. The behaviour change if I run df.replace({'value': {0:"/"}},inplace=True) in the terminal or if I run the same command in a script. In the first case, it works inside the script it doesn't.

[jbrockmendel]

This looks right to me on main. Could use a test (first check to see if one already exists)

[Arjundas08]

"Hi, I'm a new contributor and would like to try working on this 'good first issue'. Could you please assign it to me?"